NCERT Solutions Class 12 Mathematics Chapter 6 Application of Derivatives

Exercise - 6.1

1. Find the rate of change of the area of a circle with respect to its radius r when 
(i) r = 3 cm
(ii) r = 4 cm 
Solution
The area of a circle (A) with radius (r) is given by, 
A = πr²        
Now, the rate of change of the area with respect to its radius is given by, dA/dr = d/dr(πr²) = 2πr 
(i) When r = 3 cm, 
dA/dr = 2π(3) = 6π
Hence, the area of the circle is changing at the rate of 6π cm²/cm when its radius is 3 cm. 
(ii) When r = 4 cm
dA/dr = 2π(3) = 8π
Hence, the area of the circle is changing at the rate of 8π cm²/cm when its radius is 4 cm.

2. The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution
Let x be the length of a side, V be the volume , and s be the surface area of the cube. 
Then, V = x³ and S = 6x² where x is a function of time t. 
It is given that dv/dt = 8cm cm³/s  . 
Then, by using the chain rule, we have : 

Thus, when x = 12 cm, ds/dt = 32/12  cm²/s = 8/3 cm²/s. 
Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rat of 8/3 cm²/s

3. The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Solution
The area of a circle (A) with radius (r) is given by, 
A = πr²   
Now, the rate of change of area (A) with respect to time (t) is given by, 
dA/dt = d/dr(πr²) .dr/dt = 2πr dr/dt                    [By chain rule] 
It is given that, 
dr/dt = 3 cm/s 
∴ dA/dt = 2πr(3) = 6πr
Thus, when r = 10 cm, 
dA/dt = 6π(10) = 60π cm² /s 
Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60π cm²/s.

4. An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Solution
Let x be the length of a side and V be the volume of the cube. Then, 
V = x³ . 
∴ dV/dt = 3x² . dx/dt                    (By chain rule) 
It is given that, 
dx/dt = 3 cm/s 
∴ dV/dt = 3x² (3) = 9x² 
Thus, when x = 10 cm, 
dV/dt =  9(10)² = 900 cm³/s 
Hence, the volume of the cube is increasing at the rate of 900 cm³/s when the edge is 10 cm long.

5. A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Solution
The area of a circle (A) with radius (r) is given by A = πr² . 
Therefore, the rate of change of area (A) with respect to time (t) is given by, 
dA/dt = d/dt(πr² ) = d/dr(πr² ) dr/dt = 2πr dr/dt                    [By chain rule] 
It is given that dr/dt = 5 cm/s . 
Thus, when r = 8 cm, 
dA/dt = 2π (8)(5) = 80π 
Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm²/s.

6.  The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
Solution
The circumference of a circle (C) with radius (r) is given by 
C = 2πr . 
Therefore, the rate of change of circumference (C) with respect to time (t) is given by, 
dC/dt = dC/dr . dr/dt (By chain rule) 
= d/dr (2πr ) dr/dt 
= 2π . dr/dt 
It is given that dr/dt = 0.7 cm/s. 
Hence, the rate of increase of the circumference is 2π(0.7) = 1.4π cm/s.

7. The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.
Solution
Since the length (x) is decreasing at the rate of 5 cm/minute and the width(y) is increasing at the rate of 4 cm/minute, we have : 
dx/dt = -5 cm/min and dy/dt = 4 cm/min 
(a) The perimeter (p) of a rectangle is given by, 
P = 2(x + y)
∴ dP/dt = 2(dx/dt + dy/dt) = 2(-5 + 4) = -2 cm/min 
Hence, the perimeter is decreasing at the rate of 2 cm/min. 
(b) The area (A) of a rectangle is given by, 
A = x × y 
∴ dA/dt = dx/dt . y + x. dy/dt = -5y + 4x 
When x = 8 cm and y= 6 cm, dA/dt = (-5 × 6 + 4 × 8) cm²/min = 2 cm²/min 
Hence, the area of the rectangle is increasing at the rate of 2 cm²/min.

8. A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Solution
The volume of a sphere (V) with radius (r) is given by, 
V = 4/3(πr³ )
∴ Rate of change of volume (V) with respect to time (t) is given by,  
dV/dt = dV/dr . dr/dt  [By chain rule] 

9. A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
Solution
The volume of a sphere (V) with radius (r) is given by V = (4/3)πr³ . 
Rate of change of volume (V) with respect to its radius (r) is given by, 

Therefore, when radius = 10 cm, 
dV/dr = 4π(10)² = 400π
Hence, the volume of the balloon is increasing at the rate of 400π cm³/cm.

10. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
Solution
Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x m away from the wall. 
Then, by Pythagoras theorem, we have : 
x² + y²  = 25                   [Length of the ladder = 5 m]
⇒ y = √(25 - x² )
Then the rate of change of height(y) with respect to time (t) is given by, 

11. A particle moves along the curve 6y = x³ +2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Solution
The equation of the curve is given as : 
6y = x³ +2 
The rate of change of the position of the particle with respect to time (t) is given by, 

Hence, the points required on the curve are (4, 11) and (-4, -31/3).

12. The radius of an air bubble is increasing at the rate 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution
The air bubble is in the shape of a sphere. 
Now, the volume of an air bubble(V) with radius(r) is given by, 
V = (4/3)πr³ 
The rate of change of volume (V) with respect to time (t) is given by, 

It is given that dr/dt = 1/2 cm/s. 
Therefore, when r = 1 cm, 
dv/dt = 4π (1)² (1/2) = 2π cm³/s 
Hence, the rate at which the volume of the bubble increases is 2π cm³/s

13. A balloon, which always remains spherical, has a variable diameter 3/2 (2x+ 1) Find the rate of change of its volume with respect to x.
Solution
The volume of a sphere (V) with radius (r) is given by, 
V = (4/3)πr³ 
It is given that : 
Diameter = (3/2) (2x + 1) 
⇒ r = (3/4) (2x + 1) 

14. Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Solution
The volume of a cone (V) with radius (r)  and height (h) is given by, 
V = (1/3)πr³ h 
It is given that, 
h = (1/6)r
⇒ r = 6h 
∴ V = (1/3)π(6h)² h = 12πh³ 
The rate of change of volume with respect to time (t) is given by,

Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of 1/48π  cm/s. 

15. The total cost C(x) in Rupees associated with the production of x units of an item is given by C(x) = 0.007x³ – 0.003x² + 15x + 4000.
Find the marginal cost when 17 units are produced
Solution 
Marginal cost is the rate of change of total cost with respect to output. 
∴ Marginal cost (MC) = dC/dx = 0.007 (3x² ) - 0.003(2x) + 15 
= 0.021x² - 0.006x + 15 
When x = 17, MC = 0.021(17²) - 0.006 (17) + 15
= 0.021(289) - 0.006(17) + 15 
= 6.069 - 0.102 + 15 
= 20.967 
Hence, when 17 units are produced, the marginal cost is Rs. 20.967

16. The total revenue in Rupees received from the sale of x units of a product is given by
R(x) = 13x² + 26x + 15.
Find the marginal revenue when x = 7.
Solution
Marginal revenue is the rate of change of total revenue with respect to the number of units sold. 
∴ Marginal Revenue (MR) = dR/dx = 13(2x) + 26 = 26x + 26 
When x = 7, 
MR = 26(7)  + 26 = 182 + 26 = 208 
Hence, the required marginal revenue is Rs 208. 

17. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π
(B) 12π
(C) 8π
(D) 11π
Solution
The area of a circle (A) with radius (r) is given by,  
a = πr³ 
Therefore, the rate of change of the area with respect to its radius r is 
dA/dr = d/dr (πr² ) - 2πr 
∴ When r = 6 cm, 
dA/dr = 2π × 6 = 12π cm²/s 
Hence, the required rate of change of the area of a circle is 12πcm²/s . 
The correct answer is B. 

18. The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. The marginal revenue, when x = 15 is
(A) 116
(B) 96
(C) 90
(D) 126
Solution
Marginal revenue is the rate of change of total revenue with respect to the number of units sold. 
∴ Marginal Revenue (MR) =dR/dx = 3(2x) + 36 = 6x + 36 
∴ When x = 15, 
MR = 6(15) + 36 = 90 + 36 = 126 
Hence, the required marginal revenue is Rs 126. 
The correct answer is D.

Exercise 6.2

1. Show that the function given by f(x) = 3x + 17 is strictly increasing on R. 
Solution
Let x₁ and x₂ be any two numbers in R.
Then, we have : 
x₁ < x₂
⇒ 3x₁ < 3x₂
⇒ 3x₁ + 17 < 3x₂ + 17
⇒ f(x₁) < f(x₂)
Hence, f is strictly increasing on R. 
Alternate method : 
f '(x) = 3 > 0, in every interval or R. 
Thus, the function is strictly increasing on R.

2. Show that the function given by f(x) = 3x + 17 is strictly increasing on R.
Solution
Let x₁ and x₂ be any two numbers in R. 
Then, we have : 
x₁ < x₂
⇒ 2x₁ < 2x₂
⇒ e^2x₁  < e^2x₂
⇒ f(x₁) < f (x₂)
Hence, f is strictly increasing on R.

3. Show that the function given by f(x) = sin x is
(a) strictly increasing in (0,π/2)
(b) strictly decreasing in (π/2,π)
(c) neither increasing nor decreasing in (0, π)
Solution
The given function is f(x) = sin x. 
∴ f '(x) = cos x
(a) Since for each x ∊ (0, π/2), cos x > 0, we have f '(x) > 0 . 
Hence, f is strictly increasing in (0, π/2). 
(b) Since for each x ∊ (π/2, π), cos x < 0, we have f '(x) < 0. 
Hence, f is strictly decreasing in (π/2, π). 
(c) From the results obtained in (a) and (b) , it is clear that f is neither increasing nor decreasing in (0, π).

4. Find the intervals in which the function f given by f(x) = 2x² − 3x is
(a) strictly increasing
(b) strictly decreasing
Solution
The given function is f(x) = 2x² - 3x . 
f '(x) = 4x - 3 
∴ f '(x) = 0
⇒ x = 3/4 
Now, the point 3/4 divides the real line into two disjoint intervals i.e., ( - ∞, 3/4) and (3/4, ∞). 

In interval (- ∞, 3/4), f '(x) = 4x - 3 < 0. 
Hence, the given function (f) is strictly decreasing in interval (- ∞, 3/4). 
In interval (3/4, ∞), f '(x) = 4x - 3 > 0. 
Hence, the given function (f) is strictly increasing in interval (3/4 , ∞).

