RBSE Maths Class 12 Chapter 8: Application of Derivatives Important Questions and Solutions

RBSE Class 12 Maths Chapter 8 – Application of Derivatives Important questions and solutions are given here. All the important questions and solutions of this chapter have stepwise solutions. The RBSE Class 12 important questions and solutions provided at BYJU’S cover all the concepts of Class 12 and are given as per the new pattern prescribed by the board.

Chapter 8 of RBSE Class 12 contains six exercises which cover the concepts of rate of change in different situations. Enough number of practice questions are given here which cover the questions on finding the maximum, minimum value of the given functions at different points and different intervals. Also, there are some questions on finding the equation of tangent and normal for the given function.

RBSE Maths Chapter 8: Exercise 8.1 Textbook Important Questions and Solutions

Question 1: Find the rate of change of the area of a circle with respect to radius r, when r = 6 cm and r = 11 cm.

Solution:

Let A be the area of the circle.

A = πr2

Differentiating with respect to r on both sides,

dA/dr = d/dr(πr2)

dA/dr = 2πr

At r = 6,

dA/dr = 2π(6) = 12π cm2/sec

At r = 11,

dA/dr = 2π(11) = 22π cm2/sec

Question 2: The total cost C(x) rupees, associated with the production of x units of an item is given by C(x) = 0.005x3 – 0.02x2 + 30x + 5000

Find the marginal cost when 3 units are produced, here by marginal cost we mean the instantaneous rate of change of total cost at any level of output.

Solution:

Let C(x) be the cost of production of x units.

Given,

C(x) = 0.005x3 – 0.02x2 + 30x + 5000

Marginal cost (MC) = d/dx C(x)

= d/dx(0.005x3 – 0.02x2 + 30x + 5000)

= 0.005 d/dx(x3) – 0.02 d/dx(x2) + 30 d/dx(x) + d/dx(5000)

= 0.005(3x2) – 0.02 (2x) + 30 (1) + 0

= 0.015x2 – 0.04x + 30

For x = 3,

MC = 0.015 × (3)2 – 0.04 × (3) + 30

= (0.015 × 9) – 0.12 + 30

= 0.135 – 0.12 + 30

= 30.135 – 0.12

= 30.015

Therefore, the marginal cost is Rs. 30.015 or Rs. 30.02 (approx).

RBSE Maths Chapter 8: Exercise 8.2 Textbook Important Questions and Solutions

Question 3: Show that f(x) = x2 is increasing in (0, ∞) and decreasing in (-∞, 0).

Solution:

Let x1, x2 ∈ [0, ∞] such that x1 < x2.

∴ x1 < x2

⇒ x12 < x1x2 ….(i)

and x1 < x2

⇒ x1x2 < x22….(ii)

From (i) and (ii),

x1 < x2

⇒ x12< x22

⇒ f(x1) < f(x2)

∴ x1 < x2 ⇒ f(x1) < f(x2) where x1, x2 ∈ [0, ∞]

Hence, f(x) is continuously increasing in [0, ∞)

Again,

Let x1, x2 ∈ (-∞, 0) such that x1 < x2

∴ x1 < x2

⇒ x12> x1x2

[∵ – 3 < – 2 and (- 3) (- 3) = 9

(- 3) × (- 2) = 6

∴ 9 > 6

x12 > x1 x2]

Now,

x1 < x2

⇒ x1x2 > x22

[consider – 3 < – 2

(- 3) × (- 2) = 6

(- 2) × (- 2) = 4

6 > 4

∴ x1x2 > x22]

Again from (i) and (ii),

x1 < x2

⇒ x12 > x22

⇒ f(x1) > f(x2)

∴ x1 < x2 ⇒ f(x1) > f(x2)

Hence, f(x) is continuously decreasing in (-∞, 0).

Question 4: Find the least value of a when the function f(x) = x3 + ax + 5 is increasing in the interval [1, 2].

Solution:

Given function is:

f(x) = x2 + ax + 5

⇒ f'(x) = 2x + a

Function is increasing in the interval [1, 2].

x ∈ [1, 2]

⇒ 1 ≤ x ≤ 2

⇒ 2 ≤ 2x ≤ 4

⇒ 2 + a ≤ 2x + a ≤ 4 + a

⇒ 2 + a ≥ 0

⇒ a ≥ -2

Hence, the least value of a is (– 2, ∞).

