NCERT Class 12 Mathematics Chapter 6 Applications of Derivatives

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NCERT Class 12 Mathematics Chapter 6 Applications of Derivatives

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Applications of Derivatives

Chapter – 6

Exercise 6.1

1. Find the rate of change of the area of a circle with respect to its radius r when

(a) r = 3 cm. 

(b) r = 4 cm.

Ans: We know that the area of a circle, A = πr²

Therefore, the rate of change of the area with respect to its radius is given by 

= 2πr 

(a) When r = 3cm

Then, 

dA/dr = 2π(3) 

= 6π

Thus, the area is changing at the rate of 6π.

(b) When r = 4 cm

Then, 

dA/dr = 2π(4) 

= 8π

Thus, the area is changing at the rate of 8π.

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?

Ans: Let the side length, volume and surface area respectively be equal to x, V and S. 

Hence, V = x³ and S = 6x²

We have,

Therefore, 

Now, 

So, when x = 12cm 

Then,

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.

Ans: We know that A = πr²

Now, 

We have,

dr/dt = 3 cm/s

Hence, 

= 6πr

So when r = 10 cm

Then,

dA/dt = 6π(10)

= 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? 

Ans: Let the length and the volume of the cube respectively be x and V.

Hence, V = x³ 

Now,

We have,

dx/dt = 3 cm / s

Hence, 

dV/dt = 3x² (3)

= 9x²

So, when x = 10cm

Then, 

= 900 cm³ / s

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?

Ans: We know that A = πr²

Now, 

We have, 

dr/dt = 5 cm / s 

Hence, 

dA/dt = 2πr (5)

= 10πr 

So, when r = 8 cm

Then, 

dA/dt = 10π(8)

= 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? 

Ans: We know that C = 2πr

Now, 

We have, 

dr/dt = 0.7π cm / s 

Hence,

dC/dt = 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.

Ans: It is given that

(a) The perimeter of a rectangle is given by 

P = 2(x + y)

Therefore,

= 2 (- 5 + 4)

= – 2 cm / minute

(b) The area of a rectangle is given by A = xy 

Therefore,

When x = 8 cm and y = 6 cm

Then,

dA/dt = (- 5 × 6 + 4 × 8) cm² / minute

= 2 cm² / minute

8. A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Ans: We know that 

Hence,

We have,

dV/dt = 900 cm² / s

Therefore,

When radius, r =15 cm

Then,

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.

Ans: We know that 

Therefore,

When radius, r = 10 cm 

Then, 

dV/dt = 4π (10)² 

= 400π 

Thus, the volume of the balloon is increasing at the rate of 400π cm³ / s.

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?

Ans: Let the height of the wall at which the ladder is touching it be ym and the distance of its foot from the wall on the ground be xm.

Hence,

x² + y² = 5² 

⇒ y² = 25 – x²

Therefore, 

We have, 

dx/dt = 2 cm / s 

Thus, 

When x = 4 cm 

Then,

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.

Ans: The equation of the curve is 6y = x³ + 2

Differentiating with respect to time, we have

According to the question, di Henceo, 

When x = 4 

Then, 

When x = – 4

Then, 

Thus, the points on the curve are (4,11) and

12. The radius of an air bubble is increasing at the rate of 1/2 cm / s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Ans: Assuming that the air bubble is a sphere,

Therefore,

We have,

When r = 1 cm

Then, 

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.

Ans: We know that

It is given that diameter, d = 3/2 (2x + 1)

Hence, 

Therefore, 

Thus,

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?

Ans: We know that 

It is given that,

h = 1/6r

Hence, r = 6h 

Therefore,

Thus,

We have,

dV/dt = 12 cm² / s 

When h = 4 cm

Then,

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.

Ans: Marginal cost (MC) is the rate of change of the total cost with respect to the output.

Therefore,

MC = dC/dx = 0.007 (3x²) – 0.003(2x) + 15 

= 0.021x² – 0.006x + 15 

When x = 17 

Then,

MC = 0.021(17)² – 0.006(17) + 15 

= 0.021(289) – 0.006(17) + 15 

= 6.069 – 0.102 + 15 

= 20.967 

So, when 17 units are produced, the marginal cost is ₹ 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. 

Ans: Marginal revenue (MR) is the rate of change of the total revenue with respect to the number of units sold. 

Therefore,

MR = dR/dx = 13(2x) + 26

= 26 x + 26 

When, x = 7 

Then, 

MR = 26(7) + 26 

= 182 + 26 

= 208 

Thus, the marginal revenue is ₹ 208

Choose the correct answer for questions 17 and 18.

17. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is

(Α) 10π 

(Β) 12π 

(C) 8π 

(D) 11π

Ans: We know that A = πr² 

Therefore, 

When r = 6 cm 

Then,

dA/dr = 2π × 6 

= 12π cm² / s

Thus, the rate of change of the area of the circle is 12 cm² / s. 