5. Find the intervals in which the function f given by f(x) = 2x³ − 3x² − 36x + 7 is
(a) strictly increasing
(b) strictly decreasing
Solution
The given function is f(x) = 2x³ - 3x² - 36x + 7 . 
f '(x) = 6x² - 6x - 36 = 6(x² - x - 6) = 6(x + 2)(x - 3) 
∴ f '(x) = 0 ⇒ x  = -2, 3
The points x = -2 and x = 3 divide the real line into three disjoint intervals i.e., 
(- ∞, -2), (-2, 3), and (3, ∞). 

In intervals (- ∞, -2) and (3, ∞), f '(x) is positive while in interval (-2, 3), f '(x) is negative. 
Hence, the given function (f) is strictly increasing in intervals (-∞, -2) and (3, ∞), while function (f) is strictly decreasing in interval (-2, 3). 

6. Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x² + 2x − 5
(b) 10 − 6x − 2x² 
(c) -2x³ - 9x² - 12x + 1 
(d) 6 - 9x - x² 
(e) (x + 1)³ (x - 3)³ 
Solution
(a) We have, 
f(x) = x² + 2x - 5 
∴ f '(x) = 2x + 2
Now,  
f '(x) = 0
⇒ x = - 1 
Point x = -1 divides the real line into two disjoint intervals i.e., (- ∞, -1) and (-1, ∞) . 
In interval (- ∞, -1), f '(x) = 2x + 2 < 0.
∴  f is strictly decreasing in interval ( -∞. -1). 
Thus, f is strictly decreasing for x < - 1. 
In interval (-1, ∞), f '(x) = 2x + 2 > 0.
∴ f is strictly increasing in interval (-1, ∞). 
Thus, f is strictly increasing for x > -1.

(b) We have, 
f(x) = 10 - 6x - 2x² 
∴ f'(x) = -6 - 4x 
Now, 
f '(x) = 0
⇒ x = -3/2
The point x = -3//2 divides the real line into two disjoint intervals i.e., (-∞, -3/2) and (-3/2, ∞). 
In interval (-∞, -3/2) i.e., when x < -3/2, f '(x) = -6 - 4x > 0. 
∴ f is strictly increasing for x < -3/2. 
In interval (-3/2, ∞) i.e., when x > -3/2, f '(x) = -6 - 4x < 0. 
∴ f is strictly decreasing for x > -3/2. 

(c) We have, 
f(x) = -2x³ - 9x² - 12x + 1 
∴ f '(x) = -6x² - 18x - 12 = -6(x² + 3x + 2) = -6(x + 1)(x + 2)
Now, 
f '(x) = 0
⇒ x = -1 and x = -2 
Points x = -1 and x = -2 divide the real line into three disjoint intervals 
i.e., (-∞, -2), (-2, -1), and (-1, ∞). 
In intervals (-∞, -2) and (-1, ∞)i.e., when x < -2 and x > -1, 
f '(x) = -6(x + 1)(x + 2) < 0 
∴ f is strictly decreasing for x < -2 and x > -1.
Now, in interval (-2, -1) i.e., when -2 < x < - 1, f '(x) = -6(x + 1)(x + 2) > .
∴ f is strictly increasing for -2 < x < -1. 

(d) We have,  
f(x) = 6 - 9x - x² 
∴ f '(x)  = -9 - 2x 
Now, f '(x) = 0 gives x = -9/2
The point x = -9/2 divides the real line into two disjoint intervals i.e., (-∞, -9/2) and (-9/2, ∞). 
In interval (-∞, -9/2) i.e., for x < -9/2, f '(x) = -9 - 2x > 0. 
∴ f is strictly increasing for x < -9/2. 
In interval (-9/2, ∞) i.e., for x > -9/2, f '(x) = -9 - 2x < 0. 
∴ f is strictly decreasing for x > -9/2. 

(e) We have,  
f(x) = (x + 1)³ (x - 3)³ 
f '(x) = 3(x + 1)³ (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) 
Now, 
f '(x) = 0
⇒ x = -1, 3, 1 
The points x = -1, x = 1, and x = 3 divide the real line into four disjoint intervals i.e., (-∞, -1), (-1, 1), (1, 3), and (3, ∞) . 
In intervals (-∞, -1) and (-1, 1), f '(x) = 6(x + 1)² (x - 3)² (x - 1) < 0. 

7. Show that y = log (l + x) -  2x/(2 + x) , x > -1, is an increasing function of x throughout its domain. 
Solution
Consider y = log(1 + x) - 2x/(2 + x)

8. Find the values of x for y = [x(x - 2)]² is an increasing function. 
Solution
We have, 
y = [x(x - 2)]² = [x² - 2x]²
∴ dy/dx = y' = 2(x² - 2x)(2x - 2) = 4x(x - 2)(x - 1)
∴ dy/dx = 0 ⇒ x = 0, x = 2, x = 1.
The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals 
i.e., (- ∞, 0), (0, 1)(1, 2), and (2, ∞).
In intervals ( - ∞,  0) and (1, 2), dy/dx < 0.
∴ y is strictly decreasing in intervals (- ∞, 0) and (1, 2).
However, in intervals (0, 1) and (2, ∞), dy/dx > 0.
∴ y is strictly increasing for 0 < x < 1 and x > 2.

9.  Prove that y = 4 sinθ/(2 + cos θ) - θ is an increasing function of θ in (0, π/2)
Solution
We have, 
y = 4 sinθ/(2 + cos θ) - θ 

⇒ 8 cos θ + 4 = 4 + cos² θ + 4 cos θ
⇒ cos² θ - 4cos θ = 0 
⇒ cos θ (cos θ  - 4) = 0 
⇒ cos θ = 0 or cos θ  = 4 
Since cos θ ≠ 4,  cos θ  = 0. 
cos θ = 0
⇒ ፀ = π/2 
Now, 

Therefore y is strictly increasing in interval (0, π/2). 
Also, the given function is continuous at x = 0 and x = π/2 
Hence, y is increasing in interval [0, π/2). 

10. Prove that the logarithmic function is strictly increasing on (0, ∞).
Solution
The given function is f(x) = log x. 
∴ f '(x) = 1/x 
It is clear that for x > 0, f '(x) = 1/x >0. 
Hence, f(x) = log x is strictly increasing in interval (0, ∞) 

11. Prove that the function f given by f(x) = x² − x + 1 is neither strictly increasing nor strictly decreasing on (−1, 1).
Solution
The given function is f(x) =  x² − x + 1.
∴ f '(x) = 2x - 1 
Now, f '(x) = 0 ⇒ x = 1/2. 
The point 1/2 divides the interval (-1, 1) into two disjoint intervals i.e., (-1, 1/2) and (1/2, 1). 
Now, in interval (-1, 1/2), f '(x) = 2x - 1 < 0. 
Therefore, f is strictly decreasing in interval (-1, 1/2). 
However, in interval (1/2 , 1), f '(x) = 2x - 1 > 0.
Therefore, f is strictly increasing in interval (1/2, 1). 
Hence, f is neither strictly increasing nor decreasing in interval (-1, 1). 

12.Which of the following functions are strictly decreasing on (0, pi/2)?
(A) cos x
(B) cos 2x
(C) cos 3x
(D) tan x
Solution
(A) Let f₁ (x) = cos x 
∴ f '₁ (x) = - sin x 
In interval (0, π/2),  f '₁ (x) = - sin x < 0. 
∴ f ₁ (x) = cos x is strictly decreasing in interval (0, π/2). 

(B) Let f₂ (x) = cos x
∴ f '₂ (x) = - 2sin 2x 
Now, 0 < x < π/2
⇒ 0 < 2x < π
⇒ sin 2x > 0
⇒ - 2sin 2x < 0 
∴ f '₂ (x) = - 2sin 2x < 0 on (0, π/2)
∴ f₂ (x) = cos 2x is strictly decreasing in interval (0, π/2). 

(C) Let f₃ (x) = cos 3x .
f’₃ (x) = -3 sin 3x 
Now, f’₃ (x) = 0.
⇒ sin 3x = 0
⇒ 3x = π, as x ∈ (0, π/2) 
⇒ x = π/3 
The point x = π/3  divides the interval (0, π/2) into two disjoint intervals 
i.e., 0(0, π/3) and (π/3, π/2). 
Now, in interval (0, π/3), f₃ (x) = -3 sin 3x > 0 [as π/3 < x < π/2 ⇒ π < 3x < 3π/2]. 
∴ f₃ is strictly increasing in interval (π/3 , π/2) . 
Hence, f₃ is neither increasing nor decreasing in interval (0, π/2). 

(D) Let f₄ (x) = tan x. 
∴ f’₄ (x) = sec² x 
In interval (0, π/2), f’₄ (x) = sec² x > 0.
∴ f₄ is strictly increasing in interval (0, π/2). 
Therefore, functions cos x and cos 2x are strictly decreasing in (0, π/2). 
Hence, the correct answer are A and B. 

13. On which of the following intervals is the function f given by f(x) = x¹⁰⁰ + sin x –1 strictly decreasing?
(A) (0,1)
(B) (π/2,π)
(C) (0,π/2)
(D) None of these
Solution
We have, 
f(x)  = x¹⁰⁰ + sin x –1 
∴ f '(x) = 100x⁹⁹ + cos x 
In interval (0, 1), cos x > 0 and 100x⁹⁹ > 0. 
∴ f '(x) > 0. 
Thus, function f is strictly increasing in interval (0, 1). 
In interval (π/2 , π) , cos x < 0  and 100x⁹⁹ > 0. Also, 100x⁹⁹ > cos x. 
∴ f '(x)  > 0  in (π/2 , π)
Thus, function f is strictly increasing in interval (π/2 , π). 
In interval (0 , π/2), cos x > 0 and 100x⁹⁹  > 0. 
∴ 100x⁹⁹ + cos x > 0 
⇒ f '(x) > 0 on (0, π/2)
∴ f is strictly increasing in interval (0, π/2). 
Hence, function f is strictly decreasing in none of the intervals. 
The correct answer is D.

14. Find the least value of a such that the function f given by f(x) = x² + ax + 1 is strictly increasing on (1, 2).
Solution
We have, 
f(x) = x² + ax + 1 
∴ f '(x) = 2x + a 
Now, function f will be increasing in (1, 2), if f '(x) > 0 in (1, 2). 
f '(x) > 0 
⇒ 2x + a > 0 
⇒ 2x > - a 
⇒ x > -a/2 
Therefore, we have to find the least value of a such that 
x > -a/2, when x ∊(1, 2). 
⇒ x > -a/2 (when 1 < x < 2) 
Thus, the least value of a for f to be increasing on (1, 2) is given by,  
-a/2 = 1 
-a/2 = 1
⇒ a = -2
Hence, the required value of a is -2. 

