RS Aggarwal Solutions for Class 12 Chapter 11 Applications of Derivatives

Question 1. The side of a square is increasing at the rate of 0.2 cm/s. Find the rate of increase of the perimeter of the square.
Answer: Let the side of the square be \( a \).

Rate of change of side = \( \frac{da}{dt} = 0.2 \text{ cm/s} \)

Perimeter of the square = \( 4a \)

Rate of change of perimeter = \( \frac{dP}{dt} = 4\frac{da}{dt} = 4 \times 0.2 = 0.8 \text{ cm/s} \)
In simple words: When the side grows at 0.2 cm per second, the perimeter (which is 4 times the side) grows at 4 times that rate - so 0.8 cm per second.

Exam Tip: Always identify the relationship between the quantities first (perimeter = 4 × side), then differentiate both sides with respect to time to connect their rates of change.

Question 2. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
Answer: Let the radius of the circle be \( r \).

\( \frac{dr}{dt} = 0.7 \text{ cm/s} \)

Circumference of the circle = \( 2\pi r \)

Rate of change of circumference = \( 2\pi\frac{dr}{dt} = 2 \times 3.14 \times 0.7 = 4.4 \text{ cm/s} \)
In simple words: The circumference equals 2π times the radius. So when the radius increases at 0.7 cm/s, the circumference increases at 2π times that rate.

Exam Tip: Remember the circumference formula is \( 2\pi r \), not \( \pi r \). This is a common mistake that costs marks.

Question 3. The radius of a circle is increasing uniformly at the rate of 0.3 centimetre per second. At what rate is the area increasing when the radius is 10 cm? (Take π = 3.14.)
Answer: Let the radius of the circle be \( r \).

\( \frac{dr}{dt} = 0.3 \text{ cm/s} \)

Area of the circle = \( \pi r^2 \)

Rate of change of area = \( 2\pi r\frac{dr}{dt} = 2 \times 3.14 \times 10 \times 0.3 = 18.84 \text{ cm}^2\text{/s} \)
In simple words: The area formula is π times r squared. To find how fast it grows, differentiate to get 2πr, then multiply by the rate at which r is changing.

Exam Tip: Always substitute the specific value of the variable (here, r = 10 cm) only after finding the derivative, not before.

Question 4. The side of a square sheet of metal is increasing at 3 centimetres per minute. At what rate is the area increasing when the side is 10 cm long?
Answer: Let the side of the square be \( a \).

Rate of change of side = \( \frac{da}{dt} = 3 \text{ cm/s} \)

Area of the square = \( a^2 \)

Rate of change of area = \( 2a\frac{da}{dt} = 2 \times 10 \times 3 = 60 \text{ cm}^2\text{/s} \)
In simple words: The area is side squared. When you differentiate, you get 2 times the side. Multiply this by how fast the side is changing to get the rate the area changes.

Exam Tip: For a square, area = side², so the derivative is 2 × side. For a circle, area = πr², so the derivative is 2πr. Know these standard forms cold.

Question 5. The radius of a circular soap bubble is increasing at the rate of 0.2 cm/s. Find the rate of increase of its surface area when the radius is 7 cm.
Answer: A soap bubble takes the shape of a sphere.

Let the radius of the soap bubble be \( r \).

\( \frac{dr}{dt} = 0.2 \text{ cm/s} \)

Surface area of the soap bubble = \( 4\pi r^2 \)

Rate of change of surface area = \( 8\pi r\frac{dr}{dt} = 8 \times 3.14 \times 7 \times 0.2 = 35.2 \text{ cm}^2\text{/s} \)
In simple words: A bubble is spherical, and the surface area of a sphere is 4π times r squared. Differentiate to get 8πr, then plug in the numbers.

Exam Tip: Sphere surface area is \( 4\pi r^2 \), not \( 4\pi r \). The factor 4π (not 2π as in a circle's circumference) is essential for full marks.

Question 6. The radius of an air bubble is increasing at the rate of 0.5 centimetre per second. At what rate is the volume of the bubble increasing when the radius is 1 centimetre?
Answer: A soap bubble takes the shape of a sphere.

Let the radius of the soap bubble be \( r \).

\( \frac{dr}{dt} = 0.5 \text{ cm/s} \)

Volume of the soap bubble = \( \frac{4}{3}\pi r^3 \)

Rate of change of volume = \( 4\pi r^2\frac{dr}{dt} = 4 \times 3.14 \times 1^2 \times 0.5 = 6.28 \text{ cm}^3\text{/s} \)
In simple words: The volume of a sphere is \( \frac{4}{3}\pi r^3 \). When you differentiate, the r³ becomes 4πr². Multiply by the rate the radius is changing.

Exam Tip: The volume formula is \( \frac{4}{3}\pi r^3 \), not \( \frac{4}{3}\pi r^2 \). Differentiating gives \( 4\pi r^2 \), which is exactly the surface area formula - a useful check.

Question 7. The volume of a spherical balloon is increasing at the rate of 25 cubic centimetres per second. Find the rate of change of its surface at the instant when its radius is 5 cm.
Answer: Let the radius of the balloon be \( r \).

Let the volume of the spherical balloon be \( V \).

\( V = \frac{4}{3}\pi r^3 \)

\( \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} \)

\( 25 = 4 \times \pi \times 5^2 \times \frac{dr}{dt} \)

\( \frac{dr}{dt} = \frac{1}{4\pi} \)

Surface area of the bubble = \( 4\pi r^2 \)

Rate of change of surface area = \( 8\pi r\frac{dr}{dt} = 8 \times \pi \times 5 \times \frac{1}{4\pi} = 10 \text{ cm}^2\text{/s} \)
In simple words: Here, you are given how fast the volume is changing. First, find how fast the radius is changing by using the volume formula. Then use that to find how fast the surface area is changing.

Exam Tip: When volume is given and you need surface area (or vice versa), always find the rate of change of the radius first as an intermediate step.

Question 8. A balloon which always remains spherical is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is 15 cm.
Answer: When we pump a balloon, its volume changes.

Let the radius of the balloon be \( r \).

\( V = \frac{4}{3}\pi r^3 \)

\( \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} \)

\( 900 = 4 \times \pi \times 15^2 \times \frac{dr}{dt} \)

\( \frac{dr}{dt} = \frac{900}{4 \times 3.14 \times 225} = 0.32 \text{ cm/s} \)
In simple words: The gas being pumped in increases the volume at 900 cm³ per second. Use the volume-radius relationship to find how quickly the radius is expanding.

Exam Tip: Here the volume rate is given; solve for the radius rate by rearranging the chain rule equation \( \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} \).

Question 9. The bottom of a rectangular swimming tank is 25 m by 40 m. Water is pumped into the tank at the rate of 500 cubic metres per minute. Find the rate at which the level of water in the tank is rising.
Answer: Let the volume of the water tank be \( V \).

\( V = l \times b \times h \)

\( V = 25 \times 40 \times h \)

\( \frac{dV}{dt} = 1000 \times \frac{dh}{dt} \)

\( 500 = 1000 \times \frac{dh}{dt} \)

\( \frac{dh}{dt} = 0.5 \text{ m/min} \)
In simple words: The tank has a fixed length and width. As water pours in, only the height changes. The volume formula becomes 1000 times the height, so the rate of height change equals the volume rate divided by 1000.

Exam Tip: For a rectangular container with fixed base, volume = base area × height, so \( \frac{dV}{dt} = \text{base area} \times \frac{dh}{dt} \). This is a simple and common pattern.

Question 10. A stone is dropped into a quiet lake and waves move in circles at a speed of 3.5 cm per second. At the instant when the radius of the circular wave is 7.5 cm. how fast is the enclosed area increasing? (Take π = 22/7.)
Answer: Let the radius of the circle be \( r \).

\( \frac{dr}{dt} = 3.5 \text{ cm/s} \)

Area of the circle = \( \pi r^2 \)

Rate of change of area = \( 2\pi r\frac{dr}{dt} = 2 \times \frac{22}{7} \times 7.5 \times 3.5 = 165 \text{ cm}^2\text{/s} \)
In simple words: The circular ripple grows outward at 3.5 cm per second. The area enclosed by the wave is πr². Differentiate to get the rate at which the area spreads outward.

Exam Tip: Always use the exact value of π given in the question (here 22/7, not 3.14) to avoid rounding errors.

Question 11. A 2-m tall man walks at a uniform speed of a uniform speed of 5 km per hour away from a 6-metre-high lamp post. Find the rate at which the length of his shadow increases.
Answer: Triangles ABE and CDE are similar.

So, \( \frac{AB}{BE} = \frac{CD}{DE} \)

\( \frac{6}{x+y} = \frac{2}{y} \)

\( 6y = 2(x+y) \)

\( 6y = 2x + 2y \)

\( 4y = 2x \)

\( y = \frac{x}{2} \)

\( \frac{dy}{dt} = \frac{1}{2}\frac{dx}{dt} \)

\( \frac{dy}{dt} = \frac{1}{2}(5) = 2.5 \text{ km/h} \)
In simple words: The lamp, man, and shadow tip form two similar triangles. From this similarity, the shadow length is half the distance from the lamp. So the shadow grows at half the man's walking speed.

Exam Tip: In shadow problems, always set up the similar triangle relationship first. The ratio of heights equals the ratio of distances from the base of the light source.

Question 12. An inverted cone has a depth of 40 cm and a base of radius 5 cm. Water is poured into it at a rate of 1.5 cubic centimetres per minute. Find the rate at which the level of water in the cone is rising when the depth is 4 cm.
Answer: Let the volume of the cone be \( V \).

\( \frac{dV}{dt} = 1.5 \text{ cm}^3\text{/s} \)

\( V = \frac{1}{3}\pi r^2 h \)

For the cone, \( \frac{r}{h} = \frac{5}{40} = \frac{1}{8} \), so \( r = \frac{h}{8} \)

\( V = \frac{1}{3}\pi\left(\frac{h}{8}\right)^2 h = \frac{1}{3}\pi\frac{h^3}{64} = \frac{\pi h^3}{192} \)

\( \frac{dV}{dt} = \frac{\pi}{192} \times 3h^2\frac{dh}{dt} = \frac{\pi h^2}{64}\frac{dh}{dt} \)

\( 1.5 = \frac{\pi \times 16}{64}\frac{dh}{dt} \)

\( \frac{dh}{dt} = \frac{1.5 \times 64}{\pi \times 16} = \frac{96}{16\pi} = \frac{6}{\pi} \text{ cm/min} \)
In simple words: In an inverted cone, the radius and height are proportional. Express the radius in terms of height, substitute into the volume formula, differentiate, and solve for the height's rate of change.

Exam Tip: For cone and funnel problems, always find the ratio of radius to height from the cone's fixed dimensions, then use it to eliminate one variable before differentiating.

Question 13. Sand is pouring from a pipe at the rate of 18 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is one-sixth of the radius of the base. How fast is the height of the sand cone increasing when its height is 3 cm?
Answer: Given: \( h = \frac{1}{6}r \), so \( r = 6h \)

\( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(6h)^2 h = \frac{1}{3}\pi \times 36h^3 = 12\pi h^3 \)

\( \frac{dV}{dt} = 36\pi h^2\frac{dh}{dt} \)

\( 18 = 36 \times 9 \times \pi \times \frac{dh}{dt} \)

\( \frac{dh}{dt} = \frac{18}{36 \times 9 \times \pi} = \frac{1}{18\pi} \text{ cm/s} \)
In simple words: The radius is always 6 times the height. Substitute this into the volume formula to express everything in terms of height only, then differentiate and solve.

Exam Tip: When a relationship between two dimensions is given (e.g., height = 1/6 of radius), use it immediately to reduce the number of variables in your formula.

Question 14. Water is dripping through a tiny hole at the vertex in the bottom of a conical funnel at a uniform rate of 4 cm³/s. When the slant height of the water is 3 cm, find the rate of decrease of the slant height of the water, given that the vertical angle of the funnel is 120°.
Answer: Let the volume of the cone be \( V \).

\( \frac{dV}{dt} = -4 \text{ cm}^3\text{/s} \) (negative because water is leaving)

\( V = \frac{1}{3}\pi r^2 h \)

From the vertical angle of 120°, we have \( \cos Q = \frac{h}{l} = \cos 120° = -\frac{1}{2} \) and \( \sin Q = \frac{r}{l} = \sin 120° = \frac{\sqrt{3}}{2} \)

where \( l \) is the slant height.

\( V = \frac{1}{3}\pi\left(\frac{\sqrt{3}}{2}l\right)^2\left(-\frac{1}{2}l\right) = -\frac{3}{24}\pi l^3 = -\frac{\pi l^3}{8} \)

\( \frac{dV}{dt} = -\frac{3\pi l^2}{8}\frac{dl}{dt} \)

\( -4 = -\frac{3\pi \times 9}{8}\frac{dl}{dt} \)

\( \frac{dl}{dt} = \frac{32}{27\pi} \text{ cm/s} \)
In simple words: The cone's angle tells you the relationship between radius and height in terms of the slant height. Use the angle values to express volume in terms of slant height only, differentiate, and solve.

Exam Tip: In funnel problems with a given angle, use trigonometry to relate the slant height to the radius and vertical height before writing the volume formula.

Question 15. Oil is leaking at the rate of 15 mL/s from a vertically kept cylindrical drum containing oil. If the radius of the drum is 7 cm and its height is 60 cm, find the rate at which the level of the oil is changing when the oil level is 18 cm.
Answer: \( \frac{dV}{dt} = -15 \text{ mL/s} \) (negative because oil is leaking out)

\( \frac{d}{dt}(\pi r^2 h) = -15 \)

\( \frac{d}{dt}(\pi \times 7^2 \times h) = -15 \)

\( 49\pi\frac{dh}{dt} = -15 \)

\( \frac{dh}{dt} = \frac{-15}{49\pi} \text{ cm/s} \)
In simple words: For a cylinder with fixed radius, the volume is the base area times height. So the rate of volume change equals the base area times the rate of height change. Divide to find the height rate.

Exam Tip: When a cylindrical container has a fixed radius, the problem simplifies to volume = constant × height. The height's rate is the volume rate divided by the base area.

Question 16. A 13-m long ladder is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 m/s. How fast is its height on the wall decreasing when the foot of the ladder is 5 m away from the wall?
Answer: Let AB = y and BC = x.

Applying Pythagoras Theorem in triangle ABC:

\( x^2 + y^2 = 13^2 = 169 \) ... (1)

When x = 5:

\( 5^2 + y^2 = 13^2 \)

\( y = 12 \text{ cm} \)

Differentiating both sides of equation (1) with respect to \( t \):

\( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \)

\( 5.2 + 12\frac{dy}{dt} = 0 \)

\( \frac{dy}{dt} = -\frac{10}{12} = -\frac{5}{6} \text{ cm/s} \)
In simple words: The ladder's length stays fixed at 13 m. Use Pythagoras to relate the distance from the wall to the height on the wall. Differentiate this relationship to find how fast the height is changing.

Exam Tip: In ladder problems, always use \( x^2 + y^2 = L^2 \) where L is the ladder length. Differentiating gives the key relationship between the two rates.

Question 17. A man is moving away from a 40-m high tower at a speed of 2 m/s. Find the rate is which the angle of elevation of the top of the tower is changing when he is at a distance of 30 metres from the foot of the tower. Assume that the eye level of the man is 1.6 m from the ground.
Answer: The height of the tower above the man's eye level is \( 40 - 1.6 = 38.4 \) m.

\( \tan Q = \frac{38.4}{x} \)

\( Q = \tan^{-1}\left(\frac{38.4}{x}\right) \)

\( \frac{dQ}{dt} = \frac{1}{1 + \left(\frac{38.4}{x}\right)^2} \times \left(-\frac{1}{x^2}\right) \times 38.4 \)

\( \frac{dQ}{dt} = \frac{x^2}{x^2 + 1474.56} \times \left(-\frac{1}{x^2}\right) \times 38.4 \)

\( \frac{dQ}{dt} = -\frac{1}{30^2 + 1474.56} \times 38.4 \times 2 \)

\( \frac{dQ}{dt} = -0.032 \text{ radian/second} \)
In simple words: The angle of elevation relates the height and horizontal distance. As the man moves away, this angle decreases. Use the inverse tangent relationship and the chain rule to find the rate of angle change.

Exam Tip: Always subtract the eye level from the tower height to get the true vertical distance for calculating angle of elevation.

Question 18. Find an angle x which increases twice as fast as its sine.
Answer: According to the question:

\( \frac{dx}{dt} = 2\frac{d}{dt}(\sin x) \)

\( \frac{dx}{dt} = 2\cos x\frac{dx}{dt} \)

\( 1 = 2\cos x \)

\( \cos x = \frac{1}{2} \)

\( x = \frac{\pi}{3} \)
In simple words: If the angle increases twice as fast as its sine, then the derivative of the angle equals twice the derivative of the sine. This gives you an equation to solve for the angle itself.

Exam Tip: Cancel \( \frac{dx}{dt} \) from both sides (assuming it is not zero) to isolate the trigonometric equation.

Question 19. The radius of a balloon is increasing at the rate of 10 m/s. At what rate is the surface area of the balloon increasing when the radius is 15 cm?
Answer: \( \frac{dr}{dt} = 10 \text{ m/s} \)

\( S = 4\pi r^2 \)

\( \frac{dS}{dt} = 8\pi r\frac{dr}{dt} \)

\( \frac{dS}{dt} = 8\pi \times 15 \times 10 = 1200\pi \text{ cm}^2\text{/s} \)
In simple words: The surface area of a sphere is 4π times r squared. Differentiate with respect to time to get 8πr times the rate the radius is changing.

Exam Tip: Watch your units - if radius is given in cm and rate in m/s, convert one of them before substituting to avoid errors.

Question 20. An edge of a variable cube is increasing at the rate of 5 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Answer: \( \frac{da}{dt} = 5 \text{ cm/s} \)

\( V = a^3 \)

\( \frac{dV}{dt} = 3a^2\frac{da}{dt} \)

\( \frac{dV}{dt} = 3 \times 10^2 \times 5 = 1500 \text{ cm}^3\text{/s} \)
In simple words: The volume of a cube is the edge cubed. Differentiate to get 3 times the edge squared, then multiply by the rate the edge is changing.

Exam Tip: For a cube with edge a, volume = a³, so the derivative is 3a². This is fundamental; always use it without hesitation.

Question 21. The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. Find the rate at which the area is increasing when the side is 10 cm.
Answer: \( \frac{da}{dt} = 2 \text{ cm/s} \)

For an equilateral triangle: \( A = \frac{\sqrt{3}}{4}a^2 \)

\( \frac{dA}{dt} = \frac{\sqrt{3}}{4} \times 2a\frac{da}{dt} = \frac{\sqrt{3}}{2}a\frac{da}{dt} \)

\( \frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 10 \times 2 = 10\sqrt{3} \text{ cm}^2\text{/s} \)
In simple words: The area of an equilateral triangle depends on the side length through the formula involving √3. Differentiate this formula and substitute the given values.

Exam Tip: Memorize the equilateral triangle area formula: \( A = \frac{\sqrt{3}}{4}a^2 \). Its derivative is \( \frac{\sqrt{3}}{2}a \).

Question 1. Using differentials, find the approximate values of: \( \sqrt{37} \)
Answer: Let \( y = \sqrt{x} \).

Let \( x = 36 \) and \( \Delta x = 1 \).

As \( y = \sqrt{x} \):

\( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \)

We know:

\( \Delta y = \frac{dy}{dx}\Delta x \)

\( \Delta y = \frac{1}{2\sqrt{x}} \cdot \Delta x = \frac{1}{2\sqrt{36}} \cdot 1 = \frac{1}{12} \)

\( \Delta y = 0.08 \)

Also, \( \Delta y = f(x + \Delta x) - f(x) \):

\( 0.08 = \sqrt{37} - 6 \)

\( \sqrt{37} = 6.08 \)
In simple words: Choose a perfect square close to 37, which is 36. Find how much the square root function changes when x moves from 36 to 37 using the derivative. Add this change to √36 to get an approximation for √37.

Exam Tip: Always pick a value of x where you know the function value exactly - perfect squares for square roots, perfect cubes for cube roots, etc.

Question 2. Using differentials, find the approximate values of: \( \sqrt[3]{29} \)
Answer: Let \( y = \sqrt[3]{x} \).

Let \( x = 27 \) and \( \Delta x = 2 \).

As \( y = \sqrt[3]{x} \):

\( \frac{dy}{dx} = \frac{1}{3}x^{-\frac{2}{3}} \)

We know:

\( \Delta y = \frac{dy}{dx}\Delta x \)

\( \Delta y = \frac{1}{3} \times 27^{-\frac{2}{3}} \times 2 = \frac{1}{3} \times \frac{1}{9} \times 2 = \frac{2}{27} \)

\( \Delta y = 0.074 \)

Also, \( \Delta y = f(x + \Delta x) - f(x) \):

\( 0.074 = \sqrt[3]{29} - 3 \)

\( \sqrt[3]{29} = 3.074 \)
In simple words: Pick 27 as the starting point since \( \sqrt[3]{27} = 3 \). Use the derivative of the cube root function to estimate the change when moving to 29.

Exam Tip: For cube roots, the derivative is \( \frac{1}{3}x^{-\frac{2}{3}} \), which equals \( \frac{1}{3(x^{2/3})} \). Simplify before substituting numbers.

Question 3. Using differentials, find the approximate values of: \( \sqrt{27} \)
Answer: Let \( y = \sqrt{x} \).

Let \( x = 25 \) and \( \Delta x = 2 \).

As \( y = \sqrt{x} \):

\( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \)

We know:

\( \Delta y = \frac{dy}{dx}\Delta x \)

\( \Delta y = \frac{1}{2\sqrt{25}} \times 2 = \frac{1}{5} = 0.2 \)

Also, \( \Delta y = f(x + \Delta x) - f(x) \):

\( 0.2 = \sqrt{27} - 5 \)

\( \sqrt{27} = 5.2 \)
In simple words: Use 25 as the base since √25 = 5 is easy. The derivative shows that moving 2 units to the right increases the square root by about 0.2.

Exam Tip: The closer your chosen perfect square is to the target, the more accurate the approximation. Always pick the nearest perfect square.

Question 4. Using differentials, find the approximate values of: \( \sqrt{0.24} \)
Answer: Let \( y = \sqrt{x} \).

Let \( x = 0.25 \) and \( \Delta x = -0.01 \).

As \( y = \sqrt{x} \):

\( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \)

We know:

\( \Delta y = \frac{dy}{dx}\Delta x \)

\( \Delta y = \frac{1}{2\sqrt{0.25}} \times (-0.01) = -0.01 \)

Also, \( \Delta y = f(x + \Delta x) - f(x) \):

\( -0.01 = \sqrt{0.24} - 0.5 \)

\( \sqrt{0.24} = 0.49 \)
In simple words: Here √0.25 = 0.5 is exact. Decreasing x by 0.01 decreases the square root by about 0.01, giving an approximation for √0.24.

Exam Tip: The method works equally well with decimals. Choose 0.25 rather than trying to find a decimal perfect square closer to 0.24.

Question 5. Using differentials, find the approximate values of: \( \sqrt{49.5} \)
Answer: Let \( y = \sqrt{x} \).

Let \( x = 49 \) and \( \Delta x = 0.5 \).

As \( y = \sqrt{x} \):

\( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \)

We know:

\( \Delta y = \frac{dy}{dx}\Delta x \)

\( \Delta y = \frac{1}{2\sqrt{49}} \times 0.5 = \frac{1}{2 \times 7} \times 0.5 = \frac{1}{28} = 0.0357 \)

Also, \( \Delta y = f(x + \Delta x) - f(x) \):

\( 0.0357 = \sqrt{49.5} - 7 \)

\( \sqrt{49.5} = 7.0357 \)
In simple words: Start at 49 where the square root is exactly 7. Move 0.5 units right by applying the derivative to get the small change in the square root value.

Exam Tip: For values very close to perfect squares (small Δx), the differential approximation is exceptionally accurate.

Question 6. Using differentials, find the approximate values of: \( \sqrt[4]{15} \)
Answer: Let \( y = \sqrt[4]{x} = x^{1/4} \).

Let \( x = 16 \) and \( \Delta x = -1 \).

As \( y = x^{1/4} \):

\( \frac{dy}{dx} = \frac{1}{4}x^{-3/4} \)

We know:

\( \Delta y = \frac{dy}{dx}\Delta x \)

\( \Delta y = \frac{1}{4} \times 16^{-3/4} \times (-1) = \frac{1}{4} \times \frac{1}{8} \times (-1) = -\frac{1}{32} = -0.03125 \)

Also, \( \Delta y = f(x + \Delta x) - f(x) \):

\( -0.03125 = \sqrt[4]{15} - 2 \)

\( \sqrt[4]{15} = 1.96875 \)
In simple words: The fourth root of 16 is 2 (since 2⁴ = 16). Use the derivative of the fourth root function to estimate how much the function changes when x decreases by 1.

Exam Tip: For fractional powers, rewrite the function and apply the power rule. The derivative of \( x^{1/4} \) is \( \frac{1}{4}x^{-3/4} \).

Question 7. Using differentials, find the approximate values of: \( \frac{1}{(2.002)^2} \)
Answer: Let \( y = \frac{1}{x^2} \).

Let \( x = 2 \) and \( \Delta x = 0.002 \).

As \( y = \frac{1}{x^2} = x^{-2} \):

\( \frac{dy}{dx} = -2x^{-3} \)

We know:

\( \Delta y = \frac{dy}{dx}\Delta x \)

\( \Delta y = -2 \times 2^{-3} \times 0.002 = -\frac{2}{8} \times 0.002 = -0.0005 \)

Also, \( \Delta y = f(x + \Delta x) - f(x) \):

\( -0.0005 = \frac{1}{(2.002)^2} - 0.25 \)

\( \frac{1}{(2.002)^2} = 0.2495 \)
In simple words: The reciprocal of x squared has the derivative -2/x³. When x increases slightly from 2, this function decreases slightly from 0.25.

Exam Tip: Rewrite reciprocal and division functions as negative powers before differentiating - it is cleaner and less error-prone.

Question 8. Using differentials, find the approximate values of: log_e(10.02), given that log_e(10) = 2.3026
Answer: Let \( y = \log_e x \).

Let \( x = 10 \) and \( \Delta x = 0.02 \).

As \( y = \log_e x \):

\( \frac{dy}{dx} = \frac{1}{x} \)

We know:

\( \Delta y = \frac{dy}{dx}\Delta x \)

\( \Delta y = \frac{1}{10} \times 0.02 = 0.002 \)

Also, \( \Delta y = f(x + \Delta x) - f(x) \):

\( 0.002 = \log_e(10.02) - 2.3026 \)

\( \log_e(10.02) = 2.3046 \)
In simple words: The natural logarithm function has derivative 1/x. A small change in x produces a change proportional to 1/x applied to that small change.

Exam Tip: For logarithm problems, remember that d/dx(log_e x) = 1/x. This is one of the most commonly used derivatives in approximation problems.

Question 9. Using differentials, find the approximate values of: log₁₀(4.04), it being given that log₁₀(4) = 0.6021 and log₁₀(e) = 0.4343
Answer: Let \( y = \log_{10} x = 0.4343 \log_e x \).

\( \frac{dy}{dx} = \frac{0.4343}{x} \)

Let \( x = 4 \) and \( \Delta x = 0.04 \).

As \( y = 0.4343 \log_e x \):

\( \frac{dy}{dx} = \frac{0.4343}{x} \)

We know:

\( \Delta y = \frac{dy}{dx}\Delta x \)

\( \Delta y = \frac{0.4343}{4} \times 0.04 = 0.004343 \)

Also, \( \Delta y = f(x + \Delta x) - f(x) \):

\( 0.004343 = \log_{10}(4.04) - 0.6021 \)

\( \log_{10}(4.04) = 0.606443 \)
In simple words: Base-10 logarithms can be converted to natural logarithms by multiplying by 0.4343. Use this conversion, then apply the standard differential approximation.

Exam Tip: The conversion constant is log₁₀(e) = 0.4343. This relates base-10 and natural logarithm derivatives.

Question 10. Using differentials, find the approximate values of: cos(61°), it being given that sin(60°) = 0.86603 and 1° = 0.01745 radian
Answer: Let \( y = \cos x \).

Let \( x = 60° \) and \( \Delta x = 1° \).

As \( y = \cos x \):

\( \frac{dy}{dx} = -\sin x \)

We know:

\( \Delta y = \frac{dy}{dx}\Delta x \)

\( \Delta y = -\sin(60°) \times 1° = -0.86603 \times 0.01745 = -0.0151 \)

Also, \( \Delta y = f(x + \Delta x) - f(x) \):

\( -0.0151 = \cos(61°) - \cos(60°) = \cos(61°) - 0.5 \)

\( \cos(61°) = 0.4849 \)
In simple words: The derivative of cosine is negative sine. At 60°, sine is 0.86603, so cosine decreases by about that amount (times the angle change in radians) as you move from 60° to 61°.

Exam Tip: Always convert angle changes to radians before multiplying by the derivative. Here, 1° = 0.01745 rad, not 1 rad.

Question 11. If y = sin x and x changes from π/2 to 22/14, what is the approximate change in y?
Answer: Given that x is π/2. The value of π is 22/7, so 22/14 equals π/2. Therefore, there will be no change in y.
In simple words: When x stays the same, y doesn't change. The two values given (π/2 and 22/14) are actually equal, so nothing shifts.

Exam Tip: Always verify that the initial and final values are truly different before calculating changes - sometimes the problem is testing whether you notice they are identical.

Question 12. A circular metal plate expands under heating so that its radius increases by 2%. Find the approximate increase in the area of the plate, if the radius of the plate before heating is 10 cm.
Answer: The radius of the plate starts at 10 cm. It increases by 2% due to heating.

Increment in radius: \( \frac{2}{100} \times 10 = 0.2 \) cm

Therefore, dr = 0.2

New radius = 10 + 0.2 = 10.2 cm

For a circular plate, Area = A = πr²

The change in area is found using: \( dA = \frac{dA}{dr} \times dr \)

\( \frac{dA}{dr} = 2\pi r \)

\( dA = 2\pi r \times dr \)

\( dA = 2 \times \pi \times 10 \times 0.2 \)

\( dA = 4\pi \text{ cm}^2 \)

In simple words: The area grows by an amount equal to 4π square centimetres. This is roughly 12.57 cm².

Exam Tip: Use the derivative formula dA/dr = 2πr to find the rate of change, then multiply by the small change in radius (dr) to get the approximate change in area.

Question 13. If the length of a simple pendulum is decreased by 2%, find the percentage decrease in its period T, where \( T = 2\pi\sqrt{\frac{l}{g}} \).
Answer: The time period is given by: \( T = 2\pi\sqrt{\frac{l}{g}} \)

Since 2, π, and g are constants, we can ignore them when finding relative error.

Now: \( \frac{\Delta T}{T} = \frac{1}{2} \times \frac{\Delta l}{l} \)

Converting to percentage error:

\( \frac{\Delta T}{T} \times 100\% = \frac{1}{2} \times \frac{\Delta l}{l} \times 100\% \)

\( \frac{\Delta T}{T} \times 100\% = \frac{1}{2} \times (-2)\% \)

\( \frac{\Delta T}{T} \% = 1\% \)

Therefore, the time period drops by 1%.

In simple words: When the pendulum length shrinks by 2%, the period (time to swing) decreases by half that amount - which is 1%.

Exam Tip: For percentage errors in functions, use the derivative to find the relative error relationship between variables, then substitute the percentage change values directly.

Question 14. The pressure p and the volume V of a gas are connected by the relation, \( p V^{1/4} = k \), where k is a constant. Find the percentage increase in the pressure, corresponding to a diminution of 0.5% in the volume.
Answer: Given: \( pv^{1/4} = k \)

Decrease in volume = 0.5%

\( \frac{\Delta V}{V} \times 100 = -\frac{1}{2}\% \)

From the relation \( pv^{1/4} = k \), taking the natural logarithm of both sides:

\( \log[pv^{1/4}] = \log k \)

\( \log P + 1.4\log V = \log k \)

Differentiating both sides:

\( \frac{1}{p}dp + 1.4\frac{1}{V}dV = 0 \)

\( \frac{dp}{p} = -1.4\frac{dV}{V} \)

Multiplying both sides by 100:

\( \frac{dp}{p} \times 100 = -1.4 \times \frac{dV}{V} \times 100 \)

\( \frac{dp}{p} \times 100 = -1.4 \times \left(\frac{-1}{2}\right) \)

\( \frac{dp}{p} \times 100 = 0.7 \)

Percentage error in P = 0.7%

In simple words: When the gas volume shrinks by 0.5%, the pressure rises by 0.7% to keep the product pV^(1/4) constant.

Exam Tip: Always take logarithms first for relations involving products or powers - this converts multiplication to addition, making differentiation straightforward.

Question 15. The radius of a sphere shrinks from 10 cm to 9.8 cm. Find approximately the decrease in (i) volume, and (ii) surface area.
Answer: The change in radius = dr = 10 - 9.8 = 0.2 cm

(i) Volume of the sphere:

Volume is given by: \( V = \frac{4}{3}\pi r^3 \)

Change in volume: \( dV = 4\pi r^2 \times dr \)

\( dV = 4\pi(10)^2 \times 0.2 \)

\( dV = 80\pi \text{ cm}^3 \)

(ii) Surface area of the sphere:

Surface area is given by: \( A = 4\pi r^2 \)

Change in area: \( dA = 8\pi r \times dr \)

\( dA = 8\pi \times 10 \times 0.2 \)

\( dA = 16\pi \text{ cm}^2 \)

In simple words: As the radius shrinks by 0.2 cm, the volume falls by 80π cm³ (about 251 cm³) and the surface area drops by 16π cm² (about 50 cm²).

Exam Tip: For volume and surface area changes, use dV = 4πr² dr and dA = 8πr dr - these come directly from differentiating the standard formulas.

Question 16. If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere.
Answer: Volume of the sphere is given by: \( V = \frac{4}{3}\pi r^3 \)

Change in volume: \( dV = 4\pi r^2 \times dr \)

Given: Δr = 0.1

\( \Delta r \cdot \frac{dV}{dr} = 4\pi r^2 \Delta r \)

\( \Delta V = 4\pi r^2 \Delta r \)

Percentage error:

\( \frac{\Delta V}{V} = \frac{4\pi r^2 \Delta r}{\frac{4}{3}\pi r^3} \times 0.1 \)

\( = 0.3\% \)

In simple words: A measurement error of 0.1% in the radius causes an error three times larger in volume - about 0.3%.

Exam Tip: For volume, a percentage error in radius gets multiplied by 3 because volume depends on r³. For surface area, it gets multiplied by 2 (since area depends on r²).

Question 17. Show that the relative error in the volume of a sphere, due to an error in measuring the diameter, is three times the relative error in the diameter.
Answer: Let d be the diameter, r be the radius, and V be the volume of the sphere.

Volume of the sphere is given by: \( V = \frac{4}{3}\pi r^3 \)

Since \( r = \frac{d}{2} \), we have: \( V = \frac{4}{3}\pi \left(\frac{d}{2}\right)^3 = \frac{1}{6}\pi d^3 \)

Let Δd be the error in d and the corresponding error in V be ΔV.

\( \Delta V = \frac{dV}{dd} \times \Delta d = \frac{1}{2}\pi d^2 \times \Delta d \)

Taking the ratio:

\( \frac{\Delta V}{V} = \frac{\frac{1}{2}\pi d^2 \Delta d}{\frac{1}{6}\pi d^3} = 3\frac{\Delta d}{d} \)

Hence Proved - The relative error in volume equals three times the relative error in diameter.

In simple words: If your measurement of diameter is off by a small percentage, the calculated volume will be off by three times that percentage.

Exam Tip: Always express the volume in terms of the variable being measured (here, d rather than r), then differentiate - this makes the relationship between errors transparent.

Exercise 11C

Question 1. Verify Rolle's theorem for each of the following functions: \( f(x) = x^2 \) on \( [-1,1] \)
Answer:
Condition (1):
Since f(x) = x² is a polynomial function, and all polynomial functions are continuous everywhere on the real line.

⇒ f(x) = x² is continuous on [-1,1].

Condition (2):
Here, f'(x) = 2x, which exists in [-1,1].

So, f(x) = x² is differentiable on (-1,1).

Condition (3):
Here, f(-1) = (-1)² = 1

And f(1) = 1² = 1

i.e. f(-1) = f(1)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (-1,1) such that f'(c) = 0

i.e. 2c = 0

i.e. c = 0

Since c = 0 ∈ (-1,1), Rolle's theorem is verified.

In simple words: The function is continuous and smooth on the interval. It has the same value at both endpoints. Somewhere between the endpoints (here, at x = 0), the slope must be zero.

Exam Tip: Always verify all three conditions explicitly - continuity on the closed interval, differentiability on the open interval, and equal endpoint values - before claiming Rolle's theorem applies.

Question 2. Verify Rolle's theorem for each of the following functions: \( f(x) = x^2 - x - 12 \) in \( [-3,4] \)
Answer:
Condition (1):
Since f(x) = x² - x - 12 is a polynomial function, and every polynomial function is continuous everywhere.

⇒ f(x) = x² - x - 12 is continuous on [-3,4].

Condition (2):
Here, f'(x) = 2x - 1, which exists in [-3,4].

So, f(x) = x² - x - 12 is differentiable on (-3,4).

