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

In this chapter, we provide NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 pdf, free NCERT solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 book pdf download.

BoardCBSE
TextbookNCERT
ClassClass 12
SubjectMaths
ChapterChapter 6
Chapter NameApplication of Derivatives
ExerciseEx 6.4
Number of Questions Solved9
CategoryNCERT Solutions

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
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 1.1
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 1.2
tiwari academy class 12 maths Chapter 6 Application of Derivatives 1.3
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 1.4

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

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