Maharashtra Board Class 12 Maths Part 1 Chapter 4 Miscellaneous Solutions

(I) Choose The Correct Alternative:

Question 1. The equation of tangent to the curve \(y = x^2 + 4x + 1\) at (-1, -2) is
(a) \(2x - y = 0\)
(b) \(2x + y - 5 = 0\)
(c) \(2x - y - 1 = 0\)
(d) \(x + y - 1 = 0\)
Answer:
(a) \(2x - y = 0\)
In simple words: To find the tangent equation, calculate the derivative for the slope at the given point, then use the point-slope form. The given equation `2x - y = 0` represents this tangent.

🎯 Exam Tip: Remember to first find the slope of the tangent using differentiation and then apply the point-slope form \(y - y_1 = m(x - x_1)\).

Question 2. The equation of tangent to the curve \(x^2 + y^2 = 5\), where the tangent is parallel to the line \(2x - y + 1 = 0\) are
(a) \(2x - y + 5 = 0; 2x - y - 5 = 0\)
(b) \(2x + y + 5 = 0; 2x + y - 5 = 0\)
(c) \(x - 2y + 5 = 0; x - 2y - 5 = 0\)
(d) \(x + 2y + 5; x + 2y - 5 = 0\)
Answer:
(a) \(2x - y + 5 = 0; 2x - y - 5 = 0\)
In simple words: Tangents parallel to a given line will have the same slope. By finding the derivative of the curve and equating its slope to that of the given line, we can determine the points of tangency and subsequently the equations of the two possible tangent lines.

🎯 Exam Tip: When tangents are parallel to a given line, their slopes are equal. Don't forget that a circle can have two parallel tangents for a given slope.

Question 3. If the elasticity of demand \(\eta = 1\), then demand is
(a) constant
(b) inelastic
(c) unitary elastic
(d) elastic
Answer:
(c) unitary elastic
In simple words: When the elasticity of demand is exactly 1, it means the percentage change in quantity demanded is equal to the percentage change in price, indicating unitary elastic demand.

🎯 Exam Tip: Understanding elasticity concepts is crucial. Remember that \(\eta = 1\) signifies unitary elasticity, where total revenue remains unchanged with a price change.

Question 4. If \(0 < \eta < 1\), then the demand is
(a) constant
(b) inelastic
(c) unitary elastic
(d) elastic
Answer:
(b) inelastic
In simple words: If the elasticity of demand is between 0 and 1, a change in price leads to a proportionately smaller change in quantity demanded, which is known as inelastic demand.

🎯 Exam Tip: For inelastic demand, consumers are not very responsive to price changes. This typically applies to essential goods or services.

Question 5. The function \(f(x) = x^3 - 3x^2 + 3x - 100, x \in R\) is
(a) increasing for all \(x \in R, x \neq 1\)
(b) decreasing
(c) neither increasing nor decreasing
(d) decreasing for all \(x \in R, x \neq 1\)
Answer:
(a) increasing for all \(x \in R, x \neq 1\)
In simple words: To determine if a function is increasing or decreasing, we examine the sign of its first derivative. If \(f'(x) > 0\), the function is increasing. In this case, the derivative is positive except at a single point, making it increasing for almost all real numbers.

🎯 Exam Tip: Evaluate \(f'(x)\) to find intervals of increasing/decreasing. If \(f'(x) = 0\) at isolated points but \(f'(x) > 0\) around them, the function is still considered increasing.

Question 6. If \(f(x) = 3x^3 - 9x^2 - 27x + 15\), then
(a) f has maximum value 66
(b) f has minimum value 30
(c) f has maxima at \(x = -1\)
(d) f has minima at \(x = -1\)
Answer:
(c) f has maxima at \(x = -1\)
In simple words: To find maxima/minima, calculate the first derivative and set it to zero to find critical points. Then, use the second derivative test: if \(f''(x) < 0\) at a critical point, it's a maximum. In this function, a maximum occurs at \(x = -1\).

🎯 Exam Tip: The second derivative test is efficient for classifying critical points. A negative second derivative indicates a local maximum.

