## MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives

### Application of Derivatives Important Questions

**Application of Derivatives Objective Type Questions:**

Question 1.

Choose the correct answer:

Question 1.

The rate of change of the area of a circle with respect to its radius r when r = 5 is:

(a) 10 π

(b) 12 π

(c) 8 π

(d) 11 π.

Answer:

(b) 12 π

Question 2.

Tangent line of a curve y^{2} = 4x at y = x + 1 is:

(a) (1,2)

(b) (2, 1)

(c) (1, -2)

(d) (- 1, 2).

Answer:

(a) (1,2)

Question 3.

Approximate change in the volume of a cube of side x metre caused by increasing the side by 2%:

(a) 0.03 x^{3}

(b) 0.02 x^{3}

(c) 0.06 x^{3}

(d) 0.09 x^{3}

Answer:

(c) 0.06 x^{3}

Question 4.

Point on the curve x^{2} = 2y which lie minimum distance from the point (0,5):

(a) (2\(\sqrt{2}\),4)

(b) (2\(\sqrt{2}\),0)

(c) (0, 0)

(d) (2, 2).

Answer:

(a) (2\(\sqrt{2}\),4)

Question 5.

Minimum value of the function f(x) = x^{4} – x^{2} – 2x + 6:

(a) 6

(b) 4

(c) 8

(d) None of these.

Answer:

(b) 4

Question 2.

Fill in the blanks:

- The function f(x) = cosx, for 0 ≤ x ≤ π is …………………………..
- The radius of circular plate is increasing at the rate of 0.2 cm /sec. when r = 10, then the rate of change of the area of the plate is ………………………………
- The function y = x(5 – x) is maximum at x is equal to ……………………………
- The minimum value of 2x + 3y is ………………………….. when xy = 6.
- The maximum value of sin x + cos x is ……………………………
- If line y = mx+ 1 is a tangent line to the curve y
^{2}= 4x. Then value of m will be …………………………….. - Slope of the tangent to the curve y = x
^{2}at the point (1, 1) is …………………………. - Using differential the approximate value of \(\sqrt{0.6}\) is …………………………….

Answer:

- Decreasing
- 4πcm
^{2}/sec - \(\frac{5}{2}\)
- 12
- \(\sqrt{2}\)
- 1,
- 2,
- 0.8.

Question 3.

Write True/False:

- For all real values of x, the function f(x) = e
^{x}– e^{-x}is increasing function. - If the length of equal sides of an isosceles traingle be x then its maximum area will be \(\frac{1}{2}\) x
^{2} - Function f(x) = 3x
^{2}– 4x is increasing in the interval (- ∞, \(\frac{2}{3}\) ) - Function f(x) = x – cot x is always decreasing.
- Equation of the normal of the curve y = e
^{x}at point (0, 1) is x + y = l.

Answer:

- True
- True
- False
- False
- True.

**Application of Derivatives Short Answer Type Questions**

Question 1.

The radius of a circle increases at the rate of 2cm/sec. At what rate the area increases when radius is 10cm?

Solution:

Given:

\(\frac { dr }{ dt } \) = 2cm/sec

Let the area of circle be A

Then,

∵ Area of circle A = πr^{2}

= 40π sq.cm/second

Question 2.

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

Solution:

Let the radius of air bubble be r.

∴ Volume V = \(\frac{4}{3}\) πr^{3}

Question 3.

The radius of a balloon is increasing at the rate of 10 cm/sec? At what rate surface area of the balloon is increasing when radius is 15cm?

Solution:

Let r be the radius of balloon at any time t and its surface be x then

A = 4πr^{2}

Differentiating eqn.(1) w.r.t. t,

Hence, when radius of ballon is 15cm, then its area is increasing at rate of 1200π cm/sec.

Question 4.

Find those intervals in which the function f(x) = 2x^{3} – 15x^{2} + 36x + 1 is increasing or decreasing?

