Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4

Question 1.

Using differentials, find the approximate value of each of the following upto 3 places of decimal :

(i) \(\sqrt{25.3}\)

Solution:

(ii) \(\sqrt{49.5}\)

Solution:

(iii) \(\sqrt{0.6}\)

Solution:

(iv) (0.009)^{\(\frac { 1 }{ 3 }\)}

Solution:

Let y = x^{\(\frac { 1 }{ 3 }\)}

Also, let x = 0.008, ∆x = 0.001,

so, that x + ∆x = 0.009.

∴ Approximate value of ∆y = dy = 0.008.

Hence, from (1), we have :

(0.009)^{\(\frac { 1 }{ 3 }\)} – o.2 + 0.008 = 0.208 (approx.)

(v) (0.999)^{\(\frac { 1 }{ 10 }\)}

Solution:

(vi) (15)^{\(\frac { 1 }{ 4 }\)}

Solution:

Let y = ^{\(\frac { 1 }{ 4 }\)}.

Also, let x = 16, ∆x = – 1, so that

x + ∆x = 15.

∴ Approximate value of ∆y = dy = 0.03125.

Hence, from (1), we have :

(15)^{\(\frac { 1 }{ 4 }\)} = 2 – 0.03125 = 1.969 (approx)

(vii) (26)^{\(\frac { 1 }{ 3 }\)}

Solution:

Let y = x^{\(\frac { 1 }{ 3 }\)}.

Also, let x = 27, ∆x = – 1, so that

x + ∆x = 26

Now, ∆y = (x + ∆x)³-x³

= (26)^{\(\frac { 1 }{ 3 }\)} – (27)^{\(\frac { 1 }{ 3 }\)} = (26)^{\(\frac { 1 }{ 3 }\)} – 3.

⇒ (26)^{\(\frac { 1 }{ 3 }\)} = 3 + ∆y

∴ Approximate value of ∆y = dy = – 0.037037.

Hence, from (1), we have :

(26)^{\(\frac { 1 }{ 3 }\)} = 3 – 0.37037

= 2.963 (approx.).

(viii) (255)^{\(\frac { 1 }{ 4 }\)}

Solution:

Let y = f(x) = x^{\(\frac { 1 }{ 4 }\)}

Let x = 256, x + ∆x =255, so that ∆x = 255 – 256 = – 1.

Approximate value of ∆y = dy = – 0.0039.

Hence, from (1), we have :

(255)^{\(\frac { 1 }{ 4 }\)}= 4 – 0.0039 = 3.9961.

(ix) (82)^{\(\frac { 1 }{ 4 }\)}

Solution:

(x) (401)^{\(\frac { 1 }{ 2 }\)}

Solution:

(xi) (0.0037)^{\(\frac { 1 }{ 2 }\)}

Solution:

Let y = \(\sqrt{x}\)

Also, let x = 0.0036 and ∆x = 0.0001, so that x + ∆x = 0.0037

Approximate value of ∆y = dy = 0.000833.

Hence, from (1), we have :

\(\sqrt{0.0037}\) = 0.06 + 0.00083

= 0.061 (approx.)

(xii) (26.57)^{\(\frac { 1 }{ 3 }\)}

Solution:

(xiii) (81.5)^{\(\frac { 1 }{ 4 }\)}

Solution:

Let y = x^{\(\frac { 1 }{ 4 }\)}, x = 81 and ∆x = 0.5

(xiv) (3.968)^{\(\frac { 3 }{ 2 }\)}

Solution:

(xv) (32.15)^{\(\frac { 1 }{ 5 }\)}

Solution:

Question 2.

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

Solution:

Let x + ∆x = 2.01, x = 2 and ∆x = 0.01.

So, f(x + ∆x) = f(2.01) and

f(x) = f(2) = 4 . 2² + 5.2 + 2

= 16 + 10 + 2 = 28

Now, f(x) = 8x + 5.

Also, f(x + ∆x) = f(x) + ∆f(x)

= f(x) + f'(x). ∆x

= 28 + (8x + 5) x ∆x

= 28 + (16 + 5) x 0.01

= 28 + 21 x 0.01 = 28 + 0.21 =28.21.

Hence, f(2.01) = 28.21.

Question 3.

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

Solution:

Let x + ∆x = 5.001, x = 5 and ∆x = 0.001.

So, f(x) = f(5) = 5³ – 7 x 5² + 15

= 125-175 +15 = -35.

Also, f(x + ∆x) = f(x) + ∆f(x) = f(x) + f(x). ∆x

= (x³ – 7x² + 15) + (3x² – 14x) x ∆x

∴ f(5.001) = – 35 + (3 x 5² – 14 x 5) x 0.001

= – 35+ (75 – 14 x 5) x 0.001

= – 35 + 5 x 0.001

= – 35 + 0.005

= – 34.995.

Question 4.

Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.

Solution:

The side of the cube = x metres.

∴ Increase in side = 1% = 0.01 × x = 0.01 x

Volume of cube V =x³

∴ Approximate change in volume = AV

= \(\frac { dv }{ dx }\) x ∆x

= 3x² x 0.01 x = 0.03 x³ m³.

Question 5.

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

Solution:

The side of the cube = x metres.

∴ Decrease in side = 1% = 0.01x

So, Increase in side = ∆x = – 0.01 x.

Surface area of cube = 6x² m² = s (say)

Approximate change in surface area of cube = ∆s

= \(\frac { ds }{ dx }\) x ∆x

= 12x × (- 0.01 x) = – 0.12 x² m².

Question 6.

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

Solution:

Radius of the sphere = 7 m.

Error in measurement of radius = ∆r = 0.02 m.

Volume of the sphere V = \(\frac { 4 }{ 3 }\)πr³

= 4 π r² x ∆r

= 4π x 7² x 0.02 = 3.92 π m³.

Question 7.

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

Solution:

Radius of the sphere = 9 m.

Error in measurement = ∆r = 0.03 m.

Surface area of sphere S = 4 π r²

∴ ∆S = \(\frac { dS }{ dr }\) x ∆r = 8π r x ∆r dr

= 8π x 9 x 0.03

= 2.16 π m².

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:

x + ∆x = 3.02, where x = 3, ∆x = 0.02.

∆ f(x) = f(x + ∆x) – f(x)

⇒ f(x + ∆x) = f(x) + ∆ f(x)

= f(x) + f'(x) ∆x … (1)

Now, f(x) = 3x² + 15x + 5.

So, f(3) = 3 . 3² + 15.3 + 5 = 27 + 45 + 5 = 77

and f'(x) = 6x + 15.

⇒ f'(3) = 6 x 3 + 15 = 18 + 15 = 33.

Putting these values in (1), we get

∴ f(3.02) = 87 + 33 x 0.02 = 77 + 0.66 = 77.66.

∴ Part (D) is the correct answer.

Question 9.

The approximate change in volume of a cube of side x metres caused by increasing the side by 3% is

(A) 0.06 x³m³

(B) 0.6 x³m³

(C) 0.09 x³m³

(D) 0.9 x³m³

Solution:

Side of a cube = x metres.

Volume of cube = x³m³

Now, ∆x = 3% of x = 0.03 x.

Let ∆V be the change in volume.

So, ∆V = \(\frac { dv }{ dx }\) x ∆x = 3x² x ∆x

But ∆x = 0.03x.

∴ ∆V = 3x² x 0.03x = 0.09 x³m³.

∴ Part (C) is the correct answer.