(II) Fill In The Blanks:

Question 1. The slope of tangent at any point (a, b) is called as ______
Answer:
gradient
In simple words: The slope of the tangent at a specific point on a curve is also known as the gradient of the curve at that point. It measures the instantaneous rate of change.

🎯 Exam Tip: The terms "slope of tangent" and "gradient" are often used interchangeably in calculus to refer to the value of the derivative at a point.

Question 2. If \(f(x) = x^3 - 3x^2 + 3x - 100, x \in R\), then \(f''(x)\) is ______
Answer:
\(6x - 6 = 6(x - 1)\)
In simple words: To find the second derivative, you differentiate the function once to get \(f'(x)\), and then differentiate \(f'(x)\) again to get \(f''(x)\). For the given function, \(f''(x)\) simplifies to \(6(x - 1)\).

🎯 Exam Tip: Practice differentiating polynomial functions carefully to avoid errors, especially with signs and powers. Always double-check your calculations.

Question 3. If \(f(x) = \frac{7}{x} - 3, x \in R, x \neq 0\), then \(f''(x)\) is ______
Answer:
\(14x^{-3}\)
In simple words: First, write \(f(x)\) as \(7x^{-1} - 3\). Then, find the first derivative \(f'(x) = -7x^{-2}\) and then the second derivative \(f''(x) = 14x^{-3}\).

🎯 Exam Tip: For functions involving \(x\) in the denominator, convert them to negative exponents (e.g., \(1/x = x^{-1}\)) before differentiating to apply the power rule correctly.

Question 4. A rod of 108 m in length is bent to form a rectangle. If area j at the rectangle is maximum, then its dimensions are ______
Answer:
27 and 27
In simple words: For a fixed perimeter, a rectangle has the maximum area when it is a square. Since the total length (perimeter) is 108 m, each side of the square would be 108/4 = 27 m.

🎯 Exam Tip: Problems involving optimization (maximum or minimum values) often lead to geometric shapes. Remember that a square maximizes area for a given perimeter among rectangles.

Question 5. If \(f(x) = x \cdot \log x\), then its maximum value is ______
Answer:
\(-\frac{1}{e}\)
In simple words: To find the maximum value, differentiate the function, set the derivative to zero to find the critical point, and then substitute this critical point back into the original function. The critical point for this function leads to a maximum value of \(-1/e\).

🎯 Exam Tip: When working with logarithmic functions in optimization problems, ensure you correctly apply the product rule for differentiation and solve for the critical point. Also, confirm it's a maximum using the second derivative test.

(III) State Whether Each Of The Following Is True Or False:

Question 1. The equation of tangent to the curve \(y = 4xe^x\) at \(\left(-1, -\frac{1}{e}\right)\) is \(y \cdot e + 4 = 0\).
Answer:
True
In simple words: To verify this, calculate the slope of the tangent by differentiating \(y = 4xe^x\) and evaluating it at \(x = -1\). Then, use the point-slope form with the given point \(\left(-1, -\frac{1}{e}\right)\) and the calculated slope. If the resulting equation is \(y \cdot e + 4 = 0\), the statement is true.

🎯 Exam Tip: Always double-check your differentiation for complex functions involving products (like \(xe^x\)) and accurately substitute values to find the tangent equation.

Question 2. \(x + 10y + 21 = 0\) is the equation of normal to the curve \(y = 3x^2 + 4x - 5\) at (1, 2).
Answer:
False
In simple words: First, find the slope of the tangent to the curve at (1, 2) by differentiating. Then, calculate the slope of the normal (negative reciprocal). Finally, use the point-slope form for the normal line. If this doesn't match \(x + 10y + 21 = 0\), the statement is false.

🎯 Exam Tip: Remember that the slope of the normal is the negative reciprocal of the slope of the tangent. A common mistake is forgetting the negative sign or reciprocal.