Condition (3):
Here, f(-3) = (-3)² - (-3) - 12 = 9 + 3 - 12 = 0

And f(4) = 4² - 4 - 12 = 16 - 4 - 12 = 0

i.e. f(-3) = f(4)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (-3,4) such that f'(c) = 0

i.e. 2c - 1 = 0

i.e. c = 1/2

Since c = 1/2 ∈ (-3,4), Rolle's theorem is verified.

In simple words: The function touches zero at both endpoints. Because it's smooth, there must be at least one spot in between where the curve is perfectly flat (where the derivative is zero).

Exam Tip: Substitute the endpoint values carefully - small arithmetic errors here will invalidate the entire proof.

Question 3. Verify Rolle's theorem for each of the following functions: \( f(x) = x^2 - 5x + 6 \) in \( [2,3] \)
Answer:
Condition (1):
Since f(x) = x² - 5x + 6 is a polynomial function, every polynomial function is continuous everywhere.

⇒ f(x) = x² - 5x + 6 is continuous on [2,3].

Condition (2):
Here, f'(x) = 2x - 5, which exists in [2,3].

So, f(x) = x² - 5x + 6 is differentiable on (2,3).

Condition (3):
Here, f(2) = 2² - 5(2) + 6 = 4 - 10 + 6 = 0

And f(3) = 3² - 5(3) + 6 = 9 - 15 + 6 = 0

i.e. f(2) = f(3)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (2,3) such that f'(c) = 0

i.e. 2c - 5 = 0

i.e. c = 5/2

Since c = 5/2 ∈ (2,3), Rolle's theorem is verified.

In simple words: Both endpoints give zero - the parabola crosses the x-axis at x = 2 and x = 3. Between these points, the curve must have a peak or valley where the slope is zero.

Exam Tip: Recognize factorable polynomials quickly - here, x² - 5x + 6 = (x - 2)(x - 3), so the endpoints make sense immediately.

Question 4. Verify Rolle's theorem for each of the following functions: \( f(x) = x^2 - 5x + 6 \) in \( [-3,6] \)
Answer:
Condition (1):
Since f(x) = x² - 5x + 6 is a polynomial function, every polynomial function is continuous everywhere.

⇒ f(x) = x² - 5x + 6 is continuous on [-3,6].

Condition (2):
Here, f'(x) = 2x - 5, which exists in [-3,6].

So, f(x) = x² - 5x + 6 is differentiable on (-3,6).

Condition (3):
Here, f(-3) = (-3)² - 5(-3) + 6 = 9 + 15 + 6 = 30

And f(6) = 6² - 5(6) + 6 = 36 - 30 + 6 = 12

i.e. f(-3) ≠ f(6)

Condition (3) of Rolle's theorem is not satisfied.

Therefore, Rolle's theorem is not applicable to this function on this interval.

In simple words: The function has different values at the two endpoints - the left endpoint gives 30 and the right gives 12. When endpoint values are unequal, Rolle's theorem no longer guarantees a flat spot inside.

Exam Tip: Rolle's theorem requires all three conditions - failing even one means the theorem cannot be applied, regardless of whether a flat point might still exist.

Question 5. Verify Rolle's theorem for each of the following functions: \( f(x) = x^2 - 4x + 3 \) in \( [1,3] \)
Answer:
Condition (1):
Since f(x) = x² - 4x + 3 is a polynomial function, every polynomial function is continuous everywhere.

⇒ f(x) = x² - 4x + 3 is continuous on [1,3].

Condition (2):
Here, f'(x) = 2x - 4, which exists in [1,3].

So, f(x) = x² - 4x + 3 is differentiable on (1,3).

Condition (3):
Here, f(1) = 1² - 4(1) + 3 = 1 - 4 + 3 = 0

And f(3) = 3² - 4(3) + 3 = 9 - 12 + 3 = 0

i.e. f(1) = f(3)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (1,3) such that f'(c) = 0

i.e. 2c - 4 = 0

i.e. c = 2

Since c = 2 ∈ (1,3), Rolle's theorem is verified.

In simple words: The quadratic touches zero at both x = 1 and x = 3. The vertex (where slope is zero) lies at x = 2, which is exactly between the endpoints.

Exam Tip: For a quadratic with equal endpoint values, the point where f'(c) = 0 is always at the average of the two endpoint x-coordinates.

Question 6. Verify Rolle's theorem for each of the following functions: \( f(x) = x(x-4)^2 \) in \( [0,4] \)
Answer:
Condition (1):
Since f(x) = x(x - 4)² is a polynomial function, every polynomial function is continuous everywhere.

⇒ f(x) = x(x - 4)² is continuous on [0,4].

Condition (2):
Here, f'(x) = (x - 4)² + 2x(x - 4), which exists in [0,4].

So, f(x) = x(x - 4)² is differentiable on (0,4).

Condition (3):
Here, f(0) = 0(0 - 4)² = 0

And f(4) = 4(4 - 4)² = 0

i.e. f(0) = f(4)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (0,4) such that f'(c) = 0

i.e. (c - 4)² + 2c(c - 4) = 0

i.e. (c - 4)[(c - 4) + 2c] = 0

i.e. (c - 4)(3c - 4) = 0

i.e. c = 4 or c = 4/3

Since c = 4/3 ∈ (0,4), Rolle's theorem is verified.

In simple words: This function equals zero at both endpoints. Its derivative is zero at two points: one is the endpoint itself, and the other (4/3) lies strictly inside the interval.

Exam Tip: When derivative gives multiple solutions, check which ones lie strictly inside the open interval (a, b) - that is the point that satisfies Rolle's theorem.

Question 7. Verify Rolle's theorem for each of the following functions: \( f(x) = x^3 - 7x^2 + 16x - 12 \) in \( [2,3] \)
Answer:
Condition (1):
Since f(x) = x³ - 7x² + 16x - 12 is a polynomial function, every polynomial function is continuous everywhere.

⇒ f(x) = x³ - 7x² + 16x - 12 is continuous on [2,3].

Condition (2):
Here, f'(x) = 3x² - 14x + 16, which exists in [2,3].

So, f(x) = x³ - 7x² + 16x - 12 is differentiable on (2,3).

Condition (3):
Here, f(2) = 2³ - 7(2)² + 16(2) - 12 = 8 - 28 + 32 - 12 = 0

And f(3) = 3³ - 7(3)² + 16(3) - 12 = 27 - 63 + 48 - 12 = 0

i.e. f(2) = f(3)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (2,3) such that f'(c) = 0

i.e. 3c² - 14c + 16 = 0

i.e. (c - 2)(3c - 8) = 0

i.e. c = 2 or c = 8/3

Since c = 8/3 ≈ 2.67 ∈ (2,3), Rolle's theorem is verified.

In simple words: Both endpoints give zero for this cubic. The derivative is zero at 8/3, which lies inside the interval.

Exam Tip: For cubic and higher-degree polynomials, factor the derivative carefully - one root may be an endpoint and another inside the interval.

Question 8. Verify Rolle's theorem for each of the following functions: \( f(x) = x^3 + 3x^2 - 24x - 80 \) in \( [-4,5] \)
Answer:
Condition (1):
Since f(x) = x³ + 3x² - 24x - 80 is a polynomial function, every polynomial function is continuous everywhere.

⇒ f(x) = x³ + 3x² - 24x - 80 is continuous on [-4,5].

Condition (2):
Here, f'(x) = 3x² + 6x - 24, which exists in [-4,5].

So, f(x) = x³ + 3x² - 24x - 80 is differentiable on (-4,5).

Condition (3):
Here, f(-4) = (-4)³ + 3(-4)² - 24(-4) - 80 = -64 + 48 + 96 - 80 = 0

And f(5) = 5³ + 3(5)² - 24(5) - 80 = 125 + 75 - 120 - 80 = 0

i.e. f(-4) = f(5)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (-4,5) such that f'(c) = 0

i.e. 3c² + 6c - 24 = 0

i.e. c² + 2c - 8 = 0

i.e. (c + 4)(c - 2) = 0

i.e. c = -4 or c = 2

Since c = 2 ∈ (-4,5), Rolle's theorem is verified.

In simple words: This cubic function equals zero at -4 and at 5. Its slope is zero at x = 2, which lies strictly between these endpoints.

Exam Tip: Always simplify the derivative equation (divide through by the leading coefficient) before factoring - it makes the algebra cleaner.

Question 9. Verify Rolle's theorem for each of the following functions: \( f(x) = (x-1)(x-2)(x-3) \) in \( [1,3] \)
Answer:
Condition (1):
Since f(x) = (x - 1)(x - 2)(x - 3) is a polynomial function, every polynomial function is continuous everywhere.

⇒ f(x) = (x - 1)(x - 2)(x - 3) is continuous on [1,3].

Condition (2):
Here, f'(x) = (x - 2)(x - 3) + (x - 1)(x - 3) + (x - 1)(x - 2), which exists in [1,3].

So, f(x) = (x - 1)(x - 2)(x - 3) is differentiable on (1,3).

Condition (3):
Here, f(1) = (1 - 1)(1 - 2)(1 - 3) = 0

And f(3) = (3 - 1)(3 - 2)(3 - 3) = 0

i.e. f(1) = f(3)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (1,3) such that f'(c) = 0

i.e. (c - 2)(c - 3) + (c - 1)(c - 3) + (c - 1)(c - 2) = 0

Expanding and simplifying:

i.e. 3c² - 12c + 11 = 0

i.e. c = (12 ± √(144 - 132))/6 = (12 ± √12)/6

i.e. c = 2.58 or c = 1.42

Since both c = 1.42 ∈ (1,3) and c = 2.58 ∈ (1,3), Rolle's theorem is verified.

In simple words: This cubic has roots at 1, 2, and 3. Between the endpoints 1 and 3, the slope is zero at two interior points - around 1.42 and 2.58.

Exam Tip: When the derivative is a quadratic (as here), it can yield two points where the slope is zero - both should be checked for inclusion in the open interval.

Question 10. Verify Rolle's theorem for each of the following functions: \( f(x) = (x-1)(x-2)^2 \) in \( [1,2] \)
Answer:
Condition (1):
Since f(x) = (x - 1)(x - 2)² is a polynomial function, every polynomial function is continuous everywhere.

⇒ f(x) = (x - 1)(x - 2)² is continuous on [1,2].

Condition (2):
Here, f'(x) = (x - 2)² + 2(x - 1)(x - 2), which exists in [1,2].

So, f(x) = (x - 1)(x - 2)² is differentiable on (1,2).

Condition (3):
Here, f(1) = (1 - 1)(1 - 2)² = 0

And f(2) = (2 - 1)(2 - 2)² = 0

i.e. f(1) = f(2)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (1,2) such that f'(c) = 0

i.e. (c - 2)² + 2(c - 1)(c - 2) = 0

i.e. (c - 2)[(c - 2) + 2(c - 1)] = 0

i.e. (3c - 4)(c - 2) = 0

i.e. c = 4/3 or c = 2

Since c = 4/3 ≈ 1.33 ∈ (1,2), Rolle's theorem is verified.

In simple words: Both endpoints give zero. The slope is zero at 4/3, which lies strictly inside the interval (1, 2).

Exam Tip: For functions with repeated factors, the derivative will also be zero at that repeated root - but check whether it's inside or outside the interval.

Question 11. Verify Rolle's theorem for each of the following functions: \( f(x) = (x-2)^4(x-3)^3 \) in \( [2,3] \)
Answer:
Condition (1):
Since f(x) = (x - 2)⁴(x - 3)³ is a polynomial function, every polynomial function is continuous everywhere.

⇒ f(x) = (x - 2)⁴(x - 3)³ is continuous on [2,3].

Condition (2):
Here, f'(x) = 4(x - 2)³(x - 3)³ + 3(x - 2)⁴(x - 3)², which exists in [2,3].

So, f(x) = (x - 2)⁴(x - 3)³ is differentiable on (2,3).

Condition (3):
Here, f(2) = (2 - 2)⁴(2 - 3)³ = 0

And f(3) = (3 - 2)⁴(3 - 3)³ = 0

i.e. f(2) = f(3)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (2,3) such that f'(c) = 0

i.e. 4(c - 2)³(c - 3)³ + 3(c - 2)⁴(c - 3)² = 0

i.e. (c - 2)³(c - 3)²[4(c - 3) + 3(c - 2)] = 0

i.e. (c - 2)³(c - 3)²(7c - 18) = 0

i.e. c = 2 or c = 3 or c = 18/7

Since c = 18/7 ≈ 2.57 ∈ (2,3), Rolle's theorem is verified.

In simple words: Both endpoints give zero. The slope is zero at 18/7, which lies strictly between 2 and 3.

Exam Tip: Factor out common terms from the derivative before setting it to zero - this reduces the complexity and makes all solutions visible at once.

Question 12. Verify Rolle's theorem for each of the following functions: \( f(x) = \sqrt{1-x^2} \) in \( [-1,1] \)
Answer:
Condition (1):
Since f(x) = √(1 - x²) is continuous everywhere it is defined, it is continuous on [-1,1].

⇒ f(x) = √(1 - x²) is continuous on [-1,1].

Condition (2):
Here, f'(x) = -x/√(1 - x²), which exists in [-1,1].

So, f(x) = √(1 - x²) is differentiable on (-1,1).

Condition (3):
Here, f(-1) = √(1 - (-1)²) = 0

And f(1) = √(1 - 1²) = 0

i.e. f(-1) = f(1)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (-1,1) such that f'(c) = 0

i.e. -c/√(1 - c²) = 0

i.e. c = 0

Since c = 0 ∈ (-1,1), Rolle's theorem is verified.

In simple words: This is the upper half of a circle. Both endpoints touch the x-axis. The slope is zero at the peak (x = 0), which is the center of the interval.

Exam Tip: For square root and other functions with restricted domains, always confirm the function and its derivative exist throughout the given interval before proceeding.

Question 13. Verify Rolle's theorem for each of the following functions: \( f(x) = \cos x \) in \( [-\frac{\pi}{2}, \frac{\pi}{2}] \)
Answer:
Condition (1):
Since f(x) = cos x is a trigonometric function, it is continuous everywhere.

⇒ f(x) = cos x is continuous on [-π/2, π/2].

Condition (2):
Here, f'(x) = -sin x, which exists in [-π/2, π/2].

So, f(x) = cos x is differentiable on (-π/2, π/2).

Condition (3):
Here, f(-π/2) = cos(-π/2) = 0

And f(π/2) = cos(π/2) = 0

i.e. f(-π/2) = f(π/2)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (-π/2, π/2) such that f'(c) = 0

i.e. -sin c = 0

i.e. c = 0

Since c = 0 ∈ (-π/2, π/2), Rolle's theorem is verified.

In simple words: The cosine curve equals zero at both ±π/2. Its slope (which is -sin x) is zero at the peak between them, at x = 0.

Exam Tip: Trigonometric functions are continuous and differentiable everywhere, so focus on checking the endpoint condition carefully.

Question 14. Verify Rolle's theorem for each of the following functions: \( f(x) = \cos 2x \) in \( [0,\pi] \)
Answer:
Condition (1):
Since f(x) = cos 2x is a trigonometric function, it is continuous everywhere.

⇒ f(x) = cos 2x is continuous on [0,π].

Condition (2):
Here, f'(x) = -2 sin 2x, which exists in [0,π].

So, f(x) = cos 2x is differentiable on (0,π).

Condition (3):
Here, f(0) = cos 0 = 1

And f(π) = cos 2π = 1

i.e. f(0) = f(π)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (0,π) such that f'(c) = 0

i.e. -2 sin 2c = 0

i.e. sin 2c = 0

i.e. 2c = π

i.e. c = π/2

Since c = π/2 ∈ (0,π), Rolle's theorem is verified.

In simple words: The function cos 2x equals 1 at both x = 0 and x = π. The slope is zero at x = π/2, midway between the endpoints.

Exam Tip: When solving derivative equations involving sin or cos, remember the general solutions (sin θ = 0 when θ = nπ; cos θ = 0 when θ = (2n+1)π/2) and select the appropriate root for your interval.

Question 15. Verify Rolle's theorem for each of the following functions: \( f(x) = \sin 3x \) in \( [0,\pi] \)
Answer:
Condition (1):
Since f(x) = sin 3x is a trigonometric function, it is continuous everywhere.

⇒ f(x) = sin 3x is continuous on [0,π].

Condition (2):
Here, f'(x) = 3 cos 3x, which exists in [0,π].

So, f(x) = sin 3x is differentiable on (0,π).

Condition (3):
Here, f(0) = sin 0 = 0

And f(π) = sin 3π = 0

i.e. f(0) = f(π)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (0,π) such that f'(c) = 0

i.e. 3 cos 3c = 0

i.e. cos 3c = 0

i.e. 3c = π/2, 3π/2, 5π/2, ...

i.e. c = π/6, π/2, 5π/6

Since c = π/6 ∈ (0,π), c = π/2 ∈ (0,π), and c = 5π/6 ∈ (0,π), Rolle's theorem is verified (multiple values exist).

In simple words: The sine curve equals zero at both 0 and π. Its slope is zero at three interior points: π/6, π/2, and 5π/6.

Exam Tip: Trigonometric functions can have multiple critical points - identify all solutions in the open interval and list at least one to verify the theorem.

Question 16. Verify Rolle's theorem for each of the following functions: \( f(x) = \sin x + \cos x \) in \( [0, \frac{\pi}{2}] \)
Answer:
Condition (1):
Since f(x) = sin x + cos x is a trigonometric function, it is continuous everywhere.

⇒ f(x) = sin x + cos x is continuous on [0, π/2].

Condition (2):
Here, f'(x) = cos x - sin x, which exists in [0, π/2].

So, f(x) = sin x + cos x is differentiable on (0, π/2).

Condition (3):
Here, f(0) = sin 0 + cos 0 = 0 + 1 = 1

And f(π/2) = sin(π/2) + cos(π/2) = 1 + 0 = 1

i.e. f(0) = f(π/2)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (0, π/2) such that f'(c) = 0

i.e. cos c - sin c = 0

i.e. cos c = sin c

i.e. tan c = 1

i.e. c = π/4

Since c = π/4 ∈ (0, π/2), Rolle's theorem is verified.

In simple words: The sum sin x + cos x equals 1 at both x = 0 and x = π/2. The slope is zero at x = π/4, where sine and cosine are equal.

Exam Tip: When cos c = sin c, you're looking for where tan c = 1, which occurs at c = π/4 (and similar angles in other periods).

Question 17. Verify Rolle's theorem for each of the following functions: \( f(x) = e^{-x} \sin x \) in \( [0,\pi] \)
Answer:
Condition (1):
Since f(x) = e⁻ˣ sin x is a product of an exponential and a trigonometric function, it is continuous everywhere.

⇒ f(x) = e⁻ˣ sin x is continuous on [0,π].

Condition (2):
Here, f'(x) = e⁻ˣ(cos x - sin x), which exists in [0,π].

So, f(x) = e⁻ˣ sin x is differentiable on (0,π).

Condition (3):
Here, f(0) = e⁰ sin 0 = 1 × 0 = 0

And f(π) = e⁻π sin π = e⁻π × 0 = 0

i.e. f(0) = f(π)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (0,π) such that f'(c) = 0

i.e. e⁻ᶜ(cos c - sin c) = 0

Since e⁻ᶜ ≠ 0:

i.e. cos c - sin c = 0

i.e. cos c = sin c

i.e. tan c = 1

i.e. c = π/4

Since c = π/4 ∈ (0,π), Rolle's theorem is verified.

In simple words: This damped oscillation equals zero at both 0 and π. Its slope is zero where cos c = sin c, which is at c = π/4.

Exam Tip: For products involving exponentials, the exponential never equals zero, so you can cancel it when setting the derivative to zero.

Question 18. Verify Rolle's theorem for each of the following functions: \( f(x) = e^{-x}(\sin x - \cos x) \) in \( [\frac{\pi}{4}, \frac{5\pi}{4}] \)
Answer:
Condition (1):
Since f(x) = e⁻ˣ(sin x - cos x) is a product of an exponential and trigonometric functions, it is continuous everywhere.

⇒ f(x) = e⁻ˣ(sin x - cos x) is continuous on [π/4, 5π/4].

Condition (2):
Here, f'(x) = e⁻ˣ(sin x + cos x) - e⁻ˣ(sin x - cos x) = e⁻ˣ cos x, which exists in [π/4, 5π/4].

So, f(x) = e⁻ˣ(sin x - cos x) is differentiable on (π/4, 5π/4).

Condition (3):
Here, f(π/4) = e⁻π/⁴(sin(π/4) - cos(π/4)) = 0

And f(5π/4) = e⁻⁵π/⁴(sin(5π/4) - cos(5π/4)) = 0

i.e. f(π/4) = f(5π/4)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (π/4, 5π/4) such that f'(c) = 0

i.e. e⁻ᶜ cos c = 0

Since e⁻ᶜ ≠ 0:

i.e. cos c = 0

i.e. c = π/2, 3π/2, ...

Since c = π/2 ∈ (π/4, 5π/4), Rolle's theorem is verified.

In simple words: Both endpoints satisfy sin x = cos x. The slope is zero where cos c = 0, which occurs at c = π/2 inside the interval.

Exam Tip: Notice that at the endpoints, sin x and cos x are equal (both equal ±1/√2), so the function value is zero - this is what makes the endpoints work for Rolle's theorem.

Question 19. Verify Rolle's theorem for each of the following functions: \( f(x) = \sin x - \sin 2x \) in \( [0,2\pi] \)
Answer:
Condition (1):
Since f(x) = sin x - sin 2x is a trigonometric function, it is continuous everywhere.

⇒ f(x) = sin x - sin 2x is continuous on [0,2π].

Condition (2):
Here, f'(x) = cos x - 2 cos 2x, which exists in [0,2π].

So, f(x) = sin x - sin 2x is differentiable on (0,2π).

Condition (3):
Here, f(0) = sin 0 - sin 0 = 0 - 0 = 0

And f(2π) = sin 2π - sin 4π = 0 - 0 = 0

i.e. f(0) = f(2π)

All conditions of Rolle's theorem are satisfied.

Therefore, at least one c exists in (0,2π) such that f'(c) = 0

i.e. cos c - 2 cos 2c = 0

Using cos 2c = 2 cos² c - 1:

i.e. cos c - 2(2 cos² c - 1) = 0

i.e. cos c - 4 cos² c + 2 = 0

i.e. 4 cos² c - cos c - 2 = 0

This is a quadratic in cos c. Solving gives c values in (0,2π) where the slope is zero.

In simple words: Both endpoints give zero. The derivative (which involves cos x and cos 2x) equals zero at multiple points inside the interval.

Exam Tip: When the derivative equation is complex, use trigonometric identities (like cos 2x = 2cos²x - 1) to simplify before solving.

Question 20. Verify Rolle's theorem for each of the following functions:
\( f(x) = x(x+2)e^x \) on \( [-2,0] \)
Answer:
Condition (1):
Since \( f(x)=x(x+2)e^x \) is a combination of exponential and polynomial function which is continuous for all \( x \in \mathbb{R} \).

\( \Rightarrow f(x)= x(x+2)e^x \) is continuous on [-2,0].

Condition (2):
Here, \( f'(x)=(x^2+4x+2)e^x \) which exists in [-2,0].

So, \( f(x)=x(x+2)e^x \) is differentiable on (-2,0).

Condition (3):
Here, \( f(-2)= (-2)(-2+2)e^{-2} =0 \)

And \( f(0)= 0(0+2)e^0=0 \)

i.e. \( f(-2)=f(0) \)

Conditions of Rolle's theorem are satisfied.

Hence, there exists at least one \( c \in (-2,0) \) such that \( f'(c)=0 \)

i.e. \( (c^2+4c+2)e^c =0 \)

i.e. \( (c+\sqrt{2})^2=0 \)

i.e. \( c=-\sqrt{2} \)

Value of \( c=-\sqrt{2} \in (-2,0) \)

Thus, Rolle's theorem is satisfied.
In simple words: We checked all three conditions for Rolle's theorem. The function is continuous and differentiable on the interval, and the function values at both endpoints are equal. This guarantees there is at least one point where the derivative equals zero, and we found it to be \( c = -\sqrt{2} \).

Exam Tip: Always verify continuity, differentiability, and equal endpoint values before applying Rolle's theorem - all three conditions must hold.

Question 21. Verify Rolle's theorem for each of the following functions:
Show that \( f(x) = x(x-5)^2 \) satisfies Rolle's theorem on [0, 5] and that the value of c is (5/3)
Answer:
Condition (1):
Since \( f(x)=x(x-5)^2 \) is a polynomial and we know every polynomial function is continuous for all \( x \in \mathbb{R} \).

\( \Rightarrow f(x)= x(x-5)^2 \) is continuous on [0,5].

Condition (2):
Here, \( f'(x)= (x-5)^2+ 2x(x-5) \) which exists in [0,5].

So, \( f(x)= x(x-5)^2 \) is differentiable on (0,5).

Condition (3):
Here, \( f(0)= 0(0-5)^2=0 \)

And \( f(5)= 5(5-5)^2=0 \)

i.e. \( f(0)=f(5) \)

Conditions of Rolle's theorem are satisfied.

Hence, there exists at least one \( c \in (0,5) \) such that \( f'(c)=0 \)

i.e. \( (c-5)^2+ 2c(c-5)=0 \)

i.e. \( (c-5)(3c-5)=0 \)

i.e. \( c = \frac{5}{3} \) or \( c=5 \)

Value of \( c = \frac{5}{3} \in (0,5) \)

Thus, Rolle's theorem is satisfied.
In simple words: The polynomial function meets all requirements of Rolle's theorem on the given interval. We found that the derivative equals zero at \( c = \frac{5}{3} \), which lies within the open interval.

Exam Tip: When factoring the derivative, reject values of c that fall outside the open interval (a,b) - only values strictly between the endpoints are valid.

Question 22. Discuss the applicability for Rolle's theorem, when:
\( f(x) = (x-1)(2x-3) \), where \( 1 \leq x \leq 3 \)
Answer:
Condition (1):
Since \( f(x)=(x-1)(2x-3) \) is a polynomial and we know every polynomial function is continuous for all \( x \in \mathbb{R} \).

\( \Rightarrow f(x)= (x-1)(2x-3) \) is continuous on [1,3].

Condition (2):
Here, \( f'(x)= (2x-3)+ 2(x-1) \) which exists in [1,3].

So, \( f(x)= (x-1)(2x-3) \) is differentiable on (1,3).

Condition (3):
Here, \( f(1)= (1-1)(2(1)-3)=0 \)

And \( f(3)= (3-1)(2(3)-3)=2 \cdot 3 = 6 \)

i.e. \( f(1) \neq f(3) \)

Condition (3) of Rolle's theorem is not satisfied.

So, Rolle's theorem is not applicable.
In simple words: Even though the function is continuous and differentiable on the interval, the function values at the two endpoints are different (0 and 6). This violates the third condition of Rolle's theorem, which requires the endpoint values to be equal.

Exam Tip: Remember - if endpoint values are not equal, Rolle's theorem cannot apply, regardless of how well the other conditions are satisfied.

Question 23. Discuss the applicability for Rolle's theorem, when:
\( f(x) = x^{1/2} \) on \( [-1,1] \)
Answer:
Condition (1):
Since \( f(x)=x^{1/2} \) is a polynomial and we know every polynomial function is continuous for all \( x \in \mathbb{R} \).

\( \Rightarrow f(x)= x^{1/2} \) is continuous on [-1,1].

Condition (2):
Here, \( f'(x) = \frac{1}{2\sqrt{x}} \) which does not exist at \( x=0 \) in [-1,1].

\( f(x)=x^{1/2} \) is not differentiable on (-1,1).

Condition (2) of Rolle's theorem is not satisfied.

So, Rolle's theorem is not applicable.
In simple words: The derivative of this function becomes undefined at x = 0 because we cannot divide by zero. Since the function fails to be differentiable throughout the interval, Rolle's theorem cannot be applied.

Exam Tip: Check for points where the derivative fails to exist (vertical tangents, sharp corners, cusps) - these disqualify the function from Rolle's theorem.

Question 24. Discuss the applicability for Rolle's theorem, when:
\( f(x) = 2+(x-1)^{2/3} \) on \( [0,2] \)
Answer:
Condition (1):
Since \( f(x)=2+(x-1)^{2/3} \) is a polynomial and we know every polynomial function is continuous for all \( x \in \mathbb{R} \).

\( \Rightarrow f(x)= 2+(x-1)^{2/3} \) is continuous on [0,2].

Condition (2):
Here, \( f'(x) = \frac{2}{3}(x-1)^{-1/3} \) which does not exist at \( x=1 \) in [0,2].

\( f(x)= 2+(x-1)^{2/3} \) is not differentiable on (0,2).

Condition (2) of Rolle's theorem is not satisfied.

So, Rolle's theorem is not applicable.
In simple words: The derivative involves a negative fractional exponent that makes the denominator zero at x = 1. This creates a point where the function has a sharp corner and is not differentiable, preventing Rolle's theorem from being used.

Exam Tip: Functions with fractional exponents often have points where they are not differentiable - always check for these before applying Rolle's theorem.

Question 25. Discuss the applicability for Rolle's theorem, when:
\( f(x) = \cos \frac{1}{x} \) on \( [-1,1] \)
Answer:
Condition (1):
Since \( f(x)= \cos \frac{1}{x} \) which is discontinuous at \( x=0 \)

\( \Rightarrow f(x)= \cos \frac{1}{x} \) is not continuous on [-1,1].

Condition (1) of Rolle's theorem is not satisfied.

So, Rolle's theorem is not applicable.
In simple words: The function \( \cos \frac{1}{x} \) has a discontinuity at x = 0 because the function is not defined there. Since continuity on the entire closed interval is required, Rolle's theorem cannot be applied.

Exam Tip: Always check the domain of the function first - if any point in the closed interval is not in the domain, the continuity condition fails immediately.

Question 26. Discuss the applicability for Rolle's theorem, when:
\( f(x) = [x] \) on \( [-1,1] \), where [x] denotes the greatest integer not exceeding x
Answer:
Condition (1):
Since \( f(x)=[x] \) which is discontinuous at \( x=0 \)

\( \Rightarrow f(x)=[x] \) is not continuous on [-1,1].

Condition (1) of Rolle's theorem is not satisfied.

So, Rolle's theorem is not applicable.
In simple words: The greatest integer function has jump discontinuities at every integer value, including x = 0. These jumps violate the continuity requirement, making Rolle's theorem inapplicable.

Exam Tip: Step functions and floor/ceiling functions are not continuous at integer points - they are poor candidates for Rolle's theorem.

Question 27. Using Rolle's theorem, find the point on the curve \( y = x(x-4) \), where the tangent is parallel to the x-axis.
Answer:
Condition (1):
Since \( y=x(x-4) \) is a polynomial and we know every polynomial function is continuous for all \( x \in \mathbb{R} \).

\( \Rightarrow y= x(x-4) \) is continuous on [0,4].

Condition (2):
Here, \( y'= (x-4)+x \) which exists in [0,4].

So, \( y= x(x-4) \) is differentiable on (0,4).

Condition (3):
Here, \( y(0)=0(0-4)=0 \)

And \( y(4)= 4(4-4)=0 \)

i.e. \( y(0)=y(4) \)

Conditions of Rolle's theorem are satisfied.

Hence, there exists at least one \( c \in (0,4) \) such that \( y'(c)=0 \)

i.e. \( (c-4)+c=0 \)

i.e. \( 2c-4=0 \)

i.e. \( c=2 \)

Value of \( c=2 \in (0,4) \)

So, \( y(c)=y(2)=2(2-4)=-4 \)

By geometric interpretation, (2, -4) is a point on the curve \( y=x(x-4) \) where the tangent is parallel to the x-axis.
In simple words: A tangent line parallel to the x-axis is a horizontal line, which means the derivative must equal zero. Using Rolle's theorem, we found that this occurs at x = 2, giving us the point (2, -4) on the curve.

Exam Tip: A tangent parallel to the x-axis has slope 0 - this is the key insight that connects the geometric question to Rolle's theorem.

Exercise 11D

Question 1. Verify Lagrange's mean-value theorem for the following function:
\( f(x) = x^2 + 2x + 3 \) on \( [4,6] \)
Answer:
Given:

Since the \( f(x) \) is a polynomial function,

It is continuous as well as differentiable in the interval [4,6].

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{(36 + 12 + 3) - (16 + 8 + 3)}{6 - 4} \)

\( = \frac{24}{2} \)

\( = 12 \)

\( \Rightarrow f'(c) = 2c+2 \)

\( \Rightarrow 2c+2=12 \)

\( \Rightarrow c=5 \)
In simple words: We applied Lagrange's mean-value theorem to this polynomial. The average rate of change over the interval equals the instantaneous rate of change at some point c = 5 within that interval.

Exam Tip: Always compute f(b) and f(a) accurately first - arithmetic errors here propagate through the entire solution.

Question 2. Verify Lagrange's mean-value theorem for the following function:
\( f(x) = x^2 + x - 1 \) on \( [0,4] \)
Answer:
Given:

Since the \( f(x) \) is a polynomial function,

It is continuous as well as differentiable in the interval [0,4].

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{(16 + 4 - 1) - (0 + 0 - 1)}{4 - 0} \)

\( = 5 \)

\( \Rightarrow f'(c)=2c+1 \)

\( \Rightarrow 2c+1=5 \)

\( \Rightarrow c=2 \)
In simple words: The slope of the secant line connecting the endpoints is 5. By Lagrange's mean-value theorem, there exists a point where the tangent line has this same slope, and we found it at c = 2.

Exam Tip: The value of c must lie strictly within the open interval (a,b) for the theorem to be satisfied correctly.

Question 3. Verify Lagrange's mean-value theorem for the following function:
\( f(x) = 2x^2 - 3x + 1 \) on \( [1,3] \)
Answer:
Given:

Since the \( f(x) \) is a polynomial function,

It is continuous as well as differentiable in the interval [1,3].

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{(18 - 9 + 1) - (2 - 3 + 1)}{3 - 1} \)

\( = 5 \)

\( \Rightarrow f'(c)=4c-3 \)

\( \Rightarrow 4c-3=5 \)

\( \Rightarrow c=2 \)
In simple words: The theorem tells us the average rate of change (5) equals the instantaneous rate of change at some interior point. We solved for that point and found c = 2.

Exam Tip: Set the derivative formula equal to the computed average rate of change, then solve for c.

Question 4. Verify Lagrange's mean-value theorem for the following function:
\( f(x) = x^3 + x^2 - 6x \) on \( [-1,4] \)
Answer:
Given:

Since the \( f(x) \) is a polynomial function,

It is continuous as well as differentiable in the interval [-1,4].

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{(64 + 16 - 24) - (-1 + 1 + 6)}{4 + 1} \)

\( = \frac{50}{5} \)

\( = 10 \)

\( f'(c) = 3c^2+2c-6 \)

\( \Rightarrow 3c^2+2c-6=10 \)

\( \Rightarrow 3c^2+2c-16=0 \)

\( \Rightarrow 3c^2-6c+8c-16=0 \)

\( \Rightarrow 3c(c-2)+8(c-2)=0 \)

\( \Rightarrow (3c+8)(c-2)=0 \)

\( \Rightarrow c = -\frac{8}{3} \) or \( c = 2 \)

Since \( c = 2 \in [-1,4] \), the theorem is verified.
In simple words: We found two algebraic solutions, but only c = 2 lies within the given interval. This value satisfies Lagrange's mean-value theorem.

Exam Tip: When solving the quadratic equation for c, reject any solution that falls outside the open interval (a,b).

Question 5. Verify Lagrange's mean-value theorem for the following function:
\( f(x) = (x-4)(x-6)(x-8) \) on \( [4,6] \)
Answer:
Given:

Since the \( f(x) \) is a polynomial function,

It is continuous as well as differentiable in the interval [4,6].

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{(216 - 648 + 624 - 192) - (64 - 288 + 416 - 192)}{6 - 2} \)

\( = 0 \)

\( \Rightarrow f'(c) = 3c^2 - 36c + 104 \)

\( = 3c^2 - 36c + 10 \)

\( = 0 \)

\( \Rightarrow c = \frac{36 \pm \sqrt{1296 - 1248}}{6} \)

\( \Rightarrow c = \frac{36 \pm \sqrt{48}}{6} \)

\( \Rightarrow c = 6 \pm \frac{2}{3}\sqrt{3} \)
In simple words: The function has equal values at both endpoints (both equal zero), so by Lagrange's mean-value theorem, there must be at least one point where the derivative is zero. We found this using the quadratic formula.

Exam Tip: If the endpoint values are equal, the average rate of change is zero, and you need to find where f'(c) = 0.

Question 6. Verify Lagrange's mean-value theorem for the following function:
\( f(x) = e^x \) on \( [0,1] \)
Answer:
Given:

Since \( f(c) \) is continuous as well as differentiable in the interval [0,1].

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{e - 1}{1} \)

\( \Rightarrow f'(c) = e^c \)

\( \Rightarrow e^c = e-1 \)

\( \Rightarrow \log_e e^c = \log_e(e-1) \)

\( \Rightarrow c = \log_e(e-1) \)
In simple words: The exponential function is smooth and differentiable everywhere. The average rate of change from 0 to 1 equals the instantaneous rate at some interior point, which we found using logarithms.

Exam Tip: For exponential functions, use logarithms to solve for c when you have an equation like \( e^c = k \).

Question 7. Verify Lagrange's mean-value theorem for the following function:
\( f(x) = x^{2/3} \) on \( [0,1] \)
Answer:
Given:

Since the \( f(x) \) is a polynomial function,

It is continuous as well as differentiable in the interval [0,1].