Hence, the correct option 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

Ans: Marginal revenue (MR) is the rate of change of the total revenue with respect to the number of units sold. 

Therefore, 

MR = dR/dx = 3(2x) + 36 

= 6x + 36 

When, x = 15

Then, 

MR = 6(15) + 36

= 90 + 36

= 126

Thus, the marginal revenue is ₹ 126

Hence, the correct option is D.

EXERCISE 6.2

1. Show that the function given by f(x) = 3x + 17 is increasing on R. 

Ans: Let x₁ and x₂ be any two numbers in R. 

Then,

x₁ < x₂ = 3x₁ + 17 < 3x₂ + 17 = f(x₁) < f(x₂) 

Thus, f  is strictly increasing on R.

2. Show that the function given by f(x) = e²ˣ is increasing on R.

Ans: Let x₁ and x₂ be any two numbers in R. 

Then,

Thus, f  is strictly increasing on R.

3. Show that the function given by f(x) = sin x is 

(a) increasing in (0,π/2)

(b) decreasing in (π/2,π) 

(c) neither increasing nor decreasing in (0, π)

Ans: It is given that f (x) = sin x

Hence, f'(x) = cos x

(a) Here,

⇒ cos x > 0

⇒ f'(x) > 0 

Thus, f is strictly increasing in (0,π/2).

(b) Here,

⇒ cos x < 0 

⇒ f'(x) < 0

Thus, f is strictly decreasing in (π/2,π).

(c) Here, 

X∈ (0,π) 

The results obtained in (a) and (b) are sufficient to state 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) increasing.

(b) decreasing.

Ans: The given function is f(x) = 2x² – 3x 

Hence,

 f'(x) = 4x – 3

Therefore, f'(x) = 0 

⇒ x = 3/4

In

Hence, f is strictly decreasing in 

In

Hence, f is strictly increasing in

5. Find the intervals in which the function given by f(x) = 2x³ – 3x² – 36x + 7 is

(a) increasing.

(b) decreasing.

Ans: The given function is 

f(x) = 2x³ – 3x² – 36x + 7 

Hence,

f'(x) = 6x² – 6x – 36 

= 6(x² – x – 6) 

= 6(x + 2)(x – 3)

Therefore,

f'(x) = 0 

⇒ x = – 2, 3

Hence, f is strictly increasing in (- ∞,-2), (3,∞) and strictly decreasing in (-2,3)

6. Find the intervals in which the following functions are strictly increasing or decreasing:

(a) x² + 2x – 5 

Ans: f(x) = x² + 2x – 5

Hence, 

f'(x) = 2x + 2 

Therefore, 

⇒ f'(x) = 0 

 ⇒ x = – 1 

x = – 1 divides the number line into intervals (- ∞,1) and (- 1,∞) 

In (- ∞,1), f'(x) = 2x + 2 < 0

Thus f  is strictly decreasing in (-∞, 1)

In (- 1,∞), f'(x) = 2x + 2 > 0

Thus, f is strictly increasing in (-1,∞)

(b) 10 – 6x – 2x²

Ans: f(x) = 10 – 6x – 2x²

Hence,

f'(x) = – 6 – 4x 

Therefore,

⇒ f'(x) = 0 

⇒ x = – 3/2

x = – 3/2, divides the number line into two intervals (- ∞, – 3/2) and ( – 3/2, ∞)

In (- ∞, – 3/2), f'(x) = – 6 – 4x < 0

Hence, f is strictly increasing for x < – 3/2

In (- 3/2, ∞), f'(x) = – 6 – 4x > 0 

Hence, f is strictly increasing for x > – 3/2

(c) – 2x³ – 9x² – 12x + 1

Ans: f(x) = – 2x³ – 9x² – 12x + 1

Hence, 

f'(x) = – 6x² – 18x – 12 

= – 6(x² + 3x + 2) 

= – 6(x + 1)(x + 2) 

Therefore,

⇒ f'(x) = 0 

⇒ x = – 1, 2

x = – 1 and x = – 2 divide the number line into intervals (- ∞, – 2), (- 2, – 1) and (- 1, ∞),

Hence, f is strictly decreasing for x < – 2 and x > – 1

Hence, is strictly increasing in – 2 < x < – 1

(d) 6 – 9x – x²  

Ans: f(x) = 6 – 9x – x²

Hence, f'(x) = – 9 – 2x 

Therefore, 

⇒ f'(x) = 0 

⇒ x = – 9/2

Hence, f is strictly decreasing for x > – 9/2

Hence, f is strictly decreasing in x > – 9/2

(e) (x + 1)³ (x – 3)³

Ans: f(x) = (x + 1)³ (x – 3)³

Hence, 

f'(x) = 3(x + 1)² (x – 3)³ + 3(x – 3)² (x + 1)³

= 3(x + 1)² (x – 3)² [x – 3 + x + 1] 