Question 3. An absolute maximum must occur at a critical point or at an endpoint.
Answer:
True
In simple words: For a continuous function on a closed interval, the Extreme Value Theorem states that an absolute maximum (and minimum) must exist and will occur either where the derivative is zero or undefined (critical points) or at the boundaries of the interval (endpoints).

🎯 Exam Tip: When finding absolute extrema, always evaluate the function at all critical points within the interval and at both endpoints of the interval.

Question 4. The function \(f(x) = x \cdot e^{x(1-x)}\) is increasing on \(\left(-\frac{1}{2}, 1\right)\).
Answer:
True.
Hint: \[f(x) = x e^{x(1-x)} = x e^{x-x^2}\] \[ \implies f'(x) = x \frac{d}{dx} (e^{x-x^2}) + e^{x-x^2} \frac{d}{dx} (x) \] \[ = x \cdot e^{x-x^2} (1-2x) + e^{x-x^2} \times 1 \] \[ = e^{x-x^2} (x-2x^2 + 1) \] \[ = -2 e^{x-x^2} \left(x^2 - \frac{1}{2}x - \frac{1}{2}\right) \] \[ = -2 e^{x-x^2} \left[ \left(x - \frac{1}{4}\right)^2 - \frac{9}{16} \right] \] \[ \therefore f'(x) = -2 e^{x-x^2} \left[ \left(x - \frac{1}{4}\right)^2 - \frac{9}{16} \right] \quad (1) \] Now, \(x \in \left(-\frac{1}{2}, 1\right)\)
\( \implies -\frac{1}{2} < x < 1 \)
\( -\frac{1}{2} - \frac{1}{4} < x - \frac{1}{4} < 1 - \frac{1}{4} \)
\( -\frac{3}{4} < x - \frac{1}{4} < \frac{3}{4} \)
\( \implies 0 < \left(x - \frac{1}{4}\right)^2 < \frac{9}{16} \)
\( \implies \left(x - \frac{1}{4}\right)^2 - \frac{9}{16} < 0 \quad (2) \) Also, \(2e^{x-x^2} > 0\) for \(x \in \left(-\frac{1}{2}, 1\right)\)
\( \implies -2e^{x-x^2} < 0 \quad (3) \) From (2) and (3), \(-2e^{x-x^2} \left[ \left(x - \frac{1}{4}\right)^2 - \frac{9}{16} \right] > 0\)
\( \therefore f'(x) > 0 \) Hence, function f(x) is increasing on \(\left(-\frac{1}{2}, 1\right)\).
In simple words: To check if a function is increasing, we need to find its first derivative, \(f'(x)\), and see if it's positive on the given interval. The provided steps show that \(f'(x)\) is indeed greater than zero for \(x\) values between \(-1/2\) and \(1\), confirming that the function is increasing.

🎯 Exam Tip: For functions involving exponentials, be very careful with the chain rule during differentiation. The sign of the derivative determines the increasing/decreasing nature of the function.

(IV) Solve The Following:

Question 1. Find the equations of tangent and normal to the following curves:
(i) \(xy = c^2\) at \(\left(ct, \frac{c}{t}\right)\), where t is a parameter.
Solution:
\(xy = c^2\)
Differentiating both sides w.r.t. x, we get
\[ x \frac{dy}{dx} + y \frac{d}{dx} (x) = 0 \] \[ \implies x \frac{dy}{dx} + y \times 1 = 0 \] \[ \implies x \frac{dy}{dx} = -y \] \[ \implies \frac{dy}{dx} = -\frac{y}{x} \] \[ \left(\frac{dy}{dx}\right)_{\text{at } \left(ct, \frac{c}{t}\right)} = -\frac{c/t}{ct} = -\frac{1}{t^2} \] = slope of the tangent at \(\left(ct, \frac{c}{t}\right)\)
The equation of the tangent at \(\left(ct, \frac{c}{t}\right)\) is
\[ y - \frac{c}{t} = -\frac{1}{t^2} (x - ct) \]
\( \implies t^2y - ct = -x + ct \)
\( \implies x + t^2y - 2ct = 0 \)
The slope of the normal at \(\left(ct, \frac{c}{t}\right)\)
\[ \frac{-1}{\left(\frac{dy}{dx}\right)_{\text{at } \left(ct, \frac{c}{t}\right)}} = \frac{-1}{-\frac{1}{t^2}} = t^2 \] The equation of the normal at \(\left(ct, \frac{c}{t}\right)\) is
\[ y - \frac{c}{t} = t^2(x - ct) \]
\( \implies ty - c = t^3x - ct^4 \)
\( \implies t^3x - ty - c(t^4 - 1) = 0 \)
Hence, equations of tangent and normal are \(x + t^2y - 2ct = 0\) and \(t^3x - ty - c(t^4 + 1) = 0\) respectively.
In simple words: To find the tangent, differentiate the curve implicitly to get its slope, then use the point-slope formula. For the normal, the slope is the negative reciprocal of the tangent's slope, and again apply the point-slope formula.