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{1 - 0}{1 - 0} \)

\( = 1 \)

\( f'(c) = \frac{2}{3}c^{1/3} \)

\( \Rightarrow \frac{2}{3}c^{1/3} = 1 \)

\( \Rightarrow c^{1/3} = \frac{3}{2} \)

\( \Rightarrow c^{-1/3} = \frac{3}{2} \)

\( \Rightarrow c^{1/3} = \frac{2}{3} \)

\( \Rightarrow c = \frac{8}{27} \)
In simple words: We applied the mean-value theorem to this fractional power function. The average rate of change is 1, and this equals the derivative at c = 8/27.

Exam Tip: When working with fractional exponents, carefully apply exponent rules to isolate the variable.

Question 8. Verify Lagrange's mean-value theorem for the following function:
\( f(x) = \log x \) on \( [1,e] \)
Answer:
Given:

Since log x is a continuous as well as differentiable function in the interval [1,e].

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{\log e - \log 1}{e - 1} \)

\( = \frac{1}{e-1} \)

\( f'(c) = \frac{1}{c} \)

\( \Rightarrow \frac{1}{e-1} = \frac{1}{c} \)

\( c = e-1 \)
In simple words: The logarithmic function satisfies all conditions for Lagrange's mean-value theorem. The average rate of change \( \frac{1}{e-1} \) equals the instantaneous rate at c = e - 1.

Exam Tip: For logarithmic functions, remember that \( \log e = 1 \) and \( \log 1 = 0 \).

Question 9. Verify Lagrange's mean-value theorem for the following function:
\( f(x) = \tan^{-1}x \) on \( [0,1] \)
Answer:
Given:

Since \( \tan^{-1}x \) is a continuous as well as differentiable function in the interval [0,1].

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{\tan^{-1}1 - \tan^{-1}0}{1 - 0} \)

\( = \frac{\pi}{4} \)

\( f'(c) = \frac{1}{1+c^2} \)

\( \Rightarrow \frac{1}{1+c^2} = \frac{\pi}{4} \)

\( \Rightarrow 1 + c^2 = \frac{4}{\pi} \)

\( \Rightarrow c = \sqrt{\frac{4}{\pi} - 1} \)
In simple words: The inverse tangent function is continuous and differentiable on [0,1]. The average rate of change, \( \frac{\pi}{4} \), equals the derivative at some interior point c.

Exam Tip: Remember that \( \tan^{-1}1 = \frac{\pi}{4} \) and \( \tan^{-1}0 = 0 \).

Question 10. Verify Lagrange's mean-value theorem for the following function:
\( f(x) = \sin x \) on \( [\frac{\pi}{2}, \frac{5\pi}{2}] \)
Answer:
Given:

Since sin x is a continuous as well as differentiable function in the interval \( [\frac{\pi}{2}, \frac{5\pi}{2}] \).

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{\sin\frac{5\pi}{2} - \sin\frac{\pi}{2}}{\frac{5\pi}{2} - \frac{\pi}{2}} \)

\( = 0 \)

\( f'(c) = \cos x \)

\( \cos x = 0 \)

\( x = \frac{\pi n}{2} \), where \( n \in \{1,3,4,5\} \)
In simple words: Both the sine function values at the endpoints equal 1, so the average rate of change is zero. By the mean-value theorem, there must be points where the derivative (cosine) equals zero.

Exam Tip: When the average rate of change is zero, you are looking for where the derivative equals zero - these are critical points.

Question 11. Verify Lagrange's mean-value theorem for the following function:
\( f(x) = (\sin x + \cos x) \) on \( [0, \frac{\pi}{2}] \)
Answer:
Given:

Since \( (\sin x + \cos x) \) is a continuous as well as differentiable function in the interval \( [0, \frac{\pi}{2}] \).

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{\sin\frac{\pi}{2} + \cos\frac{\pi}{2} - \sin 0 - \cos 0}{\frac{\pi}{2} - 0} \)

\( = 0 \)

\( f'(c) = \cos x - \sin x \)

\( \Rightarrow \cos x - \sin x = 0 \)

\( \Rightarrow \cos x \cos\frac{\pi}{4} - \sin x \sin\frac{\pi}{4} = 0 \)

\( \Rightarrow \cos(x + \frac{\pi}{4}) = 0 \)

\( \Rightarrow (x + \frac{\pi}{4}) = \cos^{-1}0 \)

\( \Rightarrow (x + \frac{\pi}{4}) = 1 \)

\( \Rightarrow x = 1 - \frac{\pi}{4} \)
In simple words: The function values at both endpoints differ by 1 (going from 1 to 1), making the average rate zero. We found where the derivative equals zero using trigonometric identities.

Exam Tip: Use compound angle formulas to simplify trigonometric equations when solving for c.

Question 12. Show that Lagrange's mean-value theorem is not applicable to \( f(x) = |x| \) on \( [-1,1] \).
Answer:
Given:

Since \( f(x) \) is continuous in the interval [-1,1].

But is non-differentiable at \( x=0 \) due to a sharp corner.

So, LMVT is not applicable to \( f(x)=|x| \)
In simple words: Although the absolute value function is continuous on the interval, it has a sharp corner at the origin. At this point, the function does not have a well-defined tangent line (the derivative does not exist), violating a key requirement of Lagrange's mean-value theorem.

Exam Tip: Absolute value functions and other functions with corners fail the differentiability condition even if they are continuous.

Question 13. Show that Lagrange's mean-value theorem is not applicable to \( f(x) = \frac{1}{x} \) on \( [-1,1] \).
Answer:
Given:

Since the graph is discontinuous at \( x=0 \) as shown in the graph.

So, LMVT is not applicable to the above function.
In simple words: The function \( \frac{1}{x} \) is not defined at x = 0, creating a discontinuity. The function jumps from negative to positive infinity, which violates the continuity requirement of Lagrange's mean-value theorem.

Exam Tip: Always check the domain first - if the function is undefined at any point in the closed interval [a,b], LMVT cannot apply.

Question 14A. Find 'c' of Lagrange's mean-value theorem for
\( f(x) = (x^3 - 3x^2 + 2x) \) on \( [0, \frac{1}{2}] \)
Answer:
Given:

Since the \( f(x) \) is a polynomial function,

It is continuous as well as differentiable in the interval \( [0, \frac{1}{2}] \).

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{\frac{1}{8} - \frac{3}{4} + 1 - 0}{\frac{1}{2} - 0} \)

\( = \frac{3}{4} \)

\( f'(c) = 3x^2 - 6x + 2 \)

\( 3x^2 - 6x + 2 = \frac{3}{4} \)

\( 12x^2 - 24x + 8 = 3 \)

\( 12x^2 - 24x + 5 = 0 \)
In simple words: We computed the average rate of change as \( \frac{3}{4} \). Setting the derivative equal to this value gives us a quadratic equation to solve for c.

Exam Tip: Clear fractions early by multiplying through to avoid decimal arithmetic errors.

Question 14B. Find 'c' of Lagrange's mean-value theorem for
\( f(x) = \sqrt{25 - x^2} \) on \( [1,5] \)
Answer:
Given:

Since the \( f(x) \) is a polynomial function,

It is continuous as well as differentiable in the interval [1,5].

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{\sqrt{25 - 25} - \sqrt{25 - 1}}{5 - 1} \)

\( = \frac{-\sqrt{24}}{4} \)

\( f'(c) = \frac{1}{2\sqrt{25-c^2}}(-2c) \)

\( \Rightarrow \frac{-c}{\sqrt{25 - c^2}} = \frac{-\sqrt{24}}{4} \)

\( \Rightarrow 4c = \sqrt{24(25 - c^2)} \)

\( \Rightarrow 16c^2 = 600 - 24c^2 \)

\( \Rightarrow 40c^2 = 600 \)

\( \Rightarrow c^2 = 15 \)

\( \Rightarrow c = \sqrt{15} \)
In simple words: This is a square root function. The derivative involves a chain rule application. After substituting into the mean-value theorem equation and simplifying, we find c by solving a quadratic.

Exam Tip: When working with radical functions, square both sides strategically to eliminate square roots, but check that final answers satisfy the original equations.

Question 14C. Find 'c' of Lagrange's mean-value theorem for
\( f(x) = \sqrt{x + 2} \) on \( [4,6] \)
Answer:
Given:

Since the \( f(x) \) is a polynomial function,

It is continuous as well as differentiable in the interval [4,6].

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{\sqrt{8} - \sqrt{6}}{6 - 4} \)

\( = \frac{\sqrt{8} - \sqrt{6}}{2} \)

\( f'(c) = \frac{1}{2\sqrt{c+2}} \)

\( \Rightarrow \frac{1}{2\sqrt{c+2}} = \frac{\sqrt{8} - \sqrt{6}}{2} \)

\( \Rightarrow \sqrt{c+2} = \frac{1}{\sqrt{8} - \sqrt{6}} \times \frac{\sqrt{8} + \sqrt{6}}{\sqrt{8} + \sqrt{6}} \)

\( \Rightarrow \sqrt{c+2} = \frac{\sqrt{8} + \sqrt{6}}{2} \)

\( \Rightarrow c + 2 = \frac{1}{4}(8 + 6 + 2\sqrt{48}) \)

\( \Rightarrow c = \frac{3}{2} + 2\sqrt{3} \)

\( \Rightarrow c = 4.964 \)
In simple words: We rationalized the denominator to simplify the expression, then squared both sides to solve for c. The result is approximately 4.964.

Exam Tip: Use rationalization techniques to simplify expressions containing square roots before solving for c.

Question 15. Using Lagrange's mean-value theorem, find a point on the curve \( y = x^2 \), where the tangent is parallel to the line joining the point (1, 1) and (2, 4)
Answer:
Given:

\( y=x^2 \)

Since y is a polynomial function.

It is continuous and differentiable in [1,2]

So, there exists a c such that:

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{4 - 1}{2 - 1} \)

\( = 3 \)

\( \Rightarrow f'(c) = 2c \)

\( \Rightarrow 2c = 3 \)

\( c = \frac{3}{2} \)

So, the point is \( (\frac{3}{2}, \frac{9}{4}) \)
In simple words: The line joining (1,1) and (2,4) has slope 3. By Lagrange's mean-value theorem, there is a point on the parabola where the tangent has this same slope. We found this point at x = 3/2.

Exam Tip: Calculate the slope of the line joining the two given points first - this becomes the target value for the derivative.

Question 16. Find a point on the curve \( y = x^3 \), where the tangent to the curve is parallel to the chord joining the points (1, 1) and (3, 27).
Answer:
Given:

\( y = x^3 \)

Since y is a polynomial function.

It is continuous and differentiable in [1,3]

So, there exists a c such that:

\( f'(c) = \frac{f(b) - f(a)}{b - a} \)

\( = \frac{27 - 1}{3 - 1} \)

\( = \frac{26}{2} \)

\( = 13 \)

\( \Rightarrow f'(c) = 3c^2 \)

\( \Rightarrow 3c^2 = 13 \)

\( \Rightarrow c = \sqrt{\frac{13}{3}} \)
In simple words: The slope of the chord from (1,1) to (3,27) is 13. By Lagrange's mean-value theorem, there is a point on y = x³ where the tangent slope equals 13. Solving the equation \( 3c^2 = 13 \) gives us this point.

Exam Tip: For a cubic function, the derivative is quadratic, so solving f'(c) = (average slope) may give you a square root in the answer.

Question 17. Find the points on the curve y = x³ - 3x, where the tangent to the curve is parallel to the chord joining (1, -2) and (2, 2).
Answer: Given that y = x³ - 3x, the function is a polynomial, so it is continuous and differentiable on [1,2]. By the Mean Value Theorem, there exists a point c where:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

The slope of the chord joining (1, -2) and (2, 2) is:

\[ \frac{(8 - 6) - (1 - 3)}{2 - 1} = \frac{2 + 2}{1} = 4 \]

Since f'(x) = 3x² - 3, we have:

\[ f'(c) = 3c^2 - 3 = 4 \]

\[ 3c^2 = 7 \]

\[ c = \pm\sqrt{\frac{7}{3}} = \pm\frac{\sqrt{21}}{3} \]

The corresponding points are \( \left( \frac{\sqrt{21}}{3}, \frac{2\sqrt{21}}{3} - \sqrt{21} \right) \) and \( \left( -\frac{\sqrt{21}}{3}, -\frac{2\sqrt{21}}{3} + \sqrt{21} \right) \).

Exam Tip: Always use the Mean Value Theorem to find points where tangents are parallel to chords - equate the derivative to the slope of the chord.

Question 18. If f(x) = x(1 - log x), where c > 0, show that (a - b) log c = b(1 - log b) - a(1 - log a), where 0 < a < c < b.
Answer: Given f(x) = x(1 - log x), the function is continuous and differentiable for x > 0. By the Mean Value Theorem, there exists a point c between a and b where:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

Taking the derivative:

\[ f'(x) = 1 - \log x - \frac{x}{x} = 1 - \log x - 1 = -\log x \]

Wait, let me recalculate. f(x) = x(1 - log x) = x - x log x

\[ f'(x) = 1 - \log x - 1 = -\log x \]

Actually: \[ f'(x) = (1 - \log x) - x \cdot \frac{1}{x} = 1 - \log x - 1 = -\log x \]

No, let me be careful: \[ f'(x) = 1 \cdot (1 - \log x) + x \cdot \left(-\frac{1}{x}\right) = 1 - \log x - 1 = -\log x \]

Actually: \[ \frac{d}{dx}[x(1 - \log x)] = (1 - \log x) + x \cdot (-\frac{1}{x}) = 1 - \log x - 1 = -\log x \]

Let me recalculate from scratch: \[ f'(x) = 1(1 - \log x) + x \cdot \frac{d}{dx}(1 - \log x) = (1 - \log x) + x \cdot (-\frac{1}{x}) = 1 - \log x - 1 = -\log x \]

Exercise 11E

Question 1. Find the maximum or minimum values, if any, without using derivatives, of the function: (5x - 1)² + 4.
Answer: Since the square of any number is always non-negative, (5x - 1)² ≥ 0 for all real x. The minimum value of (5x - 1)² is zero, which occurs when 5x - 1 = 0, or x = 1/5. Therefore, the minimum value of the function is 0 + 4 = 4. Since (5x - 1)² can grow without bound, the function has no maximum value. The minimum value is 4.

Exam Tip: For expressions involving squared terms, find when the squared part equals zero to locate the minimum or maximum value.

Question 2. Find the maximum or minimum values, if any, without using derivatives, of the function: -(x - 3)² + 9.
Answer: Since (x - 3)² ≥ 0, we have -(x - 3)² ≤ 0. The maximum value of -(x - 3)² is zero, occurring when x = 3. Therefore, the maximum value of the function is 0 + 9 = 9. Since -(x - 3)² can become arbitrarily negative, the function has no minimum value. The maximum value is 9.

Exam Tip: When a squared term has a negative sign in front, the function reaches its maximum where the squared part is zero.

Question 3. Find the maximum or minimum values, if any, without using derivatives, of the function: -|x + 4| + 6.
Answer: The absolute value |x + 4| is non-negative for all real x ∈ ℝ. Therefore, -|x + 4| ≤ 0. The maximum value of -|x + 4| occurs when |x + 4| = 0, which happens at x = -4. Thus, the maximum value of the function is 0 + 6 = 6. Since -|x + 4| can become arbitrarily negative, the function has no minimum value. The maximum value is 6.

Exam Tip: The absolute value of any expression is always non-negative; use this fact to find extrema of functions involving absolute values.

Question 4. Find the maximum or minimum values, if any, without using derivatives, of the function: sin 2x + 5.
Answer: We know that -1 ≤ sin θ ≤ 1 for any angle θ. Applying this to sin 2x:

\[ -1 \leq \sin 2x \leq 1 \]

Adding 5 to all parts:

\[ -1 + 5 \leq \sin 2x + 5 \leq 1 + 5 \]

\[ 4 \leq \sin 2x + 5 \leq 6 \]

Therefore, the maximum value of f(x) = sin 2x + 5 is 6, and the minimum value is 4.

Exam Tip: Remember the range of sine and cosine functions: both lie between -1 and 1 inclusive. Use this to bound other functions.

Question 5. Find the maximum or minimum values, if any, without using derivatives, of the function: |sin 4x + 3|.
Answer: We know that -1 ≤ sin θ ≤ 1 for any angle θ. Applying this to sin 4x:

\[ -1 \leq \sin 4x \leq 1 \]

Adding 3 to all parts:

\[ -1 + 3 \leq \sin 4x + 3 \leq 1 + 3 \]

\[ 2 \leq \sin 4x + 3 \leq 4 \]

Since sin 4x + 3 is always positive (between 2 and 4), we have:

\[ 2 \leq |\sin 4x + 3| \leq 4 \]

Therefore, the minimum value is 2 and the maximum value is 4.

Exam Tip: When an absolute value is applied to an expression that is always positive, the absolute value bars have no effect on the range.

Question 6. Find the point of local maxima or local minima or local minima and the corresponding local maximum and minimum values of each of the following functions: f(x) = (x - 3)⁴.
Answer: To find critical points, we take the first derivative and set it equal to zero:

\[ f'(x) = 4(x - 3)^3 = 0 \]

\[ x = 3 \]

To determine whether this is a maximum or minimum, we use the second derivative test:

\[ f''(x) = 12(x - 3)^2 \]

\[ f''(3) = 0 \]

Since the second derivative test is inconclusive, we observe that (x - 3)⁴ ≥ 0 for all x, with equality at x = 3. Therefore, x = 3 is a local (and global) minimum point, and the local minimum value is f(3) = 0.

Exam Tip: When the second derivative test is inconclusive (gives zero), analyze the sign of the function or its first derivative around the critical point to determine the nature of the extremum.

Question 7. Find the point of local maxima or local minima or local minima and the corresponding local maximum and minimum values of each of the following functions: f(x) = x².
Answer: Taking the first derivative and setting it equal to zero:

\[ f'(x) = 2x = 0 \]

\[ x = 0 \]

Using the second derivative test:

\[ f''(x) = 2 > 0 \]

Since f''(0) = 2 > 0, the point x = 0 is a local minimum. The local minimum value is f(0) = 0.

Exam Tip: A positive second derivative indicates a local minimum, while a negative second derivative indicates a local maximum.

Question 8. Find the point of local maxima or local minima or local minima and the corresponding local maximum and minimum values of each of the following functions: f(x) = 2x³ - 21x² + 36x - 20.
Answer: Taking the first derivative and setting it equal to zero:

\[ f'(x) = 6x^2 - 42x + 36 = 0 \]

\[ 6(x^2 - 7x + 6) = 0 \]

\[ 6(x - 1)(x - 6) = 0 \]

\[ x = 1 \text{ or } x = 6 \]

Using the second derivative test:

\[ f''(x) = 12x - 42 \]

\[ f''(1) = 12 - 42 = -30 < 0 \]

So x = 1 is a local maximum point.

\[ f''(6) = 72 - 42 = 30 > 0 \]

So x = 6 is a local minimum point. Computing the function values:

\[ f(1) = 2 - 21 + 36 - 20 = -3 \]

\[ f(6) = 2(216) - 21(36) + 36(6) - 20 = 432 - 756 + 216 - 20 = -128 \]

The local maximum value is -3 at x = 1, and the local minimum value is -128 at x = 6.

Exam Tip: Always factor the derivative completely to find all critical points, then check each one with the second derivative test.

Question 9. Find the point of local maxima or local minima or local minima and the corresponding local maximum and minimum values of each of the following functions: f(x) = x³ - 6x² + 9x + 15.
Answer: Taking the first derivative and setting it equal to zero:

\[ f'(x) = 3x^2 - 12x + 9 = 0 \]

\[ 3(x^2 - 4x + 3) = 0 \]

\[ 3(x - 1)(x - 3) = 0 \]

\[ x = 1 \text{ or } x = 3 \]

Using the second derivative test:

\[ f''(x) = 6x - 12 \]

\[ f''(1) = 6 - 12 = -6 < 0 \]

So x = 1 is a local maximum point.

\[ f''(3) = 18 - 12 = 6 > 0 \]

So x = 3 is a local minimum point. Computing the function values:

\[ f(1) = 1 - 6 + 9 + 15 = 19 \]

\[ f(3) = 27 - 54 + 27 + 15 = 15 \]

The local maximum value is 19 at x = 1, and the local minimum value is 15 at x = 3.

Exam Tip: When you have two critical points from a cubic function, use the second derivative to classify each one without guessing.

Question 10. Find the point of local maxima or local minima or local minima and the corresponding local maximum and minimum values of each of the following functions: f(x) = x⁴ - 62x² + 120x + 9.
Answer: Taking the first derivative and setting it equal to zero:

\[ f'(x) = 4x^3 - 124x + 120 = 0 \]

\[ 4(x^3 - 31x + 30) = 0 \]

Testing x = 1: 1 - 31 + 30 = 0, so (x - 1) is a factor.

Dividing x³ - 31x + 30 by (x - 1):

\[ x^3 - 31x + 30 = (x - 1)(x^2 + x - 30) = (x - 1)(x + 6)(x - 5) \]

So the critical points are x = 1, -6, 5.

Using the second derivative test:

\[ f''(x) = 12x^2 - 124 \]

\[ f''(1) = 12 - 124 = -112 < 0 \]

So x = 1 is a local maximum.

\[ f''(-6) = 12(36) - 124 = 432 - 124 = 308 > 0 \]

So x = -6 is a local minimum.

\[ f''(5) = 12(25) - 124 = 300 - 124 = 176 > 0 \]

So x = 5 is a local minimum. Computing the function values:

\[ f(1) = 1 - 62 + 120 + 9 = 68 \]

\[ f(-6) = 1296 - 62(36) + 120(-6) + 9 = 1296 - 2232 - 720 + 9 = -1647 \]

\[ f(5) = 625 - 62(25) + 120(5) + 9 = 625 - 1550 + 600 + 9 = -316 \]

The local maximum value is 68 at x = 1. The local minimum values are -1647 at x = -6 and -316 at x = 5.

Exam Tip: For polynomials of degree 4 and higher, use polynomial division or factoring to find all critical points after taking the derivative.

Question 11. Find the point of local maxima or local minima or local minima and the corresponding local maximum and minimum values of each of the following functions: f(x) = -x³ + 12x² - 5.
Answer: Taking the first derivative and setting it equal to zero:

\[ f'(x) = -3x^2 + 24x = 0 \]

\[ -3x(x - 8) = 0 \]

\[ x = 0 \text{ or } x = 8 \]

Using the second derivative test:

\[ f''(x) = -6x + 24 \]

\[ f''(0) = 24 > 0 \]

So x = 0 is a local minimum.

\[ f''(8) = -48 + 24 = -24 < 0 \]

So x = 8 is a local maximum. Computing the function values:

\[ f(0) = -5 \]

\[ f(8) = -512 + 12(64) - 5 = -512 + 768 - 5 = 251 \]

The local maximum value is 251 at x = 8, and the local minimum value is -5 at x = 0.

Exam Tip: For cubic functions with a negative leading coefficient, the function increases then decreases as x increases, creating one local maximum and one local minimum.

Question 12. Find the point of local maxima or local minima or local minima and the corresponding local maximum and minimum values of each of the following functions: f(x) = (x - 1)(x + 2)².
Answer: Using the product rule to find the first derivative:

\[ f'(x) = (x + 2)^2 + (x - 1) \cdot 2(x + 2) = (x + 2)[(x + 2) + 2(x - 1)] = (x + 2)(3x) \]

Setting f'(x) = 0:

\[ x = 0 \text{ or } x = -2 \]

Using the second derivative test, or analyzing the sign of f'(x):

\[ f''(x) = 3(x + 2) + 3x = 6x + 6 \]

\[ f''(0) = 6 > 0 \]

So x = 0 is a local minimum.

\[ f''(-2) = -12 + 6 = -6 < 0 \]

So x = -2 is a local maximum. Computing the function values:

\[ f(0) = (-1)(4) = -4 \]

\[ f(-2) = (-3)(0) = 0 \]

The local maximum value is 0 at x = -2, and the local minimum value is -4 at x = 0.

Exam Tip: When differentiating products, apply the product rule carefully and factor the result to find critical points more easily.

Question 13. Find the point of local maxima or local minima or local minima and the corresponding local maximum and minimum values of each of the following functions: f(x) = -(x - 1)³(x + 1)².
Answer: Taking the first derivative using the product rule:

\[ f'(x) = -(x - 1)^2 \cdot 2(x + 1) - 3(x - 1)^2(x + 1)^2 = (x - 1)^2[-(2(x+1) + 3(x+1)^2)] = 0 \]

This gives us critical points at x = 1, x = -1, and x = -1/5.

By analyzing f''(x) or the sign changes of f'(x), we find:

\[ f''(1) < 0 \text{ and } f''(-1) < 0 \]

So x = 1 and x = -1 are local maxima.

\[ f''(-1/5) > 0 \]

So x = -1/5 is a local minimum. Computing the function values:

\[ f(1) = 0 \]

\[ f(-1) = 0 \]

\[ f(-1/5) = -(-6/5)^3(4/5)^2 = -\frac{3456}{3125} \]

The local maximum value is 0 at x = 1 and x = -1. The local minimum value is \( -\frac{3456}{3125} \) at x = -1/5.

Exam Tip: For functions expressed as products of factors with different powers, the repeated factors give critical points where f' = 0, and you must check each point individually.

Question 14. Find the point of local maxima or local minima or local minima and the corresponding local maximum and minimum values of each of the following functions: f(x) = x/2 + 2/x, x > 0.
Answer: Taking the first derivative and setting it equal to zero:

\[ f'(x) = \frac{1}{2} - \frac{2}{x^2} = 0 \]

\[ x^2 = 4 \]

\[ x = 2 \text{ (since } x > 0) \]

Using the second derivative test:

\[ f''(x) = \frac{4}{x^3} \]

\[ f''(2) = \frac{4}{8} = \frac{1}{2} > 0 \]

So x = 2 is a local minimum. Computing the function value:

\[ f(2) = 1 + 1 = 2 \]

The local minimum value is 2 at x = 2.

Exam Tip: For rational functions with a positive domain restriction, always check that your critical points satisfy the domain constraint.

Question 15. Find the maximum and minimum values of 2x³ - 24x + 107 on the interval [-3, 3].
Answer: To find the extrema on a closed interval, we find critical points and evaluate the function at the critical points and endpoints.

Taking the first derivative:

\[ f'(x) = 6x^2 - 24 = 0 \]

\[ x^2 = 4 \]

\[ x = \pm 2 \]

Both critical points lie in [-3, 3]. Now we evaluate f at x = -3, -2, 2, 3:

\[ f(-3) = 2(-27) - 24(-3) + 107 = -54 + 72 + 107 = 125 \]

\[ f(-2) = 2(-8) - 24(-2) + 107 = -16 + 48 + 107 = 139 \]

\[ f(2) = 2(8) - 24(2) + 107 = 16 - 48 + 107 = 75 \]

\[ f(3) = 2(27) - 24(3) + 107 = 54 - 72 + 107 = 89 \]

The maximum value is 139 at x = -2, and the minimum value is 75 at x = 2.

Exam Tip: When finding extrema on a closed interval, never forget to check the endpoints - the global extremum might occur there rather than at a critical point.

Question 16. Find the maximum and minimum values of 3x⁴ - 8x³ + 12x² - 48x + 1 on the interval [1, 4].
Answer: Taking the first derivative:

\[ f'(x) = 12x^3 - 24x^2 + 24x - 48 = 0 \]

\[ 12(x^3 - 2x^2 + 2x - 4) = 0 \]

Testing x = 2: 8 - 8 + 4 - 4 = 0, so (x - 2) is a factor.

Dividing x³ - 2x² + 2x - 4 by (x - 2):

\[ x^3 - 2x^2 + 2x - 4 = (x - 2)(x^2 + 2) \]

Since x² + 2 > 0 for all real x, the only critical point in [1, 4] is x = 2. Now we evaluate f at x = 1, 2, 4:

\[ f(1) = 3 - 8 + 12 - 48 + 1 = -40 \]

\[ f(2) = 3(16) - 8(8) + 12(4) - 48(2) + 1 = 48 - 64 + 48 - 96 + 1 = -63 \]

\[ f(4) = 3(256) - 8(64) + 12(16) - 48(4) + 1 = 768 - 512 + 192 - 192 + 1 = 257 \]

The maximum value is 257 at x = 4, and the minimum value is -63 at x = 2.

Exam Tip: For higher-degree polynomials, try testing simple integer values when finding factors of the derivative.

Question 17. Find the maximum and minimum of \( f(x) = \sin x + \frac{1}{2}\cos x \) in \( 0 \leq x \leq \frac{\pi}{2} \).
Answer: Taking the first derivative and setting it equal to zero:

\[ f'(x) = \cos x - \frac{1}{2}\sin x = 0 \]

\[ \cos x = \frac{1}{2}\sin x \]

\[ 2\cos x = \sin x \]

\[ \tan x = 2 \]

Within the interval \( [0, \pi/2] \), we find x = arctan(2), which gives us critical points. We also check the endpoints and critical points by substituting into f(x). The general solutions are \( x = \frac{\pi}{6} \) and \( x = \frac{\pi}{3} \). Using the second derivative test, we determine which is a maximum and which is a minimum. Computing the function values at key points: \[ f(\pi/6) = \frac{1}{2} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} + \frac{\sqrt{3}}{4} = \frac{2 + \sqrt{3}}{4} \] However, using more careful analysis of f'(x) = 0 gives \( \tan x = 2 \), so: \[ f'(x) = 0 \text{ gives } \cos x - \frac{1}{2}\sin x = 0 \] Within the given interval, the maximum value is \( \frac{3}{4} \) at \( x = \frac{\pi}{6} \), and the minimum value is \( \frac{1}{2} \) at \( x = \frac{\pi}{2} \).

Exam Tip: For trigonometric functions on a closed interval, evaluate at critical points found by setting f'(x) = 0 and at the endpoints to find the global maximum and minimum.

Question 18. Show that the maximum value of x^(1/x) is e^(1/e).
Answer: Let \( y = x^{1/x} \). Taking the natural logarithm of both sides:

\[ \ln y = \frac{1}{x} \ln x \]

Differentiating both sides with respect to x:

\[ \frac{1}{y} \frac{dy}{dx} = -\frac{1}{x^2}\ln x + \frac{1}{x} \cdot \frac{1}{x} = -\frac{\ln x}{x^2} + \frac{1}{x^2} = \frac{1 - \ln x}{x^2} \]

\[ \frac{dy}{dx} = y \cdot \frac{1 - \ln x}{x^2} = x^{1/x} \cdot \frac{1 - \ln x}{x^2} \]

Setting \( \frac{dy}{dx} = 0 \):

\[ 1 - \ln x = 0 \]

\[ \ln x = 1 \]

\[ x = e \]

To confirm this is a maximum, note that the sign of \( \frac{dy}{dx} \) changes from positive to negative as x increases through e. Therefore, the maximum value is:

\[ y = e^{1/e} \]

Hence, the maximum value of x^(1/x) is e^(1/e).

Exam Tip: For functions involving exponents where the variable appears in both base and exponent, take logarithms first to convert to a more manageable form.

Question 19. Show that \( \left( x + \frac{1}{x} \right) \) has a maximum and minimum, but the maximum value is less than the minimum value.
Answer: Let \( f(x) = x + \frac{1}{x} \). Taking the first derivative:

\[ f'(x) = 1 - \frac{1}{x^2} \]

Setting \( f'(x) = 0 \):

\[ 1 - \frac{1}{x^2} = 0 \]

\[ x^2 = 1 \]

\[ x = 1 \text{ or } x = -1 \]

Using the second derivative test:

\[ f''(x) = \frac{2}{x^3} \]

\[ f''(1) = 2 > 0 \]

So x = 1 gives a local minimum.

\[ f''(-1) = -2 < 0 \]

So x = -1 gives a local maximum. Computing the function values:

\[ f(1) = 1 + 1 = 2 \]

\[ f(-1) = -1 - 1 = -2 \]

The function has both a maximum and minimum. The maximum value is -2 and the minimum value is 2. Since -2 < 2, the maximum value is indeed less than the minimum value.

Exam Tip: This unusual situation (max < min) occurs because the function is defined on a disconnected domain - the critical points for x > 0 and x < 0 are different, and the function has opposite signs on these regions.

Question 20. Find the maximum profit that a company can make, if the profit function is given by \( p(x) = 41 + 24x - 18x^2 \).
Answer: To maximize the profit, we take the derivative of p(x) and set it equal to zero:

\[ \frac{dp}{dx} = 24 - 36x = 0 \]

\[ x = \frac{24}{36} = \frac{2}{3} \]

Checking the second derivative:

\[ \frac{d^2p}{dx^2} = -36 < 0 \]

Since the second derivative is negative, x = 2/3 corresponds to a maximum. Computing the maximum profit:

\[ p(2/3) = 41 + 24(2/3) - 18(2/3)^2 = 41 + 16 - 18 \cdot \frac{4}{9} = 41 + 16 - 8 = 49 \]

The maximum profit is 49.

Exam Tip: In optimization problems, always verify that the critical point is a maximum (not a minimum) using the second derivative test before reporting the answer.

Question 21. An enemy jet is flying along the curve \( y = (x^2 + 2) \). A soldier is placed at the point (3, 2). Find the nearest point between the soldier and the jet.
Answer: Let P(x, y) be a point on the curve where y = x² + 2, and the soldier is at A(3, 2). The distance from A to P is:

\[ AP = \sqrt{(x - 3)^2 + (y - 2)^2} \]

Since y = x² + 2, we have y - 2 = x², so:

\[ AP^2 = (x - 3)^2 + x^4 = z \text{ (say)} \]

To minimize this distance, we minimize AP²:

\[ \frac{dz}{dx} = 2(x - 3) + 4x^3 = 0 \]

\[ 2x - 6 + 4x^3 = 0 \]

\[ 4x^3 + 2x - 6 = 0 \]

\[ 2x^3 + x - 3 = 0 \]

Testing x = 1: 2 + 1 - 3 = 0, so x = 1 is a solution.

Using the second derivative test:

\[ \frac{d^2z}{dx^2} = 2 + 12x^2 \]

\[ \frac{d^2z}{dx^2}\bigg|_{x=1} = 2 + 12 = 14 > 0 \]

So x = 1 gives a minimum. When x = 1, y = 1 + 2 = 3. The nearest point is (1, 3).

Exam Tip: When finding the nearest point on a curve to a fixed point, it's often easier to minimize the square of the distance rather than the distance itself, to avoid dealing with square roots.

Question 22. Find the maximum and minimum values of \( f(x) = (-x + 2\sin x) \) on \( [0, 2\pi] \).
Answer: Taking the first derivative and setting it equal to zero:

\[ f'(x) = -1 + 2\cos x = 0 \]

\[ \cos x = \frac{1}{2} \]

Within [0, 2π], this gives x = π/3 and x = 5π/3. Using the second derivative test:

\[ f''(x) = -2\sin x \]

\[ f''(\pi/3) = -2 \cdot \frac{\sqrt{3}}{2} = -\sqrt{3} < 0 \]

So x = π/3 is a local maximum.

\[ f''(5\pi/3) = -2 \cdot (-\frac{\sqrt{3}}{2}) = \sqrt{3} > 0 \]

So x = 5π/3 is a local minimum. Computing the function values at critical points and endpoints:

\[ f(\pi/3) = -\frac{\pi}{3} + 2 \cdot \frac{\sqrt{3}}{2} = -\frac{\pi}{3} + \sqrt{3} \]

\[ f(5\pi/3) = -\frac{5\pi}{3} + 2 \cdot (-\frac{\sqrt{3}}{2}) = -\frac{5\pi}{3} - \sqrt{3} \]

\[ f(0) = 0 \]

\[ f(2\pi) = -2\pi \]

The maximum value is \( -\frac{\pi}{3} + \sqrt{3} \) at x = π/3, and the minimum value is \( -\frac{5\pi}{3} - \sqrt{3} \) at x = 5π/3.

Exam Tip: Always evaluate the function at critical points AND at the endpoints of the closed interval to find the absolute maximum and minimum values.

Exercise 11F

Question 1. Find two positive number whose product is 49 and the sum is minimum.
Answer: Let x and y be the two positive numbers with x > 0 and y > 0. We are given that:

\[ xy = 49 \]

From this constraint, we express y in terms of x:

\[ y = \frac{49}{x} \]

The sum to minimize is:

\[ S = x + y = x + \frac{49}{x} \]

Taking the first derivative with respect to x:

\[ \frac{dS}{dx} = 1 - \frac{49}{x^2} \]

Setting this equal to zero:

\[ 1 - \frac{49}{x^2} = 0 \]

\[ x^2 = 49 \]

\[ x = 7 \text{ (since } x > 0) \]

To confirm this is a minimum, we check the second derivative:

\[ \frac{d^2S}{dx^2} = \frac{98}{x^3} \]

\[ \frac{d^2S}{dx^2}\bigg|_{x=7} = \frac{98}{343} > 0 \]

Since the second derivative is positive, x = 7 gives a minimum. When x = 7:

\[ y = \frac{49}{7} = 7 \]

Therefore, both numbers are 7, and their sum is minimized at 14.

Exam Tip: When optimizing subject to a constraint, express one variable in terms of the other using the constraint, then reduce the problem to a single-variable optimization.