Solution:

f(x) = 2x^{3} – 15x^{2} + 36x + 1

⇒ f'(x) = 6x^{2} – 30x + 36

= 6(x^{2} – 5x + 6)

= 6(x – 2) (x – 3)

For increasing function of f(x),

f'(x) > 0

or (x – 2) (x – 3) > 0

⇒ x – 2 > 0 and x – 3 > 0

⇒ x > 2 and x < 3 ⇒ x > 3

Clearly the function is increasing in interval (3, ∞)

Again, (x – 2) (x – 3) > 0

or x – 2 < 0 and x – 3 < 0

⇒ x < 2 and x < 3

⇒ x < 2

Clearly the function is increasing in interval (-∞, 2)

Hence, the function is increasing in the interval (-∞, 2) ∪ (3, ∞)

Again for decreasing function of f(x),

f'(x) < 0

⇒ (x – 2) (x – 3) < 0

or x – 2 < 0 and x – 3 > 0

⇒ x < 2 and x > 3, which is impossible

or x – 2 > 0 and x – 3 < 0 ⇒ x > 2 and x < 3

⇒ 2 < x < 3

Hence, f(x) is decreasing function in the interval (2, 3).

Question 5.

(A) If x + y = 8, then find maximum value of xy?

Solution:

Let P = xy

⇒ x + y = 8

⇒ y = 8 – x

∴P = x(8 – x) = 8x – x^{2}

⇒ \(\frac { dP }{ dx } \) = 8 – 2x

⇒ \(\frac { d^{ 2 }P }{ dx^{ 2 } } \) = -2

For maximum or minimum value

8 – 2x = 0

⇒ x = 4

Now, x = 4 then \(\frac { d^{ 2 }P }{ dx^{ 2 } } \) = -2, which is negative

∴ at x = 4, then y = 4

Maximum value of P, when x = 4, y = 4

= 4 × 4 = 16.

(B) If x + y = 10 then find maximum value of xy?

Solution:

Solve like Q.No.5(A).

Question 6.

The radius of a circle increases at the rate of 3cm/sec. At what rate the area increases when the radius is 10cm?

Solution:

Let r be the radius and A be the area of circle then,

A = πr^{2}

\(\frac { dA }{ dt } \) = rate of change of area = ?

⇒ \(\frac { dA }{ dt } \) = 2πr \(\frac { dr }{ dt } \)

⇒ \(\frac { dA }{ dt } \) = 2π.(10).3

= 60π cm^{2}/second.

Question 7.

The volume of a cube is increasing at the rate of 9 cm^{3}/sec? If the edge of cube is 10cm then at what rate the surface area of cube is increasing? (NCERT)

Solution:

Let the edge of cube = a dm

Volufne = V= a^{3}

Surface area of cube = s = 6a^{2}

Surface area of cube

When a = 10cm, then \(\frac { ds }{ dt } \) = \(\frac { 36 }{ 10 } \) = 3.6 cm^{2}/sec.

Question 8.

(A) A man 180 cm high walks away from a lamp post at a rate of 1.2metre per second. If the height of lamp post is 4.5 metre. Find the rate at which the length of his shadow increases?

Solution:

Let AB be the lamp post of the height 4.5 m, PQ be the position of man at time t, CQ = x and BQ = y

drrf

(B) A man of height 2 metre walking away from a lamp post at a rate of 5km/ sec. If the height of the lamp post 6m. Find the rate at which the length of his shadow is increasing? (NCERT)

Solution:

Solve like Q. No.8(A).

[Ans. \(\frac{5}{8}\) km/sec.]

Question 9.

A ladder 5 metre 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 metre per second. How just is its height on the wall decreasing when the foot of the ladder is 4 metre away from the wall? (NCERT)

Solution:

Let at any time t, the bottom of the ladder be at a distance x metre from the wall and the height of the wall bey metre.

QA = xm, OB = ym, AB = 5m (given)

\(\frac { dx }{ dt } \) = 2m/sec.

In ∆OAB,

Question 10.

Find the intervals in which f(x) = 5x^{3} + 7x – 13 is increasing or decreasing?

