NCERT Solutions Class 12 maths Chapter-6 (Applications of Derivatives)Exercise 6.4

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-6 (Applications of Derivatives)Exercise 6.4 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 6.4

Question 1. Using differentials, find the approximate value of each 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

We take y = √x

Let x = 25 and △x = 0.3

△y = √(x +△x) – √x

= √25.3 – √25

= √25.3 – 5

or √25.3 = 5 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{1}{2√x}(0.3)$

= $\frac{1}{2√25}(0.3)$

= 0.3/10

= 0.03

Hence, the approximate value of√25.3 = 5 + 0.03 = 5.03

(ii)√49.5

We take y = √x

Let x = 49 and △x = 0.5

△y = √(x +△x) – √x

= √49.5 – √49

= √49.5 – 7

or √49.5 = 7 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{1}{2√x}(0.5)$

= $\frac{1}{2√49}(0.5)$

= 0.5/14

= 0.0357

Hence, the approximate value of √49.5 = 7 + 0.0357 = 7.0357

(iii) √0.6

We take y = √x

Let x = 0.64 and △x = -0.04

△y = √(x +△x) – √x

= √0.6 – √0.64

= √0.6 – 0.8

or √0.6 = 0.8 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{1}{2√x}(-0.04)$

= $\frac{1}{2√0.64}(-0.04)$

= -0.04/1.6

= -0.025

Hence, the approximate value of √0.6 = 0.8 – 0.025 = 0.775

(iv) (0.009)1/3

We take y = x1/3

Let x = 0.008 and △x = 0.001

△y = (x +△x)1/3 – x1/3

= (0.009)1/3 – (0.008)1/3

= (0.009)1/3 – 0.2

or (0.009)1/3 = 0.2 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{0.001}{3x^{\frac{2}{3}}}$

= $\frac{0.001}{3(0.008)^{\frac{2}{3}}}$

= 0.0083

Hence, the approximate value of (0.009)1/3 = 0.2 + 0.0083 = 0.2083

(v) (0.999)1/10

We take y = x1/10

Let x = 1 and △x = -0.001

△y = (x +△x)1/10 – x1/10

= (1)1/10 – (0.001)1/10

= (1)1/10 – 1

or (1)1/10 = 1 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{-0.001}{10x^{\frac{9}{10}}}$

= $\frac{-0.001}{10(1)^{\frac{9}{10}}}$

= -0.0001

Hence, the approximate value of (0.999)1/10 = 1 – 0.0001 = 0.9999

(vi) (15)1/4

We take y = x1/4

Let x = 16 and △x = -1

△y = (x +△x)1/4 – x1/4

= (15)1/4 – (16)1/4

= (15)1/4 – 2

or (15)1/4 = 2 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{-1}{4x^{\frac{3}{4}}}$

= $\frac{-1}{4(16)^{\frac{3}{4}}}$

= -1/32

Hence, the approximate value of (15)1/4 = 2 – 1/32 = 1.9687

(vii) (26)1/3

We take y = x1/3

Let x = 27 and △x = -1

△y = (x +△x)1/3 – x1/3

= (26)1/3 – (27)1/3

= (26)1/3 – 3

or (26)1/3 = 3 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{-1}{3x^{\frac{2}{3}}}$

= $\frac{-1}{3(27)^{\frac{2}{3}}}$

= -1/27

Hence, the approximate value of (26)1/3 = 3 – 1/27 = 2.962

(viii) (255)1/4

We take y = x1/4

Let x = 256 and △x = -1

△y = (x +△x)1/4 – x1/4

= (255)1/4 – (256)1/4

= (255)1/4 – 4

or (255)1/4 = 4 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{-1}{4x^{\frac{3}{4}}}$

= $\frac{-1}{4(256)^{\frac{3}{4}}}$

= -1/256

Hence, the approximate value of (255)1/4 = 4 – 1/256 = 3.996

(ix) (82)1/4

We take y = x1/4

Let x = 81 and △x = 1

△y = (x +△x)1/4 – x1/4

= (82)1/4 – (81)1/4

= (82)1/4 – 3

or (82)1/4 = 3 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{1}{4x^{\frac{3}{4}}}$

= $\frac{1}{4(81)^{\frac{3}{4}}}$

= 1/108

Hence, the approximate value of (82)1/4 = 3 + 1/108 = 3.009

(x) (401)1/2

We take y = x1/2

Let x = 400 and △x = 1

△y = (x +△x)1/2 – x1/2

= (401)1/2 – (400)1/2

= (401)1/2 – 20

or (401)1/2 = 20 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{1}{2\sqrt{x}}$