15. Let I be any interval disjoint from (−1, 1). Prove that the function f given by f(x)= x+1/x is strictly increasing on I.
Solution
We have, 
f(x) = x + 1/x
∴ f '(x) = 1 - 1/x² 
Now, 
f '(x) = 0 ⇒ 1/x² = 1 ⇒ x = ± 1 
The points x = 1 and x = -1 divide the real line in three disjoint intervals 
i.e., (-∞, -1), (-1, 1) and (1, ∞). 
In interval (-1, 1), it is observed that : 
-1 < x < 1 
⇒ x² < 1 
⇒ 1 < 1/x² , x ≠ 0 
⇒ 1 - 1/x²  < 0, x ≠ 0 
∴ f '(x) = 1 - 1/x² < 0 on (-1, 1) ~ {0}. 
∴ f is strictly decreasing on (-1, 1) ~ {0}. 
In intervals (- ∞, -1) and (1, ∞), it is observed that : 
x < -1 or 1 < x 
⇒ x² > 1 
⇒ 1 > 1/x² 
⇒ x - 1/x² > 0 
∴ f '(x) = 1 - 1/x²  > 0 on (-∞, 1) and (1, ∞). 
∴ f is strictly increasing on (- ∞, 1) and (1, ∞). 
Hence, function f is strictly increasing in interval I disjoint from (-1, 1). 
Hence, the given result is proved. 

16. Prove that the function f given by f(x) = log sin x is strictly increasing on (0, π/2) and strictly decreasing on (π/2 , π).
Solution
We have, 
f(x) = log sin x 
∴ f '(x) = (1/sin x) cos x =  cot x 
In interval (0, π/2), f '(x) = cot x > 0. 
∴ f is strictly increasing in (0, π/2). 
In interval (π/2, π) , f '(x) = cos x < 0. 
∴ f is strictly decreasing in (π/2 , π). 

17. Prove that the function f given by f(x) = log cos x is strictly decreasing on (0,π/2) and strictly increasing on (π/2,π) 
Solution
We have, 
f(x) = log cosx 

18.Prove that the function given by f (x) = x³ – 3x² + 3x – 100 is increasing in R.
Solution
We have, 
f(x) = x³ - 3x² + 3x - 100 
f '(x) = 3x² - 6x + 3 
= 3(x² - 2x + 1)
= 3(x - 1)² 
For any x ∈ R, (x - 1)²  > 0. 
Thus, f '(x) is always positive in R. 
Hence, the given function (f) is increasing in R. 

19. The interval in which y = x² e^–x is increasing is
(A) (– ∞, ∞)
(B) (– 2, 0)
(C) (2, ∞)
(D) (0, 2)
Solution
We have, 
y =  x² e^–x 
∴ dy/dx = 2xe^–x - x² e^–x = xe^–x (2 - x)
Now, dy/dx = 0.
⇒ x = 0 and x = 2 
The points x = 0 and x = 2 divide the real line into three disjoint intervals 
i.e., (-∞, 0), (0, 2), and (2, ∞). 
In intervals (-∞, 0) and (2, ∞), f '(x) < 0 as e^–x is always positive. 
∴ f is decreasing on (- ∞, 0) and (2, ∞). 
In interval (0, 2), f '(x) > 0. 
∴ f is strictly increasing on (0, 2). 
Hence, f is strictly increasing in interval (0, 2). 
The correct answer is D.

Exercise 6.3

1. Find the slope of the tangent to the curve y = 3x⁴ − 4x at x = 4.
Solution
The given curve is y = 3x⁴ − 4x.
Then, the slope of the tangent to the given curve at x = 4 is given by, 

2. Find the slope of the tangent to the curve y = (x -1)/(x - 2), x != 2 at x = 10.
Solution

3. Find the slope of the tangent to curve y = x³ − x + 1 at the point whose x-coordinate is 2.
Solution
The given curve is y = x³ - x + 1. 
∴ dy/dx = 3x² - 1. 
The slope of the tangent to a curve a (x₀ , y₀) is dy/dx . 
It is given that x₀ = 2. 
Hence, the slope of the tangent at the point where the x - coordinate is 2 is given by,  

4. Find the slope of the tangent to the curve y = x³ − 3x + 2 at the point whose x-coordinate is 3.
Solution
The given curve is y = x³ − 3x + 2 . 
∴ dy/dx = 3x² - 3 
The slope of the tangent to a curve at  (x₀ , y₀) is dy/dx]_{(x0 , y0)} .
Hence, the slope of the tangent at the point where the x - coordinate is 3 is given by, 
dy/dx]_x-3 = 3x² - 3]_x-3 = 3(3)² - 3 = 2 -3 = 24 

5. Find the slope of the normal to the curve x = acos³θ, y = asin³θ at θ = π/4 . 
Solution
It is given that x = acos³θ, and y = asin³θ . 

6. Find the slope of the normal to the curve x = 1 - a sin θ and y = b cos² θ at θ = π/2. 
Solution
It is given that x = 1 - a sin θ and y = b cos² θ . 

7. Find points at which the tangent to the curve y = x³ − 3x² − 9x + 7 is parallel to the x-axis.
Solution
The equation of the given curve is y = x³ − 3x² − 9x + 7 . 
∴ dy/dx = 3x² - 6x - 9 
Now, the tangent is parallel to the x- axis if the slope of the tangent is zero. 
∴ 3x² - 6x - 9 = 0 
⇒ x² - 2x - 3 = 0 
⇒ (x -3)(x + 1) = 0 
⇒ x = 3 or x = -1 
When x = 3, y = (3)³ - 3(3)² - 9(3) + 7 = 27 - 27 - 27 + 7 = -20. 
When x = -1, y = (-1)³ - 3(-1)² - 9(-1) + 7 = -1 - 3 + 9 + 7 = 12. 
Hence, the points at which the tangent is parallel to the x - axis are (3, -20) and (-1, 12). 

8. Find a point on the curve y = (x − 2)² at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
Solution
If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then the slope of the tangent = the slope of the chord. 
The slope of the chord is (4- 0)/(4 - 2) = 4/2 = 2. 
Now, the slope of the tangent to the given curve at a point (x, y) is given by,  
dy/dx = 2(x - 2) 
Since the slope of the tangent = slope of the chord, we have : 
2(x - 2) = 2 
⇒ x - 2 = 1
⇒ x = 3 
When x = 3, y = (3 - 2)² = 1.
Hence, the required point is (3, 1). 

9. Find the point on the curve y = x³ − 11x + 5 at which the tangent is y = x − 11.
Solution
The equation of the given curve is y =  x³ − 11x + 5 . 
The equation of the tangent to the given curve is given as y = x - 11 (which is of the form y =  mx + c). 
∴ Slope of the tangent = 1 
Now, the slope of the tangent to the given curve at the point (x, y) is given by, dy/dx = 3x² - 11
Then, we have : 
3x² - 11 = 1 
⇒ 3x² = 12 
⇒ x² = 4 
⇒ x = ± 2 
When x = 2, y = (2)³ - 11(2) + 5 = 8 - 22 + 5 = -9. 
When x = -2, y = (-2)³ - 11(-2) +5 = -8 + 22 + 5 = 19. 

Hence, the required points are (2, -9) and (-2, 19). 

10. Find the equation of all lines having slope −1 that are tangents to the curve y = 1/(x-1), x ≠ 1 . 
Solution
The equation of the given curve is y = 1/(x-1), x ≠ 1 . 
The slope of the tangents to the given curve at any point (x, y) is given by,  
dy/dx = -1/(x - 1)² 
If the slope of the tangent is -1, then we have : 
-1/(x - 1)² = -1 
⇒ (x - 1)² = 1 
⇒ x - 1 = ± 1 
⇒ x = 2, 0 
When x = 0, y = -1 and when x = 2, y = 1. 
Thus, there are two tangents to the given curve having slope -1. These are passing through the points (0, -1) and (2, 1). 
∴ The equation of the tangent through (0, -1) is given by, 
y - (-1) = -1(x - 0)
⇒ y + 1 = -x 
⇒ y + x + 1 = 0 
∴ The equation of  the tangent through (2, 1) is given by, 
y - 1 = -1(x - 2)
⇒ y - 1 = -x + 2 
⇒ y + x - 3 = 0 
Hence, the equations of the required lines are y + x + 1 = 0 and y + x - 3 = 0. 

11. Find the equation of all lines having slope 2 which are tangents to the curve y = 1/(x - 3), x ≠ 3.
Solution
The equation of the given curve is y = 1/(x - 3), x ≠ 3. 
The slope of the tangent to the given curve at any point (x, y)  is given by,  
dy/dx = -1/(x - 3)² 
If the slope of the tangent is 2, then we have : 
-1/(x- 3)² = 2 
⇒ 2(x - 3)² = -1 
⇒ (x - 3)² = -1/2 
This is not possible since the L.H.S. is positive while the R.H.S. is negative . 
Hence, there is no tangent to the given curve having slope 2. 

12. Find the equations of all lines having slope 0 which are tangent to the curve y = 1/(x² - 2x + 3) .
Solution
The slope of the tangent to the given curve at any point (x, y) is given by,  

⇒ - 2(x - 1) = 0
⇒ x = 1 
When x = 1, y = 1/(1 - 2 + 3) = 1/2. 
∴ The equation of the tangent through (1, 1/2) is given by, 
y - 1/2 = 0 (x - 1) 
⇒ y - 1/2 = 0 
⇒ y = 1/2 
Hence, the equation of the required line is y = 1/2.

13. Find points on the curve x²/9 + y²/16 =1 at which the tangent are 
(i) parallel to x - axis
(ii) parallel to y - axis 
Solution
The equation of the given curve is x²/9 + y²/16 =1 . 
On differentiating both sides with respect to x, we have : 
2x/9 + 2y/16 .dy/dx = 0 
⇒ dy/dx = -16x/9y 

(i) The tangent is parallel to the x - axis if the slope of the tangent is 0 i.e., -16x/9y = 0, which is possible if x = 0.
Then, x²/9 + y²/16 =1 for x = 0 
⇒ y² = 16
⇒ y = ± 4 
Hence, the points at which the tangents are parallel to the x - axis are (0, 4) and (0, -4). 

(ii) The tangent is parallel to the y - axis if the slope of the normal is 0, which gives 

Then, x² /9 + y² /16 = 1 for y = 0. 
⇒ x = ± 3 
Hence, the points at which the tangents are parallel to the y - axis are (3, 0) and (-3, 0). 

14. Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x⁴ − 6x³ + 13x² − 10x + 5 at (0, 5)
(ii) y = x⁴ − 6x³ + 13x² − 10x + 5 at (1, 3)
(iii) y = x³ at (1, 1) 
(iv) x² at (0, 0) 
(v) x = cos t, y = sin t at  t = π/4 
Solution
(i) The equation of the curve is y = x⁴ − 6x³ + 13x² − 10x + 5
On differentiating with respect to x, we get : 
dy/dx = 4x³ - 18x² + 26x - 10
dy/dx]_{(0, 5)} = -10
Thus, the slope of the tangent at (0, 5) is -10. The equation of the tangent is given as : 
y - 5 = -10(x - 0) 
⇒ y - 5 = -10x 
⇒ 10x + y = 5 
The slope of the normal at (0, 5) is -1/[Slope of the tangent at (0, 5)] = 1/10 . 
Therefore, the equation of the normal at (0, 5) is given as : 
y - 5 = 1/10 (x - 0) 
⇒ 10y - 50  = x 
⇒ x - 10y + 50 = 0 

(ii) The equation of the curve is y = x⁴ - 6x³ + 13x² - 10x + 5. 
On differentiating with respect to x, we get : 
dy/dx = 4x³ - 18x² + 26x - 10 
dy/dx]_{(1, 3)} = 4 - 18 + 26 - 10 = 2
Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as : 
y - 3 = 2(x - 1) 
⇒ y - 3 = 2x - 2 
⇒ y = 2x + 1 
The slope of the normal at (1, 3) is -1/[Slope of the tangent at (1, 3)] = -1/2 . 

Therefore, the equation of the normal at (1, 3) is given as : 
y - 3 = -1/2 (x - 1) 
⇒ 2y - 6 = -x + 1 
⇒ x + 2y - 7 = 0 

(iii) The equation of the curve is y = x³ . 
On differentiating with respect to x, we get :  
dy/dx = 3x² 
dy/dx]_{(1, 1)} = 3(1)² = 3 
Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as : 
y - 1 = 3(x - 1) 
⇒ y = 3x - 2 
The slope of the normal at (1, 1) is -1/[Slope of the tangent at (1, 1)] = -1/3 . 
Therefore, the equation of the normal at (1, 1) is given as : 
y - 1 = -1/3 (x - 1) 
⇒ 3y - 3 = - x + 1 
⇒ x + 3y - 4 = 0 

(iv) The equation of the curve is y = x² . 
On differentiating with respect to x, we get :  
dy/dx = 2x 
dy/dx]_{(0, 0)} = 0 
Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as : 
y - 0 = 0(x - 0) 
⇒ y = 0 
The slope of the normal at (0, 0) is -1/[Slope of the tangent at (0, 0)] = -1/0 , which is not defined. 
Therefore, the equation of the normal at (x₀ , y₀) = (0, 0) is given by 
x = x₀ = 0.

(v) The equation of the curve is x = cos t, y = sin t. 
x = cos t and y = sin t
∴ dx/dt = - sin t, dy/dt = cos t

∴ The slope of the tangent at t = π/4 is -1. 
When t = π/4 , x = 1/√2 and y = 1/√2 . 
Thus, the equation of the tangent to the given curve at t = π/4 i.e., at  [(1/√2 , 1/√2)] is 
 y - 1/√2 = -1 (x - 1/√2) . 
⇒ x + y - 1/√2 - 1/√2 = 0 
⇒ x + y - √2 = 0 
The slope of the normal at t = π/4 is -1/[Slope of the tangent at t = π/4] = 1.
Therefore, the equation of the normal to the given curve at t = π/4 i.e., at [(1/√2 , 1/√2)] is 
y - 1/√2 = 1 (x - 1/√2) 
⇒ x = y 

15. Find the equation of the tangent line to the curve y = x² − 2x + 7 which is
(a) parallel to the line 2x − y + 9 = 0.
(b) perpendicular to the line 5y - 15x = 13 
Solution
The equation of the given curve is y =  x² − 2x + 7 . 
On differentiating with respect to x, we get : 
dy/dx = 2x - 2 

(a) The equation of the line is 2x - y + 9 = 0. 
2x - y + 9 = 0 ⇒ y = 2x + 9 
This is of the form y = mx + c . 
∴ Slope of the line = 2 
If a tangent is parallel to the line 2x - y + 9 = 0, then the slope of the tangent is equal to the slope of the line. 
Therefore, we have : 
2 = 2x - 2 
⇒ 2x = 4 
⇒ x = 2 
Now, x = 2 
⇒ y = 4 - 4 + 7 = 7 
Thus, the equation of the tangent passing through (2, 7) is given by, 
y - 7 = 2(x - 2) 
⇒ y- 2x - 3 = 0 
Hence, the equation of the tangent line to the given curve (which is parallel to line 2x - y + 9 = 0 ) is y - 2x - 3 = 0.

(b) The equation of the line is 5y - 15x = 13. 
5y - 15x = 13 ⇒ y = 3x + 13/5 
This is of the form y = 2x + c .
∴ Slope of the line  = 3 
If a tangent is perpendicular to the line 5y - 15x = 13, then the slope of the tangent is -1/[slope of the line] = -1/3 . 
⇒ 2x - 2 = -1/3 
⇒ 2x = -1/3 + 2 
⇒ 2x = 5/3 
⇒ x = 5/6 
Now, x = 5/6 
⇒ y = (25/36) -(10/6) + 7 = (25 - 60 + 252)/36 = 217/36 
Thus, the equation of the tangent passing through (5/6, 217/36) is given by,  

⇒ 36y - 217 = -2(6x - 5) 
⇒ 36y - 217 = -12x + 10 
⇒ 36y + 12x - 227 = 0 
Hence, the equation of the tangent line to the given curve (which is perpendicular to line 5y - 15x = 13) is 36y + 12x - 227 = 0. 

16. Show that the tangents to the curve y = 7x³ + 11 at the points where x = 2 and x = −2 are parallel.
Solution
The equation of the given curve is y = 7x³ + 11. 
∴ dy/dx = 21x² 
The slope of the tangent to a curve at (x₀ , y₀) is dy/dx]_{(x0 ,y0)} 
Therefore, the slope of the tangent at the point where x = 2 is given by, 
dy/dx]_{x = 2} = 21(2)² = 84 
And, the slope of the tangent at the point where x = -2 is given by,  
dy/dx]_{x = -2} = 21(2)² = 84 
It is observed that the slopes of the tangents at the points where x = 2 and x = -2 are equal. 
Hence, the two tangents are parallel. 

17. Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinate of the point.
Solution
The equation of the given curve is y = x³. 
∴ dy/dx = 3x² 
The slope of the tangent at the point (x, y) is given by,  
dy/dx]_{(x, y)} = 3x² 
When the slope of the tangent is equal to the y - coordinate of the point, then y = 3x². 
Also, we have y = x³ . 
∴ 3x² = x³ 
⇒ x² (x - 3) = 0 
⇒ x = 0, x = 3 
When x = 0, they y = 0 and when x = 3, then y = 3 (3)² = 27. 
Hence, the required points are (0, 0) and (3, 27). 

18. For the curve y = 4x³ − 2x⁵, find all the points at which the tangents passes through the origin.
Solution
The equation of the given curve is y = 4x³ − 2x⁵  . 
∴ dy/dx = 12x² - 10x⁴ 
Therefore, the slope of the tangent at a point (x, y) is 12x² - 10x⁴ . 
The equation of the tangent at (x, y) is given by, 
Y - y = (12x² - 10x⁴ )(X - x) ...(1) 
When the tangent passes through the origin (0, 0), then X = Y = 0.
Therefore, equation (1) reduces to : 
-y = (12x² - 10x⁴ )(-x)
y = 12x³ - 10x⁵ 
Also, we have y = 4x³ - 2x⁵ , 
∴ 12x³ - 10x⁵ = 4x³ - 2x⁵ 
⇒ 8x⁵ - 8x³ = 0 
⇒ x⁵ - x³ = 0 
⇒ x³ (x² - 1) = 0 
⇒ x = 0, ± 1 
When x = 0, y = 4(0)³ - 2(0)⁵ = 0. 
When x = 1, y = 4(1)³ - 2(1)⁵ = 2. 
When x = -1, y = 4(-1)³ - 2(-1)⁵ = -2. 
Hence, the required points are (0, 0), (1, 2) and  (-1, -2). 

19. Find the points on the curve x² + y² − 2x − 3 = 0 at which the tangents are parallel to the x-axis.
Solution
The equation of the given curve is x² + y² − 2x − 3 = 0 . 
On differentiating with respect to x, we have : 
2x + 2y(dy/dx) - 2 = 0 
⇒ y. (dy/dx) = 1 - x 
⇒ dy/dx = (1 - x)/y 
Now, the tangents are parallel to the x - axis if the slope of the tangent is 0. 
∴ (1 - x)/y = 0
⇒ 1 - x = 0
⇒ x = 1 
But, x² + y² -2x - 3 = 0 for x = 1. 
⇒ y² = 4
⇒ y = ± 2
Hence, the points at which the tangents are parallel to the x - axis are (1, 2) and (1, -2). 

20. Find the equation of the normal at the point (am², am³) for the curve ay² = x³.
Solution
The equation of the given curve is ay² = x³ . 
On differentiating with respect to x, we have :  
2ay . (dy/dx) = 3x² 
⇒ dy/dx = 3x² /2ay 

∴ Slope of normal at (am², am³)
= -1/[slope of the tangent at (am², am³) = -2/3m 
Hence, the equation of the normal at (am², am³) is given by,  
y - am³ = -2/3m (x - am²) 
⇒ 3my - 3am⁴ = -2x + 2am² 
⇒ 2x + 3my - am² (2 + 3m²) = 0 

21. Find the equation of the normal to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
Solution
The equation of the given curve is y = x³ + 2x + 6. 
The slope of the tangent to the given curve at any point (x, y) is given by, 
dy/dx = 3x² + 2 
∴ Slope of the normal to the given curve at any point (x, y) 
= -1/[Slope of the tangent at the point (x, y)]
= -1/(3x² + 2)
The equation of the given line is x + 14y + 4 = 0. 
x + 14y + 4 ⇒ y = -(1/14)x - 4/14 (which is of the form y = mx + c) 
∴ Slope of the given line = -1/14 
If the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line. 
∴ -1/(3x² +2)   = -1/14 
⇒ 3x³ + 2 = 14 
⇒ 3x² = 12 
⇒ x² = 4 
⇒ x = ± 2 
When x = 2, y = 8 + 4 + 6 = 18. 
When x = -2, y = -8 - 4 + 6 = -6. 
Therefore, there are two normals to the given curve with slope -1/14 and passing through the points (2, 18) and (-2, -6). 
thus, the equation of the normal through (2, 18) is given by,  
y - 18 = (-1/14)(x - 2) 
⇒ 14y - 252 = -x + 2 
⇒ x + 14y - 254 = 0 
And, the equation of the normal through (-2, -6) is given by,  
y -(-6) = -1/14 (x - 2)
⇒ 14y - 252 = -x + 2 
⇒ x + 14y - 254 = 0 
And, the equation of the normal through (-2, -6) is given by,  
y - (-6)  = (-1/14) [x-(-2)] 
⇒ y + 6 = (-1/14) (x + 2) 
⇒ 14y + 84 = -x - 2 
⇒ x + 14y + 86 = 0 
Hence, the equations of the normals to the given curve (which are parallel to the given line ) are x + 14y - 254 = 0 and x + 14y + 86 = 0. 

22. Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at², 2at).
Solution
The equation of the given parabola is y² = 4ax. 
On differentiating y² = 4ax with respect to x, we have:
2y . (dy/dx) = 4a 
⇒ dy/dx = 2a/y 
∴ The slope of the tangent at (at² , 2at) is

Then, the equation of the tangent at (at² , 2at) is given by,  
y - 2at = 1/t (x - at² ) 
⇒ ty - 2at² = x - at² 
⇒ ty = x + at² 
Now, the slope of the normal at (at² , 2at) is given by,  
-1/[Slope of the tangent at (at² , 2at)] = -t 
Thus, the equation of the normal at (at² , 2at) is given as :  
y - 2at = -t(x - at² )
⇒ y - 2at = -tx + at³ 
⇒ y = -tx + 2at + at³ 

23. Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1. [Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.] 
Solution
The equations of the given curves are given as x = y² and xy = k. 
Putting x = y² in xy  = k, we get : 
y³ = k
⇒ y = k^1/3 
∴ x = k^2/3 
Thus, the point of intersection of the given curves is (k^2/3 , k^1/3 ). 
Differentiating x = y² with respect to x, we have :  
1 = 2y. (dy/dx)

⇒ dy/dx = 1/2y 

We know that two curves intersect at right angles if the tangents to the curves at the point of intersection i.e., at (k^2/3 , k^1/3 ) are perpendicular to each other. 
This implies that we should have the product of the tangents as -1. 
Thus, the given two curves cut at right angles if the product of the slopes of their respective tangents at (k^2/3 , k^1/3) is -1.

24. Find the equations of the tangent and normal to the hyperbola x²/a² - y²/b² at the point (x₀, y₀).
Solution
Differentiating x² /a² - y² /b²  = 1 with respect to x, we have :

25. Find the equation of the tangent to the curve y = √(3x-2) which is parallel to the line 4x − 2y + 5 = 0. 
Solution
The equation of the given curve is y = √(3x-2) . 
The slope of the tangent to the given curve at any point (x, y) is given by, 
dy/dx = 3/2√(3x - 2) 
The equation of the given line is 4x - 2y + 5 = 0. 
4x - 2y + 5 = 0 ⇒ y = 2x + 5/2 which is of the form y = mx + c) 
∴ Slope of the line = 2 
Now, the tangent to the given curve is parallel to the line 4x -2y - 5 = 0 if the slope of  the tangent is equal to the slope of the line. 
3/[2√(3x - 2)] = 2 
⇒ √(3x - 2) = 3/4 
⇒ 3x - 2 = 9/16 
⇒ 3x = 9/16 + 2 = 41/16 
⇒ x = 41/48 

⇒ 24y - 18 = 48x - 41 
⇒ 48x - 24y = 23 
Hence, the equation of the required tangent is  48x - 24y = 23 

26. The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is
(A) 3
(B) 1/3
(C) −3
(D) -1/3
Solution
The equation of the given curve is y = 2x² + 3sin x. 
Slope of the tangent to the given curve at x = 0 is given by, 
dy/dx]_{x = 0} = 4x + 3 cos x]_{x = 0} = 0 + 3 cos 0 = 3 
Hence, the slope of the normal to the given curve at x = 0 is 
-1/[Slope of the tangent at x = 0] = -1/3 . 
The correct answer is D. 

27. The line y = x + 1 is a tangent to the curve y² = 4x at the point
(A) (1, 2)
(B) (2, 1)
(C) (1, −2)
(D) (−1, 2)
Solution
The equation of the given curve is  y² = 4x . 
Differentiating with respect to x, we have : 
2y. dy/dx = 4 ⇒ dy/dx = 2/y 
Therefore, the slope of the tangent to the given curve at any point (x, y) is given by, 
dy/dx = 2/y 
Therefore, the slope of the tangent to the given curve at any point (x, y) is given by, 
dy/dx =2/y 
The given line is y = x + 1 (which is of the form y = mx + c)
∴ Slope of the line = 1 
The line y = x + 1 is a tangent to the given curve if the slope of the line is equal to the slope of the tangent. Also, the line must intersect the curve. 
Thus, we must have : 
2/y = 1 
⇒ y = 2 
Now,
y = x + 1
⇒ x = y- 1
⇒ x = 2 - 1 = 1 
Hence, the line y = x + 1 is a tangent to the given curve at the point (1, 2). 
The correct answer is A. 

Exercise 6.4

1. Using differentials, find the approximate value of the following up to 3 places of decimal
(i) √25.3
(ii) √49.5
(iii) √0.6
(iv) (0.009)^1/3
(v) (0.999)^1/10
(vi) (15)^1/4
(vii) (26)^1/3
(viii) (255)^1/4
(ix) (82)^1/4
(x) (401)^1/2
(xi) (0.0037)^1/2
(xii) (26.57)^1/3
(xiii) (81.5)^1/4
(xiv) (3.968)^3/2
(xv) (32.15)^1/5
Solution
(i) √25.3  
Consider y = √x . Let x = 25 and Δx = 0.3
Then,
Δy = √(x + Δx) - √x = √(25.3) - √(25) =  √25.3 - 5
⇒ √25.3 = Δy + 5 
Now, dy is approximately equal to Δy and is given by, 

(ii) √49.5
Consider y = √x . Let x = 49 and Δx = 0.5
Then,  
Δy = √(x + Δx) - √x = √(49.5) - √(49) =  √49.5 - 7
⇒ √49.5 = 7 + Δy 
Now, dy is approximately equal to Δy and is given by,  

(iii) √0.6
Consider y = √x.
Let x  = 1 and Δx = -0.4
Then,
Δy = √(x + Δx) - √x = √0.6 - 1 
⇒ √0.6 = 1 + Δy 
Now, dy is approximately equal to Δy and is given by, 

(iv) (0.009)^1/3   
Consider y = x^1/3.
Let x = 0.008 and ∆x = 0.001
Then,
Δy = (x + Δx)^1/3 - (x)^1/3 = (0.009)^1/3 - (0.008)^1/3 = (0.009)^1/3 - 0.2
⇒ (0.009)^1/3 = 0.2 + Δy 
Now, dy is approximately equal to Δy and is given by

(v) (0.999)^1/10 
Consider y = (x)^1/10.
Let x = 1 and Δx = -0.001.
Then, 
∆y = (x + ∆x)^1/10 - (x)^1/10 = (0.999)^1/10 - 1
⇒ (0.999)^1/10 = 1 + Δy 
Now, dy is approximately equal to ∆y and is given by, 

(vi) (15)^1/4 
Consider y = x^1/4.
Let x = 16 and Δx = -1.
Then,
∆y = (x + ∆x)^1/4 - (x)^1/4 = (15)^1/4 - (16)^1/4 = (15)^1/4 - 2
⇒ (15)^1/4 = 2 + Δy 
Now, dy is approximately equal to Δy and is given by, 

(vii) (26)^1/3
Consider y = (x)^1/3.
Let x = 27 and Δx = -1.
Then,
∆y = (x + ∆x)^1/3 - (x)^1/3 = (26)^1/3 - (27)^1/3 = (26)^1/3 - 3
⇒ (26)^1/3 = 3 + Δy 
Now, dy is approximately equal to ∆y and is given by,  

(viii) (255)^1/4
Consider y = (x)^1/4.
Let x = 256 and ∆x = -1 
Then,
∆y = (x + ∆x)^1/4 - (x)^1/4 = (255)^1/4 - (256)^1/4 = (255)^1/4 - 4
⇒ (255)^1/4 = 4 + Δy 
Now, dy is approximately equal to ∆y and is given by,  

(ix) (82)^1/4 
Consider y = x^1/4.
Let x = 81 and Δx = 1.
Then,
∆y = (x + ∆x)^1/4 - (x)^1/4 = (82)^1/4 - (81)^1/4 = (82)^1/4 - 3
⇒ (82)^1/4 =  Δy + 3

Now, dy is approximately equal to ∆y and is given by,  

(x) (401)^1/2
Consider y = x^1/2
Let x = 400 and ∆x = 1.
Then,
∆y = √(x + ∆x) - √x = √401 - √400 = √401 - 20 

⇒ √401 = 20 + ∆y
Now, dy is approximately equal to ∆y and is given by,  

(xi) (0.0037)^1/2 
Consider y = x^1/2
Let x = 0.0036 and ∆x = 0.0001. 
Then,
∆y = (x + Δx)^1/2 - (x)^1/2 = (0.0037)^1/2 - (0.0036)^1/2 = (0.0037)^1/2 - 0.06

⇒ (0.0037)^1/2 = 0.06 + Δy 
Now, dy is approximately equal to Δy and is given by,  

(xii) (26.57)^1/3 
Consider y = x^1/3.
Let x = 27 and Δx = -0.43 
Then,
∆y = (x + Δx)^1/3 - (x)^1/3 = (26.57)^1/3 - (27)^1/3 = (26.57)^1/3 - 3
⇒ (26.57)^1/3 = 3 + Δy 
Now, dy is approximately equal to Δy and is given by,  

(xiii) (81.5)^1/4 
Consider y = x^1/4.
Let x = 81 and Δx = 0.5 
Then,
∆y = (x + Δx)^1/4 - (x)^1/4 = (81.5)^1/4 - (81)^1/4 = (81.5)^1/4 - 3
⇒ (81.5)^1/4 = 3 + Δy 
Now, dy is approximately equal to Δy  and is given by, 

(xiv) (3.968)^3/2 
Consider y = x^3/2,
Let x = 4 and Δx = - 0.032.
Then,
∆y = (x + Δx)^3/2 - (x)^3/2 = (3.968)^3/2  - (4)^3/2  = (3.968)^3/2  - 8
⇒ (3.968)^3/2  = 8 + Δy 
Now, dy is approximately equal to ∆y and is given by,  

= -0.096

Hence, the approximate value of (3.968)^3/2 is 8 + (-0.096) = 7.904

(xv) (32.15)^1/5 
Consider y = x^1/5.
Let x = 32 and Δx = 0.15 
Then,
∆y = (x + Δx)^1/5 - (x)^1/5 = (32.15)^1/5 - (32)^1/5 = (32.15)^1/5 - 2
⇒ (32.15)^1/5 = 2 + Δy 
Now, dy is approximately equal to Δy and is given by,

2. Find the approximate value of f (2.01), where f (x) = 4x² + 5x + 2.
Solution
Let x = 2 and ∆x = 0.01 Then, we have : 
f(2.01) = f(x + ∆x) = 4(x + ∆x)² + 5(x + Δx) + 2 
Now, Δy = f(x + Δx) - f(x) 
∴ f(x + Δx) = f(x) + Δx
≈f(x) + f '(x).Δx     (as dx = Δx)
⇒ f(2.01) ≈ (4x² + 5x + 2) + (8x + 5)Δx
= [4(2)² + 5(2) + 2] + [8(2) + 5](0.01)  [as x = 2, Δx = 0.01]
= (16 + 10 + 2) + (16+5)(0.01)
= 28 + (21)(0.01) 
= 28 + 0.21 
= 28.21 
Hence, the approximate value of f(2.01) is 28.21. 