Solution:

Given:

f(x) = 5x^{2} + 7x – 13 (given)

∴ f'(x) = 10x + 7

If f(x) is increasing,

f'(x) > 0

⇒ 10x + 7 > 0

⇒ x > \(\frac{-7}{10}\)

Hence, f(x) is increasing in interval ( \(\frac{-7}{10}\) , ∞)

If f(x) os decreasing, then,

f'(x) < 0

⇒ 10x + 7 < 0

⇒ x < \(\frac{-7}{10}\)

∴ f(x) is decreasing in (-∞, \(\frac{-7}{10}\) ).

Question 11.

Find the intervals in which the function f(x) = 2x^{3} – 24x + 5 is increasing or decreasing?

Solution:

f(x) = 2x^{3} – 24x + 5

Differentiating w.r.t x,

f'(x) = 6x^{2} – 24

⇒ f'(x) = 6(x^{2} – 4)

⇒ f'(x) = 6(x + 2) (x-2) …………………. (2)

(A) If f(x) in is increasing, then

f'(x) > 0

6(x + 2)(x-2) > 0

∴ f(x), x ∈ (-∞,-2) ∪(2, ∞) is increasing.

(B) If f(x) is decreasing,

f'(x) < 0 ⇒ 6(x + 2) (x – 2) < 0

f(x), x ∈ (-2,2) decreasing.

Question 12.

Show that the function f(x) = x – cos x is always increasing?

Solution:

Given function f(x) = x – cos x

∴ f'(x) = 1 – (- sin x)

⇒ f'(x) = 1 + sin x

We know that -1 ≤ x ≤ 1

-1 + 1 ≤ 1 + sin x ≤ 1 + 1

0 ≤ 1 + sin x ≤ 2

Hence f'(x) = 1 + sin x is always positive for all values of x.

∴ f(x) = x – cos x is always increasing.

Question 13.

Find the least value of a such that the function given by f(x) = x^{2} + ax + 1 is strictly increasing on (1, 2)? (NCERT)

Solution:

Given:

f(x) = x^{2} + ax + 1

∴ f'(x) = 2x + a

When, x ∈ (1,2)

∴ 1 < x < 2

⇒ 2 < 2x < 4

⇒ 2 + a < 2x + a < A + a

⇒ 2 + a < f'(x) < 4 + a Since f(x) is increasing function ∴ f'(x) > 0

⇒ 2 + a > 0 ⇒ a > -2

Hence, the least value of a is -2.

Question 14.

Let I be any interval disjoint from [-1, 1] prove that the function f given by f(x) = x + \(\frac{1}{x}\) is strictly increasing on I? (NCERT)

Solution:

f(x) = x + \(\frac{1}{x}\)

x ∈ 1, x ∉ (-1, 1)

x < – 1 0r x > 1

⇒ x^{2} – 1 > 0

⇒ \(\frac { x^{ 2 }-1 }{ x^{ 2 } } \) > 0

⇒ f'(x) > 0, x ∈ 1

∴ f(x) is increasing on I. Proved.

Question 15.

Find the interval in which the given function is increasing or decreasing:

f(x) = x^{4} – \(\frac { x^{ 3 } }{ 3 } \)?

Solution:

f(x) = x^{4} – \(\frac { x^{ 3 } }{ 3 } \)

Hence, the interval 0 and \(\frac{1}{4}\) on X – axis is divided in three intervals

(- ∞, 0), (0, \(\frac{1}{4}\) ), ( \(\frac{1}{4}\), ∞)

In interval (- ∞, 0)

f'(x) = x^{2} (4x – 1),

⇒ f'(x) < 0 [x^{2} = +ve] [4x – 1 > 0]

⇒ f'(x) < 0

∴ In interval (- ∞, 0) the function f(x) decreasing

In interval (0, \(\frac{1}{4}\) ):

f'(x) = x^{2} (4x – 1),

⇒ f'(x) > 0 [∵ x^{2} = positive] [4x – 1 > 0]

∴ In interval (0, \(\frac{1}{4}\) ) f(x) is decreasing

In interval ( \(\frac{1}{4}\), ∞):

f'(x) = x^{2} (4x – 1), [∵ x^{2} = positive] [4x – 1 > 0]

⇒ f'(x) > 0

In interval ( \(\frac{1}{4}\), ∞) in function f(x) increasing.

