# GSEB Solutions Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4

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, âˆ†x = 0.001,
so, that x + âˆ†x = 0.009.

âˆ´ Approximate value of âˆ†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, âˆ†x = – 1, so that
x + âˆ†x = 15.

âˆ´ Approximate value of âˆ†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, âˆ†x = – 1, so that
x + âˆ†x = 26
Now, âˆ†y = (x + âˆ†x)Â³-xÂ³
= (26)$$\frac { 1 }{ 3 }$$ – (27)$$\frac { 1 }{ 3 }$$ = (26)$$\frac { 1 }{ 3 }$$ – 3.
â‡’ (26)$$\frac { 1 }{ 3 }$$ = 3 + âˆ†y

âˆ´ Approximate value of âˆ†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 + âˆ†x =255, so that âˆ†x = 255 – 256 = – 1.

Approximate value of âˆ†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 âˆ†x = 0.0001, so that x + âˆ†x = 0.0037

Approximate value of âˆ†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 âˆ†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) = 4xÂ² + 5x +2.
Solution:
Let x + âˆ†x = 2.01, x = 2 and âˆ†x = 0.01.
So, f(x + âˆ†x) = f(2.01) and
f(x) = f(2) = 4 . 2Â² + 5.2 + 2
= 16 + 10 + 2 = 28
Now, f(x) = 8x + 5.
Also, f(x + âˆ†x) = f(x) + âˆ†f(x)
= f(x) + f'(x). âˆ†x
= 28 + (8x + 5) x âˆ†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) = xÂ³ – 7xÂ² +15.
Solution:
Let x + âˆ†x = 5.001, x = 5 and âˆ†x = 0.001.
So, f(x) = f(5) = 5Â³ – 7 x 5Â² + 15
= 125-175 +15 = -35.
Also, f(x + âˆ†x) = f(x) + âˆ†f(x) = f(x) + f(x). âˆ†x
= (xÂ³ – 7xÂ² + 15) + (3xÂ² – 14x) x âˆ†x
âˆ´ f(5.001) = – 35 + (3 x 5Â² – 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.
âˆ´ Increase in side = 1% = 0.01 Ã— x = 0.01 x
Volume of cube V =xÂ³
âˆ´ Approximate change in volume = AV
= $$\frac { dv }{ dx }$$ x âˆ†x
= 3xÂ² x 0.01 x = 0.03 xÂ³ mÂ³.

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.
âˆ´ Decrease in side = 1% = 0.01x
So, Increase in side = âˆ†x = – 0.01 x.
Surface area of cube = 6xÂ² mÂ² = s (say)
Approximate change in surface area of cube = âˆ†s
= $$\frac { ds }{ dx }$$ x âˆ†x
= 12x Ã— (- 0.01 x) = – 0.12 xÂ² mÂ².

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 = âˆ†r = 0.02 m.
Volume of the sphere V = $$\frac { 4 }{ 3 }$$Ï€rÂ³
= 4 Ï€ rÂ² x âˆ†r
= 4Ï€ x 7Â² x 0.02 = 3.92 Ï€ mÂ³.

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 = âˆ†r = 0.03 m.
Surface area of sphere S = 4 Ï€ rÂ²
âˆ´ âˆ†S = $$\frac { dS }{ dr }$$ x âˆ†r = 8Ï€ r x âˆ†r dr
= 8Ï€ x 9 x 0.03
= 2.16 Ï€ mÂ².

Question 8.
If f(x) = 3xÂ² + 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 + âˆ†x = 3.02, where x = 3, âˆ†x = 0.02.
âˆ† f(x) = f(x + âˆ†x) – f(x)
â‡’ f(x + âˆ†x) = f(x) + âˆ† f(x)
= f(x) + f'(x) âˆ†x … (1)
Now, f(x) = 3xÂ² + 15x + 5.
So, f(3) = 3 . 3Â² + 15.3 + 5 = 27 + 45 + 5 = 77
and f'(x) = 6x + 15.
â‡’ f'(3) = 6 x 3 + 15 = 18 + 15 = 33.
Putting these values in (1), we get
âˆ´ f(3.02) = 87 + 33 x 0.02 = 77 + 0.66 = 77.66.
âˆ´ 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 xÂ³mÂ³
(B) 0.6 xÂ³mÂ³
(C) 0.09 xÂ³mÂ³
(D) 0.9 xÂ³mÂ³
Solution:
Side of a cube = x metres.
Volume of cube = xÂ³mÂ³
Now, âˆ†x = 3% of x = 0.03 x.
Let âˆ†V be the change in volume.
So, âˆ†V = $$\frac { dv }{ dx }$$ x âˆ†x = 3xÂ² x âˆ†x
But âˆ†x = 0.03x.
âˆ´ âˆ†V = 3xÂ² x 0.03x = 0.09 xÂ³mÂ³.
âˆ´ Part (C) is the correct answer.