= 3(x + 1)² (x – 3)² (2x – 2) 

= 6(x + 1)² (x – 3)² (x – 1) 

Therefore,

 f'(x) = 0 

= x = – 1, 3, 1

x = – 1,3,1 divides the number line into four intervals (- ∞ ,- 1), (- 1,1), (1,3) and (3, ∞) 

Hence, f is strictly decreasing in (- ∞, – 1) and (- 1,1)

Hence,f is strictly increasing in (1,3) and (3,∞)

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

Ans: It is given that y = log (1 + x) – 2x /2 + x

Therefore,

Now, 

dy/dx = 0

Hence, 

⇒ x² = 0

⇒ x = 0

Since, x > -1, x = 0 divides domain (- 1,∞) in two intervals – 1 < x < 0 and x > 0

When, – 1 < x < 0 

Then,

x < 0 ⇒ x² > 0 

x > – 1 ⇒ (2 + x) > 0

⇒ (2 + x)² > 0 

Hence,

When, x > 0

Then, 

x > 0 ⇒ x² > 0 

⇒ (2 + x)² > 0 

Hence,

Thus, f is increasing throughout the domain.

8. Find the values of x for which y = [x(x-2)]² is an increasing function.

Ans: We have,

y = [x(x – 2)]² 

= [x² – 2x]²

Therefore, 

= 2(x² – 2x)(2x – 2) 

= 4x (x – 2)(x – 1) 

Now, 

dy/dx = 0

Hence, 

⇒ 4x (x – 2)(x – 1)

⇒ x = 0, x = 2, x = 1 

x = 0, x = 1 and x = 2 divide the number line intervals (- ∞, 0), (0,1), (1,2) and (2, ∞)

Hence, y is strictly decreasing in intervals (- ∞, 0) and (1,2) 

In (0,1) and (2,∞), dy/dx > 0

Hence, y is strictly increasing in intervals (0,1) and (2,∞)

Ans: We have, 

Therefore, 

Now, 

dy/dθ = 0

Hence, 

Since, cos θ ≠ 4 

Therefore,

cos θ = θ

⇒ θ = π/2

Now, 

In interval [0, π/2], we have cos θ > 0

Also, 

4 > cos θ 

⇒ 4 – cosθ > 0 

Hence, cosθ (4 – cosθ) > 0 and also (2 + cosθ)² > 0 

Therefore,

Hence, 

dy/dθ > 0

So, y is strictly increasing in (0,π/2) and the given function is continuous at x = 0 and x = π/2

Thus, y is increasing in interval [0,π/2]

10. Prove that the logarithmic function is increasing on (0,∞)

Ans: The given function is f(x) = log x

Therefore, 

f'(x) = 1/x

For, 

x > 0, f'(x) = 1/x > 0 

Thus, the logarithmic function is strictly increasing in interval (0, ∞).

11. Prove that the function f given by f(x) = x² – x + 1 is neither strictly increasing nor decreasing on (- 1, 1).

Ans: The given function is f'(x) = x² – x + 1 

Therefore, 

f'(x) = 2x – 1

Now,

f'(x) = 0 

⇒ x = 1/2

x = 1/2 divides the interval v into (- 1,1/2) and (1/2,1)

In interval (½, 1), f'(x) = 2x – 1 > 0 

Hence, f is strictly decreasing in (- 1, 1/2)

In interval (1/2, 1), f’(x) = 2x – 1 > 0 

Hence, f is strictly increasing in (1/2,1)

Thus, f is strictly increasing nor strictly decreasing in interval (- 1,1)

12. Which of the following functions are decreasing on 0,π/2?

(A) cos x

Ans: Let f₁ (x) = cos x 

Therefore, f₁'(x) = – sin x

In interval 

Thus, cos x is strictly decreasing in (0,π/2).

(B) cos 2x

Ans: Let f₂ (x) = cos 2x 

Therefore,

Now, 

⇒ 0 < x < π/2 

⇒ 0 < 2x < π 

⇒ sin 2x >0 

⇒ – 2 sin 2x < 0

Hence, 

Thus, cos 2x is strictly decreasing in (0,π/2)

(C) cos 3x 

Ans: Let 

Therefore, 

Now,

Hence,

f₃ is strictly decreasing in (0, π/3)

In interval 

Hence,

f₃ is strictly increasing in (π/3,π/2)

Thus, cos 3x is neither increasing nor decreasing in interval (0,π/2)

(D) tan x

Ans: Let f₄ (x) = tan x

Therefore, 

In interval 

Thus, tan x is strictly increasing in (0,π/2)

Thus, the correct options are A and B.