Question 16.

The perimeter of a rectangle is 100 cm. Find the length of sides of the rectangle for maximum area?

Solution:

Let length of rectangle be x and breadth be y.

∴Perimeter of rectangle = 2(x + y)

⇒ 2x + 2y = 100

⇒ x + y = 50

Let area of rectangle,

A = xy = x(50 – x) = 50x – x^{2}, [from eqn.(1)]

\(\frac { dA }{ dx } \) = 50 – 2x

and \(\frac { d^{ 2 }A }{ dx^{ 2 } } \) = -2

For maximum or minimum of A,

\(\frac { dA }{ dx } \) = 0

⇒ 50 – 2x = 0 or or x = 25

For any value of x, \(\frac { d^{ 2 }A }{ dx^{ 2 } } \) is -ve

For x = 25, area of rectangle is maximum

Put in eqn.(1),

y = 50 – x = 50 – 25 = 25.

Question 17.

Area of a rectangle is 25 sq.cm. Find its length and breadth when its perimeter is minimum?

Solution:

Let the length and breadth of rectangle be x and y units and A be the area.

xy = 25 ……………………….. (1)

Perimeter of rectangle,

P = 2(x + y)

For maximum or minimum \(\frac { dP }{ dx } \) = 0

Which is positive for x = 5

∴ For Minimum parimeter x = cm.

y = \(\frac{25}{x}\) = \(\frac{25}{5}\) = 5 cm.

Question 18.

Find the maximum value of sin x + cos x = \(\sqrt{2}\)? (NCERT)

Solution:

Given:

f(x) = sin x + cos x …………………. (1)

∴f'(x) = cos x – sin x …………………… (2)

and f”(x) = – sinx – cos x …………………….. (3)

For maxima or minima

f'(x) = 0

∴ cos x – sin x = 0

⇒ sin x = cos x

⇒ tan x = 1

∴ x = \(\frac { \pi }{ 4 } \), \(\frac { 3\pi }{ 4 } \), \(\frac { 5\pi }{ 4 } \)

Put x = \(\frac { \pi }{ 4 } \) in eqn (3)

Hence, at x = \(\frac { \pi }{ 4 } \) the given function is maximum. In the same way for x = \(\frac { 3\pi }{ 4 } \), \(\frac { 5\pi }{ 4 } \) ………………… the function is maximum.

Put x = \(\frac { \pi }{ 4 } \) in eqn.(1),

Maximum value f ( \(\frac { \pi }{ 4 } \) ) = sin \(\frac { \pi }{ 4 } \) + cos \(\frac { \pi }{ 4 } \)

= \(\frac { 1 }{ \sqrt { 2 } } \) + \(\frac { 1 }{ \sqrt { 2 } } \) = \(\frac { 2 }{ \sqrt { 2 } } \) = \(\sqrt{2}\). Proved.

Question 19.

Two positive numbers are such that x +y = 60 and xy^{3} is maximum. Prove (NCERT)

Solution:

Given:

x + y = 60 …………………….. (1)

Let p = xy^{3}

⇒ P = (60 – y). y^{3}

⇒ P = 60y^{3} – y^{4}

⇒ \(\frac{dp}{dy}\) = 180 y^{2} – 4y^{3} …………………….. (2)

Put \(\frac{dp}{dy}\) = 0

180 y^{2} – 4y^{2} = 0

⇒ 4y^{2} (45 – y) = 0

∴ y = 45

Differentiating eqn. (2) w.r.t y,

\(\frac { d^{ 2 }p }{ dy^{ 2 } } \) = 360y – 12y^{2
}= 12 y(30 – y)

Put y = 45,

\(\frac { d^{ 2 }p }{ dy^{ 2 } } \) = 12 × 45 (30 – 45) = – ve

Hence, at y = 45, p is maximum

From eqn.(1),

x + 45 = 60

⇒ x = 15

∴ The two numbers are 15 and 45.