3. Find the approximate value of f (5.001), where f (x) = x³ − 7x² + 15.
Solution
Let x = 5 and Δx = 0.001. Then, we have :
f(5.001) = f(x + ∆x) = (x + ∆x)³ - 7(x + Δx)² + 15
Now, Δy = f(x + Δx) - f(x) 
∴ f(x + Δx) = f(x) + Δy
≈ f(x) + f '(x).Δx     (as dx = Δx)
⇒ f(5.001) ≈ (x³ - 7x²  + 15) + (3x² - 14x)Δx
= [(5)³ + 7(5)²  + 15] + [3(5)²  - 14(5)](0.001)  [x = 5, Δx = 0.001]
= (125 - 175 + 15) + (75 - 70) (0.001)
= -35 + (5)(0.001)
= -35 + 0.005
= -34.995 
Hence, the approximate value of f(5.001) is -34.995.

4. Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%.
Solution
The volume of a cube (V) of side x is given by V = x³ . 
∴ (dv/dx) Δx 
= (3x² )Δx 
= (3x² )(0.01x)  [as 1% of x is 0.01x] 
= 0.03x³ 
Hence, the approximate change in the volume of the cube is 0.03x³ m³ . 

5. Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1% . 
Solution
The surface area of a cube (S) of side x is given by S = 6x² . 
∴ dS = (dS/dx) Δx
= (12x) Δx
= (12x) (-0.01x)  [as 1% of x is 0.01 x]
= -0.12x² 
Hence, the approximate change in the surface area of the cube is -0.12x² m². 

6. If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume.
Solution
Let r be the radius of the sphere and Δ r be the error in measuring the radius. 
Then, 
r = 7m and Δr = 0.02 m 
Now; the volume V of the sphere is given by, 

= (4πr² ) ∆r 
= 4π(7)² (0.02)m³ = 3.92π m³ 
Hence, the approximate error in calculating the volume is 3.92 πm³ . 

7. If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume.
Solution
Let r be the radius of the sphere and Δr be the error in measuring the radius. 
Then, 
r = 9 m and Δr = 0.03 m 
Now, the surface area of the sphere (S) is given by, 
S = 4πr² 
∴ dS/dr = 8πr 
∴ dS = (dS/dr) Δr
= (8πr)Δr
= 8π(9)(0.03) m² 
= 2.16π m² 
Hence, the approximate error in calculating the surface area is 2.16π m² . 

8. If f (x) = 3x² + 15x + 5, then the approximate value of f (3.02) is
A. 47.66
B. 57.66
C. 67.66
D. 77.66
Solution
Let x = 3 and Δx = 0.02. Then, we have : 
f(3.02) = f(x + Δx) = 3(x + Δx)² + 15(x + ∆x) + 5 
Now, ∆y = f(x + ∆x) - f(x) 
⇒ f(x + ∆x) = f(x) + ∆y 
≈ f(x) + f '(x) ∆x      (As dx = ∆x) 
⇒ f(3.02) ≈ (3x² + 15x + 5) + (6x + 15)∆x 
= [3(3)² + 15(3) + 5] + [6(3) + 15](0.02)    [As x = 3, Δx = 0.02] 
= (27 + 45 + 5) + (18 + 15)(0.02)
= 77+ 33× 0.02
= 77 + 0.66
= 77.66
Hence, the approximate value of f(3.02) is 77.66
The correct answer is D.

9. The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
A. 0.06 x³ m³ 
B. 0.6 x³ m³
C. 0.09 x³ m³
D. 0.9 x³ m³
Solution

Therefore, option (C) is correct.

Exercise 6.5 

1. Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = (2x − 1)² + 3 
(ii) f(x) = 9x² + 12x + 2 
(iii) f(x) = -(x -1)² + 10 
(iv) g(x) = x³ + 1 

Solution

(i) The given function is f(x) = (2x - 1)² + 3.
It can be observed that (2x - 1)² ≥ 0 for every x ∊ R. 
Therefore, f(x) = (2x - 1)² + 3 ≥ 3 for every x ∊ R. 
The minimum value of f is attained when 2x - 1 = 0 
2x - 1 = 0
⇒ x = 1/2 
∴ Minimum value of f = f(1/2) = [2. (1/2) - 1]² + 3 = 3 
Hence, function f does not have a maximum value. 

(ii) The given function is f(x) = 9x² + 12x + 2 = (3x + 2)² - 2.
It can be observed that (3x + 2)² ≥ 0 for every x ∈ R. 
Therefore, f(x)  = (3x +2)² - 2 ≥ -2 for every x ∈ R. 
The minimum value of f is attained when 3x + 2 = 0 
3x + 2 = 0
⇒ x = -2/3 
∴ Minimum value of f =

Hence, function f does not have a maximum value. 

(iii) The given function is f(x) = -(x - 1)² + 10. 
It can be observed that (x - 1)² ≥ 0 for every x ∈ R.
Therefore, f(x) = -(x - 1)² + 10 ≤ 10 for every x ∈ R. 
The maximum value of f is attained when (x - 1) = 0.
(x - 1) = 0
⇒ x = 1 
∴ Maximum value of f = f(1) = -(1 - 1)² + 10 = 10 
Hence, function f does not have a minimum value. 

(iv) The given function is g(x) x³ + 1. 
Hence, function g neither has a maximum value nor a minimum value.

 

2. Find the maximum and minimum values, if any, of the functions given by
(i) f(x) = |x + 2| − 1
(ii) g(x) = -|x + 1| + 3
(iii) h(x) = sin(2x) + 5
(iv) f(x)  = |sin 4x + 3|
(v) h(x) = x+  4, x ∊ (-1, 1)

Solution

(i)  f(x) = |x + 2| − 1
We know that |x + 2| ≥ 0 for every x ∊ R.
Therefore, f(x) = |x + 2| - 1 ≥ - 1 for every x ∊ R. 
The minimum value of f is attained when |x + 2| = 0 .
|x + 2| = 0 
⇒ x = -2 
∴ Minimum value of f = f(-2) = |-2 + 2| - 1 = -1
Hence, function f does not have a maximum value. 

(ii) g(x) = -|x + 1| + 3 
We know that -|x + 1|≤ 0 for every x ∊  R 
Therefore, g(x) = -|x + 1| + 3≤ 3 for every x ∊ R. 
The maximum value of g is attained when |x + 1| = 0.
|x + 1| = 0 
⇒ x = -1 
∴ Maximum value of g = g(-1) = -|-1 + 1| + 3 = 3 
Hence, function g does not have a minimum value. 

(iii) h(x) = sin 2x + 5 
We know that - 1 ≤ sin 2x ≤ 1.
⇒ -1 + 5 ≤ sin 2x + 5 ≤ 1 + 5
⇒ 4 ≤ sin 2x + 5 ≤ 6 
Hence, the maximum and minimum values of h are 6 and 4 respectively. 

(iv) f(x) = |sin 4x + 3| 
We know that -1 ≤ sin 4x ≤ 1. 
⇒ 2 ≤sin 4x + 3 ≤4 
⇒ 2 ≤ |sin 4x + 3| ≤ 4 
Hence, the maximum and minimum values of f are 4 and  2 respectively. 

(v) h(x) = x + 1, x ∊ (-1, 1) 
Here, if a point x₀ is closest to -1, then we find x₀/2 + 1 < x₀ + 1 for all x₀ ∈ (-1, 1).
Also, if x₁ is closest to 1, then x₁ + 1 < (x₁ + 1)/2 + 1 for all x₁ ∊ (-1, 1). 
Hence, function h(x) has neither maximum nor minimum value in (-1, 1). 

 

3. Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be
(i) f(x) = x² 
(ii) g(x) = x³ - 3x 
(iii) h(x) = sin x + cos x, 0 < x < π/2 
(iv) f(x) = sin x - cos x, 0 < x < 2π
(v) f(x) = x³ - 6x²  + 9x + 15 
(vi) g(x) = x/2 + 2/x , x > 0 
(vii) g(x) = 1/(x² + 2) 
(viii) f(x) = x√(1- x) , 0 < x < 1 

Solution

(i) f(x) = x² 
∴ f '(x) = 2x 
Now, 
f '(x) = 0
⇒ x = 0 
Thus, x = 0 is the only critical point which could possible be the point of local maxima or local minima of f. 
We have f''(0) = 2, which is positive. 
Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is f(0) = 0. 

(ii) g(x) = x³ - 3x 
∴ g '(x) = 3x² - 3 
Now, 
g''(x) = 0
⇒ 3x² = 3
⇒ x = ± 1 
g''(x) = 6x 
g''(1) = 6 > 0 
g''(-1) = -6 < 0 
By second derivative test, x = 1 is a point of local minima and local minimum value of g at x = 1 is g(1) = 1³ - 3 = 1 - 3= -2. However, 
x = -1 is a point of local maxima and local maximum value of g at 
x = -1 is g(-1) = (-1)³ - 3(-1) = -1 + 3 = 2.

(iii) h(x) = sin x + cos x, 0 < x < π/2 

Therefore, by second derivative test, x = π/4 is a point of local maxima and the local maximum value of h at x = π/4 is h(π/4) = sin . (π/4) + cos.(π/4) = 1/√2 + 1/√2 = √2. 

(iv) f(x) = sin x - cos x, 0 < x < 2π
∴ f '(x) = cos x + sin x 
f '(x) = 0
⇒ cos x = -sin x
⇒ tan x = -1
⇒ x = 3π/4, 7π/4 ∊ (0, 2π)
f ''(x) = - sin x + cos x

(v) f(x) = x³ - 6x² + 9x + 15 
∴ f '(x) = 3x² - 12x + 9 
f 'x = 0 ⇒ 3(x² - 4x + 3) = 0 
⇒ 3(x - 1)(x - 3) = 0 
⇒ x = 1, 3 
Now, f ''(x) = 6x - 12 = 6(x - 2) 
f ''(1) = 6(1 - 2) = -6 < 0 
f ''(3) = 6(3 - 2) = 6 > 0 
Therefore, by second derivative test, x = 1 is a point of local maxima and the local maximum value of f at x = 1 is f(1) = 1 - 6 + 9 + 15 = 19. However, x = 3 is a point of local minima and the local minimum value of f at x = 3 is f(3) = 27 - 54 + 27 + 15 = 15. 