13. On which of the following intervals is the function f given by f(x) = x¹⁰⁰ + sin x – 1 decreasing? 

(A) (0,1) 

(B) π/2,π

(C) 0, π/2 

(D) None of these.

Ans: We have,

 f(x) = x¹⁰⁰ + sin x – 1 

Therefore,

In interval (0,1), cos x > 0 and 100x⁹⁹ > 0

Hence,

 Thus, f is strictly increasing in (0,1)

In interval 

Hence, 

Thus, f is strictly increasing in interval (π/2,π)

Now, in interval 

Hence, 

Thus, f is strictly increasing in interval (0, π/2) 

Hence, f is strictly decreasing in none of the intervals.

Thus, the correct option is D.

14. For what values of a the function f given by f(x) = x² + ax + 1 is increasing on [1, 2]? 

Ans: We have

f(x) = x² + ax + 1 

Therefore, 

Now, the function f is strictly increasing on [1,2] 

Therefore, 

⇒ 2x + a > 0 

⇒ 2x > – a 

⇒ x > – a/2

Here, we have 1 ≤ x ≤ 2 

Thus,

15. Let I be any interval disjoint from [- 1,1]. Prove that the function f given by 

f(x) = x + 1/x is increasing on I.

Ans: We have 

f(x) = x + 1/x 

 Therefore,

Now, 

The points x = 1 and x = – 1 divide the real line intervals (- ∞,1), (-1,1) and (1,∞) 

In interval (- 1,1), – 1 < x < 1

⇒ x² < 1

Therefore, 

Hence, f is strictly decreasing on (- 1,1) ~ {0} 

Now, in interval (- ∞,- 1) and (1,∞), x < – 1 or 1 < x

⇒ x² > 1

Therefore, 

Hence, f is strictly increasing on (- ∞,- 1) and (1,∞) 

Thus, f is strictly increasing in I in [- 1,1]

16. Prove that the function f given by f(x) = log sin x is increasing on (0,π/2) and  decreasing on (π/2,π).

Ans: We have 

f(x) = log sin x 

Therefore,

= cot x 

In interval 

Hence, f is strictly increasing in (0, π/2)

In interval 

Hence, f is strictly decreasing in (π/2,π).

17. Prove that the function f given by f(x) = log |cos x| is decreasing on (0,π/2) and  increasing on (3π/2, 2π)

Ans: We have

f(x) = log |cos x|

Therefore,

In interval 

Hence, 

Thus, f is strictly decreasing on (0,π/2)

In interval 

Hence,

Thus, f is strictly increasing on (3π/2,2π)

18. Prove that the function given by f(x) = x³ – 3x² + 3x – 100 is increasing in R.

Ans: We have 

f(x) = x³ – 3x² + 3x – 100 

Therefore,

f'(x) = 3x² – 6x + 3 

= 3(x² – 2x + 1) 

= 3(x – 1)²

For x ∈ R, (x – 1)²  ≥ 0 

So, f'(x) is always positive in R. 

Thus, the function 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)

Ans: We have y = x² e⁻ˣ

Therefore,

Now, 

dy/dx = 0

Hence, 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. 

Hence, f is decreasing on (- ∞,0) and (2,∞) 

In interval (0,2), f’(x) > 0 

Hence, f is strictly increasing in (0,2) 

Thus, the correct option is D.

EXERCISE 6.3

1. Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = (2x – 1)² + 3

Ans: (i) The given function is f(x) = (2x – 1)² + 3 

The minimum value of f is attained when 2x – 1 = 0

2x – 1 = 0

⇒ x = 1/2

Hence, minimum value of

Thus, the function f does not have a maximum value.

(ii) f(x) = 9x² + 12x + 2 

Ans: The given function is f(x) = 9x² + 12x + 2

It can be observed that

The minimum value of f is attained when 3x + 2 = 0 

3x + 2 = 0 

⇒ x = – 2/3 

Therefore, Minimum value of

Hence, the function f does not have a maximum value.

(iii) f(x) = – (x – 1)² + 10

Ans: The given function is 

f(x) = – (x – 1)² + 10

The maximum value of f is attained when (x – 1) = 0

(x – 1) = 0 

x = 1 

Therefore, Maximum value of 

f = f (1)

= – (1 – 1)² + 10 

= 10

Hence, the function f does not have a minimum value.

(iv) g(x) = x³ + 1

Ans: The given function is g(x) = (x)³ + 1 

Hence, function g neither has a maximum value nor a minimum value.

2. Find the maximum and minimum values, if any, of the following functions given by: 

(i) f(x) = |x + 2| – 1

Ans: The given function is f(x) = |x + 2| – 1

The minimum value of f is attained when |x + 2| = 0 

|x + 2| = 0 

⇒ x = – 2 

Therefore, Minimum value of 

f = f (- 2) 

= |- 2 + 1| – 1

= – 1 

Hence, the function f does not have a maximum value.

