NCERT Solutions for Maths Class 12 Exercise 6.4

NCERT Solutions for Maths Class 12 Exercise 6.4

 

       NCERT Solutions for Maths Class 12 Exercise 6.4

 

Hello Students! Welcome to maths-formula.com. In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 6.4.

 

You can download the PDF of NCERT Books Maths Chapter 6 for your easy reference while studying NCERT Solutions for Maths Class 12 Exercise 6.4.

 

Class 12th is a very crucial stage of your student’s life, since you take all important decisions about your career on this stage. Mathematics plays a vital role to take decision for your career because if you are good in mathematics, you can choose engineering and technology field as your career.

 

NCERT Solutions for Maths Class 12 Exercise 6.4 helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

 

All the schools affiliated with CBSE, follow the NCERT books for all subjects. You can check your syllabus from NCERT Syllabus for Mathematics Class 12.

 

NCERT Solutions for Maths Class 12 Exercise 6.4 are prepared by the experienced teachers of CBSE board. If you are preparing for JEE Mains and NEET level exams, then it will definitely make your foundation strong.

 

If you want to recall All Maths Formulas for Class 12, you can find it by clicking this link.

If you want to recall All Maths Formulas for Class 11, you can find it by clicking this link.


NCERT Solutions for Maths Class 12 Exercise 6.1

NCERT Solutions for Maths Class 12 Exercise 6.2

NCERT Solutions for Maths Class 12 Exercise 6.3

NCERT Solutions for Maths Class 12 Exercise 6.5

 

NCERT Solutions for Maths Class 12 Exercise 6.4

 

Maths Class 12 Ex 6.4 Question 1.
Using differentials, find the approximate value of each of the following up to 3 places of decimal.


Solution:



Maths Class 12 Ex 6.4 Question 2.
Find the approximate value of f(2.01), where f(x) = 4x² + 5x + 2.

Solution:
f(x + ∆x) = f(2.01) = f(2 + 0.01), f(x) = f(2) = 4 × 2² + 5 × 2 + 2 = 28
f’(x) = 8x + 5

We have, x = 2 and ∆x = 0.01

Now, f(x + ∆x) = f(x) + ∆f(x)
= f(x) + f’(x) × ∆x = 28 + (8x + 5) ∆x
= 28 + (16 + 5) × 0.01
= 28 + 21 × 0.01
= 28 + 0.21
Hence, f(2.01) = 28.21


Maths Class 12 Ex 6.4 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) = x3 – 7x2 +15

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

Hence, the approximate value of f(5.001) is –34.995.


Maths Class 12 Ex 6.4 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 (∆x) = 1% = 0.01 × x = 0.01 x
Volume of cube V = x3

dV/dx = 3x2
change in volume (∆V) = dV/dx × ∆x
                                              = 3x² × 0.01 x
                                              = 0.03 x3 m3

Hence, the approximate change in volume is 0.03 x3 m3.


Maths Class 12 Ex 6.4 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 a cube (S) = 6x² m²

dS/dx = 12x  
change in the surface area (∆S) = dS/dx × ∆x = 12x × (–0.01 x)
                                                            = –0.12 x² m²

Hence, the approximate change in surface area is –0.12 x² m².


Maths Class 12 Ex 6.4 Question 6.
If the radius of a sphere is 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 (r) = 7 m; ∆r = 0.02 m
Volume of the sphere (V) = 4/3 × πr3

dV/dr = 4/3 × π × 3r2 
∆V = dV/dr × ∆r = 4/3 × π × 3r2 × ∆r
      = 4π × 7² × 0.02
      = 3.92π m³

Hence, the approximate error in calculating the volume is 3.92π m³.


Maths Class 12 Ex 6.4 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 = 0.03 m
Surface area of sphere (S) = 4πr²
dS/dr = 4π × 2r = 8πr

∆S = dS/dr × ∆r
     = 8πr × ∆r
     = 8π × 9 × 0.03
     = 2.16 π m²

Hence, the approximate error in calculating the surface area is 2.16 π m².


Maths Class 12 Ex 6.4 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 = 3, ∆x = 0.02,
∆f(x) = f(x + ∆x) – f(x)
f(x + ∆x) = f(x) + ∆f(x) = f(x) + f’(x).∆x
Now, f(x) = 3x² + 15x + 5; f(3) = 77, f’(x) = 6x + 15
f’(3) = 33
f(3.02) = 77 + 33 × 0.02 = 77.66

Hence, the correct answer is option (D).


Maths Class 12 Ex 6.4 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 metres
Volume of a cube (V) = x³,
dV/dx = 3x²

Change in side (∆x) = 3% of x = 0.03 x
Let ∆V be the change in volume.

Then, ∆V = dV/dx × ∆x = 3x² × ∆x
But, ∆x = 0.03 x
∆V = 3x² × 0.03 x
          = 0.09 x³ m³

Hence, the correct answer is option (C).

NCERT Solutions for Maths Class 12 Exercise 6.1

NCERT Solutions for Maths Class 12 Exercise 6.2

NCERT Solutions for Maths Class 12 Exercise 6.3

NCERT Solutions for Maths Class 12 Exercise 6.5

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