(vi) g(x)  = x/2 + 2/x, x > 0 
∴ g'(x) = 1/2 - 2/x² 
Now,
g'(x) = 0 gives 2/x² = 1/2
⇒ x² = 4
⇒ x = ± 2 
Since x > 0, we take x = 2. 
Now, g''(x) = 4/x³ 
g''(2) = 4/2³ = 1/2 > 0 
Therefore, by second derivative test, x = 2 is a point of local minima and the local minimum value of g at x = 2 is g(2) = 2/2 + 2/2 = 1 + 1 = 2.

(vii) g(x) = 1/(x² + 2)

Now, for values close to x = 0 and to the left of 0, g'(x) > 0. Also, for values close to x = 0 and to the right of 0, g'(x) < 0. 
Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of g(0) is 1/(0+2) = 1/2.

Therefore, by second derivative test, x = 2/3 is a point of local maxima and the local maximum value of f at x = 2/3 is 

4. Prove that the following functions do not have maxima or minima : 
(i) f(x) = e^x 
(ii) g(x) = log x 
(iii) h(x) = x³ + x² + x + 1 
Solution 
(i)  We have, 
f(x) = e^x 
∴ f '(x) = e^x 
Now, if f '(x) = 0, then e^x = 0 . But, the exponential function can never assume 0 for any value of x. 
Therefore, there does not exist c ∊ R  such that f '(c) = 0 . 
Hence, function f does not have maxima or minima. 

(ii) We have, 
g(x) = log x
∴ g'(x) = 1/x 
Since log x is defined for a positive number x, g'(x)  > 0 for any x. 
Therefore, there does not exist c ∊ R such that g'(c) = 0 
Hence, function g does not have maxima or minima. 

(iii) We have, 
h(x) = x³ + x² + x+  1 
∴ h'(x) = 3x² + 2x + 1 
Now, 
h(x) = 0 ⇒ 3x² + 2x + 1 = 0

Therefore, there does not exist c ∈ R such that h'(c) = 0 . 
Hence, function h does not have maxima or minima. 

5. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals : 
(i) f(x) = x³ , x ∊ [-2, 2]
(ii) f(x) = sin x + cos x , x ∊ [0, π]
(iii) f(x) = 4x - (1/2)x² , x ∊ [-2, 9/2] 
(iv) f(x) = (x - 1)² + 3, x ∊ [-3, 1] 
Solution
(i) The given function is f(x) = x³ . 
∴ f '(x) = 3x² 
Now, 
f '(x) = 0 ⇒ x = 0 
Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [-2, 2]. 
f(0) = 0 
f(-2) = (-2)³ = -8
f(2) = (2)³ = 8 
Hence, we can conclude that the absolute maximum value of f on [-2, 2] is 8 occurring at x = 2 . 
Also, the absolute minimum value of f on [-2, 2] is -8 occurring at x = -2.

(ii) The given function is f(x) = sin x + cos x 
∴ f '(x) = cos x - sin x 
Now, 
f '(x) = 0 ⇒ sin x = cos x ⇒ tan x = 1 ⇒ x = π/4
Then, we evaluate the value of f at critical point x = π/4 and at the end points of the interval (0, π). 
f(π/4) = sin(π/4)+ cos(π/4) = 1/√2 + 1/√2 = 2/√2 = √2
f(0) = sin 0 + cos 0 = 0 + 1 = 1 
f(π) = sin π+ cosπ = 0 - 1 = - 1 
Hence, we can conclude that the absolute maximum value of f on (0, π) is √2 occurring at x = π/4 and the absolute minimum value of f on (0, π) is -1 occurring at x = π. 

(iii) The given function if f(x) = 4x - (1/2)x² . 
∴ f '(x) = 4 - (1/2) (2x) = 4 - x 
Now, 
f '(x) = 0 ⇒ x = 4 
Then; we evaluate the value of f at critical point x = 4 and at the end points of the interval [-2, 9/2 ]. 
f(4) = 16 - (1/2)(16) = 16 - 8 = 8 
f(-2) = -8 - (1/2) (4) = -8-2= -10
f(9/2) = 4(9/2) - (1/2)(9/2)² = 18 - 81/8 = 18 - 10.125 = 7.875 
Hence, we can conclude that the absolute maximum value of f on [-2, 9/2] is  8 occurring at x = 4 and the absolute minimum value of f on [-2, 9/2] is -10 occurring at x = -2. 

(iv) The given function is f(x) = (x-1)² + 3. 
∴ f '(x)  = 2(x - 1) 
Now, 
f '(x) = 0 ⇒ 2(x - 1) = 0 ⇒ x = 1 
Then, we evaluate the value of f at critical point x = 1 and at the end points of the interval (-3, 1). 
f(1)  = (1 - 1)² + 3 = 0 + 3 = 3 
f(-3) = (-3 - 1)² + 3 = 16 + 3 = 19 
Hence, we can conclude that the absolute maximum value of f on (-3, 1) is 19 occurring at x = -3 and the minimum value of f on (-3, 1) is 3 occurring at x = 1 . 

6. Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 − 72x− 18x². 
Solution
The profit function is given as p(x) = 41 - 72x - 18² . 
∴ p'(x) = -72 - 36x 
p''(x) = -36 
Now, 
p'(x) = 0 ⇒ x = -72/36 = -2 
Also, 
p"(-2) = -36 < 0 
By second derivative test, x = -2 is the point of local maxima of p. 
∴ Maximum profit = p(-2)
= 41 - 72(-2) - 18(-2)²  
= 41 + 144 - 72 = 113
Hence, the maximum profit that the company can make is 113 units. 

7. Find both the maximum value and the minimum value of 3x⁴ − 8x³ + 12x² − 48x + 25 on the interval [0, 3]. 
Solution
Let f(x) = 3x⁴ − 8x³ + 12x² − 48x + 25 
∴ f '(x) = 12x³ - 24x² + 24x - 48 
= 12(x³ - 2x² + 2x - 4)
= 12[x² (x - 2) + 2(x - 2)] 
= 12(x -2)(x² + 2) 
Now, f '(x) = 0 gives x = 2 or x² + 2 = 0 for which there are no real roots. 
Therefore, we consider only x = 2 ∊[0, 3].
Now, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3]. 
f(2) = 3(16) - 8(8) + 12(4) - 48(2) + 25 
= 48 - 64 + 48 - 96 + 25 
= - 39 
f(0) = 3(0) - 8(0) + 12(0) - 48(0) + 25
= 25 
f(3) = 3(81) - 8(27) + 12(9) - 48(3) + 25 
=243 - 216 + 108 - 144 + 25 = 16
 Hence, we can conclude that the absolute maximum value of f on (0, 3) is 25 occurring at x = 0 and the absolute minimum value of f at (0, 3) is  -39 occurring at x = 2. 

8. At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?
Solution
Let f(x) = sin 2x . 
∴ f'(x) = 2 cos 2x 
Now, 
f '(x) = 0 ⇒ cos 2x = 0 

Hence, we can conclude that the absolute maximum value of f on [0, 2π) is occurring at x = π/4 and x = 5π/4 .

9. What is the maximum value of the function sin x + cos x?
Solution
Let f(x) = sin x + cos x. 
∴ f '(x) = cos x - sin x 
f '(x) = 0 ⇒ sin x = cos x ⇒ tan x = 1 ⇒ x = π/4, 5π/4 ....., 
f "(x) = - sin x - cos x = -(sin x + cos x) 
Now, f "(x) will be negative when (sin x + cos x) is positive i.e., when sin x and cos x are both positive. Also, we know that sin x and cos x both are positive in the first quadrant. 

Then, f ''(x) will be negative when x ∊ (0, π/2) . 
Thus, we consider x = π/4 . 

∴ By second derivative test, f will be the maximum at x = π/4 and the maximum value of f is (π/4) = sin(π/4) + cos(π/4)  = 1/√2 × 1/√2 = 2/√2 = √2

10. Find the maximum value of 2x³ − 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [−3, −1].
Solution
Let f(x) = 2x³ − 24x + 107
∴ f '(x) = 6x² - 24 = 6(x² - 4)
Now, 
f '(x)  = 0 ⇒ 6(x² - 4) = 0 ⇒ x² = 4 ⇒ x = ± 2 
We first consider the interval (1, 3). 
Then, we evaluate the value of f at the critical point x = 2 ∊ (1, 3) and at the end points of the interval (1, 3). 
f(2) = 2(8) - 24(2) + 107 = 16 - 48 + 107 = 75 
f(1) = 2(1) - 24(1) + 107 = 2 - 24 + 107 = 85 
f(3) = 2(27) - 24(3) + 107 = 54 - 72 + 107 = 89 
Hence, the absolute maximum value of f(x) in the interval (1, 3) is 89 occurring at x = 3. 
Next we consider the interval (-3, -1). 
Evaluate the value of f at the critical point x = -2∊ (-3, -1) and at the end points of the interval (1, 3). 
f(-3) = 2(-27) - 24(-3) + 107 = -54 + 72 + 107 = 125 
f(-1) = 2(-1) - 24(-1) + 107 = -2 + 24 + 107 = 129 
f(-2) = 2(-8) - 24(-2) +107 = -16 + 48 + 107 = 139 
Hence, the absolute maximum value of f(x) in the interval (-3, -1) is 139 occurring at x = -2

11. It is given that at x = 1, the function x⁴− 62x² + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.
Solution
Let f(x) = x⁴ - 62x² + ax + 9 
∴ f '(x) = 4x³ - 124 x + a 
It is given that function f attains its maximum value on the interval (0, 2) at x = 1. 
∴ f '(1) = 0 
⇒ 4 - 124 + a = 0 
⇒ a = 120 
Hence, the value of a is 120. 

12. Find the maximum and minimum values of x + sin 2x on [0, 2π].
Solution
Let f(x) = x + sin 2x. 
∴ f '(x) = 1 + 2 cos 2x 
Now, f '(x) = 0 ⇒ cos 2x = -1/2 = - cos (π/3) = cos (π - π/3) = cos (2π/3) 
2x = 2nπ ± 2π/3 , n ∊ z 
⇒ x = nπ ± π/3, n ∊ z
⇒ x = π/3, 2π/3, 4π/3, 5π/3 ∈ (0, 2π) 
Then, we evaluate the value of f at critical points x = π/3, 2π/3, 4π/3, 5π/3 and at the end points of the interval (0, 2π). 