Question 20.

Find the slope of the tangent to curve y = x^{3} – x + 1 at the point whose x coordinate is 2? (NCERT)

Solution:

The equation of given curve,

y = x^{3} – x + 1

Slope of tangent at x = 2,

= 3 × 4 – 1 = 11.

Question 21.

Find the slope of the tangent to the curve y = \(\frac { x-1 }{ x-2 } \), x ≠ 2 at x = 10? (NCERT)

Solution:

The equation of given curve:

At x = 10 slope of tangent is:

Question 22.

Find the point at which the tangent to curve y = x^{3} – 3x^{2} – 9x + 7 is parllel to X – axis? (NCERT)

Solution:

Equation of given curve y = x^{3} – 3x^{2} – 9x + 7

\(\frac { dy }{ dx } \) = \(\frac { d }{ dx } \) (x^{3} – 3x^{2} – 9x + 7)

⇒ \(\frac { dy }{ dx } \) = 3x^{2} – 6x – 9

= 3[x^{2} – 2x – 3]

= 3[x^{2} – 3x + x – 3]

= 3[x(x – 3) + 1(x – 3)]

⇒ \(\frac { dy }{ dx } \) = 3(x – 3) (x + 1)

Tangent is parllel to X – axis.

∴\(\frac { dy }{ dx } \) = 0

3 (x – 3) (x + 1) = 0

⇒ x = 3,-1

When x = 3, then y = (3)^{3} – 3(3)^{2} – 9 × 3 + 7

y = 27 – 27 – 27 + 7

y = – 20

When x = 3, then y = (-1)^{3} – 3(-1)^{2} – 9(-1) + 7

y = – 1 – 3 + 9 + 7

y = 12

Hence, the point at which the tangent is parllel to X – axis are (3, -20) and (-1, 12).

Question 23.

Find the equation of the tangent to the parabolas y^{2} = 4ax at the point (at^{2}, 2at)? (NCERT)

Solution:

Given:

y^{2} = 4ax ……………………. (1)

Equation of tangent is:

y – y_{1} = \(\frac { dy }{ dx } \) (x – x_{1})

Where x_{1} = at^{2}, y_{1} = 2at, \(\frac { dy }{ dx } \) = \(\frac { 1 }{ t } \)

y – 2at = \(\frac { 1 }{ t } \) (x – at^{2})

⇒ yt – 2at^{2} = x – at^{2}

⇒ x – ty + at^{2} = 0.

Question 24.

Find the equation of the tangent to the curve x^{2/3} + y^{2/3} = 2 at point (1, 1)? (NCERT)

Solution:

Equation of curve:

x^{2/3} + y^{2/3} = 2

Equation of tangent at point (x_{1}, y_{1}) is:

y – 1 = – (x – 1)

⇒ y – 1 = – x + 1

⇒ x + y – 2 = 0

Question 25.

Find the equation of tangent to the curve 2y + x^{2} = 3 at point (1, 1)? (NCERT)

Solution:

Equation of given curve:

2y + x^{2} = 3

Equation of tangent at point (x_{1}, y_{1}) is:

⇒ y – 1 = x – 1

⇒ x – y = 0.

Question 26.