Question 2. Find two positive numbers whose sum is 16 and the sum of whose squares is minimum.
Answer: Let x and y be the two positive numbers where x > 0 and y > 0. We have the constraint:

\[ x + y = 16 \]

From this, y = 16 - x. We want to minimize:

\[ S = x^2 + y^2 = x^2 + (16 - x)^2 \]

Expanding:

\[ S = x^2 + 256 - 32x + x^2 = 2x^2 - 32x + 256 \]

Taking the first derivative:

\[ \frac{dS}{dx} = 4x - 32 \]

Setting this equal to zero:

\[ 4x - 32 = 0 \]

\[ x = 8 \]

Checking the second derivative:

\[ \frac{d^2S}{dx^2} = 4 > 0 \]

Since the second derivative is positive, x = 8 gives a minimum. When x = 8:

\[ y = 16 - 8 = 8 \]

Therefore, both numbers are 8, and the sum of their squares is minimized at 128.

Exam Tip: For optimization problems with linear constraints, substitute to get a function of a single variable, then apply calculus to find the extremum.

Question 3. Divide 15 into two parts such that the square of one number multiplied with the cube of the other number is maximum.
Answer: Let x and y be the two numbers such that x + y = 15. We want to maximize:

\[ P = x^2 y^3 \]

From the constraint, y = 15 - x, so:

\[ P = x^2(15 - x)^3 \]

Taking the first derivative using the product rule:

\[ \frac{dP}{dx} = 2x(15 - x)^3 + x^2 \cdot 3(15 - x)^2 \cdot (-1) \]

\[ = 2x(15 - x)^3 - 3x^2(15 - x)^2 \]

\[ = x(15 - x)^2[2(15 - x) - 3x] \]

\[ = x(15 - x)^2[30 - 2x - 3x] \]

\[ = x(15 - x)^2[30 - 5x] \]

\[ = 5x(15 - x)^2(6 - x) \]

Setting \( \frac{dP}{dx} = 0 \):

\[ x = 0 \text{ or } x = 15 \text{ or } x = 6 \]

Since we need both numbers to be positive and meaningful, x = 6 is the relevant critical point. When x = 6:

\[ y = 15 - 6 = 9 \]

We can verify this is a maximum by checking the second derivative or by noting that P = 0 at the boundaries (x = 0, 15). Therefore, we divide 15 into 6 and 9, and the maximum value of the product is \( 6^2 \cdot 9^3 = 36 \cdot 729 = 26244 \).

Exam Tip: When maximizing a product of powers subject to a sum constraint, factor out common terms from the derivative to find critical points more easily.

Question 4. Divide 8 into two positive parts such that the sum of the square of one and the cube of the other is minimum.
Answer: Let x and y represent the two numbers. Since their sum is 8, we have x + y = 8, which gives y = 8 - x. The function to minimize is S = x³ + y². Substituting y = 8 - x, we get S = x³ + (8 - x)². Taking the derivative and setting it to zero: dS/dx = 3x² + 2(8 - x)(-1) = 3x² - 2(8 - x) = 0, which simplifies to 3x² + 2x - 16 = 0. Using the quadratic formula gives x = 2 or x = -2.67. Since we need positive parts, we consider x = 2. Checking the second derivative at x = 2 gives d²S/dx² = 6x + 2 = 14 > 0, confirming a minimum. Therefore, x = 2 and y = 8 - 2 = 6 are the two positive numbers whose sum is 8 and the sum of the square of one and cube of the other is at its smallest.
In simple words: Split 8 into two pieces: 2 and 6. When you square one piece (2² = 4) and cube the other piece (6³ = 216), their sum (220) is as small as it can be.

Exam Tip: Always verify critical points with the second derivative test - a positive second derivative confirms a minimum, while negative confirms a maximum.

Question 5. Divide a into two parts such that the product of the pth power of one part and the qth power of the second part may be maximum.
Answer: Let x and y be the two numbers such that x + y = a. The function to maximize is P = x^p × y^q. Substituting y = a - x gives P = x^p × (a - x)^q. Taking the first derivative: dP/dx = x^(p-1)(a - x)^(q-1)[ap - x(p + q)]. Setting this equal to zero yields the critical point x = ap/(p + q). Using the first derivative test, we verify that the derivative changes from positive to negative as x passes through this critical point, confirming a local maximum. Therefore, the two parts are x = ap/(p + q) and y = aq/(p + q), and their product of p-th and q-th powers is maximized.
In simple words: To maximize the product of powers, divide the number a into two parts using the ratio p : q. The first part gets p/(p + q) of the total, and the second part gets q/(p + q).

Exam Tip: Remember that the first derivative test identifies local extrema by checking sign changes - positive to negative signals a maximum.

Question 6. The rate of working of an engine is given by R = 15v + 6000/v, where 0 < v < 30 and v is the speed of the engine. Show that R is the least when v = 20.
Answer: Given the rate function R = 15v + 6000/v where 0 < v < 30. Taking the first derivative with respect to v: dR/dv = 15 - 6000/v². Setting this equal to zero gives 15 = 6000/v², so v² = 400, yielding v = 20 (taking the positive root). To verify this is a minimum, we compute the second derivative: d²R/dv² = 12000/v³. At v = 20, this gives d²R/dv² = 12000/8000 = 3/2 > 0, confirming that R reaches its minimum value at v = 20.
In simple words: When the engine speed is 20, the rate of working is at its lowest point. The second derivative is positive, which proves this is truly a minimum and not a maximum.

Exam Tip: For applied optimization problems, always verify that your critical point lies within the given domain constraints (here, 0 < v < 30).

Question 7. Find the dimensions of the rectangle of area 96 cm² whose perimeter is the least. Also, find the perimeter of the rectangle.
Answer: Let x and y be the length and breadth of the rectangle with area xy = 96. From this constraint, y = 96/x. The perimeter is P = 2(x + y) = 2(x + 96/x). Taking the first derivative: dP/dx = 2(1 - 96/x²). Setting this to zero gives x² = 96, so x = √96 = 4√6. Checking the second derivative: d²P/dx² = 2(192/x³) > 0, confirming a minimum. When x = 4√6, we have y = 96/(4√6) = 24√6/24 = 4√6. Therefore, the rectangle is actually a square with sides 4√6 cm. The minimum perimeter is P = 2(4√6 + 4√6) = 16√6 cm.
In simple words: To get the smallest perimeter for a 96 cm² rectangle, make it a square with each side equal to 4√6 cm, giving a perimeter of 16√6 cm.

Exam Tip: When optimizing perimeter for a fixed area, the extremum often occurs at a square - this is a useful pattern to remember for rectangles.

Question 8. Prove that the largest rectangle with a given perimeter is a square.
Answer: Let p be the fixed perimeter and x, y be the sides of a rectangle. Then p = 2(x + y), giving y = (p - 2x)/2. The area is A = xy = x(p - 2x)/2 = (1/2)(px - 2x²). Taking the first derivative: dA/dx = (1/2)(p - 4x). Setting this to zero yields p = 4x, so x = p/4. Substituting back: y = (p - 2(p/4))/2 = p/4. Thus x = y, proving the rectangle is a square. The second derivative d²A/dx² = -2 < 0 confirms this is a maximum. Therefore, a rectangle with maximum area and fixed perimeter must be a square.
In simple words: Among all rectangles with the same perimeter, the square has the biggest area. This happens because equal sides allow the area to be as large as possible.

Exam Tip: This classic result shows that symmetric shapes (like squares) often optimize area or perimeter constraints - a principle worth remembering across geometry problems.

Question 9. Given the perimeter of a rectangle, show that its diagonal is minimum when it is a square.
Answer: Let p be the fixed perimeter and x, y be the sides. From p = 2(x + y), we get y = (p - 2x)/2. The diagonal squared is Z = D² = x² + y² = x² + ((p - 2x)/2)². Expanding: Z = x² + (p - 2x)²/4. Taking the first derivative: dZ/dx = 2x + (1/4) · 2(p - 2x)(-2) = 2x - (p - 2x)/2 = 4x - p. Setting to zero gives x = p/4. Substituting back: y = (p - 2(p/4))/2 = p/4, so x = y. The second derivative d²Z/dx² = 4 > 0 confirms this is a minimum. Therefore, the diagonal is minimized when the rectangle is a square.
In simple words: For a rectangle with fixed perimeter, the shortest diagonal happens when the rectangle becomes a square with equal sides.

Exam Tip: When working with diagonal length, remember to square it first to eliminate the square root - this simplifies differentiation significantly.

Question 10. Show that a rectangle of maximum perimeter which can be inscribed in a circle of radius a is a square of side a√2.
Answer: Let x and y be the length and breadth of the rectangle inscribed in a circle of radius a. The diagonal of the rectangle equals the diameter of the circle: x² + y² = 4a². From this constraint, y = √(4a² - x²). The perimeter is P = 2x + 2y = 2x + 2√(4a² - x²). Taking the first derivative: dP/dx = 2 - 2x/√(4a² - x²). Setting this to zero: √(4a² - x²) = x, which gives 4a² - x² = x², so x² = 2a², yielding x = a√2. Substituting back: y = √(4a² - 2a²) = √(2a²) = a√2. The second derivative is negative at this point, confirming a maximum. Therefore, the inscribed rectangle with maximum perimeter is a square with side a√2.
In simple words: When you fit a rectangle inside a circle of radius a with the maximum perimeter, that rectangle must be a square with each side measuring a√2.

Exam Tip: For problems involving inscribed or circumscribed shapes, use the geometric constraint (here, the diagonal equals the diameter) to relate the variables before optimizing.

Question 11. The sum of the perimeters of a square and a circle is given. Show that the sum of their areas is least when

Exam Tip: This optimization problem combines two shapes - set up the constraint linking their perimeters, then express total area in terms of a single variable before differentiating.

Question 11. Write the repeating decimal for each of the following and use a bar to show the repetend.
(i) \( \frac { 1 }{ 9 } \)
(ii) \( \frac { -4 }{ 3 } \)
(iii) \( \frac { 1 }{ 6 } \)
Answer: The first fraction equals 0 with 1 repeating, written as \( 0.\overline{1} \).
(ii) When you divide -4 by 3, you get -1 with 3 repeating, shown as \( -1.\overline{3} \)
(iii) \( \frac { 1 }{ 6 } \) produces 0.1 with 6 repeating indefinitely, so it's written \( 0.1\overline{6} \). Notice that only the 6 repeats - the 1 appears just once before the repeating section starts.
In simple words: When you divide the top by the bottom and get a digit (or several digits) repeating forever, put a bar over those repeating digits to show the pattern continues without end.

Exam Tip: Be careful to place the bar only over the digits that truly repeat, not over any digits that appear just once before the repetition begins.

Question 12. Show that the right triangle of maximum area that can be inscribed in a circle is an isosceles triangle.
Answer: Consider a right-angled triangle placed inside a circle such that the hypotenuse lies along the diameter. Let the radius of the circle be r, and let x and y stand for the base and height of the triangle.

Since the hypotenuse equals the diameter, we have \( 4r^2 = x^2 + y^2 \), which gives us \( y = \sqrt{4r^2 - x^2} \).

The area of the triangle is \( A = \frac{1}{2}xy = \frac{1}{2}x\sqrt{4r^2 - x^2} \).

To find when the area is largest, we differentiate with respect to x and set equal to zero. This yields \( 2r^2x = x^3 \), or \( x^2 = 2r^2 \).

From the second derivative test, we confirm this is indeed a maximum. Substituting back, we find that \( x = y = r\sqrt{2} \).

Since the base and height are equal, the triangle is isosceles (two of its sides are the same length).
In simple words: When you want a right triangle inside a circle to have the biggest possible area, the two legs (the sides that form the right angle) must be exactly equal in length, which makes it isosceles.

Exam Tip: Always use the relationship between the hypotenuse and the circle's diameter to set up your constraint, then use calculus to prove equality of the two perpendicular sides.

Question 13. Prove that the perimeter of a right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.
Answer: Let h be the fixed hypotenuse, and let x and y denote the two perpendicular sides. By the Pythagorean theorem, \( y = \sqrt{h^2 - x^2} \).

The perimeter is \( P = h + x + \sqrt{h^2 - x^2} \).

Taking the derivative with respect to x and setting it to zero yields the condition \( x = \sqrt{h^2 - x^2} \), which simplifies to \( x^2 = \frac{h^2}{2} \), so \( x = \frac{h}{\sqrt{2}} \).

The second derivative test confirms this is a maximum. Substituting back into the expression for y, we find \( y = \frac{h}{\sqrt{2}} \).

Since x equals y, the two legs of the right triangle are equal in length, making it isosceles.
In simple words: If you keep the longest side (hypotenuse) of a right triangle fixed and try to make the perimeter as large as possible, the other two sides must be equal to each other.

Exam Tip: The key insight is that setting the derivative equal to zero forces the two legs to be equal - this is the algebraic proof of the isosceles property.

Question 14. The perimeter of a triangle is 8 cm. If one of the sides of the triangle be 3 cm, what will be the other two sides for maximum area of the triangle?
Answer: We are given that the perimeter is 8 cm and one side is 3 cm. This means the sum of the other two sides is \( 8 - 3 = 5 \) cm. Let these two sides be x and y, so \( y = 5 - x \).

Using Heron's formula, the semi-perimeter is \( s = \frac{8}{2} = 4 \). The area is \( A = \sqrt{4(4-3)(4-x)(4-(5-x))} = \sqrt{4(4-x)(x-1)} \).

Squaring both sides to simplify, we get \( Z = A^2 = 4(5x - x^2 - 4) \).

Differentiating with respect to x and setting the result to zero gives \( 20 - 8x = 0 \), so \( x = 2.5 \).

The second derivative test confirms this is a maximum. Substituting back, we find \( y = 5 - 2.5 = 2.5 \).

Therefore, the other two sides are both 2.5 cm each, making the triangle isosceles with sides 3 cm, 2.5 cm, and 2.5 cm.
In simple words: When you have one side fixed at 3 cm and the perimeter locked at 8 cm, the area becomes biggest when the remaining two sides are each 2.5 cm long.

Exam Tip: Always use Heron's formula when the perimeter constraint is given, and remember that squaring the area simplifies the calculus.

Question 15. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 metres. Find the dimensions of the windows to admit maximum light through it.
Answer: Let x be the width of the rectangular base and y be its height. The semicircle has diameter x, so its radius is \( \frac{x}{2} \).

The perimeter condition states \( 10 = (x + 2y) + \pi\left(\frac{x}{2}\right) \), which simplifies to \( y = \frac{20 - 2x - \pi x}{4} \).

The total area is the rectangle plus the semicircle: \( A = xy + \frac{1}{2}\pi\left(\frac{x}{2}\right)^2 \).

After substitution and simplification, \( A = \frac{1}{8}\left[x^2(\pi - 2\pi - 4) + 40x\right] \).

Differentiating with respect to x and setting equal to zero gives \( x(4 + \pi) = 20 \), so \( x = \frac{20}{4+\pi} \).

The second derivative is negative, confirming a maximum. Substituting back yields \( y = \frac{10}{4+\pi} \).

Thus the window has width \( \frac{20}{4+\pi} \) metres and height \( \frac{10}{4+\pi} \) metres.
In simple words: To let the most light in through a window made of a rectangle with a semicircle on top, given a fixed perimeter of 10 metres, the width should be twice the height.

Exam Tip: Set up the perimeter constraint carefully, including the curved part of the semicircle, and use it to eliminate one variable before optimizing.

Question 16. A square piece of tin of side 12 cm is to be made into a box without a lid by cutting a square from each corner and folding up the flaps to form the sides. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find this maximum volume.
Answer: Let x be the side length of the square cut from each corner. After cutting and folding, the base of the box has dimensions \( (12 - 2x) \times (12 - 2x) \) and the height is x.

The volume is \( V = (12-2x)^2 \cdot x = 4x^3 - 48x^2 + 144x \).

Taking the derivative and setting it to zero: \( \frac{dV}{dx} = 12x^2 - 96x + 144 = 0 \), which simplifies to \( x^2 - 8x + 12 = 0 \).

This factors to give \( x = 6 \) or \( x = 2 \). Since \( x = 6 \) would result in zero dimensions, we reject it. So \( x = 2 \) cm.

The second derivative test confirms this is a maximum. When \( x = 2 \), the base side becomes \( 12 - 4 = 8 \) cm.

The maximum volume is \( V = 8^2 \times 2 = 128 \text{ cm}^3 \).
In simple words: Cut 2 cm squares from each corner, fold up the sides, and you get a box with an 8 cm by 8 cm base and 2 cm height - this holds the most volume of 128 cubic centimetres.

Exam Tip: Always check both critical points and reject solutions that lead to non-positive dimensions - physical reality constrains the answer.

Question 17. An open box with a square base is to be made out of a given cardboard of area c² (square) units. Show that the maximum volume of the box is \( \frac{c^3}{6\sqrt{3}} \) (cubic) units.
Answer: Let a be the side of the square base and h be the height. The surface area (base plus four sides) is \( c^2 = a^2 + 4ah \), giving \( h = \frac{c^2 - a^2}{4a} \).

The volume is \( V = a^2h = a^2 \cdot \frac{c^2 - a^2}{4a} = \frac{a(c^2 - a^2)}{4} \).

Differentiating with respect to a: \( \frac{dV}{da} = \frac{c^2 - 3a^2}{4} \). Setting this to zero gives \( c^2 = 3a^2 \), so \( a = \frac{c}{\sqrt{3}} \).

The second derivative is negative, confirming a maximum. Substituting back: \( h = \frac{c^2 - \frac{c^2}{3}}{4 \cdot \frac{c}{\sqrt{3}}} = \frac{c}{2\sqrt{3}} \).

The maximum volume is \( V = \left(\frac{c}{\sqrt{3}}\right)^2 \times \frac{c}{2\sqrt{3}} = \frac{c^2}{3} \times \frac{c}{2\sqrt{3}} = \frac{c^3}{6\sqrt{3}} \).
In simple words: When you have a fixed amount of cardboard to make an open box with a square bottom, the volume is maximized when the base side is \( \frac{c}{\sqrt{3}} \) and the height is \( \frac{c}{2\sqrt{3}} \), yielding a maximum volume of \( \frac{c^3}{6\sqrt{3}} \) cubic units.

Exam Tip: Remember that for an open box, the surface area includes only the base and four sides - no top - this changes the constraint equation.

Question 18. A cylindrical can is to be made to hold 1 litre of oil. Find the dimensions which will minimize the cost of the metal to make the can.
Answer: Let r be the radius and h be the height of the cylinder. The volume constraint is \( \pi r^2 h = 1000 \text{ cm}^3 \), giving \( h = \frac{1000}{\pi r^2} \).

The surface area (cost is proportional to this) is \( S = 2\pi r^2 + 2\pi rh \). Substituting the expression for h: \( S = 2\pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2} = 2\pi r^2 + \frac{2000}{r} \).

Differentiating with respect to r: \( \frac{dS}{dr} = 4\pi r - \frac{2000}{r^2} \). Setting this to zero gives \( 4\pi r^3 = 2000 \), so \( r^3 = \frac{500}{\pi} \), yielding \( r = \sqrt[3]{\frac{500}{\pi}} \).

The second derivative is positive, confirming a minimum. Substituting back: \( h = \frac{1000}{\pi \left(\sqrt[3]{\frac{500}{\pi}}\right)^2} \), which simplifies to \( h = 2\sqrt[3]{\frac{500}{\pi}} = 2r \).

Therefore, the minimum-cost cylinder has height equal to twice the radius, or equivalently, the height equals the diameter.
In simple words: To use the least metal when making a cylindrical can holding 1 litre, the height must be exactly twice the radius - in other words, the height should equal the diameter.

Exam Tip: For optimization with volume constraints, always express one dimension in terms of the other using the constraint, then minimize the surface area function.

Question 19. Show that the right circular cone of the least curved surface and given volume has an altitude equal to \( \sqrt{2} \) times the radius of the base.
Answer: Let the radius of the cone's base be r cm and its height be h cm. The slant height is l cm. Given that the cone's volume is fixed, we have \( V = \frac{1}{3}\pi r^2 h \), which gives us \( h = \frac{3V}{\pi r^2} \). The curved surface area is \( S = \pi rl \). Since \( l = \sqrt{h^2 + r^2} \), we can write \( S = \pi r\sqrt{h^2 + r^2} \). Substituting the expression for h: \( S = \pi r\sqrt{\left(\frac{3V}{\pi r^2}\right)^2 + r^2} \) To minimize the surface area, we square both sides to get \( Z = S^2 = \pi^2\left(\frac{9V^2}{\pi^2 r^4} + r^2\right) \). Taking the derivative with respect to r and setting it equal to zero: \( \frac{dZ}{dr} = \pi^2\left(\frac{-18V^2}{\pi^2 r^5} + 2r\right) = 0 \) This yields \( 2\pi^2 r^6 = 9V^2 \), or \( 2r^2 = h^2 \). Therefore, \( h = r\sqrt{2} \), confirming that the altitude equals \( \sqrt{2} \) times the radius of the base.
In simple words: When a cone holds a fixed amount of space inside it, its curved outer skin covers the smallest area when the height is \( \sqrt{2} \) times as long as the radius at the bottom.

Exam Tip: Always use the Pythagorean relationship for the slant height, and remember to square the surface area function before differentiating to simplify the algebra.

Question 20. Find the radius of a closed right circular cylinder of volume 100 cm³ which has the minimum total surface area.
Answer: Let the radius of the cylinder's base be r units and its height be h units. Since the volume is 100 cm³, we have \( \pi r^2 h = 100 \), giving us \( h = \frac{100}{\pi r^2} \). The total surface area of the closed cylinder is \( S = 2\pi r^2 + 2\pi rh \). Substituting the expression for h: \( S = 2\pi r^2 + 2\pi r\left(\frac{100}{\pi r^2}\right) = 2\pi r^2 + \frac{200}{r} \) To find the minimum, we differentiate with respect to r: \( \frac{dS}{dr} = 4\pi r - \frac{200}{r^2} \) Setting this equal to zero: \( 4\pi r = \frac{200}{r^2} \), which gives us \( 2\pi r = \frac{100}{r^2} \) Solving for r: \( 2\pi r^3 = 100 \), so \( r^3 = \frac{50}{\pi} \), and \( r = \sqrt[3]{\frac{50}{\pi}} \) To verify this is a minimum, we check the second derivative: \( \frac{d^2S}{dr^2} = 4\pi + \frac{400}{r^3} > 0 \) Since the second derivative is positive, this critical point gives the minimum surface area. From the condition \( 2\pi r = \frac{100}{r^2} \), we also find that \( V = 2\pi r^3 \), meaning the height equals twice the radius (or the diameter).
In simple words: To cover a cylinder holding 100 cubic centimeters with the smallest amount of material, make its height equal to its width across (the diameter). The radius works out to the cube root of \( \frac{50}{\pi} \) units.

Exam Tip: Remember that for a cylinder with fixed volume, the minimum surface area occurs when height equals the diameter - this is a key relationship worth memorizing.

Question 21. Show that the height of a closed cylinder of given volume and the least surface area is equal to its diameter.
Answer: Let r be the radius of the cylinder's base and h be its height. The surface area is given by \( S = 2\pi r^2 + 2\pi rh \). If V is the cylinder's volume, then \( V = \pi r^2 h \), which allows us to express \( h = \frac{V}{\pi r^2} \). Substituting into the surface area formula: \( S = 2\pi r^2 + 2\pi r \cdot \frac{V}{\pi r^2} = 2\pi r^2 + \frac{2V}{r} \) Taking the derivative with respect to r: \( \frac{dS}{dr} = 4\pi r - \frac{2V}{r^2} \) Setting the derivative equal to zero for critical points: \( 4\pi r = \frac{2V}{r^2} \), which simplifies to \( 2\pi r^3 = V \) This means \( \pi r^2 h = 2\pi r^3 \), so \( h = 2r \). Checking the second derivative: \( \frac{d^2S}{dr^2} = 4\pi + \frac{4V}{r^3} > 0 \), confirming this is indeed a minimum. Since h = 2r, the height equals twice the radius, which is the diameter of the cylinder.
In simple words: When you want a cylinder to hold a fixed amount of liquid while using the least wrapping material, make its height exactly equal to its width all the way across - that is, equal to the diameter.

Exam Tip: The relationship h = 2r (or h = diameter) is a standard result for cylinders with minimum surface area - always derive it carefully and show the second derivative test clearly.

Question 22. Prove that the volume of the largest cone that can be inscribed in a sphere is \( \frac{8}{27} \) of the volume of the sphere.
Answer: Let the sphere have radius a units. Consider a cone inscribed in the sphere with height h and base radius r. The apex of the cone touches the sphere, and its circular base lies within the sphere. From the geometry of the sphere, we can establish that \( r^2 = h(2a - h) \). The volume of the cone is \( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi h(2a - h) \cdot h = \frac{\pi}{3}(2ah^2 - h^3) \). To find the maximum volume, we differentiate with respect to h: \( \frac{dV}{dh} = \frac{\pi}{3}(4ah - 3h^2) \) Setting this equal to zero: \( 4ah - 3h^2 = 0 \), which gives \( h = \frac{4a}{3} \) (taking the non-zero solution). We verify this is a maximum by checking the second derivative: \( \frac{d^2V}{dh^2} = \frac{\pi}{3}(4a - 6h) = \frac{\pi}{3}(4a - 8a) = -\frac{4\pi a}{3} < 0 \) Substituting \( h = \frac{4a}{3} \) back into the radius formula: \( r^2 = \frac{4a}{3}\left(2a - \frac{4a}{3}\right) = \frac{4a}{3} \cdot \frac{2a}{3} = \frac{8a^2}{9} \) The maximum volume of the inscribed cone is: \( V_{cone} = \frac{1}{3}\pi \cdot \frac{8a^2}{9} \cdot \frac{4a}{3} = \frac{32\pi a^3}{81} \) The volume of the sphere is \( V_{sphere} = \frac{4}{3}\pi a^3 \). The ratio is: \( \frac{V_{cone}}{V_{sphere}} = \frac{\frac{32\pi a^3}{81}}{\frac{4\pi a^3}{3}} = \frac{32\pi a^3}{81} \cdot \frac{3}{4\pi a^3} = \frac{8}{27} \) Therefore, the largest inscribed cone has volume equal to \( \frac{8}{27} \) of the sphere's volume.
In simple words: When you fit the biggest cone possible inside a sphere, the cone's internal space takes up exactly 8 parts out of 27 parts of the sphere's total space.

Exam Tip: Use the Pythagorean relationship to link r and h to the sphere's radius, then optimize the volume function - this approach makes the algebra more manageable.

Question 23. Which fraction exceeds its pth power by the greatest number possible?
Answer: Let x be the required fraction, where 0 < x < 1. We want to maximize the difference \( y = x - x^p \). Differentiating with respect to x: \( \frac{dy}{dx} = 1 - px^{p-1} \) Setting this equal to zero: \( 1 = px^{p-1} \), which gives \( x^{p-1} = \frac{1}{p} \) Solving for x: \( x = \left(\frac{1}{p}\right)^{\frac{1}{p-1}} \) To verify this is a maximum, we check the second derivative: \( \frac{d^2y}{dx^2} = -p(p-1)x^{p-2} < 0 \) (since x > 0 and p > 1) This confirms that the critical point gives a maximum. The greatest number by which this fraction exceeds its pth power is: \( y = x - x^p = \left(\frac{1}{p}\right)^{\frac{1}{p-1}} - \left(\frac{1}{p}\right)^{\frac{p}{p-1}} = \left(\frac{1}{p}\right)^{\frac{1}{p-1}}\left(1 - \frac{1}{p}\right) = \left(\frac{1}{p}\right)^{\frac{1}{p-1}} \cdot \frac{p-1}{p} \) Therefore, the fraction that exceeds its pth power by the greatest amount is \( x = \left(\frac{1}{p}\right)^{\frac{1}{p-1}} \).
In simple words: Among all numbers between 0 and 1, the one that beats its own pth power by the biggest margin is \( \left(\frac{1}{p}\right)^{\frac{1}{p-1}} \). For example, if p = 2, this fraction equals \( \frac{1}{\sqrt{2}} \), which exceeds its square by \( \frac{1}{2\sqrt{2}} \).

Exam Tip: Always set the first derivative to zero and verify the sign of the second derivative - a negative second derivative confirms you have found a maximum, not a minimum.

Question 24. Find the point on the curve \( y^2 = 4x \) which is nearest to the point (2, -8).
Answer: Let P(x, y) be a point on the curve \( y^2 = 4x \). We want to minimize the distance from P to the point (2, -8). The distance squared is \( D^2 = (x - 2)^2 + (y + 8)^2 \). Expanding: \( D^2 = x^2 - 4x + 4 + y^2 + 16y + 64 \) Since the point lies on the curve, \( x = \frac{y^2}{4} \). Substituting: \( D^2 = \left(\frac{y^2}{4}\right)^2 - 4 \cdot \frac{y^2}{4} + 4 + y^2 + 16y + 64 = \frac{y^4}{16} - y^2 + y^2 + 16y + 68 = \frac{y^4}{16} + 16y + 68 \) To minimize the distance, we differentiate with respect to y: \( \frac{d(D^2)}{dy} = \frac{4y^3}{16} + 16 = \frac{y^3}{4} + 16 \) Setting this equal to zero: \( \frac{y^3}{4} + 16 = 0 \), so \( y^3 = -64 \), giving \( y = -4 \) Checking the second derivative: \( \frac{d^2(D^2)}{dy^2} = \frac{3y^2}{4} = \frac{3 \cdot 16}{4} = 12 > 0 \) This confirms that y = -4 gives a minimum distance. Substituting y = -4 into the curve equation: \( (-4)^2 = 4x \), so \( 16 = 4x \), giving \( x = 4 \) Therefore, the point on the curve nearest to (2, -8) is **(4, -4)**.
In simple words: The point on the parabola \( y^2 = 4x \) that sits closest to (2, -8) has coordinates (4, -4). This point lies directly to the right of and slightly above the target point.

Exam Tip: Always substitute the curve equation into the distance formula before differentiating - this reduces the number of variables and simplifies the calculus significantly.

Question 25. A right circular cylinder is inscribed in a cone. Show that the curved surface area of the cylinder is maximum when the diameter of the cylinder is equal to the radius of the base of the cone.
Answer: Let \( r_1 \) be the radius of the cone's base and \( h_1 \) be its height. Let r be the radius of the inscribed cylinder and h be its height. By similar triangles, we have the relationship: \( \frac{h_1 - h}{h_1} = \frac{r}{r_1} \), which gives us \( h = h_1\left(1 - \frac{r}{r_1}\right) \) The curved surface area of the cylinder is \( S = 2\pi rh \). Substituting the expression for h: \( S = 2\pi rh_1\left(1 - \frac{r}{r_1}\right) = 2\pi rh_1 - \frac{2\pi h_1 r^2}{r_1} \) To find the maximum, we differentiate with respect to r: \( \frac{dS}{dr} = 2\pi h_1 - \frac{4\pi h_1 r}{r_1} \) Setting this equal to zero: \( 2\pi h_1 = \frac{4\pi h_1 r}{r_1} \), which simplifies to \( r = \frac{r_1}{2} \) Checking the second derivative: \( \frac{d^2S}{dr^2} = -\frac{4\pi h_1}{r_1} < 0 \) This confirms that r = \( \frac{r_1}{2} \) gives a maximum. Since the diameter of the cylinder is 2r and this equals \( 2 \cdot \frac{r_1}{2} = r_1 \), the diameter of the cylinder equals the radius of the cone's base when the curved surface area is maximum.
In simple words: When you slide a cylinder up and down inside a cone, its outer curved skin has the largest area when the cylinder's width all the way across (its diameter) matches the cone's radius at the bottom.

Exam Tip: Use the similar triangles property to express one variable in terms of another - this eliminates the need for two variables and makes the optimization straightforward.

Question 26. Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.
Answer: Let the side of the square base be x and the height of the cuboid be h. Given that the volume is constant, \( V = x^2 h \), we have \( h = \frac{V}{x^2} \). The total surface area consists of two square bases and four rectangular sides: \( S = 2x^2 + 4xh \) Substituting the expression for h: \( S = 2x^2 + 4x \cdot \frac{V}{x^2} = 2x^2 + \frac{4V}{x} \) Differentiating with respect to x: \( \frac{dS}{dx} = 4x - \frac{4V}{x^2} \) Setting this equal to zero: \( 4x = \frac{4V}{x^2} \), which gives \( x^3 = V \) Checking the second derivative: \( \frac{d^2S}{dx^2} = 4 + \frac{8V}{x^3} > 0 \) This confirms that x³ = V gives a minimum surface area. Substituting back into the height formula: \( h = \frac{V}{x^2} = \frac{x^3}{x^2} = x \) Since h = x, all three dimensions of the cuboid are equal, making it a cube. Therefore, among all cuboids with a square base and fixed volume, the cube has the minimum surface area.
In simple words: If you want to wrap up a box holding a fixed amount of stuff, using the least paper, make it a perfect cube where all sides are the same length.

Exam Tip: When h = x emerges from your algebra, recognize that this means the cuboid must be a cube - state this conclusion clearly and explicitly in your final answer.

Question 27. A rectangle is inscribed in a semicircle of radius r with one of its sides on the diameter of the semicircle. Find the dimensions of the rectangle so that its area is maximum. Find also this area.
Answer: Let the rectangle have base x and height y. Since one side lies on the diameter of the semicircle and the opposite side's corners touch the semicircular arc, we use the equation of the circle. For a semicircle of radius r centered at the origin with diameter along the x-axis, a point on the arc at distance \( \frac{x}{2} \) from the center satisfies: \( r^2 = \left(\frac{x}{2}\right)^2 + y^2 \) This gives us: \( y^2 = r^2 - \frac{x^2}{4} \), so \( y = \sqrt{r^2 - \frac{x^2}{4}} \) The area of the rectangle is: \( A = xy = x\sqrt{r^2 - \frac{x^2}{4}} \) To find the maximum, we can work with \( A^2 \): \( A^2 = x^2\left(r^2 - \frac{x^2}{4}\right) = x^2 r^2 - \frac{x^4}{4} \) Differentiating with respect to x: \( \frac{d(A^2)}{dx} = 2xr^2 - x^3 \) Setting this equal to zero: \( 2xr^2 = x^3 \), which (for x > 0) gives \( x^2 = 2r^2 \), so \( x = r\sqrt{2} \) Substituting back to find y: \( y^2 = r^2 - \frac{(r\sqrt{2})^2}{4} = r^2 - \frac{r^2}{2} = \frac{r^2}{2} \), so \( y = \frac{r}{\sqrt{2}} = \frac{r\sqrt{2}}{2} \) The maximum area is: \( A = xy = r\sqrt{2} \cdot \frac{r\sqrt{2}}{2} = \frac{2r^2}{2} = r^2 \) Therefore, the rectangle has dimensions \( r\sqrt{2} \) by \( \frac{r\sqrt{2}}{2} \), and its maximum area is \( r^2 \).
In simple words: To fit the biggest rectangle inside a semicircle with one edge along the flat bottom, make the base \( r\sqrt{2} \) units wide and the height \( \frac{r\sqrt{2}}{2} \) units tall. This largest rectangle will have area \( r^2 \) square units.

Exam Tip: Working with the squared area function eliminates the square root and makes differentiation easier - remember to take the positive square root when finding y to ensure the correct physical dimension.

Question 28. Two sides of a triangle have lengths a and b and the angle between them is θ. What value of θ will maximize the area of the triangle?
Answer: We are given that two sides of a triangle measure 'a' and 'b', with an angle θ between them. We need to find the angle that produces the largest possible area. Consider triangle PQR. The area formula is: \( A = \frac{1}{2} ab \sin\theta \) ---- (1) To find the maximum or minimum, differentiate with respect to θ and set equal to zero. This works because if A(θ) reaches its peak or minimum at point c, then A'(c) = 0. Differentiating equation (1) with respect to θ: \( \frac{dA}{d\theta} = \frac{d}{d\theta}\left[\frac{1}{2} ab \sin\theta\right] \) \( \frac{dA}{d\theta} = \frac{1}{2} ab \cos\theta \) ---- (2) To find the critical point, equate equation (2) to zero: \( \frac{dA}{d\theta} = \frac{1}{2} ab\cos\theta = 0 \) \( \cos\theta = 0 \) \( \theta = \frac{\pi}{2} \) To verify this critical point gives maximum area, take the second derivative: Differentiating equation (2) with respect to θ: \( \frac{d^2A}{d\theta^2} = \frac{d}{d\theta}\left[\frac{1}{2} ab \cos\theta\right] \) \( \frac{d^2A}{d\theta^2} = -\frac{1}{2} ab \sin\theta \) ---- (3) Find the value of the second derivative at the critical point: \( \frac{d^2A}{d\theta^2}\bigg|_{\theta=\frac{\pi}{2}} = -\frac{1}{2} ab \sin\left(\frac{\pi}{2}\right) = -\frac{1}{2} ab < 0 \) Since the second derivative is negative, function A is maximum at \( \theta = \frac{\pi}{2} \). The area of the triangle reaches its maximum when \( \theta = \frac{\pi}{2} \) (or 90 degrees).
In simple words: The triangle's area becomes largest when the angle between sides a and b is a right angle (90 degrees).

Exam Tip: Remember that \( \sin(90°) = 1 \) is the maximum value of sine, making the area formula \( A = \frac{1}{2}ab \sin\theta \) largest when \( \theta = 90° \).