🎯 Exam Tip: Implicit differentiation is key for curves not explicitly defined for \(y\). Remember to use the negative reciprocal for the normal's slope.

(ii) \(y = x^2 + 4x\) at the point whose ordinate is -3.
Solution:
Let \(P(x_1, y_1)\) be the point on the curve
\(y = x^2 + 4x\), where \(y_1 = -3\)
\[ \implies y_1 = x_1^2 + 4x_1 \] \[ \implies x_1^2 + 4x_1 = -3 \] \[ \implies x_1^2 + 4x_1 + 3 = 0 \] \[ \implies (x_1 + 3)(x_1 + 1) = 0 \] \[ \implies x_1 = -3 \text{ or } x_1 = -1 \] Therefore, coordinates of the points are (-3, -3) and (-1, -3).
Differentiating \(y = x^2 + 4x\) w.r.t. x, we get
\[ \frac{dy}{dx} = \frac{d}{dx} (x^2 + 4x) = 2x + 4 \] For point (-3, -3):
\[ \left(\frac{dy}{dx}\right)_{\text{at } (-3, -3)} = 2(-3) + 4 = -2 \] = slope of the tangent at (-3, -3)
The equation of the tangent at (-3, -3) is
\[ y - (-3) = -2[x - (-3)] \]
\( \implies y + 3 = -2x - 6 \)
\( \implies 2x + y + 9 = 0 \)
The slope of the normal at (-3, -3)
\[ \frac{-1}{\left(\frac{dy}{dx}\right)_{\text{at } (-3, -3)}} = \frac{-1}{-2} = \frac{1}{2} \] The equation of the normal at (-3, -3) is
\[ y - (-3) = \frac{1}{2}[x - (-3)] \]
\( \implies 2y + 6 = x + 3 \)
\( \implies x - 2y - 3 = 0 \)
For point (-1, -3):
\[ \left(\frac{dy}{dx}\right)_{\text{at } (-1, -3)} = 2(-1) + 4 = 2 \] = slope of the tangent at (-1, -3)
The equation of the tangent at (-1, -3) is
\[ y - (-3) = 2[x - (-1)] \]
\( \implies y + 3 = 2x + 2 \)
\( \implies 2x - y - 1 = 0 \)
The slope of the normal at (-1, -3)
\[ \frac{-1}{\left(\frac{dy}{dx}\right)_{\text{at } (-1, -3)}} = -\frac{1}{2} \] The equation of the normal at (-1, -3) is
\[ y - (-3) = -\frac{1}{2}[x - (-1)] \]
\( \implies 2y + 6 = -x - 1 \)
\( \implies x + 2y + 7 = 0 \)
Hence, the equations of tangent and normal at
(i) \((-3, -3)\) are \(2x + y + 9 = 0\) and \(x - 2y - 3 = 0\)
(ii) \((-1, -3)\) are \(2x - y - 1 = 0\) and \(x + 2y + 7 = 0\)
In simple words: First, find the x-coordinates corresponding to the given ordinate \(y = -3\). This gives two points. Then, for each point, calculate the slope of the tangent using the derivative and use the point-slope form. The slope of the normal is the negative reciprocal of the tangent's slope, which is then used to find the normal's equation.

