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

Get Free NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Application of Derivatives Exercise 6.4 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Application of Derivatives Ex 6.4 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Applications of Derivatives chapter are the following:

Section Name	Topic Name
6	Applications of Derivatives
6.1	Introduction
6.2	Rate of Change of Quantities
6.3	Increasing and Decreasing Functions
6.4	Tangents and Normals
6.5	Approximations
6.6	Maxima and Minima
6.7	Maximum and Minimum Values of a Function in a Closed Interval
6.8	Summary

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

Ex 6.4 Class 12 Maths Question 1.
Using differentials, find the approximate value of each of the following up to 3 places of decimal.
(i) $\sqrt { 25.3 }$
(ii) $\sqrt { 49.5 }$
(iii) $\sqrt { 0.6 }$
(iv) ${ \left( 0.009 \right) }^{ \frac { 1 }{ 3 } }$
(v) ${ \left( 0.999 \right) }^{ \frac { 1 }{ 10 } }$
(vi) ${ \left( 15 \right) }^{ \frac { 1 }{ 4 } }$
(vii) ${ \left( 26 \right) }^{ \frac { 1 }{ 3 } }$
(viii) ${ \left( 255 \right) }^{ \frac { 1 }{ 4 } }$
(ix) ${ \left( 82 \right) }^{ \frac { 1 }{ 4 } }$
(x) ${ \left( 401 \right) }^{ \frac { 1 }{ 2 } }$
(xi) ${ \left( 0.0037 \right) }^{ \frac { 1 }{ 2 } }$
(xii) ${ \left( 26.57 \right) }^{ \frac { 1 }{ 3 } }$
(xiii) ${ \left( 81.5 \right) }^{ \frac { 1 }{ 4 } }$
(xiv) ${ \left( 3.968 \right) }^{ \frac { 3 }{ 2 } }$
(xv) ${ \left( 32.15 \right) }^{ \frac { 1 }{ 5 } }$
Solution:
(i) y + ∆y = $\sqrt { 25.3 }$
= $\sqrt { 25+0.3 }$
= $\sqrt { x+\Delta x }$
∴ x = 25
∆x = 0.3
⇒ y = √x
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 1

tiwari academy class 12 maths Chapter 6 Application of Derivatives 1.3

Ex 6.4 Class 12 Maths 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.

Ex 6.4 Class 12 Maths 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.

Ex 6.4 Class 12 Maths 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 = $\frac { dv }{ dx }$ × ∆x
= 3x² × 0.01 x
= 0.03 x³ m³

Ex 6.4 Class 12 Maths 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
∴ $\frac { ds }{ dx }$ × ∆x = 12x × (- 0.01 x)
= – 0.12 x² m².

Ex 6.4 Class 12 Maths 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 = $\frac { 4 }{ 3 } \pi { r }^{ 3 }$
$\Delta V=\frac { dV }{ dr } \times \Delta r=\frac { 4 }{ 3 } .\pi .3{ r }^{ 2 }\times \Delta r$
= 4π × 7² × 0.02
= 3.92 πm³

Ex 6.4 Class 12 Maths 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 = $\frac { ds }{ dr }$ × ∆r
= 8πr × ∆r
= 8π × 9 × 0.03
= 2.16 πm².

Ex 6.4 Class 12 Maths 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

Ex 6.4 Class 12 Maths 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 = $\frac { dv }{ dx }$ x ∆x = 3x² × ∆x
But, ∆x = 0.03 x
⇒ ∆v = 3x² x 0.03 x
= 0.09 x³m³

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

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