Question 29. Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius \( 5\sqrt{3} \) cm is (500π) cm³.
Answer: We are given a sphere with radius \( 5\sqrt{3} \) cm, and we must find the maximum volume of a cylinder that fits inside it. Let us set: - Radius of the sphere be R units - Volume of the inscribed cylinder be V - Height of the inscribed cylinder be h - Radius of the cylinder be r Using the Pythagorean theorem, where the diameter AC = 2R, diameter of cylinder base AB = 2r, and height BC = h: \( AC^2 = AB^2 + BC^2 \) \( 4R^2 = 4r^2 + h^2 \) ---- (1) The volume of the cylinder is: \( V = \pi r^2 h \) Substituting (1) into the volume formula: \( V = \pi h \left(\frac{1}{4}[4R^2 - h^2]\right) \) \( V = \frac{\pi}{4}(4R^2h - h^3) \) ---- (2) To find maximum or minimum, differentiate with h and set equal to zero: \( \frac{dV}{dh} = \frac{d}{dh}\left[\frac{\pi}{4}(4R^2h - h^3)\right] \) \( \frac{dV}{dh} = \frac{4R^2\pi}{4} - \frac{\pi}{4}(3h^2) \) \( \frac{dV}{dh} = R^2\pi - \frac{3h^2\pi}{4} \) ---- (3) Equate to zero to find critical point: \( \frac{dV}{dh} = R^2\pi - \frac{3h^2\pi}{4} = 0 \) \( 3h^2\pi = 4R^2\pi \) \( h^2 = \frac{4R^2}{3} = \frac{4(5\sqrt{3})^2}{3} = \frac{4(75)}{3} = 100 \) \( h = 10 \) cm Verify this is a maximum by taking the second derivative: \( \frac{d^2V}{dh^2} = \frac{d}{dh}\left[R^2\pi - \frac{3h^2\pi}{4}\right] \) \( \frac{d^2V}{dh^2} = 0 - \frac{3(2h)\pi}{4} = -2h\pi \) ---- (4) At h = 10: \( \frac{d^2V}{dh^2}\bigg|_{h=10} = -2(10)\pi = -20\pi < 0 \) Since the second derivative is negative, volume V is maximum at h = 10 cm. Substitute h into equation (1): \( r^2 = \frac{1}{4}[4(5\sqrt{3})^2 - (10)^2] \) \( r^2 = \frac{1}{4}[4(75) - 100] = \frac{1}{4}[300 - 100] = \frac{200}{4} = 50 \) Calculate the maximum volume: \( V = \pi(50)(10) = 500\pi \text{ cm}^3 \) Therefore, the maximum volume of the inscribed cylinder is \( 500\pi \) cm³.
In simple words: When a cylinder sits inside a sphere, its volume becomes largest at a specific height. Using calculus, we find this maximum volume is exactly \( 500\pi \) cubic centimetres.

Exam Tip: Always verify the second derivative is negative for a maximum volume problem - a positive second derivative would indicate a minimum instead.

Question 30. A square tank of capacity 250 cubic meters has to be dug out. The cost of the land is Rs. 50 per square metre. The cost of digging increases with the depth and for the whole tank, it is Rs. (400 × h²), where h metres is the depth of the tank. What should be the dimensions of the tank so that the cost is minimum?
Answer: We are given the tank holds 250 cubic metres, land costs Rs. 50 per square metre, and digging costs Rs. (400 × h²). We need to find dimensions that minimize total cost. Let us consider: - Side of the tank is x metres - Cost formula: \( C = 50x^2 + 400h^2 \) ---- (1) - Volume: \( V = x^2h = 250 \), so \( h = \frac{250}{x^2} \) ---- (2) Substitute (2) into (1): \( C = 50x^2 + 400\left(\frac{250}{x^2}\right)^2 \) \( C = 50x^2 + \frac{400 \times 62500}{x^4} \) ---- (3) Differentiate with respect to x: \( \frac{dC}{dx} = \frac{d}{dx}\left[50x^2 + \frac{400 \times 62500}{x^4}\right] \) \( \frac{dC}{dx} = 50(2x) + \frac{25000000 \times (-4)}{x^5} \) \( \frac{dC}{dx} = 100x - \frac{10^8}{x^5} \) ---- (4) Equate to zero for critical point: \( 100x - \frac{10^8}{x^5} = 0 \) \( x^6 = 10^6 \) \( x = 10 \) m Verify with second derivative: \( \frac{d^2C}{dx^2} = \frac{d}{dx}\left[100x - \frac{10^8}{x^5}\right] \) \( \frac{d^2C}{dx^2} = 100 + \frac{10^8 \times 5}{x^6} \) ---- (5) At x = 10: \( \frac{d^2C}{dx^2}\bigg|_{x=10} = 100 + \frac{10^8 \times 5}{(10)^6} = 100 + 500 = 600 > 0 \) Since the second derivative is positive, cost C is minimum at x = 10. Substitute x into equation (2): \( h = \frac{250}{(10)^2} = \frac{250}{100} = 2.5 \) m The cost is minimum when the tank dimensions are length = 10 m and depth = 2.5 m.
In simple words: To dig a tank with the least expense, make the square base 10 metres on each side and dig it 2.5 metres deep.

Exam Tip: For practical optimization problems, always check that the second derivative is positive (for minimum cost) and substitute back to find all unknowns.

Question 31. A square piece of tin of side 18 cm is to be made into a box without the top, by cutting a square piece from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find the maximum volume of the box.
Answer: We start with an 18 cm square sheet and cut squares of side x from each corner, then fold up the sides to form an open box. We need to find x that produces maximum volume. Let us consider: - 'x' is the length and breadth of the piece removed from each corner - After folding, the base dimensions become (18 - 2x) - The height of the resulting box is x The volume is: \( V = (18-2x)^2 \times x \) \( V = (324 + 4x^2 - 72x) \times x \) \( V = 4x^3 - 72x^2 + 324x \) ---- (1) Differentiate with respect to x: \( \frac{dV}{dx} = 12x^2 - 144x + 324 \) ---- (2) Equate to zero: \( 12x^2 - 144x + 324 = 0 \) \( x^2 - 12x + 27 = 0 \) Using the quadratic formula: \( x = \frac{12 \pm \sqrt{144 - 108}}{2} = \frac{12 \pm \sqrt{36}}{2} = \frac{12 \pm 6}{2} \) \( x = 9 \text{ or } x = 3 \) Since x = 9 is not viable (18 - 2(9) = 0), we have x = 3 cm. Verify with second derivative: \( \frac{d^2V}{dx^2} = 24x - 144 \) ---- (3) At x = 3: \( \frac{d^2V}{dx^2}\bigg|_{x=3} = 24(3) - 144 = 72 - 144 = -72 < 0 \) Since the second derivative is negative, volume V is maximum at x = 3 cm. Base side: \( 18 - 2(3) = 12 \) cm Maximum volume: \( V = (12)^2 \times 3 = 144 \times 3 = 432 \) cm³ The maximum volume of the box is 432 cm³, achieved by cutting 3 cm squares from each corner, resulting in a box with a 12 cm base and 3 cm height.
In simple words: Cut 3 cm squares from each corner and fold up. This creates a box with sides 12 cm and height 3 cm, holding 432 cubic centimetres - the most volume possible.

Exam Tip: Always check both solutions from the quadratic equation against the physical constraints - here, x cannot be so large that 18 - 2x becomes zero or negative.

Question 32. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be least when the depth of the tank is half of its width.
Answer: We must show that for a square-base open tank holding a fixed volume, construction cost is minimized when depth equals half the width. Let us consider: - Side of the tank is x metres - Height is h metres - Volume: \( V = x^2h \) - Surface area: \( S = x^2 + 4xh \) (includes base only, no top) - Let Rs. P be the price per square metre From volume constraint: \( h = \frac{V}{x^2} \) ---- (1) Cost of construction: \( C = (x^2 + 4xh)P \) ---- (2) Substitute (1) into (2): \( C = \left[x^2 + 4x \cdot \frac{V}{x^2}\right]P = \left[x^2 + \frac{4V}{x}\right]P \) ---- (3) Differentiate with respect to x: \( \frac{dC}{dx} = P\left[2x + \frac{4V \times (-1)}{x^2}\right] = P\left[2x - \frac{4V}{x^2}\right] \) ---- (4) Equate to zero: \( 2x - \frac{4V}{x^2} = 0 \) \( x^3 = 2V \) Verify with second derivative: \( \frac{d^2C}{dx^2} = P\left[2 - \frac{4V \times (-2)}{x^3}\right] = P\left[2 + \frac{8V}{x^3}\right] \) ---- (5) Since \( \frac{d^2C}{dx^2} > 0 \), the function C is minimum at \( x = \sqrt[3]{2V} \). Substitute x into equation (1): \( h = \frac{V}{(\sqrt[3]{2V})^2} = \frac{V}{(2V)^{2/3}} = \frac{V \times (2V)^{-2/3}}{1} = (2V)^{1/3} \times (2V)^{-2/3} = (2V)^{-1/3} = \frac{1}{2} \sqrt[3]{2V} \) Therefore: \( h = \frac{1}{2}x \) This proves that cost is minimum when the depth equals half the width.
In simple words: When building an open square tank to hold a set amount of water, material costs least if you make it only half as deep as it is wide.

Exam Tip: This is a classic optimization proof - set up the cost function, differentiate, equate to zero, and show the relationship between variables at the critical point.

Question 33. A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum.
Answer: A 36 cm wire is divided into two parts - one forms a square and the other forms an equilateral triangle. We must find the split that minimizes the combined area. Let us consider: - Perimeter of the square is x cm - Perimeter of equilateral triangle is (36 - x) cm - Side of square: \( \frac{x}{4} \) - Side of triangle: \( \frac{36-x}{3} \) Sum of areas: \( A = \left(\frac{x}{4}\right)^2 + \frac{\sqrt{3}}{4}\left(\frac{36-x}{3}\right)^2 \) \( A = \frac{x^2}{16} + \frac{\sqrt{3}}{4}\left(12 - \frac{x}{3}\right)^2 \) \( A = \frac{x^2}{16} + \frac{\sqrt{3}}{4}\left(144 + \frac{x^2}{9} - 8x\right) \) \( A = \frac{x^2}{16} + \frac{\sqrt{3}}{4}\left(144 + \frac{x^2}{9} - 8x\right) \) ---- (1) Differentiate with respect to x: \( \frac{dA}{dx} = \frac{2x}{16} + \frac{\sqrt{3}}{4}\left(\frac{2x}{9} - 8\right) \) ---- (2) Equate to zero: \( \frac{2x}{16} + \frac{\sqrt{3}}{4}\left(\frac{2x}{9} - 8\right) = 0 \) After solving (detailed algebra steps): \( x = \frac{144\sqrt{3}}{(9 + 4\sqrt{3})} \) cm Verify with second derivative: \( \frac{d^2A}{dx^2} = \frac{1}{8} + \frac{\sqrt{3}}{4}\left(\frac{2}{9}\right) > 0 \) Since the second derivative is positive, the sum of areas is minimum. The lengths of each piece are: - Wire for square: \( x = \frac{144\sqrt{3}}{(9 + 4\sqrt{3})} \) cm - Wire for triangle: \( 36 - x = 36 - \frac{144\sqrt{3}}{(9 + 4\sqrt{3})} = \frac{324}{(9 + 4\sqrt{3})} \) cm To minimize total area, divide the wire using these calculated lengths.
In simple words: Split the wire using the formula values above - this creates a square and triangle whose combined area is as small as possible.

Exam Tip: For combined optimization of two shapes, always express both in terms of a single variable and minimize the sum of their areas.

Question 34. Find the largest possible area of a right-angled triangle whose hypotenuse is 5 cm.
Answer: We need to find the maximum area of a right triangle with a fixed hypotenuse of 5 cm. Let us consider: - Base of triangle is 'a' cm - Perpendicular side is 'b' cm - Hypotenuse is 5 cm By the Pythagorean theorem: \( 25 = a^2 + b^2 \) \( b^2 = 25 - a^2 \) ---- (1) Area of the triangle: \( A = \frac{1}{2}ab \) Squaring both sides: \( A^2 = \frac{1}{4}a^2b^2 \) Substitute (1): \( Z = A^2 = \frac{1}{4}a^2(25 - a^2) \) ---- (2) Differentiate with respect to a: \( \frac{dZ}{da} = \frac{1}{4}[25(2a) - 4a^3] = \frac{1}{4}[25a - a^3] \) ---- (3) Equate to zero: \( 25a - a^3 = 0 \) \( a\left(\frac{25}{2} - a^2\right) = 0 \) Since a ≠ 0: \( a = \frac{5}{\sqrt{2}} \) Verify with second derivative: \( \frac{d^2Z}{da^2} = \frac{1}{4}[25 - 3a^2] \) ---- (4) At \( a = \frac{5}{\sqrt{2}} \): \( \frac{d^2Z}{da^2}\bigg|_{a=\frac{5}{\sqrt{2}}} = \frac{25}{2} - 3\left(\frac{25}{2}\right) = -25 < 0 \) Since the second derivative is negative, area is maximum. Substitute into equation (1): \( b^2 = 25 - \frac{25}{2} = \frac{25}{2} \) \( b = \frac{5}{\sqrt{2}} \) Maximum area: \( A = \frac{1}{2} \times \frac{5}{\sqrt{2}} \times \frac{5}{\sqrt{2}} = \frac{1}{2} \times \frac{25}{2} = \frac{25}{4} \) cm² The largest possible area is \( \frac{25}{4} \) cm² (or 6.25 cm²), achieved when both legs equal \( \frac{5}{\sqrt{2}} \) cm (meaning the triangle is an isosceles right triangle).
In simple words: For a right triangle with hypotenuse 5 cm, the biggest area you can get is when both legs are equal length - it becomes an isosceles right triangle with area 6.25 square centimetres.

Exam Tip: For a right triangle with fixed hypotenuse, maximum area occurs when the two legs are equal (45-45-90 triangle), making calculations simpler.

Exercise 11G

Question 1. Show that the function \( f(x) = 5x - 2 \) is a strictly increasing function on R.
Answer: The domain of the function is R. Finding the derivative: \( f'(x) = 5 \) Since the derivative is greater than 0, the function is strictly increasing throughout its domain, which is R.
In simple words: The derivative is always 5 (positive), so the function always goes upward - it never stops increasing.

Exam Tip: If the derivative is positive everywhere in the domain, the function is strictly increasing; there are no exceptions.

Question 2. Show the function \( f(x) = -2x + 7 \) is a strictly decreasing function on R.
Answer: The domain of the function is R. Finding the derivative: \( f'(x) = -2 \) Since the derivative is less than 0, the function is strictly decreasing throughout its domain, which is R.
In simple words: The derivative is always -2 (negative), so the function always goes downward - it continuously decreases.

Exam Tip: A negative derivative everywhere means the function graph always slopes downward with no exceptions.

Question 3. Prove that \( f(x) = ax + b \), where a and b are constants and a > 0, is a strictly increasing function on R.
Answer: The domain of the function is R. Finding the derivative: \( f'(x) = a \) We are given that a > 0. Since the derivative is positive, the function is strictly increasing throughout its domain, which is R.
In simple words: When the constant a is positive, the derivative is always positive, making the entire function increase without stopping.

Exam Tip: For linear functions, the sign of the coefficient determines whether the function is increasing or decreasing across the entire domain.

Question 4. Prove that the function \( f(x) = e^{2x} \) is strictly increasing on R.
Answer: The domain of the function is R. Finding the derivative: \( f'(x) = 2e^x \) We know that \( e^x \) is always positive for all real values of x. Therefore: \( f'(x) > 0 \) This proves the function is strictly increasing throughout its domain.
In simple words: Since \( e^x \) is always positive and we multiply by 2, the derivative never becomes zero or negative - the function always rises.

Exam Tip: Exponential functions with positive bases are always positive, making their derivatives positive and ensuring the functions are strictly increasing.

Question 5. Show that the function \( f(x) = x^2 \) is
(a) strictly increasing on [0, ∞[
(b) strictly decreasing on [0, ∞[
(c) neither strictly increasing nor strictly decreasing on R
Answer: The domain of the function is R. \( f'(x) = 2x \) For x > 0: \( f'(x) > 0 \), which means the function is increasing. For x < 0: \( f'(x) < 0 \), which means the function is decreasing. Therefore, the function is neither strictly increasing nor strictly decreasing across R. (a) The function is strictly increasing on [0, ∞[ because the derivative is positive for all x > 0. (c) The function is neither strictly increasing nor strictly decreasing on R because it decreases for negative x-values and increases for positive x-values.
In simple words: The parabola goes down on the left side (negative numbers) and up on the right side (positive numbers), so it cannot be called strictly increasing or decreasing across the whole real line.

Exam Tip: A function can be increasing on one interval and decreasing on another - always check the sign of the derivative on each interval separately.

Question 6. Show that the function \( f(x) = |x| \) is
(a) strictly increasing on ]0, ∞[
(b) strictly decreasing on] - ∞, 0[
Answer: For x > 0: The absolute value function becomes: \( f(x) = x \) Therefore: \( f'(x) = 1 \), which is less than 0. Actually, when x > 0, we have f'(x) = 1 > 0, so the function is strictly increasing. For x < 0: The absolute value function becomes: \( f(x) = -x \) Therefore: \( f'(x) = -1 \), which is greater than 0. Actually, when x < 0, we have f'(x) = -1 < 0, so the function is strictly decreasing. (a) The function is strictly increasing on ]0, ∞[ because f'(x) = 1 > 0. (b) The function is strictly decreasing on] - ∞, 0[ because f'(x) = -1 < 0.
In simple words: The absolute value function goes upward on the right side of zero and downward on the left side, like a V-shape.

Exam Tip: When working with absolute values, remember to separate the cases (positive and negative arguments) and check the derivative in each region.

Question 7. Prove that the function \( f(x) = \log_e x \) is strictly increasing on ]0, ∞[.
Answer: We have: \( f(x) = \ln(x) \) The derivative is: \( f'(x) = \frac{1}{x} \) For x > 0, the derivative is positive, meaning the function is strictly increasing. The function is strictly increasing on ]0, ∞[.
In simple words: The natural logarithm always rises as x gets larger - for any two positive numbers where the first is smaller, the logarithm of the first is always less than the logarithm of the second.

Exam Tip: The natural logarithm is only defined for positive x-values, and its derivative \( \frac{1}{x} \) is always positive in that domain.

Question 8. Prove that the function \( f(x) = \log_a x \) is strictly increasing on ]0, ∞[ when a > 1 and strictly decreasing on ]0, ∞[ when 0 < a < 1.
Answer: Consider \( f(x) = \log_a x \). The domain is x > 0. The derivative is: \( f'(x) = \frac{1}{x \ln(a)} \) For a > 1: \( \ln(a) > 0 \) Therefore: \( f'(x) > 0 \), which means the function is strictly increasing. For 0 < a < 1: \( \ln(a) < 0 \) Therefore: \( f'(x) < 0 \), which means the function is strictly decreasing.
In simple words: When the base is bigger than 1, logarithms increase. When the base is between 0 and 1, logarithms decrease as x grows.

Exam Tip: The sign of \( \ln(a) \) determines whether the logarithmic function increases or decreases - check this before concluding.

Question 9. Prove that \( f(x) = 3^x \) is strictly increasing on R.
Answer: Consider \( f(x) = 3^x \). The domain of the function is R. The derivative is: \( f'(x) = 3^x \ln(3) \) The value \( 3^x \) is always positive for all real x. The value \( \ln(3) \) is also positive. Therefore: \( f'(x) > 0 \) for all x in R. This proves the function is strictly increasing throughout its domain.
In simple words: Exponential functions with base 3 always rise - as you move right on the graph, the values get bigger and bigger.

Exam Tip: All exponential functions with base greater than 1 are strictly increasing on their entire domain.

Question 10. Show that \( f(x) = x^3 - 15x^2 + 75x - 50 \) is increasing on R.
Answer: Take \( f(x) = x^3 - 15x^2 + 75x - 50 \). The domain of this function is R.
\( f'(x) = 3x^2 - 30x + 75 = 3(x^2 - 10x + 25) = 3(x - 5)(x - 5) = 3(x - 5)^2 \)
\( f'(x) = 0 \) when \( x = 5 \).

For \( x < 5 \), we have \( f'(x) > 0 \), and for \( x > 5 \), we have \( f'(x) > 0 \). Since the derivative is positive throughout R (though it equals 0 at just the single point \( x = 5 \)), the function is increasing on R.
In simple words: When you take the derivative and check its sign, it stays positive everywhere. Even though it touches zero at one spot, the function still goes up overall.

Exam Tip: A function can be increasing even if the derivative is zero at isolated points - what matters is that the derivative doesn't turn negative anywhere.

Question 11. Show that \( f(x) = \left(x - \frac{1}{x}\right) \) is increasing for all \( x \in \mathbb{R} \), where \( x \neq 0 \).
Answer: Take \( f(x) = x - \frac{1}{x} \). The domain of the function is \( \mathbb{R} - \{0\} \).
\( f'(x) = 1 + \frac{1}{x^2} \)

For all \( x \in \mathbb{R} \), \( f'(x) > 0 \). Since the derivative is always positive on its domain, the function is increasing.
In simple words: The derivative stays greater than zero everywhere it's defined, so the function always rises as you move forward.

Exam Tip: Check that the function is defined on the entire stated domain - exclude points where the denominator is zero.

Question 12. Show that \( f(x) = \left(\frac{1}{x} + 5\right) \) is decreasing for all \( x \in \mathbb{R} \), where \( x \neq 0 \).
Answer: Take \( f(x) = \frac{1}{x} + 5 \). The domain of the function is \( \mathbb{R} - \{0\} \).
\( f'(x) = -\frac{1}{x^2} \)

For all \( x \), we have \( f'(x) < 0 \). Since the derivative is always negative, the function is decreasing.
In simple words: The derivative is negative everywhere, which means the function always drops as you move forward.

Exam Tip: A negative derivative means the function always decreases on that domain, excluding any points where it's undefined.

Question 13. Show that \( f(x) = \frac{1}{(1 + x^2)} \) is decreasing for all \( x \geq 0 \).
Answer: Take \( f(x) = \frac{1}{(1 + x^2)} \).
\( f'(x) = -\frac{2x}{(1 + x^2)^2} \)

For \( x \geq 0 \), the derivative \( f'(x) \) is negative (since the numerator is negative and the denominator is always positive). Therefore, the function is decreasing for \( x \geq 0 \).
In simple words: The derivative is negative whenever x is zero or positive, so the function always falls in this range.

Exam Tip: Pay careful attention to the domain - here the function decreases only for non-negative values, not for all real numbers.

Question 14. Show that \( f(x) = \left(x^3 + \frac{1}{x^3}\right) \) is decreasing on \( [-1, 1] \).
Answer: Take \( f(x) = x^3 + x^{-3} \).
\( f'(x) = 3x^2 - 3x^{-4} = 3\left(x^2 - \frac{1}{x^4}\right) = 3\left(\frac{x^6 - 1}{x^4}\right) \)

This can be factored as:
\( f'(x) = 3\left(\frac{(x^3 - 1)(x^3 + 1)}{x^4}\right) = \frac{3(x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)}{x^4} \)

The roots of \( f'(x) \) are at \( x = 1 \) and \( x = -1 \). From the sign analysis, \( f'(x) < 0 \) in the interval \( (-1, 1) \). Therefore, \( f(x) \) is decreasing on \( [-1, 1] \).
In simple words: When you compute the derivative and check its sign between -1 and 1, it comes out negative, confirming the function drops across this range.

Exam Tip: Use a sign chart or test points in each region to determine where the derivative is negative - this directly tells you where the function decreases.

Question 15. Show that \( f(x) = \frac{x}{\sin x} \) is increasing on \( \left[0, \frac{\pi}{2}\right] \).
Answer: Take \( f(x) = \frac{x}{\sin x} \).
\( f'(x) = \frac{\sin x - x\cos x}{\sin^2 x} = \frac{\cos x(\tan x - x)}{\sin^2 x} \)

On the interval \( \left[0, \frac{\pi}{2}\right] \), we have \( \cos x > 0 \) and \( \sin^2 x > 0 \). Also, \( \tan x > x \) for all \( x \) in this range (except at \( x = 0 \) where they are equal). Therefore, \( f'(x) > 0 \) throughout the interval, so the function is increasing on \( \left[0, \frac{\pi}{2}\right] \).
In simple words: The derivative stays positive on this interval because both the cosine and the numerator factor are positive, making the entire expression rise.

Exam Tip: For trigonometric functions, identify the sign of each factor separately - the inequality \( \tan x > x \) is a useful fact for this interval.

Question 16. Prove that the function \( f(x) = \log(1 + x) - \frac{2x}{x + 2} \) is increasing for all \( x > -1 \).
Answer: Take \( f(x) = \log(1 + x) - \frac{2x}{x + 2} \).
\( f'(x) = \frac{1}{1 + x} - \frac{4}{(x + 2)^2} = \frac{(x + 2)^2 - 4(1 + x)}{(1 + x)(x + 2)^2} = \frac{x^2}{(1 + x)(x + 2)^2} \)

For \( x > -1 \), we have \( f'(x) > 0 \) (the numerator is non-negative and strictly positive for \( x \neq 0 \), while the denominator is always positive). Therefore, the function is increasing for all \( x > -1 \).
In simple words: After simplifying the derivative, you find it's always positive in this domain, which means the function consistently rises.

Exam Tip: When combining rational and logarithmic derivatives, find a common denominator - this often reveals a cleaner factored form that makes the sign obvious.

Question 17. Let I be an interval disjoint from \( [-1, 1] \). Prove that the function \( f(x) = \left(x + \frac{1}{x}\right) \) is strictly increasing on I.
Answer: Take \( f(x) = x + \frac{1}{x} \).
\( f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2} = \frac{(x - 1)(x + 1)}{x^2} \)

\( f'(x) < 0 \) in the interval \( [-1, 1] \) (where the function is decreasing). However, \( f'(x) > 0 \) on \( (-\infty, -1) \cup (1, \infty) \), meaning the function is increasing on these intervals. Since I is disjoint from \( [-1, 1] \), it must be contained in one of these regions, so \( f(x) \) is strictly increasing on I.
In simple words: The derivative is positive whenever you're outside the interval from -1 to 1, so the function rises throughout region I.

Exam Tip: Understand that a function can increase on some intervals and decrease on others - always check the derivative's sign on the specific interval in question.

Question 18. Show that \( f(x) = \frac{(x - 2)}{(x + 1)} \) is increasing for all \( x \in \mathbb{R} \), except at \( x = -1 \).
Answer: Take \( f(x) = \frac{x - 2}{x + 1} \).
\( f'(x) = \frac{3}{(x + 1)^2} \)

\( f'(x) \) is not defined at \( x = -1 \), and for all \( x \in \mathbb{R} - \{-1\} \), we have \( f'(x) > 0 \). Since the derivative is positive everywhere it is defined, the function is increasing on its entire domain.
In simple words: The derivative is always positive except where it's undefined, so the function always climbs wherever it exists.

Exam Tip: Always state the exceptions clearly - points where the function or its derivative is undefined must be excluded from the increasing/decreasing intervals.

Question 19. Find the intervals on which the function \( f(x) = (2x^2 - 3x) \) is (a) strictly increasing (b) strictly decreasing.
Answer: Take \( f(x) = 2x^2 - 3x \).
\( f'(x) = 4x - 3 \)
\( f'(x) = 0 \) at \( x = \frac{3}{4} \)

(a) The function is strictly increasing for \( x \in \left[\frac{3}{4}, \infty\right) \).

(b) The function is strictly decreasing for \( x \in \left(-\infty, \frac{3}{4}\right] \).
In simple words: The function reaches its turning point at x = 3/4. Before that it falls, and after that it rises.

Exam Tip: Set the derivative equal to zero to find critical points - these divide the domain into regions where the function either increases or decreases uniformly.

Question 20. Find the intervals on which the function \( f(x) = 2x^3 - 3x^2 - 36x + 7 \) is (a) strictly increasing (b) strictly decreasing.
Answer: Take \( f(x) = 2x^3 - 3x^2 - 36x + 7 \).
\( f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x - 3)(x + 2) \)
\( f'(x) = 0 \) at \( x = 3 \) and \( x = -2 \).

(a) \( f'(x) > 0 \) for \( x \in (-\infty, -2] \cup [3, \infty) \), so the function is strictly increasing on these intervals.

(b) \( f'(x) < 0 \) for \( x \in [-2, 3] \), so the function is strictly decreasing on this interval.
In simple words: The function has two turning points at -2 and 3. It rises before -2, drops between -2 and 3, then rises again after 3.

Exam Tip: Always use a sign chart with the factored form - test a point in each region to confirm whether the derivative is positive or negative.

Question 21. Find the intervals on which the function \( f(x) = 6 - 9x - x^2 \) is (a) strictly increasing (b) strictly decreasing.
Answer: Take \( f(x) = 6 - 9x - x^2 \).
\( f'(x) = -(2x + 9) \)
\( f'(x) = 0 \) at \( x = -\frac{9}{2} \)

(a) \( f'(x) > 0 \) for \( x \in \left(-\infty, -\frac{9}{2}\right] \), so the function is strictly increasing on this interval.

(b) \( f'(x) < 0 \) for \( x \in \left[-\frac{9}{2}, \infty\right) \), so the function is strictly decreasing on this interval.
In simple words: The function reaches a peak at x = -9/2, rises before that point, and falls after it.

Exam Tip: For a quadratic, the vertex marks the transition between increasing and decreasing - the linear derivative makes this transition clear.

Question 22. Find the intervals on which the function \( f(x) = \left(x^4 - \frac{x^3}{3}\right) \) is (a) increasing (b) decreasing.
Answer: Take \( f(x) = x^4 - \frac{x^3}{3} \).
\( f'(x) = 4x^3 - x^2 = x^2(4x - 1) \)
\( f'(x) = 0 \) for \( x = 0 \) and \( x = \frac{1}{4} \).

(a) The function is increasing for \( x \in \left[\frac{1}{4}, \infty\right) \).

(b) The function is decreasing for \( x \in \left(-\infty, \frac{1}{4}\right] \).
In simple words: The critical points are at 0 and 1/4. Even though 0 is a critical point, the function only starts rising after 1/4.

Exam Tip: A repeated root in the factored derivative (like \( x^2 \) here) means the derivative touches zero but does not change sign at that point - the function continues in the same direction.

Question 23. Find the intervals on which the function \( f(x) = x^3 - 12x^2 + 36x + 17 \) is (a) increasing (b) decreasing.
Answer: Take \( f(x) = x^3 - 12x^2 + 36x + 17 \).
\( f'(x) = 3x^2 - 24x + 36 = 3(x^2 - 8x + 12) = 3(x - 6)(x - 2) \)
\( f'(x) = 0 \) at \( x = 2 \) and \( x = 6 \).

(a) The function is increasing for \( x \in (-\infty, 2) \cup (6, \infty) \).

(b) The function is decreasing for \( x \in [2, 6] \).
In simple words: The function has a local peak at x = 2 and a local valley at x = 6. It rises to 2, dips from 2 to 6, then rises again.

Exam Tip: The sign of the derivative determines monotonicity - use test points between critical points to quickly identify which regions have positive vs. negative derivatives.

Question 24. Find the intervals on which the function \( f(x) = \left(x^3 - 6x^2 + 9x + 10\right) \) is (a) increasing (b) decreasing.
Answer: Take \( f(x) = x^3 - 6x^2 + 9x + 10 \).
\( f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 3)(x - 1) \)
\( f'(x) = 0 \) at \( x = 1 \) and \( x = 3 \).

(a) The function is increasing for \( x \in (-\infty, 1) \cup (3, \infty) \).

(b) The function is decreasing for \( x \in [1, 3] \).
In simple words: Two turning points at 1 and 3 divide the graph into three regions - the function climbs, then falls, then climbs again.

Exam Tip: Once you factor the derivative, the interval endpoints are immediately visible - organize your answer by listing the increasing and decreasing regions clearly.

Question 25. Find the intervals on which the function \( f(x) = \left(6 + 12x + 3x^2 - 2x^3\right) \) is (a) increasing (b) decreasing.
Answer: Take \( f(x) = -2x^3 + 3x^2 + 12x + 6 \).
\( f'(x) = -6x^2 + 6x + 12 = -6(x^2 - x - 2) = -6(x - 2)(x + 1) \)
\( f'(x) = 0 \) at \( x = -1 \) and \( x = 2 \).

(a) The function is increasing for \( x \in [-1, 2] \).

(b) The function is decreasing for \( x \in (-\infty, -1) \cup (2, \infty) \).
In simple words: The negative leading coefficient flips the typical cubic shape - the function drops, rises in the middle, then drops again.

Exam Tip: Be careful with the sign of the leading coefficient when factoring - a negative sign reverses which intervals are increasing vs. decreasing.

Question 26. Find the intervals on which the function \( f(x) = 2x^3 - 24x + 5 \) is (a) increasing (b) decreasing.
Answer: Take \( f(x) = 2x^3 - 24x + 5 \).
\( f'(x) = 6x^2 - 24 = 6(x^2 - 4) = 6(x - 2)(x + 2) \)
\( f'(x) = 0 \) at \( x = -2 \) and \( x = 2 \).

(a) The function is increasing for \( x \in (-\infty, -2) \cup (2, \infty) \).

(b) The function is decreasing for \( x \in [-2, 2] \).
In simple words: The derivative factors as a difference of squares, giving two turning points at -2 and 2. The function falls between them and rises elsewhere.

Exam Tip: A simple difference of squares in the derivative (like \( x^2 - 4 \)) factors immediately - use this to quickly find critical points.

Question 27. Find the intervals on which the function \( f(x) = (x - 1)(x - 2)^2 \) is (a) increasing (b) decreasing.
Answer: Take \( f(x) = (x - 1)(x - 2)^2 = x^3 - 5x^2 + 8x - 4 \).
\( f'(x) = 3x^2 - 10x + 8 = (3x - 4)(x - 2) \)
\( f'(x) = 0 \) at \( x = \frac{4}{3} \) and \( x = 2 \).

(a) The function is increasing for \( x \in \left(-\infty, \frac{4}{3}\right) \cup (2, \infty) \).

(b) The function is decreasing for \( x \in \left[\frac{4}{3}, 2\right] \).
In simple words: The original function has a squared factor at x = 2, but the critical points lie at 4/3 and 2. The derivative changes sign between them.

Exam Tip: Even if the original function has a repeated root, the derivative may have distinct critical points - always solve the derivative equation directly.

Question 28. Find the intervals on which the function \( f(x) = \left(x^4 - 4x^3 + 4x^2 + 15\right) \) is (a) increasing (b) decreasing.
Answer: Take \( f(x) = x^4 - 4x^3 + 4x^2 + 15 \).
\( f'(x) = 4x^3 - 12x^2 + 8x = 4x(x^2 - 3x + 2) = 4x(x - 1)(x - 2) \)
\( f'(x) = 0 \) at \( x = 0 \), \( x = 1 \), and \( x = 2 \).

(a) The function is increasing for \( x \in (0, 1) \cup (2, \infty) \).

(b) The function is decreasing for \( x \in (-\infty, 0] \cup [1, 2] \).
In simple words: Three turning points divide the curve into regions - the function falls, rises, falls, then rises again.

Exam Tip: With four critical regions, use a sign chart or test points to track the derivative's sign - this avoids sign errors.

Question 29. Find the intervals on which the function \( f(x) = 2x^3 + 9x^2 + 12x + 15 \) is (a) increasing (b) decreasing.
Answer: Take \( f(x) = 2x^3 + 9x^2 + 12x + 15 \).
\( f'(x) = 6x^2 + 18x + 12 = 6(x^2 + 3x + 2) = 6(x + 2)(x + 1) \)
\( f'(x) = 0 \) at \( x = -2 \) and \( x = -1 \).

(a) The function is decreasing for \( x \in [-2, -1] \).

(b) The function is increasing for \( x \in (-\infty, -2) \cup (-1, \infty) \).
In simple words: The function climbs initially, dips between -2 and -1, then continues climbing.

Exam Tip: When critical points are negative, take extra care with your inequalities - test a point in each interval to verify your answer.

Question 30. Find the intervals on which the function \( f(x) = x^4 - 8x^3 + 22x^2 - 24x + 21 \) is (a) increasing (b) decreasing.
Answer: Take \( f(x) = x^4 - 8x^3 + 22x^2 - 24x + 21 \).
\( f'(x) = 4x^3 - 24x^2 + 44x - 24 = 4(x^3 - 6x^2 + 11x - 6) = 4(x - 3)(x - 1)(x - 2) \)
\( f'(x) = 0 \) at \( x = 1 \), \( x = 2 \), and \( x = 3 \).

(a) The function is decreasing for \( x \in (-\infty, 1] \cup [2, 3] \).

(b) The function is increasing for \( x \in (1, 2) \cup (3, \infty) \).
In simple words: Three critical points create an alternating pattern - the function falls, rises, falls, then rises as you move left to right.

Exam Tip: For a quartic with three critical points, always construct a sign chart showing the + - + - pattern that emerges from the three linear factors.

Question 31. Find the intervals on which the function \( f(x) = 3x^4 - 4x^3 - 12x^2 + 5 \) is (a) increasing (b) decreasing.
Answer: Take \( f(x) = 3x^4 - 4x^3 - 12x^2 + 5 \).
\( f'(x) = 12x^3 - 12x^2 - 24x = 12x(x^2 - x - 2) = 12x(x + 1)(x - 2) \)
\( f'(x) = 0 \) at \( x = -1 \), \( x = 0 \), and \( x = 2 \).

(a) The function is increasing for \( x \in (-1, 0) \cup (2, \infty) \).

(b) The function is decreasing for \( x \in (-\infty, -1] \cup [0, 2] \).
In simple words: Three critical points at -1, 0, and 2 divide the curve into regions where it alternates between falling and rising.

Exam Tip: When a critical point is at x = 0, include it carefully in your sign analysis - test nearby points to confirm the derivative's behavior.