🎯 Exam Tip: Always find all possible points on the curve that satisfy the given condition (e.g., specific x or y coordinate) as there might be multiple tangents and normals.

(iii) \(x = \frac{1}{t}, y = t - \frac{1}{t}\), at \(t = 2\).
Solution:
When \(t = 2\), \(x = \frac{1}{2}\) and \(y = 2 - \frac{1}{2} = \frac{3}{2}\)
Hence, the point P at which we want to find the equations of tangent and normal is \(\left(\frac{1}{2}, \frac{3}{2}\right)\).
Now, \(x = \frac{1}{t}, y = t - \frac{1}{t}\)
Differentiating x and y w.r.t. t, we get
\[ \frac{dx}{dt} = \frac{d}{dt} \left(\frac{1}{t}\right) = -\frac{1}{t^2} \] \[ \frac{dy}{dt} = \frac{d}{dt} \left(t - \frac{1}{t}\right) = 1 + \frac{1}{t^2} \] \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1 + 1/t^2}{-1/t^2} = -(t^2 + 1) \] \[ \left(\frac{dy}{dx}\right)_{\text{at } t=2} = -(2^2 + 1) = -5 \] = slope of the tangent at \(t = 2\)
The equation of the tangent at \(\left(\frac{1}{2}, \frac{3}{2}\right)\) is
\[ y - \frac{3}{2} = -5 \left(x - \frac{1}{2}\right) \]
\( \implies y - \frac{3}{2} = -5x + \frac{5}{2} \)
\( \implies 5x + y - 4 = 0 \)
The slope of the normal at \(t = 2\)
\[ \frac{-1}{\left(\frac{dy}{dx}\right)_{\text{at } t=2}} = \frac{-1}{-5} = \frac{1}{5} \] The equation of the normal at \(\left(\frac{1}{2}, \frac{3}{2}\right)\) is
\[ y - \frac{3}{2} = \frac{1}{5} \left(x - \frac{1}{2}\right) \]
\( \implies 10\left(y - \frac{3}{2}\right) = 2\left(x - \frac{1}{2}\right) \)
\( \implies 10y - 15 = 2x - 1 \)
\( \implies 2x - 10y + 14 = 0 \)
\( \implies x - 5y + 7 = 0 \)
Hence, the equations of tangent and normal are \(5x + y - 4 = 0\) and \(x - 5y + 7 = 0\) respectively.
In simple words: For parametric equations, first calculate \(dx/dt\) and \(dy/dt\), then find \(dy/dx\) by dividing them. Evaluate \(dy/dx\) at the given parameter value \(t\) to get the tangent's slope. Then use the point and slopes to find the equations of both the tangent and the normal.

🎯 Exam Tip: Remember the formula for \(dy/dx\) in parametric equations: \(dy/dx = (dy/dt) / (dx/dt)\). Always convert the parameter \(t\) to the corresponding \((x, y)\) coordinates for the point of tangency/normality.

(iv) \(y = x^3 - x^2 - 1\) at the point whose abscissa is -2.
Solution:
\(y = x^3 - x^2 - 1\)
\[ \implies \frac{dy}{dx} = \frac{d}{dx} (x^3 - x^2 - 1) = 3x^2 - 2x - 0 = 3x^2 - 2x \] \[ \left(\frac{dy}{dx}\right)_{\text{at } x=-2} = 3(-2)^2 - 2(-2) = 16 \] = slope of the tangent at \(x = -2\)
When \(x = -2\), \(y = (-2)^3 - (-2)^2 - 1 = -13\)
Therefore, the point P is (-2, -13).
The equation of the tangent at (-2, -13) is
\[ y - (-13) = 16[x - (-2)] \]
\( \implies y + 13 = 16x + 32 \)
\( \implies 16x - y + 19 = 0 \)
The slope of the normal at \(x = -2\)
\[ \frac{-1}{\left(\frac{dy}{dx}\right)_{\text{at } x=-2}} = -\frac{1}{16} \] The equation of the normal at (-2, -13) is
\[ y - (-13) = -\frac{1}{16}[x - (-2)] \]
\( \implies 16y + 208 = -x - 2 \)
\( \implies x + 16y + 210 = 0 \)
Hence, equations of tangent and normal are \(16x - y + 19 = 0\) and \(x + 16y + 210 = 0\) respectively.
In simple words: First, use the given x-coordinate to find the corresponding y-coordinate, yielding the point of tangency. Then, differentiate the function to find the slope of the tangent at that point. Use the point and slope to form the tangent equation. For the normal, use the negative reciprocal of the tangent's slope with the same point.

