NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4.

• Application of Derivatives Class 12 Ex 6.1
• Application of Derivatives Class 12 Ex 6.2
• Application of Derivatives Class 12 Ex 6.3
• Application of Derivatives Class 12 Ex 6.5

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 6
Chapter Name	Application of Derivatives
Exercise	Ex 6.4
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4

Question 1.
Using differentials, find the approximate value of each of the following up to 3 places of decimal.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
(xiii)
(xiv)
(xv)
Solution:
(i) y + ∆y =
=
=
∴ x = 25
∆x = 0.3
⇒ y = √x

Question 2.
Find the approximate value of f (2.01), where f (x) = 4x² + 5x + 2
Solution:
f(x+∆x) = f(2.01), f(x) = f (2) = 4.2² + 5.2 + 2 = 28,
f’ (x) = 8x + 5 Now, f(x + ∆x) = f(x) + ∆f(x)
= f(x) + f’ (x) • ∆x = 28 + (8x + 5) ∆x
= 28 + (16 + 5) x 0.01
= 28 + 21 x 0.01
= 28 + 0.21
Hence,f(2 x 01)
= 28 x 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,
f(x) = f(5) = – 35
f(x + ∆x) = f(x) + ∆f(x) = f(x) + f'(x).∆x
= (x³ – 7x² + 15) + (3x² – 14x) × ∆x
f(5.001) = – 35 + (3 × 5² – 14 × 5) × 0.001
⇒ f (5.001) = – 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 meters.
Increase in side = 1% = 0.01 × x = 0.01 x
Volume of cube V= x³
∴ ∆v = × ∆x
= 3x² × 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 m;
Decrease in side = 1% = 0.01 x
Increase in side = ∆x = – 0.01 x
Surface area of cube = 6x² m² = S
∴ × ∆x = 12x × (- 0.01 x)
= – 0.12 x² m².

Question 6.
If the radius of a sphere is measured as 7m with an error of 0.02 m, then find the approximate error in calculating its volume.
Solution:
Radius of the sphere = 7m : ∆r = 0.02 m.
Volume of the sphere V =

= 4π × 7² × 0.02
= 3.92 πm³

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: ∆r = 0.03m
Surface area of sphere S = 4πr²
∆s = × ∆r
= 8πr × ∆r
= 8π × 9 × 0.03
= 2.16 πm².

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:
(d) x + ∆x = 3.02, where x=30, ∆x=.02,
∆f(x) = f(x + ∆x) – f(x)
⇒ f(x + ∆x) = f(x) + ∆f(x) = f(x) + f’ (x)∆x
Now f(x) = 3×2 + 15x + 5; f(3) = 77, f’ (x) = 6x + 15
f’ (3) = 33
∴ f (3.02) = 87 + 33 x 0 02 = 77.66

Question 9.
The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
(a) 0.06 x³ m³
(b) 0.6 x³ m³
(c) 0.09 x³ m³
(d) 0.9 x³ m³
Solution:
(c) Side of a cube = x meters
Volume of cube = x³,
for ∆x. ⇒ 3% of x = 0.03 x
Let ∆v be the change in v0l. ∆v = x ∆x = 3x² × ∆x
But, ∆x = 0.03 x
⇒ ∆v = 3x² x 0.03 x
= 0.09 x³m³

 

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