(ii) g(x) = – |x + 1| + 3

Ans: The given function is g(x) = |- x + 1| + 3

The maximum value of g is attained when |x + 1| = 0 

|x + 1| = 0 

⇒x = – 1 

Therefore, maximum value of 

g = g(- 1)

= – | – 1 + 1| + 3 

= 3 

Hence, the function g does not have a minimum value.

(iii) h(x) = sin (2x) + 5

Ans: The given function is h(x) = sin (2x) + 5

We know that 

Therefore,

Hence, the maximum and minimum values of h are 6 and 4, respectively.

(iv) f(x) = |sin 4x + 3|

Ans: The given function is f(x) = |sin 4x + 3|

We know that -1 ≤ sin 4x ≤ 1 

Therefore,

⇒ 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)

Ans: The given function is 

h(x) = x + 1,x € (- 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

Ans: f(x) = x²

Therefore,

Now,

⇒ x = 0

Thus, x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.

Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is (0) = 0.

(ii) g(x) = x³ – 3x

Ans: g(x) = x³ – 3x

Therefore,

Now,

Also,

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

Ans: h(x) = sin x + cos x,0 < x < π/2

Therefore,

Now,

Hence,

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 

(iv) f(x) = sin x – cos x, 0 < x < 2π

Ans: f(x) = sin x – cos x, 0 < x < 2π

Therefore,

Also, 

Hence,

Therefore, by second derivative test, x = (3π/4) is a point of local maxima and the local

maximum value of f at x = (3π/4) is

However, x = (7π/4) is a point of local minima and the local minimum value of f at x = (7π/4) is 

(v) f(x) = x³ – 6x² + 9x + 15

Ans: f(x) = x³ – 6x² + 9x + 15

Therefore, 

Now, 

Also,

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

Ans: 

Therefore,

Since, x > 0, we take x = 2 

Hence,

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

Ans: 

Therefore,

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) = 1/0 + 2

= 1/2

Ans: 

Therefore,

Now,

Also, 

Hence,

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

4. Prove that the following functions do not have maxima or minima: 

(i) f(x) = eˣ

Ans: f(x) = eˣ 

Therefore, 

f’(x) = eˣ

Now,

f’(x) = 0

= eˣ = 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) g(x) = log x

Ans: g(x) = log x 

Therefore, 

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) h(x) = x³ + x² + x + 1

Ans: h(x) = x³ + x² + x + 1

Therefore,

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: 

Ans: The given function is f(x) = x³

Therefore, 

f'(x) = 3x² 

Now,

f'(x) = 0 

= 3x² = 0

Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [- 2, 2) 

Therefore,

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.

Ans: The given function is f(x) = sin x + cos x

Therefore, 

f'(x) = cos x – sin x

Now, 

Then, we evaluate the value of f at critical point x = π/4 and at the end points of the interval [0,π].

Therefore, 

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

Also, the absolute minimum value off on [0,π] is – 1 occurring at x = π.

Ans: The given function is

Therefore,

Now, 

Then, we evaluate the value of f at critical point x = 4 and at end points of the interval

[- 2,9/2]

Therefore,

Hence, we can conclude that the absolute maximum value of f on [- 2, 9/2] is 8 occurring at x = 4

Also, the absolute minimum value of f on [ – 2, 9/2] is – 10 occurring at x = – 2

Ans: The given function is f(x) = (x – 1)² + 3

Therefore,

Then, we evaluate the value of f at critical point x = 1 and at end points of the interval [- 3,1]

f(1) = (1 – 1)³ + 3 

= 3

f(- 3) = (- 3 – 1)² + 3

= 16 + 3 

= 19

Hence, we can conclude that the absolute maximum value of on [- 3,1] is 19 occurring at x = – 3

Also, the absolute 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 

Ans: The profit function is given as 

p(x) = 41 – 72x – 18x² 

Differentiating w.r.t x, we get 

p(x) = – 72 – 36x 

Putting p(x) = 0

⇒ – 72 – 36x = 0 

⇒ 36x = – 72 

⇒ x = – 2 

x = – 2 is a critical point.

p'(x) = – 72 – 36x 

Differentiating w.r.t x 

p” (x) = – 36 

Since P” (x) < 0 

x = – 2 is a point of maxima,

p(x) = 41 – 72x – 18x²

Maximum profit 

= p(- 2) 

= 41 – 72(- 2) – 18(- 2)²

= 41 + 144 – 18(4)

= 41 + 144 – 72

= 113 

Hence, the maximum profit is 113

7. Find both the maximum value and the minimum value of 

Ans: 

Therefore, 

Now, 

Now, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0.3]

Therefore,

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? 