Hence, we can conclude that the absolute maximum value of f(x) in the interval (0, 2π) is 2π occurring at x = 2π and the absolute minimum value of f(x) in the interval (0, 2π) is 0 occurring at x = 0. 

13. Find two numbers whose sum is 24 and whose product is as large as possible.
Solution
Let one number be x. Then, the other number is (24 - x). 
Let p(x) denote the product of the two numbers. Thus, we have : 
p(x) = x(24 - x) = 24x - x² 
∴ p'(x) = 24 - 2x 
p"(x) = -2 
Now,
p'(x) = 0 ⇒ x = 12 
Also, p"(12) = -2 < 0 
∴ By second derivative test, x = 12 is the point of local maxima of P. Hence, the product of the numbers is the maximum when the numbers are  12 and 24 - 12 = 12. 

14. Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.
Solution
The two numbers are x and y such that x + y = 60 . 
⇒ y = 60 - x 
Let f(x) = xy³ . 
⇒ f(x) = x (60 - x)³ 
∴ f '(x) = (60 - x)³ - 3x(60 - x)² 
= (60 - x)² [60 - x -3x]
=(60 - x)² (60 - 4x) 
And, f "(x) = -2(60 - x)(60 - 4x) - 4(60 - x)² 
= -2(60 - x)[60 - 4x + 2(60 - x)]
= -2(60 - x)(180 - 6x)
= - 12(60 - x)(30 - x) 
Now, f '(x) = 0 ⇒ x = 60 or x  = 15 
When x = 60, f "(x) = 0. 
When x = 15, f "(x) = -12(60 - 15)(30 - 15) = -12 × 45 × 15 < 0. 
∴ By second derivative test, x = 15 is a point of local maxima of f. Thus, function xy³ is maximum when x = 15 and y = 60 - 15 = 45. 
Hence, the required numbers are  15 and 45. 

15. Find two positive numbers x and y such that their sum is 35 and the product x²y⁵ is a maximum
Solution
Let one number be x. Then, the other number is y = (35 - x). 
Let p(x) =  x²y⁵ . Then, we have : 

x = 0, x = 35 , x = 10 
When x = 35, P '(x) = p(x) = 0 and y = 35 - 35 = 0. This will make the product x²y⁵ equal to 0. 
When x = 0, y = 35 - 0 = 35 and the product x²y² will be 0. 
∴ x = 0 and x = 35 cannot be the possible values of x. 
When x = 10, we have: 
P"(x) = 7(35 - 10)³ (6 ×100 - 120 × 10 + 350) 
= 7(25)³ (-250) < 0 
∴ By second derivative test, P(x) will be the maximum when x = 10 and y =35 - 10 = 25. 
Hence, the required numbers are 10 and  25.

16. Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Solution
Let one number be x. Then, the other number is (16 - x). 
Let the sum of the cubes of these numbers be denoted by S(x) . Then, 

Now, S"(8) = 6(8) + 6(16 - 8) = 48 + 48 = 96 > 0 
∴ By second derivative test, x = 8 is the point of local minima of S. 
Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and  16 - 8 = 8. 

17. A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
Solution
Let the side of the square to be cut off be x cm. The, the length and the breadth of the box will be (18 -2x) cm each and the height of the box is x cm. 
Therefore, the volume V(x)  of the box is given by, 
V(x) = x(18 - 2x)² 
∴ V'(x) = (18 - 2x)² - 4x(18 - 2x)
= (18 - 2x)[18 - 2x - 4x]
= (18 - 2x)(18 - 6x)
= 6 × 2(9 - x)(3 -x) 
= 12(9 - x)(3 - x) 
And, V"(x) = 12[-(9 - x) - (3 - x)] 
= -12(9 - x + 3 - x) 
= -12(12 - 2x) 
= -24(6x - x) 
Now, V'(x) = 0 ⇒ x = 9 or x = 3 
If x = 9, then the length and the breadth will become 0.  
∴ x ≠ 9. 
⇒ x = 3 . 
Now, V"(3) = -24(6 - 3) = -72 < 0 
∴ By second derivative test, x = 3 is the point of maxima of V. 
Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible. 

18. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square 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 the maximum possible?
Solution
Let the side of the square to be cut off be x cm. Then, the height of the box is x, the length is 45 - 2x, and the breadth is 24 - 2x. 
Therefore, the volume V(x) of the box is given by,  
V(x) = x(45 - 2x)(24 - 2x) 
= x(1080 - 90x  48x + 4x² )
= 4x³ - 138x² + 1080x 
∴ V'(x) = 12x² - 276x + 1080 
= 12(x² - 23x + 90)
= 12(x - 18)(x - 5)
V"(x) = 24x - 276 = 12(2x - 23) 
Now, V'(x) = 0 ⇒ x = 18 and x = 5 
It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet. 
Thus, x cannot be equal to 18. 
∴ x = 5 
Now, V"(5) = 12(10 - 23) = 12(-13) = -156 < 0 
∴ By second derivative test, x = 5 is the point of maxima. 
Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm. 

19. Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Solution
Let a rectangle of length l and breadth b be inscribed in the given circle of radius a. 
Then, the diagonal passes through the centre and is of length 2a cm. 

∴ By the second derivative test, when l = √2a, then the area of the rectangle is the maximum. 
Since l  = b =  √2a, the rectangle is a square. 
Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area. 

20. Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.
Solution
Let r and h be the radius and height of the cylinder respectively. 
Then, the surface area (S) of the cylinder is given by, 
S  = 2πr² + 2πrh 

Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter. 

21. Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?
Solution
Let r and h be the radius and height of the cylinder respectively. 
Then, Volume (V) of the cylinder is given by, 
V = πr² h = 100  (given) 
∴ h = 100/πr² 
Surface area (S) of the cylinder is given by, 

Hence, the required dimensions of the can which has the minimum surface area is given by radius = (50/π)^1/3 cm and height = 2(50/π)^1/3 cm. 

22. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Solution
Let a piece of length l be cut from the given wire to make a square. 
Then, the other piece of wire to be made into a circle is of length (28 - l) m. 
Now, side of square = l/4. 
Let r be the radius of the circle. The, 2πr = 28 - l ⇒ r = 1/2π (28 - l). 
The combined areas of the square and the circle (A) is given by, 
A = (side of the square)²  + πr² 

∴ By second derivative test, the area (A) is the minimum when l = 112/(π + 4) . 
Hence, the combined area is the minimum when the length of the wire in making the square is  112/(π + 4) cm while the length of the wire in making the circle is 28 - 112/(π + 4)  = 28π/(π + 4) cm. 

23. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.
Solution
Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R. 

⇒ 4R² (R² - r² )= (3r² - 2R² )² 
⇒ 4R⁴ - 4R² r² = 9r⁴ + 4R⁴ - 12r² R² 
⇒ 9r⁴ = 8R² r² 
⇒ r² = (8/9)R² 
When r² = (8/9)R² , then d² V/dr²  < 0. 
∴ By second derivative test, the volume of the cone is the maximum when r² = (8/9)R² . 

Hence, the volume of the largest cone that can be inscribed in the sphere is 8/27 
the volume of the sphere. 

24. Show that the right circular cone of least curved surface and given volume has an altitude equal to 2 times the radius of the base.
Solution
Let r and h be the radius and the height (altitude) of the cone respectively. 
Then, the volume (V) of the cone is given as : 
V = (1/3)πr² h ⇒ h = 3V/πr² 
The surface area (S) of the cone is given by, 
S = πrl (where l is the slant height ) 

Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to √2 times the radius of the base . 

25. Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^-1 √2. 
Solution
Let θ be the semi - vertical angle of the cone. 
It is clear that θ ∊ [0, π/2]. 
Let r, h, and l be the radius, height, and the slant height of the cone respectively. 
The slant height of the cone is given as constant. 

Now, dV/dθ = 0 
⇒ sin³ θ = 2 sinθ cos² θ
⇒ tan² θ = 2 
⇒ tan θ = √2 
⇒ θ = tan^-1 √2 
Now, when θ = tan^-1 √2, then tan² θ = 2 or sin² θ = 2 cos² θ . 
Then, we have : 

∴ By second derivative test, the volume (V) is the maximum when θ = tan^-1 √2. 
Hence, for a given slant height, the semi - vertical angle of the cone of the maximum volume is tan^-1 √2. 

26. Show that semi-vertical angle of right circular cone of given surface area and maximum volume is Sin^-1 (1/3) .
Solution
Total Surface Area of the cone  = S = πr(l + r) ...(1) 
[Where r and l are radius and slant height of the cone respectively]
Volume of cone = V = (1/3)πr² h
V² = (1/9)π² r⁴ h² 
V² = (1/9)π² r⁴ (l² - r² )
Using (1), we have 
V² = (1/9)π² r⁴ [(s/πr - r)² - r² ] 
Then by solving further we get - 
V² = (1/9) [S(S r² - 2πr⁴ )]
P = V² 
differentiating P with respect to r we gwt 
dp/dr = (1/9)[S (2Sr - 8πr³ )] 
equate  dp/dr = 0 
S = 4πr² 
Differentiating again with respect to r we find d² P/dr² < 0 therefore P is  maximum when S = 4πr² 
Differentiating again with respect to r we find that d² p/dr²  < 0 therefore P is maximum when S = 4πr² 
Again therefore, V is maximum when S = 4πr² 
πr(l + r) = 4πr² 
Thus l = 3r 
sin θ = r/l = 1/3 
θ = sin^-1 (1/3) 

27. The point on the curve x² = 2y which is nearest to the point (0, 5) is
(A) (2√2,4)
(B) (2√2,0)
(C) (0, 0)
(D) (2, 2)
Solution
The given curve is x² = 2y. 
For each value of x, the position of the point will be (x, x² /2). 
The distance d(x) between the points (x, x² /2) and (0, 5) is given by, 

When, x = 0 , then d" (x) = 36(-8)/6³ < 0. 
When, x = ± 2√2, d"(x) > 0. 
∴ By second derivative test, d(x) is the minimum at x = ± 2√2. 
When x = ± 2√2, y = (2√2)² /2 = 4. 
Hence, the point on the curve x² = 2y which is nearest to the point (0, 5) is (  ± 2√2, 4). 
The correct answer is A. 

28. For all real values of x, the minimum value of (1 - x + x² )/(1 + x + x² ) is 
(A) 0 
(B) 1 
(C) 3 
(D)1/3 
Solution

∴ By second derivative test, f is the minimum at x = 1 and the minimum value is given by f(1) = (1 - 1 + 1)/(1 + 1 + 1) = 1/3. 
The correct answer is D. 

29. The maximum value of [x(x - 1) + 1]^1/3 , 0 ≤ x ≤ 1 is 
(A) (1/3)^1/3 
(B) 1/2 
(C) 1 
(D) 0 
Solution
Let f(x) = [x(x - 1) + 1]^1/3