= $\frac{1}{2\sqrt{400}}$

= 1/40

Hence, the approximate value of (401)1/2 = 20 + 1/40 = 20.025

(xi) (0.0037)1/2

We take y = x1/2

Let x = 0.0036 and △x = 0.0001

△y = (x +△x)1/2 – x1/2

= (0.0037)1/2 – (0.0036)1/2

= (0.0037)1/2 – 0.06

or (0.0037)1/2 = 0.06 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{0.0001}{2\sqrt{x}}$

= $\frac{0.0001}{2\sqrt{0.0036}}$

= 0.0001/0.12

Hence, the approximate value of (0.0037)1/2 = 0.06 + 0.0001/0.12 = 0.0608

(xii) (26.57)1/3

We take y = x1/3

Let x = 27 and △x = -0.43

△y = (x +△x)1/3 – x1/3

= (26.57)1/3 – (27)1/3

= (26.57)1/3 – 3

or (26.57)1/3 = 3 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{-0.43}{3x^{\frac{2}{3}}}$

= $\frac{-0.43}{3(27)^{\frac{2}{3}}}$

= -0.43/27

Hence, the approximate value of (26.57)1/3 = 3 – 0.43/27 = 2.984

(xiii) (81.5)1/4

We take y = x1/4

Let x = 81 and △x = 0.5

△y = (x +△x)1/4 – x1/4

= (81.5)1/4 – (81)1/4

= (81.5)1/4 – 3

or (81.5)1/4 = 3 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{0.5}{4x^{\frac{3}{4}}}$

= $\frac{0.5}{4(81)^{\frac{3}{4}}}$

= 0.5/108

Hence, the approximate value of (81.5)1/4 = 3 + 0.5/108 = 3.004

(xiv) (3.968)3/2

We take y = x3/2

Let x = 4 and △x = -0.032

△y = (x +△x)3/2 – x3/2

= (3.968)3/2 – (4)3/2

= (3.968)3/2 – 8

or (3.968)3/2 = 8 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{(-0.032)(3)\sqrt{x}}{2}$

= $\frac{(-0.032)(3)\sqrt{4}}{2}$

= -0.096

Hence, the approximate value of (3.968)3/2 = 8 – 0.096 = 7.904

(xv) (32.15)1/5

We take y = x1/5

Let x = 32 and △x = 0.15

△y = (x +△x)1/5 – x1/5

= (32.15)1/5 – (32)1/5

= (32.15)1/5 – 2

or (32.15)1/5 = 2 + △y

Here, dy is approximately equal to △y

dy = (dy/dx)△x

= $\frac{0.15}{5x^{\frac{4}{5}}}$

= $\frac{0.15}{5(32)^{\frac{4}{5}}}$

= 0.15/80

Hence, the approximate value of (32.15)1/5 = 2 + 0.15/80 = 2.0018

Question 2. Find the approximate value of f(2.01), where f(x) = 4x²+ 5x + 2.

Solution:

Find the approximate value of f(2.01)

So, f(2.01) = f(2 + 0.01)

x = 2 and △x = 0.01

Given that f(x) = y = 4x²+ 5x + 2 -(1)

On differentiating we get

dy/dx = 8x + 5

dy = (8x + 5)dx = (8x + 5)△x

Now changing x = x + Δx and y = y +Δy in eq(1)

f(x + Δx) = y + Δy

f(2.01) = y + Δy -(2)

f(2 + .01) = y + Δy

since

then Δy = dy and Δx = dx

From eq(2), we get

f(2.01) = (4x²+ 5x + 2) + (8x + 5)△x

= (4(2)²+ 5(2) + 2) + (8(2) + 5)(0.01)

= 16 + 10 + 2 + 0.16 + 0.05

= 28.21

Question 3. Find the approximate value of f(5.001), where f(x) = x³– 7x²+ 15.

Solution:

Find the approximate value of f(5.001)

So, f(5.001) = f(5 + .001)

x = 5 and △x = 0.01,

Given that f(x) = y = x³– 7x²+ 15 -(1)

On differentiating we get

dy/dx = 3x²– 14x

dy = (3x²– 14x) dx = (3x²– 14x) △x

Now changing x = x + Δx and y = y +Δy in eq(1)

f(x + Δx) = y + Δy

f(5.001) = y + Δy -(2)

f(5 + .001) = y + Δy

since

then Δy = dy and Δx = dx

From eq(2), we get

f(5.001) = (x³– 7x²+ 15) + (3x²– 14x) △x

= ((5)³– 7(5)²+ 15) + (3(5)²– 14(5)) (0.001)

= 125 – 175 + 15 + 0.075 + 0.07

= -34.995

Question 4. Find the approximate change in the volume of a cube of side x meters caused by increasing the side by 1%.