Question 32. Find the intervals on which the function \( f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11 \) is (a) increasing (b) decreasing.
Answer: Take \( f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11 \).
\( f'(x) = \frac{12x^3 - 24x^2 - 60x + 72}{10} = 1.2(x^3 - 2x^2 - 5x + 6) = 1.2(x - 1)(x - 3)(x + 2) \)
\( f'(x) = 0 \) at \( x = -2 \), \( x = 1 \), and \( x = 3 \).

(a) The function is increasing for \( x \in (-2, 1) \cup (3, \infty) \).

(b) The function is decreasing for \( x \in (-\infty, -2] \cup [1, 3] \).
In simple words: The fractional coefficients simplify when you take the derivative - the factorization reveals three critical points that govern the function's shape.

Exam Tip: Factor out constants and coefficients from the derivative early - this keeps the algebraic manipulation cleaner and the factorization more obvious.

Exercise 11H

Question 1. Find the slope of the tangent to the curve
(i) \( y = (x^3 - x) \) at \( x = 2 \)
(ii) \( y = (2x^2 + 3\sin x) \) at \( x = 0 \)
(iii) \( y = (\sin 2x + \cot x + 2)^2 \) at \( x = \frac{\pi}{2} \)
Answer:
(i) \( \frac{dy}{dx} = 3x^2 - 1 \)
\( \frac{dy}{dx} \) at \( (x = 2) = 11 \)

(ii) \( \frac{dy}{dx} = 4x + 3\cos x \)
\( \frac{dy}{dx} \) at \( (x = 0) = 3 \)

(iii) \( \frac{dy}{dx} = 2(\sin 2x + \cot x + 2)(2\cos 2x - \csc^2 x) \)
\( \frac{dy}{dx} \) at \( \left(x = \frac{\pi}{2}\right) = 2(0 + 0 + 2)(-2 - 1) = -12 \)
In simple words: Take the derivative of each curve, then plug in the given x-value to find the slope at that point.

Exam Tip: For composite functions, apply the chain rule carefully - each layer of the function must be differentiated.

Question 2. Find the equations of the tangent and the normal to the given curve at the indicated point for \( y = x^3 - 2x + 7 \) at \( (1, 6) \).
Answer: Compute the slope: \( m = \frac{dy}{dx} = 3x^2 - 2 \). At \( (1, 6) \), \( m = 1 \).

Using the point-slope form for the tangent line:
\( y - 6 = 1(x - 1) \)
\( x - y + 5 = 0 \)

For the normal line, the slope is \( -1 \):
\( y - 6 = -1(x - 1) \)
\( x + y - 7 = 0 \)
In simple words: Find the slope at the point using the derivative, then use point-slope form to write the tangent line. The normal is perpendicular, so its slope is the negative reciprocal.

Exam Tip: Always verify that your point lies on the original curve before writing the tangent or normal - this catches arithmetic errors early.

Question 5. Find the equations of the tangent and the normal to the given curve \( y^2 = 4ax \) at \( \left(\frac{a}{m^2}, \frac{2a}{m}\right) \).
Answer: Start with implicit differentiation: \( 2y\frac{dy}{dx} = 4a \), so \( \frac{dy}{dx} = \frac{2a}{y} \).

At the point \( \left(\frac{a}{m^2}, \frac{2a}{m}\right) \), the slope is \( m = \frac{2a}{\frac{2a}{m}} = m \).

Tangent line: \( y - \frac{2a}{m} = m\left(x - \frac{a}{m^2}\right) \)
Simplifying: \( m^2x - my + a = 0 \)

Normal line (slope \( -\frac{1}{m} \)): \( y - \frac{2a}{m} = -\frac{1}{m}\left(x - \frac{a}{m^2}\right) \)
Simplifying: \( m^2x + m^3y - 2am^2 - a = 0 \)
In simple words: For an implicit curve, differentiate both sides with respect to x, solve for dy/dx, then substitute the point to get the slope.

Exam Tip: With parametric or implicit curves, the derivative formula may involve the parameter or both variables - always substitute the point coordinates to get the actual slope.

Question 6. Find the equations of the tangent and the normal to the given curve \( y = x^3 \) at \( P(1, 1) \).
Answer: Compute the slope: \( m = \frac{dy}{dx} = 3x^2 \). At \( (1, 1) \), \( m = 3 \).

Tangent line: \( y - 1 = 3(x - 1) \)
\( y = 3x - 2 \)

Normal line (slope \( -\frac{1}{3} \)): \( y - 1 = -\frac{1}{3}(x - 1) \)
\( x + 3y = 4 \)
In simple words: The tangent touches the curve with slope 3, while the normal is perpendicular with slope -1/3.

Exam Tip: For simple power functions, the derivative is straightforward - focus on correctly applying the point-slope form.

Question 7. Find the equations of the tangent and the normal to the given curve \( y^2 = 4ax \) at \( (at^2, 2at) \).
Answer: Using implicit differentiation: \( 2y\frac{dy}{dx} = 4a \), so \( \frac{dy}{dx} = \frac{2a}{y} \).

At \( (at^2, 2at) \), the slope is \( m = \frac{2a}{2at} = \frac{1}{t} \).

Tangent line: \( y - 2at = \frac{1}{t}(x - at^2) \)
Simplifying: \( x - ty + at^2 = 0 \)

Normal line (slope \( -t \)): \( y - 2at = -t(x - at^2) \)
Simplifying: \( tx + y = at^3 + 2at \)
In simple words: The slope at any point on this parabola is the reciprocal of the parameter t. Use this to construct both tangent and normal equations.

Exam Tip: For parametric curves, the parameter t often appears in both the slope formula and the final equation - don't eliminate it unless the problem asks for specific numeric values.

Question 8. Find the equations of the tangent and the normal to the given curve at the indicated point.
Answer: (This question lacks specific function and point details. Please provide the curve equation and the point coordinates to complete this solution.)

Exam Tip: Ensure all given information is included when stating a tangent-normal problem - curve, point, and domain restrictions.

Question 9. Find the equations of the tangent and the normal to the given curve \( 16x^2 + 9y^2 = 144 \) at \( (2, y_1) \), where \( y_1 > 0 \).
Answer: First, find \( y_1 \): Substitute \( x = 2 \) into the ellipse equation:
\( 16(4) + 9y_1^2 = 144 \)
\( 64 + 9y_1^2 = 144 \)
\( y_1 = \frac{4\sqrt{5}}{3} \)

Using implicit differentiation: \( 32x + 18y\frac{dy}{dx} = 0 \), so \( \frac{dy}{dx} = -\frac{16x}{9y} \).

At \( \left(2, \frac{4\sqrt{5}}{3}\right) \), the slope is \( m = -\frac{32}{12\sqrt{5}} = -\frac{8\sqrt{5}}{15} \).

Tangent line: \( y - \frac{4\sqrt{5}}{3} = -\frac{8\sqrt{5}}{15}\left(x - 2\right) \)
Simplifying: \( 8x + 3\sqrt{5}y = 36 \)

Normal line (slope \( \frac{3\sqrt{5}}{8} \)): \( y - \frac{4\sqrt{5}}{3} = \frac{3\sqrt{5}}{8}\left(x - 2\right) \)
Simplifying: \( 9\sqrt{5}x - 24y + 14\sqrt{5} = 0 \)
In simple words: Substitute the x-coordinate into the ellipse equation to find y, then use implicit differentiation to get the slope at that point.

Exam Tip: When a point has an unknown coordinate, always substitute the known coordinate first to find the missing one before proceeding.

Question 10. Find the equations of the tangent and the normal to the given curve \( y = x^4 - 6x^3 + 13x^2 - 10x + 5 \) at the point where \( x = 1 \).
Answer: Find the y-coordinate: \( y(1) = 1 - 6 + 13 - 10 + 5 = 3 \). The point is \( (1, 3) \).

Compute the slope: \( m = \frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10 \). At \( x = 1 \), \( m = 4 - 18 + 26 - 10 = 2 \).

Tangent line: \( y - 3 = 2(x - 1) \)
\( 2x - y + 1 = 0 \)

Normal line (slope \( -\frac{1}{2} \)): \( y - 3 = -\frac{1}{2}(x - 1) \)
\( x + 2y - 7 = 0 \)
In simple words: Substitute x = 1 to find both the y-coordinate and the slope, then apply point-slope form for both lines.

Exam Tip: Always calculate y using the original function, not the derivative - this is a common mistake that leads to incorrect line equations.

Question 11. Find the equation of the tangent to the curve \( \sqrt{x} + \sqrt{y} = a \) at \( \left(\frac{a^2}{4}, \frac{a^2}{4}\right) \).
Answer: Using implicit differentiation: \( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 0 \), so \( \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \).

At \( \left(\frac{a^2}{4}, \frac{a^2}{4}\right) \), the slope is \( m = -1 \).

Tangent line: \( y - \frac{a^2}{4} = -1\left(x - \frac{a^2}{4}\right) \)
Simplifying: \( 2(x + y) = a^2 \)
In simple words: Differentiate implicitly to find that the slope is -1 at the given point, then write the tangent equation.

Exam Tip: When a point has equal coordinates (like here where both are a²/4), the slope often simplifies nicely - use this as a check on your work.

Question 12. Show that the equation of the tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at \( (x_1, y_1) \) is \( \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \).
Answer: Using implicit differentiation on the hyperbola equation: \( \frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0 \), so \( \frac{dy}{dx} = \frac{b^2x}{a^2y} \).

At the point \( (x_1, y_1) \), the slope is \( m = \frac{b^2x_1}{a^2y_1} \).

Using point-slope form:
\( y - y_1 = \frac{b^2x_1}{a^2y_1}(x - x_1) \)

Multiply through by \( a^2y_1 \):
\( a^2y_1y - a^2y_1^2 = b^2x_1x - b^2x_1^2 \)

Rearrange: \( b^2x_1x - a^2y_1y = b^2x_1^2 - a^2y_1^2 \)

Since \( (x_1, y_1) \) lies on the hyperbola, \( \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1 \), which means \( b^2x_1^2 - a^2y_1^2 = a^2b^2 \).

Therefore: \( \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \).
In simple words: Use implicit differentiation to find the slope, apply point-slope form, then simplify using the constraint that the point lies on the hyperbola.

Exam Tip: For conic sections, knowing that the point satisfies the original equation allows you to eliminate constants and reach the elegant final form.

Question 13. Find the equation of the tangent to the curve \( y = (\sec^{-1} x - \tan^{-1} x) \) at \( x = \frac{\pi}{3} \).
Answer: Compute the slope: \( \frac{dy}{dx} = \frac{1}{|x|\sqrt{x^2 - 1}} - \frac{1}{1 + x^2} \).

At \( x = \frac{\pi}{3} \), evaluate: \( y = \sec^{-1}\left(\frac{\pi}{3}\right) - \tan^{-1}\left(\frac{\pi}{3}\right) \approx 7 \) (exact value depends on numeric evaluation).

The slope at \( x = \frac{\pi}{3} \) is \( m = 16\sqrt{3} \).

Tangent line: \( y - 7 = 16\sqrt{3}\left(x - \frac{\pi}{3}\right) \)
Simplifying: \( 3y - 48\sqrt{3}x + 16\sqrt{3}\pi - 21 = 0 \)
In simple words: Differentiate the inverse trigonometric functions, substitute x = π/3, then use point-slope form.

Exam Tip: Inverse trig derivatives are easily confused - double-check the formulas before substituting into the slope.

Question 14. Find the equation of the normal to the curve \( y = (\sin 2x + \cot x + 2)^2 \) at \( x = \frac{\pi}{2} \).
Answer: Compute the slope using the chain rule: \( \frac{dy}{dx} = 2(\sin 2x + \cot x + 2)(2\cos 2x - \csc^2 x) \).

At \( x = \frac{\pi}{2} \), \( \sin 2x = 0 \), \( \cot x = 0 \), so the first factor is \( (0 + 0 + 2) = 2 \).
The second factor: \( 2\cos(\pi) - \csc^2\left(\frac{\pi}{2}\right) = 2(-1) - 1 = -3 \).
Therefore, \( m = 2 \cdot 2 \cdot (-3) = -12 \).

Find y: \( y\left(\frac{\pi}{2}\right) = (0 + 0 + 2)^2 = 4 \).

Normal line (slope \( \frac{1}{12} \)): \( y - 4 = \frac{1}{12}\left(x - \frac{\pi}{2}\right) \)
Simplifying: \( 24y - 2x + \pi - 96 = 0 \)
In simple words: For a composite function, use the chain rule to find the derivative, evaluate at the given point, then construct the normal with reciprocal slope.

Exam Tip: At special angles like π/2, trig values simplify dramatically - use this to avoid unnecessary calculation.

Question 15. Show that the tangents to the curve \( y = 2x^3 - 4 \) at the points \( x = 2 \) and \( x = -2 \) are parallel.
Answer: Compute the derivative: \( \frac{dy}{dx} = 6x^2 \).

At \( x = 2 \): \( m = 6(4) = 24 \).
At \( x = -2 \): \( m = 6(4) = 24 \).

Since both tangent lines have the same slope, they are parallel.
In simple words: Lines are parallel if they have equal slopes. Here, the derivative gives the same value at both points.

Exam Tip: For even-power polynomials, the derivative often yields equal slopes at opposite x-values - check for this symmetry.

Question 16. Find the equation of the tangent to the curve \( x^2 + 3y = 3 \), where the tangent is parallel to the line \( y - 4x + 5 = 0 \).
Answer: The line \( y - 4x + 5 = 0 \) has slope 4, so we need a tangent with slope 4.

Using implicit differentiation: \( 2x + 3\frac{dy}{dx} = 0 \), so \( \frac{dy}{dx} = -\frac{2x}{3} \).

Set this equal to 4: \( -\frac{2x}{3} = 4 \), which gives \( x = -6 \).

Find y from the curve equation: \( 36 + 3y = 3 \), so \( y = -11 \).

Tangent line at \( (-6, -11) \): \( y + 11 = 4(x + 6) \)
Simplifying: \( 4x - y + 13 = 0 \)
In simple words: Find the slope of the given line, set the curve's derivative equal to that slope, solve for the point, then write the tangent equation.

Exam Tip: When a tangent must be parallel to a given line, extract the slope from the line equation first - this is your target slope for the derivative.

Question 17. At what point on the curve \( x^2 + y^2 - 2x - 4y + 1 = 0 \) is the tangent parallel to the y-axis?
Answer: When a tangent is parallel to the y-axis, its slope becomes undefined, which means the denominator equals zero. Differentiating the curve with respect to x, we get \( 2x + 2y\frac{dy}{dx} - 2 - 4\frac{dy}{dx} = 0 \), which simplifies to \( \frac{dy}{dx} = \frac{-(2x - 2)}{(2y - 4)} = \frac{-1}{0} \). For this to be undefined, the denominator must be zero: \( 2y - 4 = 0 \Rightarrow y = 2 \). Substituting \( y = 2 \) into the curve equation gives \( x^2 + 4 - 2x - 8 + 1 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow x = 3 \) or \( x = -1 \). Therefore, the required points are \( (-1, 2) \) and \( (3, 2) \).
In simple words: When the tangent line is vertical (parallel to the y-axis), the slope is undefined. Find where the denominator of the slope becomes zero, substitute that y-value back into the curve, and solve for x.

Exam Tip: A vertical tangent occurs when \( \frac{dy}{dx} \) becomes undefined (denominator = 0). Always verify both x-values satisfy the original curve equation.

Question 18. Find the point on the curve \( x^2 + y^2 - 2x - 3 = 0 \) where the tangent is parallel to the x-axis.
Answer: When a tangent is parallel to the x-axis, its slope equals zero. Differentiating implicitly: \( 2x + 2y\frac{dy}{dx} - 2 = 0 \). Setting \( \frac{dy}{dx} = 0 \) gives \( 2x - 2 = 0 \Rightarrow x = 1 \). Substituting into the curve: \( 1 + y^2 - 2 - 3 = 0 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2 \). Therefore, the required points are \( (1, 2) \) and \( (1, -2) \).
In simple words: A horizontal tangent has slope zero. Set \( \frac{dy}{dx} = 0 \), solve for x, then find the corresponding y-values from the curve.

Exam Tip: Horizontal tangents occur where \( \frac{dy}{dx} = 0 \). Always find both y-values (positive and negative) when solving a quadratic equation.

Question 19. Prove the tangent to the curve \( y = x^2 - 5x + 6 \) at the point (2, 0) and (3, 0) are at right angles.
Answer: Two tangent lines are perpendicular when their slopes satisfy the condition \( m_1 m_2 = -1 \). Finding the derivative: \( \frac{dy}{dx} = 2x - 5 \). At point (2, 0): \( m_1 = 2(2) - 5 = -1 \). At point (3, 0): \( m_2 = 2(3) - 5 = 1 \). Computing the product: \( m_1 m_2 = (-1)(1) = -1 \). Since the product of the slopes equals -1, the tangents at these two points are perpendicular.
In simple words: Find the slope of the tangent at each point using the derivative. If the product of the two slopes equals -1, the tangents are perpendicular to each other.

Exam Tip: For perpendicularity, always check that \( m_1 m_2 = -1 \) exactly. Verify that both points actually lie on the curve before computing slopes.

Question 20. Find the point on the curve \( y = x^2 + 3x + 4 \) at which the tangent passes through the origin.
Answer: A tangent that passes through the origin has equation \( y = mx \), where m is the slope. Let the point of tangency be \( (x_1, y_1) \). Since this point lies on the curve: \( y_1 = x_1^2 + 3x_1 + 4 \) - (1). The slope at this point is \( m = 2x_1 + 3 \). The tangent line equation is \( y_1 = (2x_1 + 3)x_1 \) - (2). Comparing equations (1) and (2): \( x_1^2 + 3x_1 + 4 = (2x_1 + 3)x_1 \Rightarrow x_1^2 - 4 = 0 \Rightarrow x_1 = 2 \) or \( x_1 = -2 \). When \( x_1 = 2 \): \( y_1 = 4 + 6 + 4 = 14 \). When \( x_1 = -2 \): \( y_1 = 4 - 6 + 4 = 2 \). Therefore, the required points are \( (2, 14) \) and \( (-2, 2) \).
In simple words: A tangent through the origin has the form y = mx. Set this equal to the curve's tangent equation at some point, then solve to find where they match.

Exam Tip: For tangents through a fixed point, equate the tangent-line equation with the point-slope form at the point of contact and solve for the x-coordinate of that contact point.

Question 21. Find the point on the curve \( y = x^3 - 11x + 5 \) at which the equation of tangent is \( y = x - 11 \).
Answer: The line \( y = x - 11 \) has slope 1. Finding the derivative of the curve: \( \frac{dy}{dx} = 3x^2 - 11 \). Setting the slope equal to 1: \( 3x^2 - 11 = 1 \Rightarrow 3x^2 = 12 \Rightarrow x = 2 \) or \( x = -2 \). At \( x = 2 \): curve gives \( y = 8 - 22 + 5 = -9 \); tangent line gives \( y = 2 - 11 = -9 \). Both match. At \( x = -2 \): curve gives \( y = -8 + 22 + 5 = 19 \); tangent line gives \( y = -2 - 11 = -13 \). These do not match, so this point is rejected. Therefore, the final answer is \( (2, -9) \).
In simple words: Set the derivative equal to the slope of the given line. For each x-value found, check that both the curve and the tangent line pass through the same point.

Exam Tip: Always verify that the point satisfies both the curve equation and the tangent line equation. A point that works for only one equation must be discarded.

Question 22. Find the equation of the tangents to the curve \( 2x^2 + 3y^2 = 14 \) parallel to the line \( x + 3y = 4 \).
Answer: The line \( x + 3y = 4 \) can be rewritten as \( y = -\frac{1}{3}x + \frac{4}{3} \), so its slope is \( -\frac{1}{3} \). Differentiating the curve implicitly: \( 4x + 6y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{2x}{3y} \). Setting this equal to \( -\frac{1}{3} \): \( -\frac{2x}{3y} = -\frac{1}{3} \Rightarrow \frac{2x}{3y} = \frac{1}{3} \Rightarrow 2x = y \). Substituting into the curve equation: \( 2x^2 + 3(2x)^2 = 14 \Rightarrow 2x^2 + 12x^2 = 14 \Rightarrow 14x^2 = 14 \Rightarrow x = \pm 1 \). At \( x = 1 \): \( y = 2 \). At \( x = -1 \): \( y = -2 \). Using point-slope form \( y - b = m(x - a) \): At (1, 2): \( y - 2 = -\frac{1}{3}(x - 1) \Rightarrow 3y + x = 7 \). At (-1, -2): \( y + 2 = -\frac{1}{3}(x + 1) \Rightarrow 3y + x = -7 \).
In simple words: Extract the slope from the given line. Use implicit differentiation to find where the curve's tangent has that same slope, then write the tangent equations using point-slope form.

Exam Tip: Always simplify your final tangent equations to the form ax + by = c for clarity. Double-check that both points lie on the original curve.

Question 23. Find the equation of the tangent to the curve \( x^2 + 2y = 8 \) which is perpendicular to the line \( x - 2y + 1 = 0 \).
Answer: The line \( x - 2y + 1 = 0 \) has slope \( \frac{1}{2} \). For a perpendicular line, the slope is \( -\frac{1}{-1/m} = -2 \). Differentiating the curve: \( 2x + 2\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -x \). Setting \( -x = -2 \): \( x = \frac{1}{2} \). From the curve: \( \frac{1}{4} + 2y = 8 \Rightarrow y = \frac{31}{8} \). Using point-slope form at \( (-\frac{1}{2}, \frac{31}{8}) \): \( y - \frac{31}{8} = -2(x + \frac{1}{2}) \Rightarrow 8y - 31 = -16x - 8 \Rightarrow 16x + 8y = 23 \).
In simple words: Find the slope of the perpendicular line (negative reciprocal). Use the derivative to locate where the curve's tangent has that slope, then form the tangent equation.

Exam Tip: The slope of a perpendicular line is the negative reciprocal. If two slopes are \( m_1 \) and \( m_2 \), then perpendicularity means \( m_1 \cdot m_2 = -1 \).

Question 24. Find the point on the curve \( y = 2x^2 - 6x - 4 \) at which the tangent is parallel to the x-axis.
Answer: A tangent parallel to the x-axis has slope zero. Finding the derivative: \( \frac{dy}{dx} = 4x - 6 \). Setting it to zero: \( 4x - 6 = 0 \Rightarrow x = \frac{3}{2} \). Substituting into the curve: \( y = 2 \cdot \frac{9}{4} - 6 \cdot \frac{3}{2} - 4 = \frac{9}{2} - 9 - 4 = -\frac{17}{2} \). Therefore, the required point is \( \left(\frac{3}{2}, -\frac{17}{2}\right) \).
In simple words: A horizontal tangent has slope zero. Set the derivative equal to zero and solve for x, then find the y-value on the curve.

Exam Tip: For horizontal tangents, always set \( \frac{dy}{dx} = 0 \) and solve. Verify that the resulting point lies on the original curve.

Question 25. Find the point on the parabola \( y = (x - 3)^2 \) where the tangent is parallel to the chord joining the point (3, 0) and (4, 1).
Answer: The slope of the chord joining (3, 0) and (4, 1) is \( m = \frac{1 - 0}{4 - 3} = 1 \). Finding the derivative of the parabola: \( \frac{dy}{dx} = 2(x - 3) \). Setting it equal to 1: \( 2(x - 3) = 1 \Rightarrow x - 3 = \frac{1}{2} \Rightarrow x = \frac{7}{2} \). When \( x = \frac{7}{2} \): \( y = \left(\frac{7}{2} - 3\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \). Therefore, the required point is \( \left(\frac{7}{2}, \frac{1}{4}\right) \).
In simple words: Find the slope of the chord by using the two given points. Then find where the curve's tangent has the same slope using the derivative.

Exam Tip: A chord connecting two points has slope \( \frac{y_2 - y_1}{x_2 - x_1} \). A tangent parallel to this chord must have the same slope.

Question 26. Show that the curves \( x = y^2 \) and \( xy = k \) cut at right angles if \( 8k^2 = 1 \).
Answer: Two curves cut orthogonally (at right angles) when the product of their slopes at the intersection point equals -1. Finding the intersection point: from \( x = y^2 \) and \( xy = k \), we have \( y^2 \cdot y = k \Rightarrow y^3 = k \Rightarrow y = k^{1/3} \) and \( x = k^{2/3} \). For the first curve \( x = y^2 \): \( \frac{dy}{dx} = \frac{1}{2y} \). At the intersection: \( m_1 = \frac{1}{2k^{1/3}} \). For the second curve \( xy = k \): \( y + x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \). At the intersection: \( m_2 = -\frac{k^{1/3}}{k^{2/3}} = -\frac{1}{k^{1/3}} \). Computing the product: \( m_1 m_2 = \frac{1}{2k^{1/3}} \cdot \left(-\frac{1}{k^{1/3}}\right) = -\frac{1}{2k^{2/3}} \). For orthogonal intersection: \( -\frac{1}{2k^{2/3}} = -1 \Rightarrow \frac{1}{2k^{2/3}} = 1 \Rightarrow k^{2/3} = \frac{1}{2} \Rightarrow k^2 = \frac{1}{8} \Rightarrow 8k^2 = 1 \).
In simple words: Find the common point of the two curves. Calculate the slope of each curve at that point. If their product equals -1, the curves intersect at right angles.

Exam Tip: For orthogonal intersection, verify that \( m_1 m_2 = -1 \) exactly. Always find the intersection point first before computing slopes.

Question 27. Show that the curves \( xy = a^2 \) and \( x^2 + y^2 = 2a^2 \) touch each other.
Answer: Two curves touch (are tangent to) each other when they share a common point and have the same tangent line at that point (the angle between them is zero). Finding intersection points: from \( xy = a^2 \), we have \( y = \frac{a^2}{x} \). Substituting into the second curve: \( x^2 + \frac{a^4}{x^2} = 2a^2 \Rightarrow x^4 + a^4 = 2a^2x^2 \Rightarrow x^4 - 2a^2x^2 + a^4 = 0 \Rightarrow (x^2 - a^2)^2 = 0 \Rightarrow x = \pm a \). At \( x = a \): \( y = a \). At \( x = -a \): \( y = -a \). For \( xy = a^2 \): \( \frac{dy}{dx} = -\frac{a^2}{x^2} \). At (a, a): \( m_1 = -1 \). At (-a, -a): \( m_1 = -1 \). For \( x^2 + y^2 = 2a^2 \): \( 2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y} \). At (a, a): \( m_2 = -1 \). At (-a, -a): \( m_2 = -1 \). Since the slopes match at both points, the curves are tangent and therefore touch each other.
In simple words: Two curves touch when they meet at a point and have identical slopes there. Find intersection points, compute the slope of each curve at those points, and verify the slopes are equal.

Exam Tip: "Touching" means tangency - the curves have the same slope at the point of contact. "Cutting orthogonally" means perpendicular slopes (product = -1).

Question 28. Show that the curves \( x^3 - 3xy^2 + 2 = 0 \) and \( 3x^2y - y^3 - 2 = 0 \) cut orthogonally.
Answer: Two curves cut orthogonally when the product of their slopes at the intersection point equals -1. Finding the intersection: adding the two equations gives \( x^3 - 3xy^2 + 2 + 3x^2y - y^3 - 2 = 0 \Rightarrow x^3 - y^3 - 3xy^2 + 3x^2y = 0 \Rightarrow (x - y)^3 = 0 \Rightarrow x = y \). Substituting into the first equation: \( x^3 - 3x^3 + 2 = 0 \Rightarrow -2x^3 + 2 = 0 \Rightarrow x = 1 \). Therefore, the intersection point is (1, 1). For the first curve, differentiating implicitly: \( 3x^2 - 3y^2 - 6xy\frac{dy}{dx} = 0 \). At (1, 1): \( 3 - 3 - 6\frac{dy}{dx} = 0 \Rightarrow m_1 = 0 \). For the second curve: \( 6xy + 3x^2\frac{dy}{dx} - 3y^2\frac{dy}{dx} = 0 \). At (1, 1): \( 6 + 3\frac{dy}{dx} - 3\frac{dy}{dx} = 0 \) is unclear, so use \( 3x^2\frac{dy}{dx} - 3y^2\frac{dy}{dx} = -6xy \Rightarrow \frac{dy}{dx}(3x^2 - 3y^2) = -6xy \). At (1, 1): the denominator is zero, so \( m_2 \) is undefined. Since \( m_1 = 0 \) (horizontal) and \( m_2 \) is undefined (vertical), the tangents are perpendicular, confirming orthogonal intersection at 90°.
In simple words: Find where the curves meet. Calculate the slope at that point for each curve. If one slope is zero and the other is undefined (or their product is -1), the curves are perpendicular.

Exam Tip: When one slope is 0 and the other undefined, the curves are automatically orthogonal. Always verify the intersection point satisfies both original equations.

Question 29. Find the equation of tangent to the curve \( x = (\theta + \sin\theta) \), \( y = (1 + \cos\theta) \) at \( \theta = \frac{\pi}{4} \).
Answer: For parametric curves, the slope is \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \). Finding the derivatives: \( \frac{dx}{d\theta} = 1 + \cos\theta \) and \( \frac{dy}{d\theta} = -\sin\theta \). At \( \theta = \frac{\pi}{4} \): \( \frac{dx}{d\theta} = 1 + \frac{1}{\sqrt{2}} \) and \( \frac{dy}{d\theta} = -\frac{1}{\sqrt{2}} \). Therefore, \( m = \frac{-1/\sqrt{2}}{1 + 1/\sqrt{2}} = \frac{-1}{\sqrt{2} + 1} = \frac{-1(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{-(\sqrt{2} - 1)}{2 - 1} = -({\sqrt{2} - 1}) = 1 - \sqrt{2} \). At \( \theta = \frac{\pi}{4} \): \( x = \frac{\pi}{4} + \frac{1}{\sqrt{2}} \) and \( y = 1 + \frac{1}{\sqrt{2}} \). Using point-slope form: \( y - (1 + \frac{1}{\sqrt{2}}) = (1 - \sqrt{2})(x - (\frac{\pi}{4} + \frac{1}{\sqrt{2}})) \). Simplifying yields \( y = (1 - \sqrt{2})x + \frac{(\sqrt{2} - 1)\pi}{4} + 2 \).
In simple words: For parametric equations, find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), then compute the slope as their ratio. Evaluate at the given parameter value and use point-slope form.

Exam Tip: Always simplify the slope expression fully. Rationalizing denominators with surds makes the final tangent equation cleaner.

Question 30. Find the equation of the tangent at \( t = \frac{\pi}{4} \) for the curve \( x = \sin 3t \), \( y = \cos 2t \).
Answer: For parametric curves, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Finding derivatives: \( \frac{dx}{dt} = 3\cos 3t \) and \( \frac{dy}{dt} = -2\sin 2t \). At \( t = \frac{\pi}{4} \): \( \frac{dx}{dt} = 3\cos \frac{3\pi}{4} = 3 \cdot (-\frac{1}{\sqrt{2}}) = -\frac{3}{\sqrt{2}} \) and \( \frac{dy}{dt} = -2\sin \frac{\pi}{2} = -2 \). Therefore, \( m = \frac{-2}{-3/\sqrt{2}} = \frac{2\sqrt{2}}{3} \). At \( t = \frac{\pi}{4} \): \( x = \sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}} \) and \( y = \cos \frac{\pi}{2} = 0 \). Using point-slope form: \( y - 0 = \frac{2\sqrt{2}}{3}(x - \frac{1}{\sqrt{2}}) \Rightarrow y = \frac{2\sqrt{2}}{3}x - \frac{2}{3} \Rightarrow 4x - 3\sqrt{2}y - 2\sqrt{2} = 0 \).
In simple words: Differentiate x and y with respect to the parameter. Form the slope as the ratio of derivatives. Evaluate at the given parameter and write the tangent using point-slope form.

Exam Tip: For parametric tangents, always simplify to avoid errors. Multiply through by appropriate factors to clear square roots from coefficients when presenting the final answer.

Question 1. Mark (√) against the correct answer: If \( y = 2^x \) then \( \frac{dy}{dx} = ? \)
(a) \( x(2^{x-1}) \)
(b) \( \frac{2^x}{(\log 2)} \)
(c) \( 2^x(\log 2) \)
(d) None of these
Answer: (c) \( 2^x(\log 2) \)
In simple words: Start with \( y = 2^x \). Taking the natural logarithm of both sides gives \( \log y = x \log 2 \). Differentiating with respect to x yields \( \frac{1}{y}\frac{dy}{dx} = \log 2 \), so \( \frac{dy}{dx} = y \log 2 = 2^x \log 2 \).

Exam Tip: For exponential functions of the form \( a^x \), the derivative is \( a^x \log a \). Memorizing this rule saves time in multiple-choice questions.

Question 2. Mark (√) against the correct answer: If \( y = 10^x \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{10^x}{\log 10} \)
(b) \( 10^x \log 10 \)
(c) \( x \cdot 10^{x-1} \)
(d) None of these
Answer: (b) \( 10^x \log 10 \)
In simple words: The derivative of \( a^x \) is \( a^x \log a \) where log denotes the natural logarithm (or common logarithm, the base choice must match context). Here, \( \frac{d}{dx}(10^x) = 10^x \log 10 \).

Exam Tip: Do not confuse the power rule (\( x^n \to nx^{n-1} \)) with the exponential rule (\( a^x \to a^x \log a \)). In option (c), the exponent is the variable, not the base.

Question 3. Mark (√) against the correct answer: If \( y = \log_{10} x \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{1}{x} \)
(b) \( \frac{1}{x}(\log 10) \)
(c) \( \frac{1}{x(\log 10)} \)
(d) None of these
Answer: (c) \( \frac{1}{x(\log 10)} \)
In simple words: To differentiate \( y = \log_{10} x \), convert using the change-of-base formula: \( y = \frac{\log x}{\log 10} \). Then \( \frac{dy}{dx} = \frac{1}{\log 10} \cdot \frac{1}{x} = \frac{1}{x \log 10} \).

Exam Tip: The derivative of \( \log_a x \) is \( \frac{1}{x \log a} \). Avoid the common mistake of forgetting the logarithm of the base in the denominator.

Question 4. Mark (√) against the correct answer: If \( y = x^x \) then \( \frac{dy}{dx} = ? \)
(a) \( x^x \log x \)
(b) \( x^x(1 + \log x) \)
(c) \( x(1 + \log x) \)
(d) None of these
Answer: (b) \( x^x(1 + \log x) \)
In simple words: Let \( y = x^x \). Taking logarithms: \( \log y = x \log x \). Differentiating both sides with respect to x: \( \frac{1}{y}\frac{dy}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1 \). Therefore, \( \frac{dy}{dx} = y(1 + \log x) = x^x(1 + \log x) \).

Exam Tip: For functions of the form \( f(x)^{g(x)} \) where both base and exponent are variables, always use logarithmic differentiation.

Question 5. Mark (√) against the correct answer: If \( y = x^{\sin x} \) then \( \frac{dy}{dx} = ? \)
(a) \( (\sin x) \cdot x^{(\sin x - 1)} \)
(b) \( (\sin x \cos x) \cdot x^{(\sin x - 1)} \)
(c) \( x^{\sin x}\left(\frac{\sin x + x \log x \cdot \cos x}{x}\right) \)
(d) None of these
Answer: (c) \( x^{\sin x}\left(\frac{\sin x + x \log x \cdot \cos x}{x}\right) \)
In simple words: Let \( y = x^{\sin x} \). Taking logarithms: \( \log y = \sin x \log x \). Differentiating with respect to x: \( \frac{1}{y}\frac{dy}{dx} = \cos x \log x + \sin x \cdot \frac{1}{x} = \frac{\cos x \log x + \sin x / x}{1} \). Therefore, \( \frac{dy}{dx} = x^{\sin x}\left(\frac{\sin x + x \log x \cos x}{x}\right) \).

Exam Tip: When both the base and exponent are functions of x, logarithmic differentiation is required. Simplify the final expression carefully.

Question 6. Mark (√) against the correct answer: If \( y = x^{\sqrt{x}} \) then \( \frac{dy}{dx} = ? \)
(a) \( \sqrt{x} \cdot x^{(\sqrt{x}-1)} \)
(b) \( \frac{x^{\sqrt{x}} \log x}{2\sqrt{x}} \)
(c) \( x^{\sqrt{x}}\left(\frac{2 + \log x}{2\sqrt{x}}\right) \)
(d) None of these
Answer: (c) \( x^{\sqrt{x}}\left(\frac{2 + \log x}{2\sqrt{x}}\right) \)
In simple words: Let \( y = x^{\sqrt{x}} \). Taking logarithms: \( \log y = \sqrt{x} \log x \). Differentiating: \( \frac{1}{y}\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \log x + \sqrt{x} \cdot \frac{1}{x} = \frac{\log x}{2\sqrt{x}} + \frac{1}{\sqrt{x}} = \frac{\log x + 2}{2\sqrt{x}} \). Thus, \( \frac{dy}{dx} = x^{\sqrt{x}} \cdot \frac{\log x + 2}{2\sqrt{x}} \).

Exam Tip: Combine fraction terms carefully. Check by substituting simple values (like x = 1) to ensure reasonableness.

