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 y2 = 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 x3
(b) 0.02 x3
(c) 0.06 x3
(d) 0.09 x3
Answer:
(c) 0.06 x3

MP Board Solutions

Question 4.
Point on the curve x2 = 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) = x4 – x2 – 2x + 6:
(a) 6
(b) 4
(c) 8
(d) None of these.
Answer:
(b) 4

Question 2.
Fill in the blanks:

  1. The function f(x) = cosx, for 0 ≤ x ≤ π is …………………………..
  2. 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 ………………………………
  3. The function y = x(5 – x) is maximum at x is equal to ……………………………
  4. The minimum value of 2x + 3y is ………………………….. when xy = 6.
  5. The maximum value of sin x + cos x is ……………………………
  6. If line y = mx+ 1 is a tangent line to the curve y2 = 4x. Then value of m will be ……………………………..
  7. Slope of the tangent to the curve y = x2 at the point (1, 1) is ………………………….
  8. Using differential the approximate value of \(\sqrt{0.6}\) is …………………………….

Answer:

  1. Decreasing
  2. 4πcm2/sec
  3. \(\frac{5}{2}\)
  4. 12
  5. \(\sqrt{2}\)
  6. 1,
  7. 2,
  8. 0.8.

MP Board Solutions

Question 3.
Write True/False:

  1. For all real values of x, the function f(x) = ex – e-x is increasing function.
  2. If the length of equal sides of an isosceles traingle be x then its maximum area will be \(\frac{1}{2}\) x2
  3. Function f(x) = 3x2 – 4x is increasing in the interval (- ∞, \(\frac{2}{3}\) )
  4. Function f(x) = x – cot x is always decreasing.
  5. Equation of the normal of the curve y = ex at point (0, 1) is x + y = l.

Answer:

  1. True
  2. True
  3. False
  4. False
  5. 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,
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 1
∵ Area of circle A = πr2
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 2
= 40π sq.cm/second

MP Board Solutions

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 = \(\frac{4}{3}\) πr3
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 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 = 4πr2
Differentiating eqn.(1) w.r.t. t,
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 4
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) = 2x3 – 15x2 + 36x + 1 is increasing or decreasing?
Solution:
f(x) = 2x3 – 15x2 + 36x + 1
⇒ f'(x) = 6x2 – 30x + 36
= 6(x2 – 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).

MP Board Solutions

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 – x2
⇒ \(\frac { dP }{ dx } \) = 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 = πr2
\(\frac { dA }{ dt } \) = rate of change of area = ?
⇒ \(\frac { dA }{ dt } \) = 2πr \(\frac { dr }{ dt } \)
⇒ \(\frac { dA }{ dt } \) = 2π.(10).3
= 60π cm2/second.

MP Board Solutions

Question 7.
The volume of a cube is increasing at the rate of 9 cm3/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= a3
Surface area of cube = s = 6a2
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 5
Surface area of cube
MP Board Class 12th Maths Important Questions ChapterMP Board Class 12th Maths Important Questions Chapter 6 Application of Derivati 6 img oo
When a = 10cm, then \(\frac { ds }{ dt } \) = \(\frac { 36 }{ 10 } \) = 3.6 cm2/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,
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 8

Question 10.
Find the intervals in which f(x) = 5x3 + 7x – 13 is increasing or decreasing?
Solution:
Given:
f(x) = 5x2 + 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 (-∞, \(\frac{-7}{10}\) ).

MP Board Solutions

Question 11.
Find the intervals in which the function f(x) = 2x3 – 24x + 5 is increasing or decreasing?
Solution:
f(x) = 2x3 – 24x + 5
Differentiating w.r.t x,
f'(x) = 6x2 – 24
⇒ f'(x) = 6(x2 – 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.

MP Board Solutions

Question 13.
Find the least value of a such that the function given by f(x) = x2 + ax + 1 is strictly increasing on (1, 2)? (NCERT)
Solution:
Given:
f(x) = x2 + 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 + \(\frac{1}{x}\)
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 9
x ∈ 1, x ∉ (-1, 1)
x < – 1 0r x > 1
⇒ x2 – 1 > 0
⇒ \(\frac { x^{ 2 }-1 }{ x^{ 2 } } \) > 0
⇒ f'(x) > 0, x ∈ 1
∴ f(x) is increasing on I. Proved.

MP Board Solutions

Question 15.
Find the interval in which the given function is increasing or decreasing:
f(x) = x4 – \(\frac { x^{ 3 } }{ 3 } \)?
Solution:
f(x) = x4 – \(\frac { x^{ 3 } }{ 3 } \)
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 10
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 10a
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) = x2 (4x – 1),
⇒ f'(x) < 0 [x2 = +ve] [4x – 1 > 0]
⇒ f'(x) < 0
∴ In interval (- ∞, 0) the function f(x) decreasing
In interval (0, \(\frac{1}{4}\) ):
f'(x) = x2 (4x – 1),
⇒ f'(x) > 0 [∵ x2 = positive] [4x – 1 > 0]
∴ In interval (0, \(\frac{1}{4}\) ) f(x) is decreasing
In interval ( \(\frac{1}{4}\), ∞):
f'(x) = x2 (4x – 1), [∵ x2 = positive] [4x – 1 > 0]
⇒ f'(x) > 0
In interval ( \(\frac{1}{4}\), ∞) in function f(x) increasing.

