Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Ex 8(e) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Exercise 8(e)

Question 1.

Determine the differentials in each of the following cases.

(i) y = x^{3} – 1

Solution:

y= x^{3} – 1

Then dy = 3x^{2} dx

(ii) y = sin^{2} x

Solution:

y = sin^{2} x

Then dy = 2 sin x cos x dx = sin 2x dx

(iii) y = \(\frac{1+\sqrt{x}}{1-\sqrt{x}}\)

Solution:

(iv) z = cos 2t – 2 cot t

Solution:

z = cos 2t – 2 cot t

dz = (-2 sin 2t + 2 cosec^{2} t) dt

(v) r = \(\frac{4}{1+\sin \theta}\)

Solution:

r = \(\frac{4}{1+\sin \theta}\)

dz = \(\frac{-4 \cos \theta}{(1+\sin \theta)^2}\) . dθ

= \(-\frac{4 \cos \theta}{(1+\sin \theta)^2}\) dθ

(vi) x^{2}y = 2

Solution:

⇒ y = \(\frac{2}{x^2}\)

dy = \(\frac{4}{x^3}\)

(vii) xy^{2} + yx^{2} = 1

Solution:

⇒ dx . y^{2} + x . 2y dy + dy . x^{2} + y . 2x dx = 0

⇒ (2 xy + x^{2}) dy = – (y^{2} + 2xy) dx

⇒ dy = –\(\frac{y^2+2 x y}{x^2+2 x y}\)dx

Question 2.

Find δf and df when

(i) f(x) = 2x^{2} – 1, x = 1, δx = 0.02

Solution:

f(x) = 2x^{2} – 1, x = 1, δx = 0.02

Then δf = f(x + δx) – f(x)

= f(1.02) – f(1)

= 2 (1.02)^{2} – 1 – (2 – 1)

= 2.0808 – 2 = 0.0808

Again df = 4x dx

= 4 × 1 × 0.02 = 0.08

(ii) f(x) = √x, x = 16, δx = 0.3

Solution:

f(x) = √x, x = 16, δx = 0.3

(iii) f(x) = (x + 1)^{3}, x = 8, δx = 0.04

Solution:

f(x) = (x + 1)^{3}, x = 8, δx = 0.04

δf = f(x + δx) – f(x)

= (9.04)^{3} – (8 +1)^{3}

= (9.04)^{3} – 9^{3} = 9.7632

Again df = 3(x + 1)^{2} dx = 3 × 9^{2} × 0.04

= 243 × 0.04 = 9.72

(iv) f(x) = In (1 + x), x = 1, δx = 0.04

Solution:

f(x) = In (1 + x), x = 1, δx = 0.04

δf = f(x + 8x) – f(x)

= In (1 + x + 5x) – ln (1 + x)

= In (2.04) – In (2) = 0.0198

Again df = \(\frac{1}{1+x}\)dx = \(\frac{1}{2}\) × 0.04 = 0.02

Question 3.

Find approximate values of the following:

(i) \(\sqrt[3]{28}\)

Solution:

(ii) \(\sqrt[6]{63}\)

Solution:

(iii) \(\sqrt{48 \cdot 96}\)

Solution:

(iv) (1.99)^{7}

Solution:

Let y = x^{7}

Then dy = 7x^{6} dx

⇒ δy = 7x^{6} δx

⇒ (x + δx)^{7} – x^{7} = 7x^{6} δx

(x + δx)^{7} = x^{7} + 7x^{6} δx

Put x = 2 and δx = -0.01

Then (1.99)^{7} = 2^{7} – 7 × 2^{6} × 0.01

= 128 – 7 × 64 × 0.01

= 128 – 4.48 = 123.52

(v) 2^{3.02}

Solution:

Let y = 2^{x}

Then dy = 2^{x} In 2 dx

⇒ δy = 2^{x} In 2 . δx

⇒ 2^{x + δx} – 2^{x} = 2^{x} In 2 . δx

⇒ 2^{x + δx} = 2^{x} + 2^{x} In 2 δx

Then 2^{3.02 }= 2^{3} + 2^{3} In 2 × 0.02

= 8 + 8 In 2 × 0.02 = 8.1109

(vi) sin 59°

Solution:

Let y = sin x

Then dy = cos x dx ⇒ δy = cos x . δx

⇒ sin (x + δx) – sin x = cos x × δx

⇒ sin (x + δx) = sin x + cos x × δx

Put x = 60°, δx = -1°

Then sin 59° = sin 60° + cos 60° x – \( \frac{\pi}{180} \)

\(\frac{\sqrt{3}}{2}\) – \(\frac{1}{2}\) × \(\frac{\pi}{180}\)

[ ∵ 1° = \(\frac{\pi}{180}\) radian = 0.85729

Question 4.

find the percentage of error in calculation of the surface area of a spherical balloon of diameter 14.02 m. if the true diameter is 14m.

Solution:

Let S be the surface area of a spherical balloon of radius V.

Then S = 4πr^{2}

Then dS = 8πr dr

Question 5.

Find approximately the difference between the volumes of two cubes of sides 3 cm and 3.04 cm.

Solution:

Let V be the volume of a cube of side x cm.

Then V = x^{3} ⇒ dV = 3x^{2} dx

Here x = 3 cm and dx = (3.04 – 3) cm = 0.04 cm

Thus dV = 3 × 9 × 0.04 = 1.08 cm^{3}

∴ Difference of two volumes is 1.08 cm^{3}.

Question 6.

The height of a regular cone is 3 times the radius of its base. The radius of the base was wrongly measured to be 5 cm. whereas its true radius is 4.88 cm. Find the relative error in measuring the curved surface area of the cone.

Ans.

Consider a regular cone of height h and radius of its base x.

Then h = 3x.

If S is the area of the curved surface of the cone

then S = πxl

Where l is the slant height of the cone.

= 0.049