🎯 Exam Tip: Always calculate the full coordinates \((x,y)\) of the point of interest before proceeding. Be careful with signs when substituting negative values into polynomials and differentiation results.

Question 2. Find the equation of the normal to the curve \(y = \sqrt{x-3}\) which is perpendicular to the line \(6x + 3y - 4 = 0\).
Solution:
Let \(P(x_1, y_1)\) be the foot of the required normal to the curve \(y = \sqrt{x-3}\)
Differentiating \(y = \sqrt{x-3}\) w.r.t. x, we get
\[ \frac{dy}{dx} = \frac{d}{dx} (x-3)^{1/2} = \frac{1}{2} (x-3)^{-1/2} \frac{d}{dx} (x-3) \] \[ = \frac{1}{2 \sqrt{x-3}} \times (1-0) = \frac{1}{2 \sqrt{x-3}} \] \[ \left(\frac{dy}{dx}\right)_{\text{at } (x_1, y_1)} = \frac{1}{2 \sqrt{x_1-3}} \] = slope of the tangent at \(P(x_1, y_1)\)
The slope of the normal at \(P(x_1, y_1)\) is
\[ m_1 = \frac{-1}{\left(\frac{dy}{dx}\right)_{\text{at } (x_1, y_1)}} = \frac{-1}{\frac{1}{2 \sqrt{x_1-3}}} = -2\sqrt{x_1-3} \] The slope of the line \(6x + 3y - 4 = 0\) is
\[ m_2 = -\frac{6}{3} = -2 \] Since, the normal at \(P(x_1, y_1)\) is perpendicular to the line \(6x+3y-4=0\), \(m_1m_2 = -1\), i.e. \(m_1 = -\frac{1}{m_2}\)
\[ m_1 = -\frac{1}{-2} = \frac{1}{2} \] So, we have
\[ -2\sqrt{x_1-3} = \frac{1}{2} \]
\( \implies \sqrt{x_1-3} = -\frac{1}{4} \)
Squaring both sides:
\[ x_1-3 = \frac{1}{16} \]
\( \implies x_1 = \frac{1}{16} + 3 = \frac{1+48}{16} = \frac{49}{16} \] Since, \((x_1, y_1)\) lies on the curve \(y = \sqrt{x-3}\)
\[ \implies y_1 = \sqrt{x_1-3} \] \[ \implies y_1 = \sqrt{\frac{49}{16} - 3} = \sqrt{\frac{49-48}{16}} = \sqrt{\frac{1}{16}} = \pm \frac{1}{4} \] Therefore, the coordinates of the point P are \(\left(\frac{49}{16}, \frac{1}{4}\right)\) or \(\left(\frac{49}{16}, -\frac{1}{4}\right)\).
And the slope of the normal is \(m_1 = \frac{1}{2}\).
The equation of the normal at \(\left(\frac{49}{16}, \frac{1}{4}\right)\) is
\[ y - \frac{1}{4} = \frac{1}{2} \left(x - \frac{49}{16}\right) \]
\( \implies 2\left(y - \frac{1}{4}\right) = x - \frac{49}{16} \)
\( \implies 2y - \frac{1}{2} = x - \frac{49}{16} \)
\( \implies x - 2y - \frac{41}{16} = 0 \)
\( \implies 16x - 32y - 41 = 0 \)
The equation of the normal at \(\left(\frac{49}{16}, -\frac{1}{4}\right)\) is
\[ y - \left(-\frac{1}{4}\right) = \frac{1}{2} \left(x - \frac{49}{16}\right) \]
\( \implies y + \frac{1}{4} = \frac{1}{2} \left(x - \frac{49}{16}\right) \)
\( \implies 2y + \frac{1}{2} = x - \frac{49}{16} \)
\( \implies x - 2y - \frac{57}{16} = 0 \)
i.e. \(16x - 32y - 57 = 0\)
Hence, the equation of the normals are \(16x - 32y - 41 = 0\) and \(16x - 32y - 57 = 0\).
In simple words: First, find the slope of the given line. Since the normal is perpendicular to this line, its slope will be the negative reciprocal. Next, differentiate the curve to find the general slope of the tangent, and thus the general slope of the normal. Equate the two normal slopes to find the point(s) on the curve, then use these points and the normal's slope to write its equation.