Ans: Let f(x) = sin 2x 

Therefore,

Now, we evaluate the value of f at critical point x = π/4, 3π/4, 5π/4, 7π/4 and at the end points of the interval [0,2π]

Therefore,

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?

Ans: Let f(x) = sin x + cos x

Therefore, 

f'(x) = cos x – sin x

Now, 

Hence, 

Now, 

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.

By second derivative test, f will be the maximum at x = π/4 and the maximum value of f is

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]

Ans: Let f(x) = 2x³ – 24x + 107

Therefore,

Now,

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],

Hence, 

Thus, the absolute maximum value of f(x) in the interval [1,3] is 89 occurring at x = 3 

Next, we first consider the interval [- 3,- 1] 

Then, we evaluate the value of f at the critical point x = – 2 € [- 3, – 1] and at the end points of the interval [- 3,- 1], 

Hence,

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 attains its maximum value, on the interval [0, 2]. Find the value of a.

Ans: 

It is given that function f attains its maximum value on the interval [0.2] at x = 1

Thus, the value of a = 120

12. Find the maximum and minimum values of x + sin 2x on [0, 2π).

Ans: Let f(x) = x + sin 2x 

Therefore,

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 [0,2]. 

Hence,

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.

Ans: Let one number be x.

Then, the other number be (24 – x)

Let P(x) denote the product of the two numbers.

Thus, we have: 

P(x) = x(24 – x) 

= 24x – x² 

By second derivative test, x = 12 is the point of local maxima of P.

Thus, the numbers are 12 and (24 – 12) = 12

Hence, the product of the numbers is the maximum when the numbers are 12 each.

14. Find two positive numbers x and y such that x + y = 60 and xy³ is maximum. 

Ans: The two numbers are x and y such that x + y = 60

Therefore, 

= y = 60 – x 

Let, f(x) = xy³

f(x) = x(60 – x)³

Therefore,

Now,

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 X² y⁵ is a maximum.

Ans: Let one number be x.

Then, the other number is y = (35 – x) 

Therefore,

Now, 

Hence, x = 0 and x = 35 cannot be the possible values of x. 

When, x = 10 

Then,

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. 

Ans: Let one number be x. 

Then, the other number be (16 – x) 

Let the sum of the cubes of these numbers be denoted by S(x) 

Then.

 S(x) = x³ + (16 – x)³

Therefore,

Now,

Also,

By second derivative test, x = 8 is the point of local minima of S.

Thus, the numbers are 8 and (16 – 8) = 8

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 each.

17. A square piece of tin of side 18 cm is to be 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.

Ans: Let the side of the square to be cut off be x cm. 

Then, the length and the breadth of the box will be (18 – 2x)cm each and the height of the box be x cm. 

Therefore, the volume V (x) of the box is given by. 

V(x) = x(18 – 2x)²

Hence,

Now, 

If, x = 9 then the length and the breadth will become 0. 

Hence, x ≠ 9

When, x = 3

Then, 

By second derivative test, x = 3 is the point of local 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 maximum?

Ans: Let the side of the square to be cut off be x cm. 

Then, the height of the box is x cm, the length is (45 – 2x) cm and the breadth is (24 – 2x) cm 

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

Hence,

Now,

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.

When,

x = 5

Then,

By second derivative test, x = 5 is the point of local 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.

Ans: 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.

Now, by applying the Pythagoras theorem, we have:

Area of triangle, 

Therefore,

Now,

dA/dl = 0

Hence,

Thus,

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 its height is equal to the diameter of the base.

Ans: 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

Therefore, 

Let V be the volume of the cylinder. 

Then,

Now,

By second derivative test, the volume is the maximum when r² = S/6π.

Now, when r² = S/6π,

Then,

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?

Ans: Let r and h be the radius and height of the cylinder respectively. 

Then, volume (V) of the cylinder is given by,

V = πr²h = 100

= h = 100/πr²

Surface area (S) is given by:

S = 2πr² + 2πrh 

= 2πr² + 200/r

Hence, 

Now, 

By second derivative test, the surface area is the minimum when the radius of the cylinder is 

Hence, the required dimensions of the can which has the minimum surface area is given by

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?

Ans: 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 is l/4 

Let r be the radius of the circle. 

Then. 2πr = 28 – l

⇒ r = l/2π (28 – l)

The combined areas of the square and the circle, (A) is given by,

Now, 

By second derivative test, the area (A) is the minimum when 

Hence, the combined area is the minimum when the length of the wire in making the square is

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.

Ans: Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.

Let V be the volume of the cone.

Then, 

Therefore, 

Now,

Then,

Therefore,

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 time the radius of the base.

Ans: Let r and h be the radius and height of the cone, respectively. 