MP Board Solutions

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 – x2, [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.

MP Board Solutions

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)
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 11
For maximum or minimum \(\frac { dP }{ dx } \) = 0
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 12
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)
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 13
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 xy3 is maximum. Prove (NCERT)
Solution:
Given:
x + y = 60 …………………….. (1)
Let p = xy3
⇒ P = (60 – y). y3
⇒ P = 60y3 – y4
⇒ \(\frac{dp}{dy}\) = 180 y2 – 4y3 …………………….. (2)
Put \(\frac{dp}{dy}\) = 0
180 y2 – 4y2 = 0
⇒ 4y2 (45 – y) = 0
∴ y = 45
Differentiating eqn. (2) w.r.t y,
\(\frac { d^{ 2 }p }{ dy^{ 2 } } \) = 360y – 12y2
= 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 = x3 – x + 1 at the point whose x coordinate is 2? (NCERT)
Solution:
The equation of given curve,
y = x3 – x + 1
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 15
Slope of tangent at x = 2,
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 16
= 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:
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 17
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 17a
At x = 10 slope of tangent is:
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 18

Question 22.
Find the point at which the tangent to curve y = x3 – 3x2 – 9x + 7 is parllel to X – axis? (NCERT)
Solution:
Equation of given curve y = x3 – 3x2 – 9x + 7
\(\frac { dy }{ dx } \) = \(\frac { d }{ dx } \) (x3 – 3x2 – 9x + 7)
⇒ \(\frac { dy }{ dx } \) = 3x2 – 6x – 9
= 3[x2 – 2x – 3]
= 3[x2 – 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).

MP Board Solutions

Question 23.
Find the equation of the tangent to the parabolas y2 = 4ax at the point (at2, 2at)? (NCERT)
Solution:
Given:
y2 = 4ax ……………………. (1)
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 19
Equation of tangent is:
y – y1 = \(\frac { dy }{ dx } \) (x – x1)
Where x1 = at2, y1 = 2at, \(\frac { dy }{ dx } \) = \(\frac { 1 }{ t } \)
y – 2at = \(\frac { 1 }{ t } \) (x – at2)
⇒ yt – 2at2 = x – at2
⇒ x – ty + at2 = 0.

Question 24.
Find the equation of the tangent to the curve x2/3 + y2/3 = 2 at point (1, 1)? (NCERT)
Solution:
Equation of curve:
x2/3 + y2/3 = 2
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 20
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 20a
Equation of tangent at point (x1, y1) is:
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 21
y – 1 = – (x – 1)
⇒ y – 1 = – x + 1
⇒ x + y – 2 = 0

Question 25.
Find the equation of tangent to the curve 2y + x2 = 3 at point (1, 1)? (NCERT)
Solution:
Equation of given curve:
2y + x2 = 3
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 22
Equation of tangent at point (x1, y1) is:
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 23
⇒ 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 1st curve:
x = cos t
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 24
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 (x1, y1) is:
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 25

(B) Find the equation of tangent and normal to the curve 16x2 + 9y2 = 145 at point (x1, y1), where x1 = 2 and y1 > 0? (CBSE 2018)
Solution:
Given:
16x2 + 9y2 = 145 ……………………. (1)
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 26
Put x = 2 in eqn.(1), we get
16.(2)2 +9y2 =145
⇒ 64 + 9y2 =145
⇒ 9y2 = 145 – 64 = 81
⇒ y2 =9, [y ≠ -3 ∵ y1 > 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),
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 27
and equation of normal of curve (1) at point (2, 3) is,
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 28

Question 27.
Use differential to approximate (25)1/3? (NCERT)
Solution:
Let y = x1/3
Where, x = 27 and ∆x = -2
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 29
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 29a
∆y is appromimately equal to dy.
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 30
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
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 31
∆y is appromimately equal to b,
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 32
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 = x1/4
Where, x = 16 and ∆x = -1
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 33
∆y is appromimately equal to dy,
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 34
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 = x1/3
Where, x = 27 and ∆x = -1
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 35
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 35a
∆y is appromimately equal to dy
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 36
Approximate value of (26)1/3
= ∆y + 3
= 3 – 0.037 = 2.963

MP Board Solutions

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
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 37
= (8πr) × 0.03
= 8π × 9 × 0.03 = 2.16 π cm2
The approximate error in caluculating the surface area is = 2.16 π cm2

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 = \(\frac{4}{3}\) πr3
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 38
The approximate error in calculating the volume is 3.92π m3

MP Board Solutions

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 = x3, ∆x = 1% of x = \(\frac{x}{100}\)
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 39
Change in volume ∆V = ( \(\frac { dV }{ dx } \) ) ∆x
MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives img 40
Approximate change in volume = 0.03 x3m3.

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 = x3
\(\frac { dv }{ dx } \) = \(\frac { d }{ dx } \) x3 = 3x2
dV = ( \(\frac { dv }{ dx } \) ) ∆x = 3x2 × 0.02 x = 0.06x3 m3
Thus the approximate change in voume = 0.06 x3m3.

MP Board Class 12 Maths Important Questions