RBSE Maths Chapter 8: Exercise 8.3 Textbook Important Questions and Solutions

Question 5: Find the equation of all lines having slope 2 and being tangent to the curve y + 2/(x – 3) = 0.

Solution:

Given curve is:

y + 2/(x – 3) = 0

y = -2/(x – 3) ….(i)

Differentiating with respect to x on both sides,

dy/dx = -2 d/dx(1/x – 3)

dy/dx = 2/(x – 3)2

Given that, slope is 2.

Thus, 2/(x – 3)2 = 2

(x – 3)2 = 2/2

(x – 3)2 = 1

x – 3 = ±1

x = ±1 + 3

x = 2, 4

Substituting x = 2 in (i),

y = -2/(2 – 3) = 2

Thus, the point is (2, 2).

Substituting x = 4 in (i),

y = -2/(4 – 3) = -2

Thus, the point is (4, -2)

Now, the tangent at (2, 2).

y – 2 = 2(x – 2)

y – 2 = 2x – 4

2x – y – 2 = 0

And the tangent at (4, -2) is:

y – (-2) = 2(x – 4)

y + 2 = 2x – 8

2x – y – 10 = 0

Therefore, the required equations are 2x – y – 2 = 0 and 2x – y – 10 = 0.

Question 6: Find the equation of the tangent and normal to the given curves at the indicated points.

y2 = 4ax at x = a

Solution:

Given curve is:

y2 = 4ax ….(i)

Substituting x = 1 in (i),

y2 = 4a(a)

y2 = 4a2

y = ±2a

Thus, the points of contact are (a, 2a) and (a, -2a).

Differentiating (i) with respect to x,

d/dx(y2) = d/dx(4ax)

2y (dy/dx) = 4a

dy/dx = 2a/y

At the point (a, 2a)

(dy/dx)(a, 2a) = 2a/2a = 1

The equation of tangent at (a, 2a) is:

y – 2a = 1(x – a)

x – y + a = 0

The equation of normal at (a, 2a) is:

y – 2a = -1(x – a)

x + y – 3a = 0

Therefore, at point (a, 2a), the equation of tangent is x – y + a = 0 and the equation of normal is x + y – 3a = 0.

RBSE Maths Chapter 8: Exercise 8.4 Textbook Important Questions and Solutions

Question 7: Using differentials, find the approximate value of the following number.

(3.968)3/2

Solution:

Let x = 4 and y = x3/2

y = (4)3/2 = (22)3/2 = 23 = 8

Δx = 3.968 – 4 = 0.032

Consider ,y = x3/2

Differentiating with respect to x,

dy/dx = (3/2)x3/2 – 1

dy/dx = (3/2)x1/2

Thus, dy = (dy/dx)Δx

= (3/2) × x1/2 × (-0.032)

dy = (3/2) × (4)1/2 × (-0.032)

= (3/2) × 2 × (-0.032)

= 3 × (-0.032)

dy = -0.096 = Δy

Therefore, (3.968)3/2 = y + Δy

= 8 + (-0.096)

= 7.904

Hence, the approximate value of (3.968)3/2 is 7.904.

Question 8: Prove that the approximation percentage error in calculating the volume of a cubical box is almost three times the approximate percentage error in calculating the edge of the cube.

Solution:

Let x be the length of the edge of the cubical box and V be its volume.

V = x3

Differentiating with respect to x,

dV/dx = d/dx(x3)

dV/dx = 3x2

ΔV = (dV/dx)Δx

ΔV = 3x2 Δx

ΔV/V = (3x2 Δx)/V

ΔV/V = (3x2 Δx)/x3

ΔV/V = 3(Δx/x)

(ΔV/V) × 100 = 3[(Δx/x) × 100]

RBSE Maths Chapter 8: Exercise 8.5 Textbook Important Questions and Solutions

Question 9: Find the maximum and minimum value of the following function if any, in the given interval.

2x3 – 24x + 107; x ∈ [1, 3]

Solution:

Let the given function be y = 2x3 – 24x + 107 and x ∈ [1, 3]

dy/dx = 6x2 – 24

For maxima or minima

dy/dx = 0

⇒ 6x2 – 24 = 0

⇒ x = ± 2. (given x ∈ [1, 3])

∴ x = 2

Now,

y1 = 2(1)3 – 24(1) + 107

= 2 – 24 + 107

= 85

y2 = 2(2)3 – 24(2) +107

= 16 – 48 + 107

= 75

y3 = 2(3)3 – 24(3) + 107

= 54 – 72 + 107

= 89

Hence, the maximum value of the given function is 89 at x = 3 whereas minimum value is 75 at x = 2.