Then, the volume (V) of the cone is given by,

The surface area (S) of the cone is given by, 

S = πrl, where l is the slant height 

Hence,

Therefore,

Now,

Thus, it can be easily verified that when

By second derivative test, the surface area of the cone is the least when

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

Ans: Let θ be the semi-vertical angle of the cone. 

It is clear that

Leth 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.

Therefore, 

Now, 

By second derivative test, the volume (V) is the maximum when θ = tan √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) Choose the correct answer in Questions 27 and 29.

Ans: Let r be the radius, l be the slant height and h be the height of the cone of given surface area S. 

Also, let a be the semi-vertical angle of the cone.

Then, 

Let V be  the volume of the cone.

Then, 

Differentiating (2) with respect to r, we get

For maximum or minimum, put dV/dr = 0

Differentiating again with respect to r, we get

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)

Ans: The given curve is x² = 2y

Let us denote PA² by Z 

Then, 

Z = y² – 8y + 25 

Differentiating both sides with respect to y, we get 

dZ/dy = 2y – 8

For maxima or minima, we have

dZ/dy = 0

= 2y – 8 = 0 

= 2y = 8

= y = 4

Also,

d²Z/dy² = 2

Now, 

Thus, the correct option is A.

(A) 0 

(B) 1 

(0) 3 

(D) 1/3

Ans: 

Therefore,

Now,

Also,

Hence, 

Also,

By second derivative test, f is the minimum at x = 1 and the minimum value is given by.

Thus, the correct option is D.

(B) 1/2

(C) 1

(D) 0

Ans: 

Then, we evaluate the value of f at critical point  x = 1/2 and at the end points of the interval [0,1] i.e., at x = 0 and x = 1.

Hence, we can conclude that the maximum value of f in the interval [0,1] is 1.

Thus, the correct option is C.

Miscellaneous Exercise on Chapter 6

1. Show that the function given by f(x) = log x/x has maximum at x = e.

Ans: 

Now, 

Also,

Now, 

Therefore, by second derivative test, f is the maximum at x = e.

2. The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Ans: Let ∆ABC be isosceles where BC is the base of fixed length b.

Also, let the length of the two equal sides of ∆ABC be a.

Now, in ∆ADC, by applying the Pythagoras theorem, we have:

Area of triangle,

The rate of change of the area with respect to time (t) is given by,

It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second. 

Therefore,

da/dr = – 3 cm/s

Hence,

Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of – √3bcm²/s.

3. Find the intervals in which the function f given by f(x) = 4sin x – 2x – x cos x/2+cos x is 

(i) increasing.

(ii) decreasing.

Ans: 

4. Find the intervals in which the function f given by

(i) increasing.

(ii) decreasing.

Ans: We have

f(x) = x³ + 1/x³

Therefore,

Now,

Now, the points x = 1 and x = – 1 divide the real line into three disjoint intervals i.e., (- ∞, – 1) (-1,1) and (1, ∞). 

In intervals (- ∞, – 1) and (1,∞ ) i.e., when x < – 1 and x > 1, f'(x) > 0.

Thus, when x < – 1 and x >1, f is increasing. 

In interval (-1, 1) i,e., when – 1 < x < 1, f’(x) < 0

Thus, when – 1 < x < 1, f is decreasing.

5. Find the e maximum area of an isosceles triangle inscribed in the ellipse x²/a² + y²/ b² = 1 with its vertex at one end of the major axis.

Ans: 

Let the major axis be along the x-axis.

Let ABC be the triangle inscribed in the ellipse where vertex C is at (a,0) 

Since the ellipse is symmetrical with respect to the x – axis and y – axis, we can assume the coordinates of A to be (x₁, y₁) and the coordinates of B to be (- x₁ – y₁)

Therefore, 

Now, 

dA/dx₁ = 0

Hence, 

Now,

Also when, 

x₁ = a/2

Then,

Thus, the area is the maximum when x₁ = a/2 

Hence, Maximum area of the triangle is given by,

6. A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³ If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?

Ans: Let l, b and h represent the length, breadth, and height of the tank respectively. 

Then, we have height, h = 2m and volume of the tank, V = 8m³

Volume of the tank

V = lbh

= 8 = l × b × 2

= lb = 4

= b = 4/l

Now, area of the base, lb = 4 

Area of the 4 walls,

However, the length cannot be negative. Therefore, we have l = 4 

Hence,

b = 4/l

= 4/2

= 2

Now,

When, l = 2 

Then,

Thus, by second derivative test, the area is the minimum when l = 2 

We have l = b = h = 2 

Therefore, 

Cost of building the base in ₹ is 

70 × (lb) = 70(4) 

= 280 

Cost of building the walls in ₹ is 

2h(l + b) × 45 = 2 × 2(2 + 2) × 45 

= 720 

Required total cost in ₹ is 

280 + 720 = 1000 

Thus, the total cost of the tank will be ₹ 1000.