🎯 Exam Tip: Be careful when solving for \(x_1\) and \(y_1\). Sometimes, squaring both sides of an equation can introduce extraneous solutions, so always check the original equation or square root constraints.

Question 3. Show that the function \(f(x) = \frac{x-2}{x+1}, x \neq -1\) is increasing.
Solution:
\(f(x) = \frac{x-2}{x+1}\)
\[ \implies f'(x) = \frac{d}{dx} \left(\frac{x-2}{x+1}\right) \] \[ = \frac{(x+1)\frac{d}{dx}(x-2) - (x-2)\frac{d}{dx}(x+1)}{(x+1)^2} \] \[ = \frac{(x+1)(1-0) - (x-2)(1+0)}{(x+1)^2} \] \[ = \frac{x+1 - x+2}{(x+1)^2} = \frac{3}{(x+1)^2} \] Given \(x \neq -1\)
\[ \implies x+1 \neq 0 \] \[ \implies (x+1)^2 > 0 \] \[ \implies \frac{3}{(x+1)^2} > 0 \] \[ \therefore f'(x) > 0\), for all \(x \in R, x \neq -1 \] Hence, the function f is increasing for all \(x \in R\), where \(x \neq -1\).
In simple words: To show a function is increasing, we need to prove that its first derivative is always positive. By differentiating \(f(x)\) using the quotient rule, we find \(f'(x) = \frac{3}{(x+1)^2}\). Since the square of any non-zero real number is positive, and 3 is positive, \(f'(x)\) is always positive (for \(x \neq -1\)), meaning the function is increasing.

🎯 Exam Tip: When using the quotient rule for differentiation, ensure you remember the formula and apply it correctly: \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\). A positive derivative implies an increasing function.

Question 4. Show that the function \(f(x) = \frac{3}{x} + 10, x \neq 0\) is decreasing.
Solution:
\(f(x) = \frac{3}{x} + 10\)
\[ \implies f'(x) = \frac{d}{dx} (3x^{-1} + 10) = 3(-x^{-2}) + 0 = -\frac{3}{x^2} \] Given \(x \neq 0\)
\[ \implies x^2 > 0 \] \[ \implies -\frac{3}{x^2} < 0 \] \[ \therefore f'(x) < 0\), for all \(x \in R, x \neq 0 \] Hence, the function f is decreasing for all \(x \in R\), where \(x \neq 0\).
In simple words: To prove a function is decreasing, its first derivative must always be negative. For \(f(x) = \frac{3}{x} + 10\), the derivative is \(f'(x) = -\frac{3}{x^2}\). Since \(x^2\) is always positive (for \(x \neq 0\)), \(\frac{3}{x^2}\) is positive, making \(-\frac{3}{x^2}\) always negative. Thus, the function is decreasing.

🎯 Exam Tip: Converting fractions to negative exponents (e.g., \(3/x = 3x^{-1}\)) often simplifies differentiation. A negative derivative directly indicates a decreasing function.