(A) Find the equation of tangent to the cure x = cos t, y = sin at t = \(\frac { \pi }{ 4 } \)? (NCERT)

Solution:

Equation of 1^{st} curve:

x = cos t

x = cos t = cos \(\frac { \pi }{ 4 } \) = \(\frac { 1 }{ \sqrt { 2 } } \)

y = sin t = sin \(\frac { \pi }{ 4 } \) = \(\frac { 1 }{ \sqrt { 2 } } \)

Equation of tangent at (x_{1}, y_{1}) is:

(B) Find the equation of tangent and normal to the curve 16x^{2} + 9y^{2} = 145 at point (x_{1}, y_{1}), where x_{1} = 2 and y_{1} > 0? (CBSE 2018)

Solution:

Given:

16x^{2} + 9y^{2} = 145 ……………………. (1)

Put x = 2 in eqn.(1), we get

16.(2)^{2} +9y^{2} =145

⇒ 64 + 9y^{2} =145

⇒ 9y^{2} = 145 – 64 = 81

⇒ y^{2} =9, [y ≠ -3 ∵ y_{1} > 0]

⇒ y = 3

At point (2, 3)

\(\frac { dy }{ dx } \) = – \(\frac { 16 }{ 9 } \). \(\frac { 2 }{ 3 } \) = – \(\frac { 32 }{ 27 } \)

Equation of tangent of curve (1) at point (2, 3),

and equation of normal of curve (1) at point (2, 3) is,

Question 27.

Use differential to approximate (25)^{1/3}? (NCERT)

Solution:

Let y = x^{1/3}

Where, x = 27 and ∆x = -2

∆y is appromimately equal to dy.

Approximately value of (25)^{1/3}

= 3 + ∆y

= 3 – 0.074 = 2.926.

Question 28.

Use differential to approximate \(\sqrt { 36.6 } \)? (NCERT)

Solution:

Let y = \(\sqrt{x}\), where x = 36 and ∆x = 0.6

∆y is appromimately equal to b,

Approximate value of \(\sqrt { 36.6 } \)

= ∆y + 6

= 0.05 + 6 = 6.05

Question 29.

Use differential to approximate (15)^{1/4}? (NCERT)

Solution:

Let y = x^{1/4}

Where, x = 16 and ∆x = -1

∆y is appromimately equal to dy,

Approximately value of (15)^{1/4}

∆y + 2 = 2 – 0.031 = 1.969.

Question 30.

Use differential to approximate (26)^{1/3}? (NCERT)

Solution:

Let y = x^{1/3}

Where, x = 27 and ∆x = -1

∆y is appromimately equal to dy

Approximate value of (26)^{1/3}

= ∆y + 3

= 3 – 0.037 = 2.963

Question 31.

If the radius of the sphere is measured as 9cm with an error 0.03 cm then find the approximate error in calculating its surface area? (NCERT)

Solution:

Let the radius of circle = r

Given:

r = 9 cm, ∆r = 0.03 cm

Area of sphere

= (8πr) × 0.03

= 8π × 9 × 0.03 = 2.16 π cm^{2}

The approximate error in caluculating the surface area is = 2.16 π cm^{2}

Question 32.

If the radius of a sphere is measured as 7m with an error of 0.02m then find the approximate error in calculating its volume?

Solution:

Let the radius of sphere = r

and ∆r be the error in measuring the radius.

r = 7m, ∆r = 0.02 m (given)

Volume of sphere V = \(\frac{4}{3}\) πr^{3}

The approximate error in calculating the volume is 3.92π m^{3}

Question 33.

Find the approximate change in the volume V of a cube of sides x metre caused by increasing the side by 1%? (NCERT)

Solution:

Let x be the sides of cube.

Volume of cube V = x^{3}, ∆x = 1% of x = \(\frac{x}{100}\)

Change in volume ∆V = ( \(\frac { dV }{ dx } \) ) ∆x

Approximate change in volume = 0.03 x^{3}m^{3}.

Question 34.

Find the approximate change in voume V of a cube of side x metre caused by increasing the side by 2%

Solution:

Let x be the side of cube,

∆x = 2% of x = \(\frac { x\times 2 }{ 100 } \) = 0.02 x

Volume of cube V = x^{3}

\(\frac { dv }{ dx } \) = \(\frac { d }{ dx } \) x^{3} = 3x^{2}

dV = ( \(\frac { dv }{ dx } \) ) ∆x = 3x^{2} × 0.02 x = 0.06x^{3} m^{3}

Thus the approximate change in voume = 0.06 x^{3}m^{3}.