7. The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Ans: Let r be the radius of the circle and a be the side of the square. 

Then, we have:

The sum of the areas of the circle and the square (A) is given by.

Hance,

Now,

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

8. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening. 

Ans: Let x and y be the length and breadth of the rectangular window. 

Radius of the semicircular opening be x/2 

It is given that the perimeter of the window is 10m. 

Therefore,

Area of the window (A) is given by,

Therefore,

Now, 

Therefore, by second derivative test, the area is the maximum when length is 20/π + 4 m

Now,

Hence, the required dimensions of the window to admit maximum light is given by length

9. A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle.

Show that the minimum length of the hypotenuse is

Ans: Let ∆ABC be right-angled at B, AB = x, BC = y and ∠C = θ. 

Also, let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and BC respectively,

Now,

Thus,

By second derivative test the length of the hypotenuse is the maximum when

10. Find the points at which the function f given by f(x) = (x – 2)⁴ (x + 1)³ has

(i) local maxima.

(ii) local minima.

(iii) point of inflexion. 

Ans:  The given function is f(x) = (x – 2)⁴ (x + 1)³

Therefore,

Now,

Thus, x = 2/7 is the point of local maxima. 

Now, for values of x close to 2 and to the left of 2.f'(x) < 0

Also, for values of x close to 2 and to the right of 2,f'(x) > 0 

Thus, x = 2 is the point of local minima. 

Now, as the value of x varies through – 1, f'(x) does not change its sign. 

Thus, x = – 1 is the point of inflexion.

11. Find the absolute maximum and minimum values of the function f given by f(x) = cos² x + sin x, x ∈ [0, π]

Ans: We have f(x) = cos² x + sin x

Therefore, 

Now,

Now, evaluating the value of f at critical points x = π/6, π/2 and at the end points of the interval

[0,π] i.e., at x = 0 and x = π, 

we have:

Hence, the absolute maximum value of f is 5/4 occurring at x = π/6 and the absolute minimum value of f is 1 occurring at 

12. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3.

Ans: A sphere of fixed radius (r) is given. Let R and h be the radius and the height of the cone respectively.

The volume (V) of the cone is given by

Now, from the right ∆BCD, we have:

Therefore,

Now,

Also,

Then, height of the cone 

Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius = 4r/3.

13. Let f be a function defined on [a, b] such that f'(x) > 0, for all x ∈ (a, b). Then prove that f is an increasing function on (a, b).

Ans: Let x₁, x₂ ∈ (a,b) such that x₁ > x₂ 

Consider the sub-interval [x₁, x₂]

Since f(x) is differentiable on (a,b) and [x₁, x₂] ⊂ (a,b). 

Therefore f(x) is continuous on [x₁, x₂] and differentiable on (x₁, x₂) 

By the Lagrange’s mean value theorem, there exists C ∈ (x₁, x₂) such that

Since, f'(x) > 0 for all x ∈ (a,b), so in particular,

Since, x₁, x₂ are arbitrary points in (a,b). 

Therefore, x₁, < x₂ 

= f(x₁) < f(x₂) for all x₁, x₂ ∈ (a,b)

Hence, f(x) is increasing on (a,b).

14. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3 Also find the maximum volume.

Ans: A sphere of fixed radius (R) is given.

Let r and h be the radius and the height of the cylinder respectively.

From the given figure, we have h = 2√R² – r² 

The volume (V) of the cylinder is given by,

V = πr²h

= 2πr² √R² – r² 

Therefore,

Now,

Also,

Hence, the volume of the cylinder is the maximum when the height of the cylinder is 2R√3.

15. Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle a is one-third that of the cone and the greatest volume of cylinder is 4/27 πh³ tan² a.

Ans: The given right circular cone of fixed height h and semi-vertical angle a can be drawn as:

Here, a cylinder of radius R and height H is inscribed in the cone.

Then, ∠GAO = a, OG = r, OA = h, OE = R and CE = H. 

We have, r = h tan a 

Now, since ∆AOG is similar to ∆CEG, we have:

Now, the volume (V) of the cylinder is given by,

Now,

Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.

Thus, the maximum volume of the cylinder can be obtained as:

Hence, the given result is proved

16. A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of 

(A) 1 m/h. 

(B) 0.1 m/h.

(C) 1.1 m/h.

(D) 0.5 m/h.

Ans: Let r be the radius of the cylinder. 

Then, volume (V) of the cylinder is given by, 

V = πr²h 

= π (10)² h

= 100πh

Differentiating with respect to time (t), we have:

 dV/ dt =100π dh/dt

The tank is being filled with wheat at the rate of 314m³/h 

dV/dt = 314m³/h

Thus, we have: 

Hence, the depth of wheat is increasing at the rate of Im/h. 

Thus, the correct option is A.