Question 5. If \(x + y = 3\), show that the maximum value of \(x^2y\) is 4.
Solution:
Given \(x + y = 3\)
\[ \implies y = 3 - x \] Substituting y in \(x^2y\), we get
\[ \implies x^2y = x^2(3 - x) = 3x^2 - x^3 \] Let \(f(x) = 3x^2 - x^3\)
Then
\[ f'(x) = \frac{d}{dx} (3x^2 - x^3) = 3 \times 2x - 3x^2 = 6x - 3x^2 \] And
\[ f''(x) = \frac{d}{dx} (6x - 3x^2) = 6 \times 1 - 3 \times 2x = 6 - 6x \] Now, set \(f'(x) = 0\) to find critical points:
\[ 6x - 3x^2 = 0 \]
\( \implies 3x(2 - x) = 0 \)
\( \implies x = 0 \text{ or } x = 2 \)
Check \(f''(x)\) at these critical points:
For \(x = 0\): \(f''(0) = 6 - 0 = 6 > 0\)
\[ \implies f\) has minimum value at \(x = 0 \] For \(x = 2\): \(f''(2) = 6 - 12 = -6 < 0\)
\[ \implies f\) has maximum value at \(x = 2 \] When \(x = 2\), then \(y = 3 - 2 = 1\)
\[ \implies \text{maximum value of } x^2y = (2)^2(1) = 4\).
In simple words: First, express \(x^2y\) in terms of a single variable, say \(x\), using the constraint \(x + y = 3\). Then, differentiate this new function with respect to \(x\), find the critical points by setting the derivative to zero, and use the second derivative test to confirm the maximum. Finally, substitute the \(x\) value that gives the maximum back into \(x^2y\) to find its value.

🎯 Exam Tip: Optimization problems often involve using a constraint equation to reduce the expression to be maximized or minimized to a single variable. The second derivative test is efficient for confirming maxima or minima.

Question 6. Examine the function f for maxima and minima, where \(f(x) = x^3 - 9x^2 + 24x\).
Solution:
\(f(x) = x^3 - 9x^2 + 24x\)
\[ \implies f'(x) = \frac{d}{dx} (x^3 - 9x^2 + 24x) = 3x^2 - 9 \times 2x + 24 \times 1 = 3x^2 - 18x + 24 \] And
\[ f''(x) = \frac{d}{dx} (3x^2 - 18x + 24) = 3 \times 2x - 18 \times 1 + 0 = 6x - 18 \] Set \(f'(x) = 0\) to find critical points:
\[ 3x^2 - 18x + 24 = 0 \] Divide by 3:
\[ \implies x^2 - 6x + 8 = 0 \] Factorize:
\[ \implies (x - 2)(x - 4) = 0 \] \[ \implies \text{the roots of } f'(x) = 0 \text{ are } x_1 = 2 \text{ and } x_2 = 4\).
Now use the second derivative test:
(a) For \(x = 2\): \(f''(2) = 6(2) - 18 = 12 - 18 = -6 < 0\)
\[ \implies \text{by the second derivative test, f has maximum at } x = 2 \text{ and maximum value of f at } x = 2 \] Maximum value: \(f(2) = (2)^3 - 9(2)^2 + 24(2) = 8 - 36 + 48 = 20\).
(b) For \(x = 4\): \(f''(4) = 6(4) - 18 = 24 - 18 = 6 > 0\)
\[ \implies \text{by the second derivative test, f has minimum at } x = 4 \text{ and minimum value of f at } x = 4 \] Minimum value: \(f(4) = (4)^3 - 9(4)^2 + 24(4) = 64 - 144 + 96 = 16\).
In simple words: To find maxima and minima, first calculate the first derivative \(f'(x)\) and find its roots (critical points). Then, calculate the second derivative \(f''(x)\). Evaluate \(f''(x)\) at each critical point: if \(f''(x) < 0\), it's a maximum; if \(f''(x) > 0\), it's a minimum. Finally, substitute these x-values into the original function to find the maximum/minimum values.

🎯 Exam Tip: Clearly show all steps, from finding critical points to applying the second derivative test. State both the location (x-value) and the actual maximum/minimum value (f(x)) in your final answer.