Question 10: Prove that the function f(x) = x/(1 + x tan x) has maximum value at x = cos x.

Solution:

Let the given function be:

y = (1 + x tan x)/x

y = (1/x) + tan x

Differentiating with respect to x,

dy/dx = (-1/x2) + sec2x

Again differentiating with respect to x,

d2y/dx2 = (2/x3) + 2 sec2x tan x

For maxima or minima, dy/dx = 0

(-1/x2) + sec2x = 0

⇒ (1/x2) = (1/cos2x)

⇒ x2 = cos2x

⇒ x = cos x

At x = cos x,

d2y/dx2 = (2/cos3x) + 2 sec2(cos x) tan x (cos x) = +ve

Thus, at x = cos x, (1 + x tan x)/x is minimum

Therefore, at x = cos x, x/(1 + x tan x) is maximum.

Hence proved.

RBSE Maths Chapter 8: Exercise 8.6 Textbook Important Questions and Solutions

Question 11: The sum of perimeter of a square and circumference of a circle is given. Prove that the sum of their areas will be minimum if the side of the square is equal to the diameter of the circle.

Solution:

Let x be the side of the square and r be the radius of the circle.

Circumference of circle = 2πr

Perimeter of square = 4x

Sum of both perimeters = 4x + 2πr = k ….(i)

Area of circle = πr2

Area of square = x2

∴ Sum of areas = A = πr2 + x2 ….(ii)

From (i),

r = (k – 4x)/2π

From (ii),

A = π[(k – 4x)/2π]2 + x2

A = [π(k – 4x)2/4π2] + x2

A = [(k – 4x)2/4π] + x2

Differentiating with respect to x,

dA/dx = [2(k – 4x)(-4)/4π] + 2x

dA/dx = [-2(k – 4x)/π] + 2x ….(iii)

For minimum area, dA/dx = 0

[-2(k – 4x)/π] + 2x = 0

-2(k – 4x) = -2πx

k – 4x = πx

k = πx + 4x

k = (π + 4)x

x = k/(π + 4)

Differentiating (iii) with respect to x,

d2A/dx2 = (8/π) + 2

For x = k/(π + 4), d2A/dx2 = 2 + (8/π) > 0

Thus, at x = k/(π + 4), the sum of the area is minimum.

r = [(k – 4)(k/ π + 4)]/2π

= (πk + 4k – 4k)/ 2π(π + 4)

r = πk/2π(π + 4)

r = k/2(π + 4)

Now,

x/r = [k/(π + 4)] [2(π + 4)/k]

x/r = 2

x = 2r

Therefore, the side of the square is equal to the diameter of the circle when their sum of areas is minimum.

Question 12: The expense for a steamer per hour is proportional to the cube of its velocity. If the velocity of stream is x km/hr, then prove that the maximum velocity of streamer per hour will be (2/3)x when the steamer runs against the direction of stream.

Solution:

Let u km/hr be the velocity of the steamer and S km be the distance covered.

According to the given,

Speed of water current = x km/hr

Relative velocity of steamer = (u – x) k/h

Time take to cover the distance = T = S/(u – x) hour

Given expense per hour = Ku3 (where K is constant)

Total expenses E = Ku3 . S/(u – x)

Differentiating with respect to u,

dE/du = K.S[(u – x)3u2 – u3(1 – 0)] / (u – x)2

dE/du = [K.S u2(2u – 3x)]/ (u – x)2

For maxima or minima,

dE/du = 0

[K.S u2(2u – 3x)]/ (u – x)2 = 0

2u – 3x = 0

2u = 3x

u = (3/2)x

Hence, (d2E/du2)u = (3/2)x = +ve

Therefore at u = (3/2)x, the expenses will be minimum and maximum at u = (⅔)x.

RBSE Maths Chapter 8: Additional Important Questions and Solutions

Question 1: The radius of a cylinder is r and height is h, then find the rate of change in surface area of the cylinder with respect to radius.

Solution:

Given,

Radius of cylinder = r

Height of cylinder = h

Rate of change in the surface area with respect to r is dS/dr.

Surface area of cylinder = S = 2πr2 + 2πrh

Differentiating with respect to r,

dS/dr = (2π) d/dr (r2) + 2πh d/dr (r)

= 2π(2r) + 2πh

= 4πr + 2πh

Therefore, the rate of change is 4πr + 2πh.

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