Question 7. Mark (√) against the correct answer: If \( y = e^{\sin\sqrt{x}} \) then \( \frac{dy}{dx} = ? \)
(a) \( e^{\sin\sqrt{x}} \cdot \cos\sqrt{x} \)
(b) \( \frac{e^{\sin\sqrt{x}} \cos\sqrt{x}}{2\sqrt{x}} \)
(c) \( \frac{e^{\sin\sqrt{x}}}{2\sqrt{x}} \)
(d) None of these
Answer: (b) \( \frac{e^{\sin\sqrt{x}} \cos\sqrt{x}}{2\sqrt{x}} \)
In simple words: Use the chain rule repeatedly: \( \frac{dy}{dx} = e^{\sin\sqrt{x}} \cdot \cos\sqrt{x} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sin\sqrt{x}} \cos\sqrt{x}}{2\sqrt{x}} \).

Exam Tip: For composite functions involving exponentials, apply the chain rule step by step: outermost function first, then work inward.

Question 8. Mark (√) against the correct answer: If \( y = (\tan x)^{\cot x} \) then \( \frac{dy}{dx} = ? \)
(a) \( \cot x \cdot (\tan x)^{\cot x - 1} \cdot \sec^2 x \)
(b) \( -(\tan x)^{\cot x} \cdot \csc^2 x \)
(c) \( (\tan x)^{\cot x} \cdot \csc^2 x(1 - \log \tan x) \)
(d) None of these
Answer: (c) \( (\tan x)^{\cot x} \cdot \csc^2 x(1 - \log \tan x) \)
In simple words: Let \( y = (\tan x)^{\cot x} \). Taking logarithms: \( \log y = \cot x \log(\tan x) \). Differentiating: \( \frac{1}{y}\frac{dy}{dx} = -\csc^2 x \log(\tan x) + \cot x \cdot \frac{\sec^2 x}{\tan x} = \csc^2 x(1 - \log \tan x) \). Therefore, \( \frac{dy}{dx} = (\tan x)^{\cot x} \cdot \csc^2 x(1 - \log \tan x) \).

Exam Tip: When the exponent is a trigonometric function, logarithmic differentiation becomes essential. Take care with signs when differentiating negative derivatives like \( -\csc^2 x \).

Question 9. Mark (√) against the correct answer: If \( y = (\sin x)^{\log x} \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{(\sin x)^{\log x}}{x}(x \cos x - \log x \sin x) \)
(b) \( (\sin x)^{\log x}\left(\frac{\log x}{x} + \frac{\cos x}{sin x}\right) \)
(c) \( (\sin x)^{\log x}\left(\frac{\log x \cos x + \sin x}{x \sin x}\right) \)
(d) None of these
Answer: (c) \( (\sin x)^{\log x}\left(\frac{\log x \cos x + \sin x}{x \sin x}\right) \)
In simple words: Let \( y = (\sin x)^{\log x} \). Taking logarithms: \( \log y = \log x \log(\sin x) \). Differentiating: \( \frac{1}{y}\frac{dy}{dx} = \frac{1}{x} \log(\sin x) + \log x \cdot \frac{\cos x}{\sin x} = \frac{\log x \cos x + \sin x / x}{sin x} \). Simplifying and multiplying by y gives the result.

Exam Tip: Carefully combine logarithmic terms. Always find a common denominator before finalizing the answer.

Question 10. Mark (√) against the correct answer: If \( y = \sin(x^x) \) then \( \frac{dy}{dx} = ? \)
(a) \( x^x \cos(x^x) \)
(b) \( x^x \cos x^x(1 + \log x) \)
(c) \( x^x \cos x^x \log x \)
(d) None of these
Answer: (b) \( x^x \cos x^x(1 + \log x) \)
In simple words: Let \( u = x^x \), so \( y = \sin u \). By the chain rule: \( \frac{dy}{dx} = \cos u \cdot \frac{du}{dx} = \cos(x^x) \cdot \frac{d}{dx}(x^x) \). From logarithmic differentiation, \( \frac{d}{dx}(x^x) = x^x(1 + \log x) \). Therefore, \( \frac{dy}{dx} = x^x \cos(x^x)(1 + \log x) \).

Exam Tip: Always use the chain rule for composite functions. Identify the "outer" function (sine) and the "inner" function (\( x^x \)), then combine derivatives.

Question 11. Mark (√) against the correct answer: If \( y = \sqrt{x \sin x} \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{(x \cos x + \sin x)}{2\sqrt{x \sin x}} \)
(b) \( \frac{1}{2}(x \cos x + \sin x) \cdot \sqrt{x \sin x} \)
(c) \( \frac{1}{2\sqrt{x \sin x}} \)
(d) None of these
Answer: (a) \( \frac{(x \cos x + \sin x)}{2\sqrt{x \sin x}} \)
In simple words: Rewrite \( y = (x \sin x)^{1/2} \). By the chain rule: \( \frac{dy}{dx} = \frac{1}{2}(x \sin x)^{-1/2} \cdot \frac{d}{dx}(x \sin x) \). Now, \( \frac{d}{dx}(x \sin x) = \sin x + x \cos x \). Therefore, \( \frac{dy}{dx} = \frac{\sin x + x \cos x}{2\sqrt{x \sin x}} \).

Exam Tip: For square root derivatives, use the power rule with exponent 1/2, then apply the product rule to the expression under the square root.

Question 12. Mark (√) against the correct answer: If \( e^{x+y} = xy \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{x(1-y)}{y(x-1)} \)
(b) \( \frac{y(1-x)}{x(y-1)} \)
(c) \( \frac{(x-xy)}{(xy-y)} \)
(d) None of these
Answer: (b) \( \frac{y(1-x)}{x(y-1)} \)
In simple words: Taking logarithms of \( e^{x+y} = xy \): \( x + y = \log x + \log y \). Differentiating implicitly: \( 1 + \frac{dy}{dx} = \frac{1}{x} + \frac{1}{y}\frac{dy}{dx} \). Rearranging: \( \frac{dy}{dx}\left(1 - \frac{1}{y}\right) = \frac{1}{x} - 1 \Rightarrow \frac{dy}{dx} \cdot \frac{y-1}{y} = \frac{1-x}{x} \Rightarrow \frac{dy}{dx} = \frac{y(1-x)}{x(y-1)} \).

Exam Tip: For implicit equations, take logarithms first to simplify before differentiating. Always collect all \( \frac{dy}{dx} \) terms on one side.

Question 13. Mark (√) against the correct answer: If \( (x + y) = \sin(x + y) \) then \( \frac{dy}{dx} = ? \)
(a) -1
(b) 1
(c) \( \frac{1 - \cos(x + y)}{\cos^2(x + y)} \)
(d) None of these
Answer: (a) -1
In simple words: Differentiating implicitly: \( 1 + \frac{dy}{dx} = \cos(x + y) \cdot (1 + \frac{dy}{dx}) \). This gives \( (1 + \frac{dy}{dx})(1 - \cos(x + y)) = 0 \). Either \( \cos(x + y) = 1 \) or \( \frac{dy}{dx} = -1 \). When \( \cos(x + y) = 1 \), we have \( x + y = 2n\pi \) (so \( x + y = \sin(2n\pi) = 0 \Rightarrow y = -x \)), which also gives \( \frac{dy}{dx} = -1 \). In both cases, the answer is -1.

Exam Tip: When implicit differentiation yields a product equal to zero, consider all possible cases before choosing the final answer. Verify consistency with the original equation.

Question 14. Mark (√) against the correct answer in the following: If \( \sqrt{x} + \sqrt{y} = \sqrt{a} \) then \( \frac{dy}{dx} = ? \)
(a) \( -\frac{\sqrt{x}}{\sqrt{y}} \)
(b) \( -\frac{1}{2} \cdot \frac{\sqrt{y}}{\sqrt{x}} \)
(c) \( -\frac{\sqrt{y}}{\sqrt{x}} \)
(d) None of these
Answer: (c) \( -\frac{\sqrt{y}}{\sqrt{x}} \)
In simple words: When you have an equation with square roots, differentiate both sides with respect to x. The derivative comes out to the negative ratio of the square roots.

Exam Tip: Apply implicit differentiation carefully by treating both \( \sqrt{x} \) and \( \sqrt{y} \) as functions of x; the chain rule is essential here.

Question 15. Mark (√) against the correct answer in the following: If \( x^y = y^x \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{(y - x \log y)}{(x - y \log x)} \)
(b) \( \frac{y(y - x \log y)}{x(x - y \log x)} \)
(c) \( \frac{y(y + x \log y)}{x(x + y \log x)} \)
(d) none of these
Answer: (b) \( \frac{y(y - x \log y)}{x(x - y \log x)} \)
In simple words: Take the natural logarithm of both sides of the equation, then differentiate using implicit differentiation. Simplify to find the derivative.

Exam Tip: When using logarithmic differentiation on equations of the form \( a^b = c^d \), remember to apply the chain rule to each logarithmic term.

Question 16. Mark (√) against the correct answer in the following: If \( x^p y^q = (x + y)^{p+q} \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{x}{y} \)
(b) \( \frac{y}{x} \)
(c) \( \frac{x^{p-1}}{y^{q-1}} \)
(d) none of these
Answer: (b) \( \frac{y}{x} \)
In simple words: Apply logarithms to simplify the given equation, then use the product and chain rules to differentiate. The final result reduces to a simple ratio.

Exam Tip: Logarithmic differentiation is particularly useful for equations involving products and powers; always expand the logarithm of products as sums of logarithms.

Question 17. Mark (√) against the correct answer in the following: If \( y \log_a x = x \log_a y \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{y(y - x \log_a y)}{x(x - y \log_a x)} \)
(b) \( \frac{y(y + x \log_a y)}{x(x + y \log_a x)} \)
(c) \( \frac{dy}{dx} = \frac{y(y - x \log_a y)}{x(x - y \log_a x)} \)
(d) None of these
Answer: (a) \( \frac{y(y - x \log_a y)}{x(x - y \log_a x)} \)
In simple words: Differentiate both sides of the given equation implicitly with respect to x, then rearrange to isolate \( \frac{dy}{dx} \).

Exam Tip: When logarithms with different bases appear, use the change of base formula and the logarithmic product rule to simplify before differentiating.

Question 18. Mark (√) against the correct answer in the following: If \( y = \cos^2 x^3 \) then \( \frac{dy}{dx} = ? \)
(a) \( -3x^2 \sin \)
(b) \( -3x^2 \sin^2 x^3 \)
(c) \( -3x^2 \cos^2 \left( 2x^3 \right) \)
(d) none of these
Answer: (c) \( -3x^2 \sin(2x^3) \)
In simple words: Use the chain rule twice - once for the square and once for the cube inside. Then apply the double-angle identity to simplify the trigonometric expression.

Exam Tip: Always identify composite functions and apply the chain rule from the outermost function inward; recognize when trigonometric identities can simplify your final answer.

Question 19. Mark (√) against the correct answer in the following: If \( y = x^2 \sin\frac{1}{x} \) then \( \frac{dy}{dx} = ? \)
(a) \( x \sin\frac{1}{x} - \cos\frac{1}{x} \)
(b) \( -\cos\frac{1}{x} + 2x \sin\frac{1}{x} \)
(c) \( -x \sin\frac{1}{x} + \cos\frac{1}{x} \)
(d) None of these
Answer: (b) \( -\cos\frac{1}{x} + 2x \sin\frac{1}{x} \)
In simple words: Apply the product rule to differentiate the product of \( x^2 \) and \( \sin\frac{1}{x} \). The chain rule is needed for the sine term.

Exam Tip: When a product involves a composite trigonometric function, the chain rule produces a chain of derivatives; organize your work carefully to avoid sign errors.

Question 20. Mark (√) against the correct answer in the following: If \( y = \log \left( x + \sqrt{x^2 + a^2} \right) \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{1}{2\left(x + \sqrt{x^2 + a^2}\right)} \)
(b) \( \frac{-1}{\sqrt{x^2 + a^2}} \)
(c) \( \frac{1}{\sqrt{x^2 + a^2}} \)
(d) none of these
Answer: (c) \( \frac{1}{\sqrt{x^2 + a^2}} \)
In simple words: Use the chain rule on the logarithmic function. The numerator simplifies when you differentiate the expression inside the square root.

Exam Tip: After differentiating a logarithm containing a nested radical, look for algebraic simplifications that cancel common terms in the numerator and denominator.

Question 21. Mark (√) against the correct answer in the following: If \( y = \log \left( \frac{1 + \sqrt{x}}{1 - \sqrt{x}} \right) \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{1}{\sqrt{x}(1 - x)} \)
(b) \( \frac{-1}{x\left(1 - \sqrt{x}\right)^2} \)
(c) \( \frac{-\sqrt{x}}{2(1 - \sqrt{x})} \)
(d) none of these
Answer: (a) \( \frac{1}{\sqrt{x}(1 - x)} \)
In simple words: Apply the chain rule to the logarithm, then use the quotient rule to differentiate the argument. Simplify by finding a common denominator.

Exam Tip: When logarithms contain fractions, use the property \( \log\frac{a}{b} = \log a - \log b \) to separate the terms before differentiating.

Question 22. Mark (√) against the correct answer in the following: If \( y = \log \left( \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2} - x} \right) \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{2}{\sqrt{1 + x^2}} \)
(b) \( \frac{2\sqrt{1 + x^2}}{x^2} \)
(c) \( \frac{-2}{\sqrt{1 + x^2}} \)
(d) none of these
Answer: (a) \( \frac{2}{\sqrt{1 + x^2}} \)
In simple words: Break the logarithm into a difference of two logarithms, then differentiate each part. The denominator simplifies through algebraic cancellation.

Exam Tip: For complex logarithmic expressions, separate numerator and denominator using logarithm laws first; this makes differentiation more straightforward.

Question 23. Mark (√) against the correct answer in the following: If \( y = \sqrt{\frac{1 + \sin x}{1 - \sin x}} \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{1}{2} \sec^2 \left( \frac{\pi}{4} - \frac{\pi}{2} \right) \)
(b) \( \frac{1}{2} \csc ec^2 \left( \frac{\pi}{4} - \frac{\pi}{2} \right) \)
(c) \( \frac{1}{2} \csc ec^2 \left( \frac{\pi}{4} - \frac{\pi}{2} \right) \cot \left( \frac{\pi}{4} - \frac{\pi}{2} \right) \)
(d) none of these
Answer: (b) \( \frac{1}{2} \csc ec^2 \left( \frac{\pi}{4} - \frac{\pi}{2} \right) \)
In simple words: Simplify the fraction inside the square root using half-angle identities, then apply the chain rule to the resulting trigonometric expression.

Exam Tip: Recognize half-angle and sum-to-product formulas that transform complex trigonometric fractions; this often leads to much simpler derivatives.

Question 24. Mark (√) against the correct answer in the following: If \( y = \sqrt{\frac{\sec x - 1}{\sec x + 1}} \) then \( \frac{dy}{dx} = ? \)
(a) \( \sec^2 x \)
(b) \( \frac{1}{2} \sec^2 \frac{x}{2} \)
(c) \( \frac{1}{2} \csc ec^2 \frac{x}{2} \)
(d) none of these
Answer: (c) \( \frac{1}{2} \sec^2 \frac{x}{2} \)
In simple words: Convert secant to cosine in the numerator and denominator, apply half-angle identities, then simplify to a tangent function before differentiating.

Exam Tip: When square roots of trigonometric fractions appear, use half-angle conversions to transform the expression into a simpler trigonometric function.

Question 25. Mark (√) against the correct answer in the following: If \( y = \tan^{-1} \left( \frac{1 - \cos x}{\sin x} \right) \) then \( \frac{dy}{dx} = ? \)
(a) 1
(b) -1
(c) \( \frac{1}{2} \)
(d) \( \frac{-1}{2} \)
Answer: (c) \( \frac{1}{2} \)
In simple words: Apply half-angle identities to simplify the argument of the inverse tangent, which reduces it to \( \frac{x}{2} \). Then differentiate.

Exam Tip: For inverse trigonometric functions, simplify the argument first using identities; the derivative often becomes much simpler after simplification.

Question 26. Mark (√) against the correct answer in the following: If \( y = \tan^{-1} \left( \frac{\cos x + \sin x}{\cos x - \sin x} \right) \) then \( \frac{dy}{dx} = ? \)
(a) 1
(b) -1
(c) \( \frac{1}{2} \)
(d) \( \frac{-1}{2} \)
Answer: (a) 1
In simple words: Divide both the numerator and denominator by cosine x to convert the fraction into a tangent expression. Use the tangent addition formula to simplify the result.

Exam Tip: When a fraction of trigonometric functions appears in an inverse function, dividing by a common trigonometric term often reveals hidden angles.

Question 27. Mark (√) against the correct answer in the following: If \( y = \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right) \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{1}{2} \)
(b) \( \frac{-1}{2} \)
(c) 1
(d) -1
Answer: (b) \( \frac{-1}{2} \)
In simple words: Apply half-angle and complementary angle identities to rewrite the argument as an inverse tangent of a half-angle. Differentiate the simplified form.

Exam Tip: Expressions of the form \( \frac{\cos x}{1 + \sin x} \) often simplify to half-angles or complementary angles; try rewriting in terms of \( \frac{x}{2} \) or \( \frac{\pi}{4} - \frac{x}{2} \).

Question 28. Mark (√) against the correct answer in the following: If \( y = \tan^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}} \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{1}{2} \)
(b) \( \frac{-1}{2} \)
(c) \( \frac{1}{(1 + x^2)} \)
(d) none of these
Answer: (a) \( \frac{1}{2} \)
In simple words: Use the half-angle formula to simplify the square root expression to a tangent of a half-angle. Then apply the inverse tangent property and differentiate.

Exam Tip: Half-angle identities \( 1 - \cos x = 2\sin^2\frac{x}{2} \) and \( 1 + \cos x = 2\cos^2\frac{x}{2} \) are key for simplifying such expressions.

Question 29. Mark (√) against the correct answer in the following: If \( y = \tan^{-1} \left( \frac{a \cos x - b \sin x}{b \cos x + a \sin x} \right) \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{a}{b} \)
(b) \( \frac{-b}{a} \)
(c) 1
(d) -1
Answer: (d) -1
In simple words: Divide the numerator and denominator by b times cosine x. Introduce a parameter for the ratio and use the tangent subtraction formula. Simplify to find the derivative.

Exam Tip: When the inverse tangent argument contains constants multiplied by sine and cosine, parametrize using inverse tangent of the ratio; the tangent subtraction formula then applies.

Question 30. Mark (√) against the correct answer in the following: If \( y = \sin^{-1} \left( 3x - 4x^3 \right) \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{3}{\sqrt{1 - x^2}} \)
(b) \( \frac{-4}{\sqrt{1 - x^2}} \)
(c) \( \frac{3}{\sqrt{1 + x^2}} \)
(d) none of these
Answer: (a) \( \frac{3}{\sqrt{1 - x^2}} \)
In simple words: Recognize that the argument \( 3x - 4x^3 \) is the triple angle formula for sine. Substitute \( x = \sin\theta \) to transform the expression into \( \sin 3\theta \), then simplify and differentiate.

Exam Tip: Identify trigonometric triple angle formulas in inverse function arguments; substituting a trigonometric variable often reveals a simpler composition.

Question 31. Mark (√) against the correct answer in the following: If \( y = \cos^{-1} \left( 4x^3 - 3x \right) \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{3}{\sqrt{1 - x^2}} \)
(b) \( \frac{-3}{\sqrt{1 - x^2}} \)
(c) \( \frac{4}{\sqrt{1 - x^2}} \)
(d) \( \frac{4}{(3x^2 - 1)} \)
Answer: (b) \( \frac{-3}{\sqrt{1 - x^2}} \)
In simple words: The argument is the triple angle formula for cosine. Substitute \( x = \cos\theta \) to convert it to \( \cos 3\theta = 3\theta \) where \( \theta = \cos^{-1}x \). Differentiate the simplified form.

Exam Tip: The cosine triple angle formula \( \cos 3\theta = 4\cos^3\theta - 3\cos\theta \) paired with inverse cosine simplifies derivatives significantly.

Question 32. Mark (√) against the correct answer in the following: If \( y = \tan^{-1} \left( \frac{\sqrt{a} + \sqrt{x}}{1 - \sqrt{ax}} \right) \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{1}{(1 + x)} \)
(b) \( \frac{1}{\sqrt{x}(1 + x)} \)
(c) \( \frac{2}{\sqrt{x}(1 + x)} \)
(d) \( \frac{1}{2\sqrt{x}(1 + x)} \)
Answer: (b) \( \frac{1}{\sqrt{x}(1 + x)} \)
In simple words: Set \( \sqrt{a} = \tan A \) and \( \sqrt{x} = \tan B \). Use the tangent addition formula to rewrite the argument as \( \tan(A + B) \). Then differentiate the resulting inverse tangent.

Exam Tip: When the inverse tangent argument has a structure resembling the tangent addition formula, introduce auxiliary angles to reveal the hidden composition.

Question 33. Mark (√) against the correct answer in the following: If \( y = \cos^{-1} \left( \frac{x^2 - 1}{x^2 + 1} \right) \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{2}{(1 + x^2)} \)
(b) \( \frac{-2}{(1 + x^2)} \)
(c) \( \frac{2x}{(1 + x^2)} \)
(d) none of these
Answer: (b) \( \frac{-2}{(1 + x^2)} \)
In simple words: Set \( x = \tan\theta \), so the argument becomes the double angle formula \( \cos 2\theta \). This simplifies to \( -2\theta = -2\tan^{-1}x \). Differentiate to obtain the result.

Exam Tip: The double angle formula \( \cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} \) appears frequently in inverse trigonometric derivatives; recognize it to simplify your work.

Question 34. Mark (√) against the correct answer in the following: If \( y = \tan^{-1} \left( \frac{1 + x^2}{1 - x^2} \right) \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{2x}{(1 + x^4)} \)
(b) \( \frac{-2x}{(1 + x^4)} \)
(c) \( \frac{x}{(1 + x^4)} \)
(d) none of these
Answer: (b) \( \frac{-2x}{(1 + x^4)} \)
In simple words: Set \( x^2 = \tan\theta \), so the argument simplifies using the double angle formula. This gives \( -2\theta = -2\tan^{-1}(x^2) \). Differentiate with the chain rule.

Exam Tip: When the argument of an inverse function is a function of \( x^2 \), use a substitution like \( x^2 = \tan\theta \) to reveal double-angle structures more clearly.

Question 35. Mark (√) against the correct answer in the following: If \( y = \tan^{-1} \left( -\sqrt{x} \right) \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{-1}{(1 + x)} \)
(b) \( \frac{2}{\sqrt{(1 + x)}} \)
(c) \( \frac{-1}{2\sqrt{x}(1 + x)} \)
(d) none of these
Answer: (c) \( \frac{-1}{2\sqrt{x}(1 + x)} \)
In simple words: Apply the chain rule to the inverse tangent function. Differentiate the argument \( -\sqrt{x} \) to obtain \( \frac{-1}{2\sqrt{x}} \). Substitute into the derivative formula for inverse tangent.

Exam Tip: Always use the chain rule when the argument of an inverse trigonometric function is itself a composite expression; track each layer carefully.

Question 36. Mark (√) against the correct answer in the following: If \( y = \cos^{-1} x^3 \) then \( \frac{dy}{dx} = ? \)
(a) \( \frac{-1}{\sqrt{1 - x^6}} \)
(b) \( \frac{-3x^2}{\sqrt{1 - x^6}} \)
(c) \( \frac{-3x^2}{\sqrt{1 - x^6}} \)
(d) none of these
Answer: (b) \( \frac{-3x^2}{\sqrt{1 - x^6}} \)
In simple words: Apply the chain rule to inverse cosine. Differentiate the inner function \( x^3 \) to get \( 3x^2 \). Substitute into the derivative formula for inverse cosine, noting that the denominator involves \( 1 - (x^3)^2 = 1 - x^6 \).

Exam Tip: For composite inverse trigonometric functions, be careful with the denominator of the derivative formula - it involves the square of the entire argument, not just part of it.

Question 36. If \( y = \cos^{-1}(x^3) \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{-3x^2}{\sqrt{1-x^6}} \)
(B) \( \frac{-3x^2}{\sqrt{1-x^6}} \)
(C) \( \frac{-3}{x^2\sqrt{1-x^6}} \)
(D) none of these
Answer: (A) \( \frac{-3x^2}{\sqrt{1-x^6}} \)
In simple words: Take the derivative of the inverse cosine function. Apply the chain rule - first differentiate the inverse cosine part, then multiply by the derivative of what's inside the parentheses.

Exam Tip: Remember the derivative formula for \( \cos^{-1}(u) \) is \( \frac{-1}{\sqrt{1-u^2}} \times \frac{du}{dx} \). Always apply the chain rule carefully when the argument is not just x.

Question 37. Mark (√) against the correct answer in the following: If \( y = \tan^{-1}(\sec x + \tan x) \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{1}{2} \)
(B) \( \frac{-1}{2} \)
(C) 1
(D) none of these
Answer: (A) \( \frac{1}{2} \)
In simple words: Simplify the expression inside the inverse tangent first using trigonometric identities. Once you reduce it to a simpler form, the derivative becomes straightforward.

Exam Tip: Use half-angle identities and the identity \( \cos^2\theta + \sin^2\theta = 1 \) to simplify \( \sec x + \tan x \) before differentiating. This makes the calculation much faster.

Question 38. Mark (√) against the correct answer in the following: If \( y = \cot^{-1}\left(\frac{1-x}{1+x}\right) \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{-1}{(1+x^2)} \)
(B) \( \frac{1}{(1+x^2)} \)
(C) \( \frac{-1}{(1+x^2)^{3/2}} \)
(D) none of these
Answer: (B) \( \frac{1}{(1+x^2)} \)
In simple words: Substitute \( x = \tan\theta \) to convert the inverse cotangent into a simpler form. Once simplified, differentiation is direct.

Exam Tip: Trigonometric substitution is highly useful here. Setting \( x = \tan\theta \) converts the fraction into a recognizable inverse function form, which then becomes easy to differentiate.

Question 39. Mark (√) against the correct answer in the following: If \( y = \sqrt{\frac{1+x}{1-x}} \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{2}{(1-x)^2} \)
(B) \( \frac{x}{(1-x)^{3/2}} \)
(C) \( \frac{1}{(1-x)^{3/2} \cdot (1+x)^{1/2}} \)
(D) none of these
Answer: (C) \( \frac{1}{(1-x)^{3/2} \cdot (1+x)^{1/2}} \)
In simple words: Use the substitution \( x = -\cos\theta \) to rewrite the expression under the square root. This transforms it into a trigonometric form that can be simplified and then differentiated.

Exam Tip: The substitution method is essential here. By substituting \( x = -\cos\theta \), the nested radicals simplify into a tangent function, making differentiation manageable. Do not skip the intermediate simplification steps.

Question 40. Mark (√) against the correct answer in the following: If \( y = \sec^{-1}\left(\frac{x^2+1}{x^2-1}\right) \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{-2}{(1+x^2)} \)
(B) \( \frac{2}{(1+x^2)} \)
(C) \( \frac{-1}{(1+x^2)} \)
(D) none of these
Answer: (A) \( \frac{-2}{(1+x^2)} \)
In simple words: Begin by converting the secant inverse to a tangent inverse using algebraic manipulation. Once expressed as an inverse tangent, the derivative follows a standard pattern.

Exam Tip: Converting between different inverse trigonometric forms using identities can drastically simplify the problem. Always look for ways to rewrite the argument into a more manageable form before differentiating.

Question 41. Mark (√) against the correct answer in the following: If \( y = \sec^{-1}\left(\frac{1}{2x^2-1}\right) \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{-2}{(1+x^2)} \)
(B) \( \frac{-2}{(1-x^2)} \)
(C) \( \frac{-2}{\sqrt{1+x^2}} \)
(D) none of these
Answer: (B) \( \frac{-2}{(1-x^2)} \)
In simple words: Start by setting \( x = \cos\theta \). This substitution will transform the inverse secant into a cleaner inverse cosine expression. After simplification, differentiate with respect to x using the chain rule.

Exam Tip: Proper choice of substitution is critical. The substitution \( x = \cos\theta \) directly relates to the double angle formula \( 2\cos^2\theta - 1 \), which appears in your problem. This connection makes the simplification efficient.

Question 42. Mark (√) against the correct answer in the following: If \( y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{1}{(1+x^2)} \)
(B) \( \frac{2}{(1+x^2)} \)
(C) \( \frac{1}{2(1+x^2)} \)
(D) none of these
Answer: (C) \( \frac{1}{2(1+x^2)} \)
In simple words: Substitute \( x = \tan\theta \). This transforms the square root expression into a secant function. Simplify further using half-angle formulas to reduce the argument to a tangent of a half angle. Then differentiate.

Exam Tip: Half-angle substitutions can be powerful in inverse trigonometric problems. After setting \( x = \tan\theta \), look for opportunities to apply half-angle identities - they often turn complicated expressions into simple ones.

Question 43. Mark (√) against the correct answer in the following: If \( y = \sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{-1}{2\sqrt{1-x^2}} \)
(B) \( \frac{1}{2\sqrt{1-x^2}} \)
(C) \( \frac{1}{2\sqrt{1+x^2}} \)
(D) none of these
Answer: (B) \( \frac{1}{2\sqrt{1-x^2}} \)
In simple words: Let \( x = \cos(2\theta) \). Use the identities for \( 1 + \cos(2\theta) \) and \( 1 - \cos(2\theta) \) to express the sum under the inverse sine as a single sine value. Then differentiate.

Exam Tip: When you see sums of square roots, immediately think of sum-to-product identities or double angle formulas. These often collapse complex nested expressions into manageable forms. Always check your substitution by working through the algebra carefully.

Question 44. Mark (√) against the correct answer in the following: If \( x = at^2, y = 2at \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{1}{t} \)
(B) \( \frac{-1}{t^2} \)
(C) \( \frac{-2}{t} \)
(D) none of these
Answer: (A) \( \frac{1}{t} \)
In simple words: When x and y are both given as functions of a parameter t, use the parametric differentiation formula: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Calculate each derivative separately and then divide.

Exam Tip: Parametric differentiation is faster than eliminating the parameter. Always use \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) for parametric equations. Be careful with signs and simplify fractions fully before your final answer.

Question 45. Mark (√) against the correct answer in the following: If \( x = a\sec\theta, y = b\tan\theta \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{b}{a}\sec\theta \)
(B) \( \frac{b}{a}\operatorname{cosec}\theta \)
(C) \( \frac{b}{a}\cot\theta \)
(D) none of these
Answer: (C) \( \frac{b}{a}\cot\theta \)
In simple words: Apply parametric differentiation using \( \theta \) as the parameter. Differentiate x and y separately with respect to \( \theta \). Then compute the ratio \( \frac{dy/d\theta}{dx/d\theta} \).

Exam Tip: For trigonometric parametric equations, keep the parameter (angle) in your answer unless specifically asked to eliminate it. The answer in terms of the parameter is often simpler and cleaner than converting back to Cartesian form.

Question 46. Mark (√) against the correct answer in the following: If \( x = a\cos^2\theta, y = b\sin^2\theta \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{-a}{b} \)
(B) \( \frac{-a}{b}\cot\theta \)
(C) \( \frac{-b}{a} \)
(D) none of these
Answer: (C) \( \frac{-b}{a} \)
In simple words: Compute \( \frac{dx}{d\theta} = -2a\cos\theta\sin\theta \) and \( \frac{dy}{d\theta} = 2b\sin\theta\cos\theta \). Form the ratio \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \). The sine and cosine terms cancel out, leaving a simple constant ratio.

Exam Tip: When trigonometric terms cancel during parametric differentiation, it often signals that the derivative is independent of the parameter. This is a good check on your work - if the answer still contains \( \theta \), recalculate carefully.

Question 47. Mark (√) against the correct answer in the following: If \( x = \theta(\cos\theta + \sin\theta) \) and \( y = a(\sin\theta - \theta\cos\theta) \), then \( \frac{dy}{dx} = ? \)
(A) cot θ
(B) tan θ
(C) a cot θ
(D) a tan θ
Answer: (B) tan θ
In simple words: Differentiate both x and y with respect to \( \theta \). For x, use the product rule along with the sum. For y, apply the product rule to \( \theta\cos\theta \). After simplification, divide the derivatives to get \( \frac{dy}{dx} \).

Exam Tip: Product rule is essential here. Take your time expanding \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) fully before simplifying. Cancellation of terms often occurs, revealing the clean final form.

Question 48. Mark (√) against the correct answer in the following: If \( y = x^{x^{x^{...}}} \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{y}{x(1-\log x)} \)
(B) \( \frac{y^2}{x(1-\log x)} \)
(C) \( \frac{y}{x(1-y\log x)} \)
(D) none of these
Answer: (A) \( \frac{y}{x(1-\log x)} \)
In simple words: Notice that the exponent itself equals y. Rewrite as \( y = x^y \). Take the natural log of both sides and differentiate implicitly. Solve for \( \frac{dy}{dx} \) by rearranging the resulting equation.

Exam Tip: Infinite power towers have a self-referential structure - the exponent is the entire expression. Always exploit this by setting the whole expression equal to y, then taking logs. Implicit differentiation is the key technique here.

Question 49. Mark (√) against the correct answer in the following: If \( y = \sqrt{x + \sqrt{x + \sqrt{x + ...}}} \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{1}{(2y-1)} \)
(B) \( \frac{1}{(y^2-1)} \)
(C) \( \frac{2y}{(y^2-1)} \)
(D) none of these
Answer: (A) \( \frac{1}{(2y-1)} \)
In simple words: The infinite nested radical means the inner part also equals y. Write \( y = \sqrt{x + y} \). Square both sides to get \( y^2 = x + y \). Differentiate this relation with respect to x and solve for \( \frac{dy}{dx} \).

Exam Tip: Infinite nested radicals are self-similar - what is inside the outer radical (after the first x) is the entire expression y. This self-similarity allows you to set up an equation in y and x, which you can then differentiate implicitly to find the derivative.

Question 50. Mark (√) against the correct answer in the following: If \( y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + ...}}} \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{\sin x}{(2y-1)} \)
(B) \( \frac{\cos x}{(y-1)} \)
(C) \( \frac{\cos x}{(2y-1)} \)
(D) none of these
Answer: (C) \( \frac{\cos x}{(2y-1)} \)
In simple words: By the self-similarity of the infinite nested radical, set \( y = \sqrt{\sin x + y} \). Square both sides to obtain \( y^2 = \sin x + y \). Differentiate both sides with respect to x and isolate \( \frac{dy}{dx} \).

Exam Tip: When differentiating \( y^2 = \sin x + y \), the left side gives \( 2y\frac{dy}{dx} \) and the right side gives \( \cos x + \frac{dy}{dx} \). Rearranging correctly is crucial - collect all \( \frac{dy}{dx} \) terms on one side before factoring.

Question 51. Mark (√) against the correct answer in the following: If \( y = e^x + e^{x^x - x} \), then \( \frac{dy}{dx} = ? \)
(A) \( \frac{1}{(1-y)} \)
(B) \( \frac{y}{(1-y)} \)
(C) \( \frac{y}{(y-1)} \)
(D) none of these
Answer: (B) \( \frac{y}{(1-y)} \)
In simple words: Rewrite the given equation as \( y = e^{x+y} \). Take the natural log of both sides to get \( \log y = (x + y) \log e = x + y \). Differentiate implicitly with respect to x. Rearrange to isolate \( \frac{dy}{dx} \).

Exam Tip: The exponent structure here is tricky - it equals the entire expression y. Once you recognize this self-referential form and rewrite it as \( y = e^{x+y} \), apply logarithmic differentiation. Careful handling of the \( \frac{dy}{dx} \) term during rearrangement is essential.

Question 52. Mark (√) against the correct answer in the following: The value of k for which \( f(x) = \begin{cases} \frac{\sin 5x}{3x}, & \text{if } x \ne 0 \\ k, & \text{if } x = 0 \end{cases} \) is continuous at x = 0 is
(A) \( \frac{5}{3} \)
(B) 0
(C) \( \frac{3}{5} \)
(D) \( \frac{5}{3} \)
Answer: (A) \( \frac{5}{3} \)
In simple words: For continuity at x = 0, the limit of f(x) as x approaches 0 must equal f(0) = k. Compute \( \lim_{x \to 0} \frac{\sin 5x}{3x} \) using the standard limit \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \). Rewrite the fraction to apply this identity and find k.

Exam Tip: Always use the standard limit formula \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) by manipulating the given expression. Multiply and divide by appropriate factors (here, 5x) to match the standard form, then evaluate.

Question 53. Mark (√) against the correct answer in the following: Let \( f(x) = \begin{cases} x \sin \frac{1}{x}, & \text{if } x \ne 0 \\ 0, & \text{when } x = 0 \end{cases} \). Then, which of the following is the true statement?
(A) f(x) is not defined at x = 0
(B) \( \lim_{x \to 0} f(x) \) does not exist
(C) f(x) is continuous at x = 0
(D) f(x) is discontinuous at x = 0
Answer: (C) f(x) is continuous at x = 0
In simple words: Compute the left and right-hand limits of \( x\sin\frac{1}{x} \) as x approaches 0. Since \( \sin\frac{1}{x} \) is bounded between -1 and 1, the product \( x\sin\frac{1}{x} \) approaches 0 as x approaches 0 (by the squeeze theorem). Thus f(0) = 0 matches the limit, confirming continuity.

Exam Tip: Bounded oscillations multiplied by a vanishing factor always approach zero. The squeeze theorem is invaluable here - \( -|x| \le x\sin\frac{1}{x} \le |x| \), and both bounds approach 0, forcing the middle expression to also approach 0.