Solution:

Volume of cube = x3

V = x3

On differentiating we get

dV/dx = 3x2

Given that Δx = 0.0

dV = $(\frac{dV}{dx}).Δx=3x.Δx$

dV = (3x2)(-0.01x)

= 0.03x3m3

Thus, the approximate change in volume is 0.03x3m3

Question 5.Find the approximate change in the surface area of a cube of side x meters caused by decreasing the side by 1%.

Solution:

Surface area of cube = 6x2

S = 6x2

On differentiating we get

dS/dx = 12x

Given that Δx = -0.01[-ve sign because of decrease]

dS = $(\frac{dS}{dx}).Δx=12x.Δx$

dS = (12x)(-0.01x)

= -0.12x2m2

Thus, the approximate change in volume is -0.12x2m2

Question 6. If the radius of a sphere is measured as 7m with an error of 0.02m, then find the approximate error in calculating its value.

Solution:

Let r be the radius of the sphere and △r be the error in measuring the radius,

then r = 7m and △r = 0.02m.

Now the volume V of the sphere is given by,

v = 4/3πr2

On differentiating we get

$\frac{dv}{dr}=4πr^2$

Therefore, dv = (4πr2)Δr = 4π(7)2.(0.02)

= 12.30m3

Thus, the approximate error in calculating the volume is 12.30m3

Question 7. If the radius of a sphere is measured as 9m with an error of 0.03m, then find the approximate error in calculating its surface area.

Solution:

Let r be the radius of the sphere and Δr be the error in measuring the radius.

Then r = 9m and Δr = 0.03m.

Now the surface area S of the sphere is given by,

S = 4πr²

On differentiating we get

dS/dr = 8πr

Therefore, dS = (8πr)Δr

= 8π(9).(0.03)

= 2.16πm²

Thus, the approximate error in measuring the surface area is 2.16πm².

Question 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:

Given: f(x) = y = 3x²+ 15x + 5 -(1)

f(3.02) = f(3 + 0.2)

So, x = 3 and Δx = 0.02

On differentiating eq(1) we get

f'(x) = y = 6x+ 15

dy = (6x + 15) dx = (6x + 15)Δx

Now changing x = x + Δx and y = y +Δy in eq(1)

f(x + Δx) = y + Δy

f(3.02) = y + Δy -(2)

f(3 + 0.2) = y + Δy

since

then Δy = dy and Δx = dx

From eq(2), we get

f(3.02) = (3x²+ 15x + 5) + (6x + 15)Δx

f(3.02) = (3x²+ 15x + 5) + (6x + 15)(0.02)

= (3(3)²+ 15(3) + 5) + (6(3)+ 15)(0.02)

= 27 + 45 + 5 + 0.36 + 0.3

= 77.66

Hence the correct option is 77.66

Question 9. The approximate change in the volume of a cube of side x meters caused by increasing the side by 3% is

(A) 0.06x³ m³ (B) 0.6x³ m³ (C)0.09x³ m³ (D)0.9x³ m³

Solution:

We have

Volume of cube = v = x3

On differentiating we get

$\frac{dv}{dr}=3x^2$

We have

Volume of cube = v = x3

On differentiating we get

Given that the side increasing 3%, so Δx = 0.03x

$dv=(\frac{dv}{dr}).Δx=3x^2.(0.03x)$

$dv=x^2(0.03x)=0.09x^3m^3$

Thus, the approximate change is 0.09x3m3

The correct option is (C)

NCERT Solutions Class 12 maths Chapter-6 (Applications of Derivatives)Exercise 6.4

NCERT Solutions Class 12 maths Chapter-6 (Applications of Derivatives)Exercise 6.4

Exercise 6.4

Chapter-6 (Applications of Derivatives)

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NCERT Solutions Class 12 maths Chapter-6 (Applications of Derivatives)Exercise 6.4

NCERT Solutions Class 12 maths Chapter-6 (Applications of Derivatives)Exercise 6.4

Exercise 6.4

Chapter-6 (Applications of Derivatives)

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