Question 55. Mark (√) against the correct answer in the following: Let \( f(x) = x^{3/2} \). Then, \( f'(0) = ? \)
(A) \( \frac{3}{2} \)
(B) \( \frac{1}{2} \)
(C) does not exist
(D) none of these
Answer: (C) does not exist
In simple words: Differentiate to get \( f'(x) = \frac{3}{2}x^{1/2} = \frac{3}{2\sqrt{x}} \). As x approaches 0 from the right, the denominator approaches 0, making the derivative approach infinity. Therefore, the derivative is undefined at x = 0.

Exam Tip: When the derivative formula contains a root in the denominator (like \( \frac{1}{\sqrt{x}} \)), be alert that the derivative diverges to infinity at x = 0. Even though the function itself is continuous there, the derivative fails to exist due to this infinite slope.

Question 56. Mark (√) against the correct answer in the following: The function \( f(x) = |x|, \, x \in \mathbb{R} \) is
(A) continuous but not differentiable at x = 0
(B) differentiable but not continuous at x = 0
(C) neither continuous nor differentiable at x = 0
(D) none of these
Answer: (A) continuous but not differentiable at x = 0
In simple words: The absolute value function is continuous everywhere - it has no breaks or gaps. However, at x = 0, the graph has a sharp corner where the slope abruptly changes from -1 (left side) to +1 (right side). This corner point prevents the derivative from existing at x = 0, even though the function is continuous there.

Exam Tip: Graphing the function helps tremendously here. Continuity means no breaks; differentiability means no sharp corners or cusps. The absolute value function is a classic example of a function that is continuous but has a non-differentiable point at the corner.

Question 57. The function \( f(x) = \begin{cases} 1+x, & \text{when } x \leq 2 \\ 5-x, & \text{when } x > 2 \end{cases} \) is
(A) continuous as well as differentiable at x = 2
(B) continuous but not differentiable at x = 2
(C) differentiable but not continuous at x = 2
(D) none of these
Answer: To determine continuity, the left-hand limit should match the right-hand limit and the function value at the point.

At x = 2:

L.H.L = \( \lim_{x \to 2^-}(1+x) = 3 \)

R.H.L = \( \lim_{x \to 2^+}(5-x) = 3 \)

f(2) = 1 + 2 = 3

Since L.H.L = R.H.L = f(2) = 3, the function is continuous at x = 2.

For differentiability:

\( f'(2^-) = \lim_{x \to 2^-} \frac{f(x)-f(2)}{x-2} = \lim_{h \to 0} \frac{(1+2-h)-3}{-h} = \lim_{h \to 0} \frac{-h}{-h} = 1 \)

\( f'(2^+) = \lim_{x \to 2^+} \frac{f(x)-f(2)}{x-2} = \lim_{h \to 0} \frac{(5-(2+h))-3}{h} = \lim_{h \to 0} \frac{-h}{h} = -1 \)

Since f'(2-) ≠ f'(2+), the function is not differentiable at x = 2.

In simple words: The function is continuous at x = 2 because the left and right limits both equal 3. However, it is not differentiable because the left and right derivatives have different values (1 and - 1).

Exam Tip: Always verify continuity first by checking that the left limit, right limit, and function value are all equal. Then separately check differentiability by comparing the left and right derivatives - they must match for the function to be differentiable.

Question 58. If \( f(x) = \begin{cases} kx+5, & \text{when } x \leq 2 \\ x+1, & \text{when } x > 2 \end{cases} \) is continuous at x = 2, then k = ?
(A) 2
(B) -2
(C) 3
(D) -3
Answer: For continuity at x = 2, the left and right limits must be equal to the function value.

L.H.L = \( \lim_{x \to 2^-}(kx+5) = k(2)+5 = 2k+5 \)

R.H.L = \( \lim_{x \to 2^+}(x+1) = 2+1 = 3 \)

Since the function is continuous at x = 2:

\( 2k+5 = 3 \)

\( 2k = -2 \)

\( k = -1 \)

In simple words: For the function to have no jump or break at x = 2, the left side expression and right side expression must give the same value. Setting them equal gives k = -1.

Exam Tip: When finding constants for continuity, equate the left-hand limit to the right-hand limit - this creates a simple algebraic equation for the unknown parameter.

Question 59. If the function \( f(x) = \begin{cases} \frac{1-\cos 4x}{8x^2}, & x \neq 0 \\ k, & x = 0 \end{cases} \) is continuous at x = 0, then k = ?
(A) 1
(B) 2
(C) \( \frac{1}{2} \)
(D) \( \frac{-1}{2} \)
Answer: Given that f(x) is continuous at x = 0, we need to find k by calculating the limit.

Using the identity \( 1-\cos 4x = 2\sin^2 2x \):

\( f(x) = \lim_{x \to 0} \frac{2\sin^2 2x}{8x^2} = \lim_{x \to 0} \frac{2\sin^2 2x}{2 \times 4x^2} = \lim_{x \to 0} \left(\frac{\sin 2x}{2x}\right)^2 \)

Using the standard limit \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \):

\( f(x) = \left(1\right)^2 = 1 \)

Therefore, k = 1

In simple words: To make the function continuous at x = 0, the constant k must equal the limit of the expression as x approaches 0. Using the double angle identity and the standard sine limit, this value turns out to be 1.

Exam Tip: For continuity at a point where the function is defined by a limiting case, always use trigonometric identities and standard limits like \( \lim \frac{\sin u}{u} = 1 \) to find the required constant.

Question 60. If the function \( f(x) = \begin{cases} \frac{\sin^2 ax}{x^2}, & \text{when } x \neq 0 \\ k, & \text{when } x = 0 \end{cases} \) is continuous at x = 0, then k = ?
(A) a
(B) a²
(C) -2
(D) -4
Answer: For continuity at x = 0, we must evaluate the limit of the expression as x approaches 0.

\( f(x) = \lim_{x \to 0} \frac{\sin^2 ax}{x^2} = \lim_{x \to 0} \frac{\sin^2 ax}{x^2} \times \frac{a^2}{a^2} = \lim_{x \to 0} \left(\frac{\sin ax}{ax}\right)^2 \times a^2 \)

Applying the standard limit \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \):

\( f(x) = (1)^2 \times a^2 = a^2 \)

Therefore, k = a²

In simple words: To find k, compute the limit as x approaches 0 by rewriting the expression as a squared sine ratio times a squared constant. The sine ratio limit gives 1, leaving just a².

Exam Tip: When finding constants for functions involving \( \frac{\sin u}{u} \), always factor out constants appropriately so the limit form \( \frac{\sin u}{u} = 1 \) can be applied directly.

Question 61. If the function \( f(x) = \begin{cases} \frac{k\cos x}{\pi-2x}, & \text{when } x \neq \frac{\pi}{2} \\ 3, & \text{when } x = \frac{\pi}{2} \end{cases} \) is continuous at \( x = \frac{\pi}{2} \), then the value of k is
(A) 3
(B) -3
(C) -5
(D) 6
Answer: Given that f(x) is continuous at x = π/2, the left-hand limit must equal the function value.

\( \text{L.H.L} = \lim_{x \to \frac{\pi}{2}^-} \frac{k\cos x}{\pi-2x} \)

Let \( x = \frac{\pi}{2} - h \) where h → 0:

\( \lim_{h \to 0} \frac{k\cos(\frac{\pi}{2}-h)}{\pi-2(\frac{\pi}{2}-h)} = \lim_{h \to 0} \frac{k\sin h}{2h} = k \lim_{h \to 0} \frac{\sin h}{2h} = k \cdot \frac{1}{2} = \frac{k}{2} \)

Since the function is continuous:

\( \frac{k}{2} = 3 \)

\( k = 6 \)

In simple words: At x = π/2, the numerator \( \cos(\frac{\pi}{2}) \) equals 0, creating an indeterminate form. By substituting \( x = \frac{\pi}{2} - h \) and using \( \cos(\frac{\pi}{2}-h) = \sin h \), the limit becomes k/2, which must equal 3.

Exam Tip: For continuous functions with indeterminate forms, use substitution to transform the limit into a recognizable standard form, then apply known limits like \( \lim \frac{\sin h}{h} = 1 \).

Question 62. At x = 2, \( f(x) = |x| \) is
(A) continuous but not differentiable
(B) differentiable but not continuous
(C) continuous as well as differentiable
(D) none of these
Answer: Given the modulus function f(x) = |x|, we examine its properties at x = 2.

The graph of f(x) = |x| shows a V-shape with the vertex at the origin. At x = 2, we are on the right branch where |x| = x, so f(2) = 2.

The function is continuous at x = 2 because the left and right limits both equal 2. On the right branch (x > 0), f(x) = x, which is a straight line with slope 1. At x = 2, the function transitions smoothly.

Similarly, the function is differentiable at x = 2 with f'(2) = 1 on the right branch.

However, the modulus function is neither continuous nor differentiable at x = 0 (the vertex), where the slope changes from - 1 to 1 abruptly. Since the question asks about x = 2, which is away from the critical point, the function is both continuous and differentiable at x = 2.

In simple words: The absolute value function forms a V-shape. At x = 2, you are on the straight-line portion of this V, so the function is smooth and has a well-defined slope. It is both continuous and differentiable at this point.

Exam Tip: The modulus function |x| is continuous everywhere but differentiable everywhere except at x = 0. Always identify the critical point (the vertex) when analyzing piecewise-defined functions.

Question 63. If \( f(x) = \begin{cases} \frac{x^2-2x-3}{x+1}, & \text{when } x \neq -1 \\ k, & \text{when } x = -1 \end{cases} \) is continuous at x = -1, then k = ?
(A) 4
(B) -4
(C) -3
(D) 2
Answer: For continuity at x = -1, find the limit by factoring the numerator.

\( f(x) = \lim_{x \to -1} \frac{x^2-2x-3}{x+1} \)

Factor the numerator: \( x^2-2x-3 = (x+1)(x-3) \)

\( f(x) = \lim_{x \to -1} \frac{(x+1)(x-3)}{x+1} = \lim_{x \to -1} (x-3) = -1-3 = -4 \)

Therefore, k = -4

In simple words: The fraction simplifies after cancelling the common factor (x+1). Once simplified, substitute x = -1 to get -4.

Exam Tip: When a rational function has a removable discontinuity, always factor the numerator to cancel the problematic denominator term before evaluating the limit.

Question 64. The function \( f(x) = x^3+6x^2+15x-12 \) is
(A) strictly decreasing on ℝ
(B) strictly increasing on ℝ
(C) increasing in \( (-\infty,-2) \) and decreasing in \( (2,\infty) \)
(D) none of these
Answer: To determine monotonicity, find the derivative and analyze its sign.

\( f'(x) = 3x^2+12x+15 \)

Rewrite by completing the square:

\( f'(x) = 3(x^2+4x+5) = 3(x^2+4x+4+1) = 3[(x+2)^2+1] = 3(x+2)^2+3 \)

Since \( (x+2)^2 \geq 0 \) for all real x, we have \( 3(x+2)^2+3 \geq 3 > 0 \) for all x ∈ ℝ.

Therefore, f'(x) > 0 for all x ∈ ℝ, which means the function is strictly increasing on ℝ.

In simple words: The derivative is always positive (never zero or negative) because it is the sum of a squared term and a positive constant. This guarantees the function always rises as x increases.

Exam Tip: When the derivative can be written as a sum of a perfect square and a positive constant, it is always positive, making the function strictly increasing everywhere.

Question 65. The function \( f(x) = 4-3x+3x^2-x^3 \) is
(A) decreasing on ℝ
(B) increasing on ℝ
(C) strictly decreasing on ℝ
(D) strictly increasing on ℝ
Answer: Rewrite the function and find its derivative.

\( f(x) = -x^3+3x^2-3x+4 \)

\( f'(x) = -3x^2+6x-3 = -3(x^2-2x+1) = -3(x-1)^2 \)

Since \( (x-1)^2 \geq 0 \) for all real x, we have \( f'(x) = -3(x-1)^2 \leq 0 \) for all x ∈ ℝ.

Moreover, f'(x) = 0 only when x = 1. For all x ≠ 1, f'(x) < 0, meaning the function is strictly decreasing wherever the derivative is negative. At x = 1, the function has a horizontal tangent but continues decreasing overall.

Therefore, the function is strictly decreasing on ℝ.

In simple words: The derivative is written as a negative constant times a perfect square. This product is always zero or negative, with only one point (x = 1) where it equals zero. The function never increases.

Exam Tip: A function is strictly decreasing if its derivative is always non-positive and is strictly negative except at isolated points. A single point where f'(x) = 0 does not break strict monotonicity if f'(x) ≤ 0 everywhere.

Question 66. The function \( f(x) = 3x+\cos 3x \) is
(A) increasing on ℝ
(B) decreasing on ℝ
(C) strictly increasing on ℝ
(D) strictly decreasing on ℝ
Answer: Find the derivative to determine the behavior of the function.

\( f'(x) = 3-3\sin 3x = 3(1-\sin 3x) \)

Since \( \sin 3x \) varies between -1 and 1:

When \( \sin 3x = -1 \), f'(x) = 3(1-(-1)) = 6

When \( \sin 3x = 1 \), f'(x) = 3(1-1) = 0

In all cases, \( 1-\sin 3x \geq 0 \), so f'(x) ≥ 0 for all x ∈ ℝ. Since f'(x) ranges from 0 to 6 and equals zero only at isolated points, the function is increasing on ℝ.

In simple words: The term (1 - sin 3x) ranges from 0 to 2, making the derivative always non-negative. The function never decreases, so it is increasing across all real numbers.

Exam Tip: For trigonometric functions, always recall that sine and cosine are bounded between -1 and 1, which constrains the derivative and helps determine monotonicity.

Question 67. The function \( f(x) = x^3+6x^2+9x+3 \) is decreasing for
(A) 1 < x < 3
(B) x > 1
(C) x < 1
(D) x < - 1 or x > - 3
Answer: To find where the function is decreasing, set the derivative equal to zero and analyze the sign of f'(x).

\( f'(x) = 3x^2+12x+9 = 0 \)

\( 3(x^2+4x+3) = 0 \)

\( 3(x+1)(x+3) = 0 \)

Critical points: x = -1 and x = -3

Testing the sign of f'(x) in each interval:

For x > -1: f'(x) > 0 (function increasing)

For x < -3: f'(x) > 0 (function increasing)

For -3 < x < -1: f'(x) < 0 (function decreasing)

Therefore, the function is decreasing in the interval (-3, -1).

In simple words: The critical points split the number line into regions. Check the sign of the derivative in each region. Between the two critical points, the derivative is negative, making the function decrease there.

Exam Tip: Always mark critical points on a number line and test the sign of the derivative in each interval created by these points. This quickly identifies where the function increases and decreases.

Question 68. The function \( f(x) = x^3-27x+8 \) is increasing when
(A) |x| < 3
(B) |x| > 3
(C) -3 < x < 3
(D) none of these
Answer: Find the critical points by setting the derivative to zero.

\( f'(x) = 3x^2-27 = 0 \)

\( 3(x^2-9) = 0 \)

\( 3(x-3)(x+3) = 0 \)

Critical points: x = 3 and x = -3

Analyzing the sign of f'(x):

For x > 3: f'(x) > 0 (function increasing)

For x < -3: f'(x) > 0 (function increasing)

For -3 < x < 3: f'(x) < 0 (function decreasing)

Therefore, the function is increasing when x > 3 or x < -3, which can be written as |x| > 3.

In simple words: Outside the interval [-3, 3] the function increases. Inside this interval, between the two critical points, the function decreases. The condition |x| > 3 captures both regions where the function rises.

Exam Tip: Use the notation |x| > a to indicate "both x > a and x < -a" - this is a compact way to express two separate intervals where a function behaves the same way.

Question 69. \( f(x) = \sin x \) is increasing in
(A) \( \left(\frac{\pi}{2}, \pi\right) \)
(B) \( \left(\pi, \frac{3\pi}{2}\right) \)
(C) \( (0, \pi) \)
(D) \( \left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \)
Answer: To find where sin x is increasing, analyze its derivative.

\( f'(x) = \cos x \)

For the function to be increasing, f'(x) = cos x > 0.

The cosine function is positive in the intervals where x ∈ \( \left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \), \( \left(\frac{3\pi}{2}, \frac{5\pi}{2}\right) \), and so on.

Within the standard interval given in option (D), cos x > 0 for all x in \( \left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \).

Therefore, sin x is increasing in \( \left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \).

In simple words: Sine increases wherever its derivative (cosine) is positive. Cosine is positive in the interval from -π/2 to π/2, so sine increases throughout this range.

Exam Tip: Memorize the intervals where trigonometric derivatives are positive - cos x > 0 and sin x > 0 have well-defined standard intervals that repeat every 2π.

Question 70. \( f(x) = \frac{2x}{\log x} \) is increasing in
(A) (0, 1)
(B) (1, e)
(C) (e, ∞)
(D) (-∞, e)
Answer: Find the derivative using the quotient rule.

\( f'(x) = \frac{2 \log x - 2x \cdot \frac{1}{x}}{(\log x)^2} = \frac{2\log x - 2}{(\log x)^2} = \frac{2(\log x - 1)}{(\log x)^2} \)

Set f'(x) = 0 to find critical points:

\( \log x - 1 = 0 \)

\( \log x = 1 \)

\( x = e \)

Analyzing the sign of f'(x):

For x > e: log x > 1, so the numerator is positive. The denominator \( (\log x)^2 \) is always positive. Thus f'(x) > 0 and the function is increasing.

For 0 < x < e and x ≠ 1: log x < 1, so the numerator is negative. Thus f'(x) < 0 and the function is decreasing.

Therefore, f(x) is increasing in (e, ∞).

In simple words: The derivative changes sign at x = e. For larger values (x > e), the logarithm exceeds 1, making the numerator positive and the function increasing. For smaller values, the function decreases.

Exam Tip: When analyzing rational functions with logarithms, identify where the numerator equals zero, then test intervals to determine where the derivative is positive or negative.

Question 71. \( f(x) = (\sin x - \cos x) \) is decreasing in
(A) \( \left(0, \frac{3\pi}{4}\right) \)
(B) \( \left(\frac{3\pi}{4}, \frac{7\pi}{4}\right) \)
(C) \( \left(\frac{7\pi}{4}, 2\pi\right) \)
(D) none of these
Answer: Find where the function is decreasing by analyzing f'(x).

\( f'(x) = \cos x + \sin x \)

Multiply and divide by √2 to rewrite:

\( f'(x) = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x\right) = \sqrt{2}\left(\sin\frac{\pi}{4}\cos x + \cos\frac{\pi}{4}\sin x\right) = \sqrt{2}\sin\left(\frac{\pi}{4} + x\right) \)

For the function to be decreasing, f'(x) < 0:

\( \sin\left(\frac{\pi}{4} + x\right) < 0 \)

Sine is negative when its argument is in \( (\pi, 2\pi) \):

\( \pi < \frac{\pi}{4} + x < 2\pi \)

\( \frac{3\pi}{4} < x < \frac{7\pi}{4} \)

Therefore, the function is decreasing in \( \left(\frac{3\pi}{4}, \frac{7\pi}{4}\right) \).

In simple words: Rewrite the derivative using the sine addition formula. The sine of an angle is negative in the third and fourth quadrants, corresponding to the interval (π, 2π). Solve the inequality to find the x-values where this occurs.

Exam Tip: For sums of sines and cosines, use the formula \( a\sin\theta + b\cos\theta = \sqrt{a^2+b^2}\sin(\theta + \phi) \) where \( \tan\phi = \frac{b}{a} \). This transforms the expression into a single sine function, making the analysis straightforward.

Question 72. \( f(x) = \frac{x}{\sin x} \) is
(A) increasing in (0, 1)
(B) decreasing in (0, 1)
(C) increasing in \( \left(0, \frac{1}{2}\right) \) and decreasing in \( \left(\frac{1}{2}, 1\right) \)
(D) none of these
Answer: Analyze the behavior using the quotient rule to find f'(x).

\( f'(x) = \frac{\sin x - x\cos x}{\sin^2 x} \)

Within the interval (0, 1):

In this small interval, sin x is increasing (since cos x > 0) and x is increasing. The term sin x grows while the term x cos x also grows, but sin x grows faster initially. For small positive x, sin x - x cos x > 0, so the numerator is positive throughout (0, 1).

Since the denominator \( \sin^2 x > 0 \) for x ∈ (0, 1), we have f'(x) > 0 throughout this interval.

Therefore, the function is increasing in (0, 1).

In simple words: The derivative's numerator (sin x - x cos x) stays positive in the interval (0, 1), making the derivative positive. The function rises as x increases from 0 to 1.

Exam Tip: For quotient derivatives, focus on the sign of the numerator when the denominator is always positive. In small intervals like (0, 1), carefully track which component dominates.

Question 73. \( f(x) = x^x \) is decreasing in the interval
(A) (0, e)
(B) \( \left(0, \frac{1}{e}\right) \)
(C) (0, 1)
(D) none of these
Answer: Find the derivative of this exponential function.

\( f(x) = x^x \)

\( f'(x) = (log x + 1) x^x \)

Set f'(x) = 0 to find critical points. Since \( x^x > 0 \) always for x > 0:

\( \log x + 1 = 0 \)

\( \log x = -1 \)

\( x = \frac{1}{e} \)

Analyzing the sign:

For \( x > \frac{1}{e} \): log x > -1, so f'(x) > 0 (function increasing)

For \( 0 < x < \frac{1}{e} \): log x < -1, so f'(x) < 0 (function decreasing)

Therefore, the function is decreasing in \( \left(0, \frac{1}{e}\right) \).

In simple words: The function x^x has its minimum at x = 1/e. For smaller positive values (before 1/e), the function decreases. For larger values (after 1/e), it increases.

Exam Tip: For functions of the form x^x, use logarithmic differentiation: take the natural log of both sides before differentiating, then solve for f'(x). Always check that the base x > 0 for the logarithm to be defined.

Question 74. \( f(x) = x^2 e^{-x} \) is increasing in
(A) (-2, 0)
(B) (0, 2)
(C) (2, ∞)
(D) (-∞, ∞)
Answer: Use the product rule to find the derivative.

\( f'(x) = 2x e^{-x} - x^2 e^{-x} = e^{-x}(2x - x^2) = -e^{-x}(x^2 - 2x) \)

Set f'(x) = 0. Since \( e^{-x} > 0 \) always:

\( x^2 - 2x = 0 \)

\( x(x - 2) = 0 \)

Critical points: x = 0 and x = 2

Analyzing the sign. Note the negative factor in front:

For x > 2: \( (x^2 - 2x) > 0 \), so f'(x) < 0 (function decreasing)

For 0 < x < 2: \( (x^2 - 2x) < 0 \), so f'(x) > 0 (function increasing)

For x < 0: \( (x^2 - 2x) > 0 \), so f'(x) < 0 (function decreasing)

Therefore, the function is increasing in (0, 2).

In simple words: The exponential factor e^(-x) is always positive. The negative sign in front flips the behavior: where (x² - 2x) is negative, the overall derivative becomes positive. This occurs between the critical points 0 and 2.

Exam Tip: When a negative sign appears in a derivative, invert your analysis - regions where the inner expression is negative become regions where the derivative is positive.

Question 75. \( f(x) = \sin x - kx \) is decreasing for all x ∈ ℝ when
(A) k < 1
(B) k ≤ 1
(C) k > 1
(D) k ≤ 1
Answer: For the function to be decreasing everywhere, its derivative must be non-positive for all x ∈ ℝ.

\( f'(x) = \cos x - k \)

For f to decrease everywhere:

\( f'(x) \leq 0 \) for all x ∈ ℝ

\( \cos x - k \leq 0 \)

\( \cos x \leq k \)

Since cos x oscillates between -1 and 1, the maximum value of cos x is 1. For the inequality \( \cos x \leq k \) to hold for all x, we need:\br />
\( k \geq 1 \)

Therefore, the function is decreasing when k ≥ 1.

In simple words: The cosine term varies between -1 and 1. For the derivative to stay non-positive everywhere, k must be at least as large as the largest value cosine can reach, which is 1.

Exam Tip: When analyzing inequalities with bounded periodic functions, consider their maximum and minimum values to determine constraints on parameters.

Question 76. \( f(x) = (x+1)^3(x-3)^3 \) is increasing in
(A) (-∞, 1)
(B) (-1, 3)
(C) (3, ∞)
(D) (1, ∞)
Answer: Use the product rule to find the derivative.

\( f'(x) = 3(x+1)^2(x-3)^3 + 3(x-3)^2(x+1)^3 = 3(x+1)^2(x-3)^2[(x-3) + (x+1)] \)

\( = 3(x+1)^2(x-3)^2(2x - 2) = 6(x+1)^2(x-3)^2(x - 1) \)

The squared factors are always non-negative. Critical points come from:\br />
\( x = -1, x = 1, x = 3 \)

Analyzing the sign of the linear factor (x - 1):\br />
For x > 1: (x - 1) > 0, so f'(x) > 0 (function increasing)

For x < 1: (x - 1) < 0, so f'(x) < 0 (function decreasing)

Therefore, the function is increasing in (1, ∞).

In simple words: After factoring, the derivative is a product of squared terms (always non-negative) and a linear term (x - 1). The sign is determined entirely by this linear factor, which is positive when x > 1.

Exam Tip: When derivatives contain squared factors, these do not affect the sign analysis - focus on the unsquared factors to determine where the derivative is positive or negative.

Question 77. \( f(x) = [x(x-3)]^2 \) is increasing in
(A) (0, ∞)
(B) (-∞, 0)
(C) (1, 3)
(D) \( \left(0, \frac{3}{2}\right) \cup (3, \infty) \)
Answer: Find the derivative using the chain rule.

\( f'(x) = 2[x(x-3)] \cdot (2x - 3) = 2x(x-3)(2x - 3) \)

Critical points: x = 0, x = 3/2, x = 3

Analyzing the sign on each interval:

For \( x > 3 \): all three factors are positive, so f'(x) > 0 (function increasing)

For \( \frac{3}{2} < x < 3 \): x > 0, (x-3) < 0, (2x-3) > 0, so f'(x) < 0 (function decreasing)

For \( 0 < x < \frac{3}{2} \): x > 0, (x-3) < 0, (2x-3) < 0, so f'(x) > 0 (function increasing)

For \( x < 0 \): x < 0, (x-3) < 0, (2x-3) < 0, so f'(x) < 0 (function decreasing)

Therefore, the function is increasing in \( \left(0, \frac{3}{2}\right) \cup (3, \infty) \).

In simple words: Test the sign of each factor (x, x-3, and 2x-3) in each interval created by the critical points. Where an even number of factors are negative, the product is positive and the function increases.

Exam Tip: Create a sign chart with all critical points marked on a number line, then determine the sign of each factor in each interval. Multiply the signs to find where f'(x) is positive.

Question 78. If \( f(x) = kx^3-9x^2+9x+3 \) is increasing for every real number x, then
(A) k > 3
(B) k ≥ 3
(C) k < 3
(D) k ≤ 3
Answer: For the function to be increasing everywhere, f'(x) must be non-negative for all x ∈ ℝ.

\( f'(x) = 3kx^2-18x+9 = 3(kx^2 - 6x + 3) \)

For f'(x) ≥ 0 for all x, the quadratic \( kx^2 - 6x + 3 \) must be non-negative for all x.

This requires:\br />
1. k > 0 (parabola opens upward)

2. Discriminant D ≤ 0: \( (-6)^2 - 4(k)(3) \leq 0 \)

\( 36 - 12k \leq 0 \)

\( 12k \geq 36 \)

\( k \geq 3 \)

Therefore, the function is increasing for all x when k ≥ 3. Since the question asks for the minimum such k, the answer is k ≥ 3, but if only one option must be chosen for the inequality itself, k > 3 strictly ensures f'(x) > 0 everywhere.

In simple words: The derivative is a quadratic that must never be negative. For this, the parabola must open upward (k > 0) and have no real roots (non-positive discriminant). The discriminant condition gives k ≥ 3.

Exam Tip: When a polynomial derivative must have constant sign, analyze the discriminant of the resulting quadratic. For a quadratic to be always positive, its discriminant must be negative (no real roots) and its leading coefficient must be positive.

Question 79. \( f(x) = \frac{x}{(x^2-1)} \) is increasing in
(A) (-1, 1)
(B) (-1, ∞)
(C) \( (-\infty, -1) \cup (1, \infty) \)
(D) none of these
Answer: Find the derivative using the quotient rule.

\( f'(x) = \frac{(x^2-1) - x(2x)}{(x^2-1)^2} = \frac{x^2-1-2x^2}{(x^2-1)^2} = \frac{-x^2-1}{(x^2-1)^2} \)

Analyze the sign of f'(x). The denominator is always positive. The numerator \( -x^2-1 \) is always negative since \( x^2 \geq 0 \).

Thus f'(x) < 0 wherever the function is defined (x ≠ ±1), meaning the function is decreasing on each connected interval of its domain.

However, examining the function's behavior on separate intervals:\br />
For x > 1 and x < -1, the function is decreasing within each interval.\br />
The function is decreasing on (-∞, -1), (-1, 1), and (1, ∞).

None of the standard single intervals show the function increasing.

In simple words: The derivative is negative everywhere the function is defined. The squared denominator is always positive, and the numerator is always negative, making f'(x) always negative. The function never increases.

Exam Tip: For rational functions, always check both the domain restrictions and the sign of the derivative separately on each continuous piece of the domain.

Question 80. The least value of k for which \( f(x) = x^2+kx+1 \) is increasing on (1, 2) is
(A) -2
(B) -1
(C) 1
(D) 2
Answer: For the function to be increasing on (1, 2), the derivative must be non-negative throughout this interval.

\( f'(x) = 2x+k \)

For f'(x) ≥ 0 on (1, 2):

\( 2x+k \geq 0 \)

\( k \geq -2x \)

Since x ranges over (1, 2), the expression -2x ranges over (-4, -2). The maximum value of -2x in this interval approaches -2 (when x approaches 1).

To ensure k ≥ -2x for all x ∈ (1, 2), we need:\br />
\( k \geq -2(1) = -2 \)

The least value of k is -2.

In simple words: The derivative 2x + k must stay non-negative. As x increases from 1 to 2, the term 2x increases, making -2x more negative. The critical constraint comes at x = 1, where we need k ≥ -2.

Exam Tip: For increasing conditions on a finite interval, find where the derivative is smallest and ensure that even at that point, f'(x) ≥ 0. This determines the minimum value of the parameter.

Question 81. \( f(x) = \frac{x}{(x^2-1)} \) is increasing in
(A) (-1, 1)
(B) (-1, ∞)
(C) \( (-\infty, -1) \cup (1, \infty) \)
(D) none of these
Answer: Find the derivative using the quotient rule.

\( f'(x) = \frac{(x^2-1) - x(2x)}{(x^2-1)^2} = \frac{x^2-1-2x^2}{(x^2-1)^2} = \frac{-x^2-1}{(x^2-1)^2} = \frac{-(x^2+1)}{(x^2-1)^2} \)

The denominator is always positive. The numerator \( -(x^2+1) \) is always negative since \( x^2+1 > 0 \) for all real x.

Therefore, f'(x) < 0 wherever f(x) is defined (x ≠ ±1), meaning the function is decreasing on each interval of its domain.\br />
For x > 1: f(x) is decreasing

For -1 < x < 1: f(x) is decreasing

For x < -1: f(x) is decreasing

The function is never increasing on any interval.

In simple words: The derivative is negative everywhere. The numerator is a negative expression (the negative of x² + 1), and the denominator is always positive. This makes f'(x) always negative, so the function always decreases.

Exam Tip: When a derivative is uniformly negative (or positive) across all allowed values, the function maintains that monotonicity consistently on each continuous piece of its domain.

Question 82. \( f(x) = |x| \) has
(A) minimum at x = 0
(B) maximum at x = 0
(C) neither a maximum nor a minimum at x = 0
(D) none of these
Answer: Analyze the absolute value function to determine its extrema.

\( f(x) = |x| = \begin{cases} -x, & x < 0 \\ x, & x \geq 0 \end{cases} \)

At x = 0, f(0) = 0.

For any x ≠ 0, |x| > 0.

Therefore, f(0) = 0 is the smallest value the function attains. This makes x = 0 a global minimum.

The function has no maximum since |x| grows without bound as x → ±∞.

In simple words: The absolute value of any number is either zero or positive. The smallest possible value is zero, which occurs only at x = 0. All other inputs give larger outputs, confirming a minimum at the origin.

Exam Tip: For absolute value functions, recognize that |x| ≥ 0 for all x, with equality only at x = 0. This immediately identifies x = 0 as a global minimum.

Question 83. When x is positive, the minimum value of \( f(x) = x^x \) is
(A) e^e
(B) \( e^{1/e} \)
(C) \( e^{-1/e} \)
(D) \( \left(\frac{1}{e}\right) \)
Answer: Find the critical points and determine the minimum value for positive x.

Let \( f(x) = x^x \). Taking the natural logarithm:

\( \ln f(x) = x \ln x \)

Differentiating:

\( \frac{f'(x)}{f(x)} = \ln x + 1 \)

\( f'(x) = x^x(\ln x + 1) \)

Set f'(x) = 0. Since \( x^x > 0 \) for x > 0:

\( \ln x + 1 = 0 \)

\( \ln x = -1 \)

\( x = \frac{1}{e} \)

The second derivative test confirms this is a minimum. The minimum value is:

\( f\left(\frac{1}{e}\right) = \left(\frac{1}{e}\right)^{1/e} = e^{-1/e} \)

In simple words: Taking logarithms converts the problem to finding when ln x = -1, which gives x = 1/e. At this point, the function value is (1/e) raised to the power 1/e, which equals e^(-1/e).

Exam Tip: For functions of the form x^x, always use logarithmic differentiation to find critical points. The expression ln x + 1 determines where extrema occur.

Question 83. Mark (√) against the correct answer in the following: The maximum value of \( \left(\frac{\log x}{x}\right) \) is
(a) \( \frac{1}{e} \)
(b) \( \frac{2}{e} \)
(c) e
(d) 1
Answer: (a) \( \frac{1}{e} \)
In simple words: To find the maximum value, differentiate the expression, set it equal to zero, and solve for x. When x equals e, you get the highest point on the curve, which gives a maximum value of 1/e.

Exam Tip: Always check the second derivative to confirm whether a critical point is a maximum or minimum - a positive second derivative indicates a minimum, while a negative one indicates a maximum.

Question 84. Mark (√) against the correct answer in the following: \( f(x) = \cos ec \, x \) in (-π, 0) has a maxima at
(a) x = 0
(b) \( x = -\frac{\pi}{4} \)
(c) \( x = -\frac{\pi}{3} \)
(d) \( x = -\frac{\pi}{2} \)
Answer: (d) \( x = -\frac{\pi}{2} \)
In simple words: Check each option by plugging in the x-value and calculating the function value. The option that gives the highest (least negative) function value is where the maximum occurs.

Exam Tip: When working with trigonometric functions on restricted intervals, always verify that your answer lies within the given domain and that you are comparing function values correctly.

Question 85. Mark (√) against the correct answer in the following: If x > 0 and xy = 1, the minimum value of (x + y) is
(a) -2
(b) 1
(c) 2
(d) none of these
Answer: (c) 2
In simple words: Since xy = 1, you can write y as 1/x. Substitute this into (x + y) to get an expression in terms of x alone, then use calculus to find where this expression reaches its smallest value.

Exam Tip: When a constraint like xy = 1 is given, use it to eliminate one variable and reduce the problem to a single-variable optimization task.

Question 86. Mark (√) against the correct answer in the following: The minimum value of \( \left(x^2 + \frac{250}{x}\right) \) is
(a) 0
(b) 25
(c) 50
(d) 75
Answer: (d) 75
In simple words: Differentiate the expression with respect to x and set the result equal to zero to find the critical point. Substitute this x-value back into the original expression to calculate the minimum.

Exam Tip: After finding a critical point, always substitute it back into the original function to find the actual minimum or maximum value - do not stop at finding the critical point itself.

Question 87. Mark (√) against the correct answer in the following: The minimum value of \( f(x) = 3x^4 - 8x^3 - 48x + 25 \) on [0, 3] is
(a) 16
(b) 25
(c) -39
(d) none of these
Answer: (c) -39
In simple words: Find all critical points within the interval by setting the derivative equal to zero. Then evaluate the function at these critical points as well as at the endpoints of the interval. The smallest value among these results is the minimum on the closed interval.

Exam Tip: For optimization on a closed interval, never forget to check the endpoint values - the absolute minimum may occur at an endpoint rather than at an interior critical point.

Question 88. Mark (√) against the correct answer in the following: The maximum value of \( f(x) = (x - 2)(x - 3)^2 \) is
(a) \( \frac{7}{3} \)
(b) 3
(c) \( \frac{4}{27} \)
(d) 0
Answer: (c) \( \frac{4}{27} \)
In simple words: Expand the product, then find the derivative and set it equal to zero to locate critical points. Use the second derivative test to identify which critical point gives a maximum. Finally, substitute that x-value into the original function to get the maximum value.

Exam Tip: When a function is given as a product, you may expand it first for easier differentiation, or apply the product rule - choose whichever method you find clearer.

Question 89. Mark (√) against the correct answer in the following: The least value of \( f(x) = \left(e^x + e^{-x}\right) \) is
(a) -2
(b) 0
(c) 2
(d) none of these
Answer: (c) 2
In simple words: The function \( e^x + e^{-x} \) is always symmetric and always positive. It reaches its lowest point where the derivative equals zero, which happens to be at x = 0. Substituting x = 0 gives 1 + 1 = 2.

Exam Tip: Recognize that \( e^x + e^{-x} \) represents the hyperbolic cosine function (cosh x), which has a well-known minimum value of 2 occurring at x = 0 - this can save time if you remember such standard results.