Exercise 6.4 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE PDF Download

 Page 1


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
 
 
Exercise 6.4                                                            Page No: 216 
 
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)  
Consider,  ……….(1) and then 
  
On differentiating equation (1) w.r.t. x, we get 
Page 2


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
 
 
Exercise 6.4                                                            Page No: 216 
 
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)  
Consider,  ……….(1) and then 
  
On differentiating equation (1) w.r.t. x, we get 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
  =  
  ……….(2) 
Now, given expression can be written as, 
 
Here,  and , then  
=  
  
Since,  and  is approximately equal to  and  respectively. 
 From equation (2), 
 
 = 0.03 
Hence, approximately value of  is 5 + 0.03 = 5.03. 
(ii)  
Consider,  ……….(1) 
On differentiating equation (1) w.r.t. x, we get 
  =  
  ……….(2) 
Now, from equation (1),  
 
Page 3


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
 
 
Exercise 6.4                                                            Page No: 216 
 
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)  
Consider,  ……….(1) and then 
  
On differentiating equation (1) w.r.t. x, we get 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
  =  
  ……….(2) 
Now, given expression can be written as, 
 
Here,  and , then  
=  
  
Since,  and  is approximately equal to  and  respectively. 
 From equation (2), 
 
 = 0.03 
Hence, approximately value of  is 5 + 0.03 = 5.03. 
(ii)  
Consider,  ……….(1) 
On differentiating equation (1) w.r.t. x, we get 
  =  
  ……….(2) 
Now, from equation (1),  
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
 
=  
Here,  and , then 
 
=  
  
Since,  and  is approximately equal to  and  respectively. 
 From equation (2),  
= 0.0357 
So, approximately value of  is 7 + 0.0357 = 7.0357. 
(iii)  
Consider,  ……….(1) 
On differentiating equation (1) w.r.t. x, we get 
  =  
  ……….(2) 
Now, from equation (1),  
=  
Here,  and , then 
 
Page 4


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
 
 
Exercise 6.4                                                            Page No: 216 
 
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)  
Consider,  ……….(1) and then 
  
On differentiating equation (1) w.r.t. x, we get 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
  =  
  ……….(2) 
Now, given expression can be written as, 
 
Here,  and , then  
=  
  
Since,  and  is approximately equal to  and  respectively. 
 From equation (2), 
 
 = 0.03 
Hence, approximately value of  is 5 + 0.03 = 5.03. 
(ii)  
Consider,  ……….(1) 
On differentiating equation (1) w.r.t. x, we get 
  =  
  ……….(2) 
Now, from equation (1),  
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
 
=  
Here,  and , then 
 
=  
  
Since,  and  is approximately equal to  and  respectively. 
 From equation (2),  
= 0.0357 
So, approximately value of  is 7 + 0.0357 = 7.0357. 
(iii)  
Consider,  ……….(1) 
On differentiating equation (1) w.r.t. x, we get 
  =  
  ……….(2) 
Now, from equation (1),  
=  
Here,  and , then 
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
=  
  
Since,  and  is approximately equal to  and  respectively. 
 From equation (2),  =  
Therefore, approximately value of  is 0.8 – 0.025 = 0.775. 
(iv)  
Consider,  ……….(1) 
On differentiating equation (1) w.r.t. x, we get 
  =  
  ……….(2) 
Now, from equation (1),  
=  
Here,  and , 
then  
=  
  
Since,  and  is approximately equal to  and  respectively. 
Page 5


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
 
 
Exercise 6.4                                                            Page No: 216 
 
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)  
Consider,  ……….(1) and then 
  
On differentiating equation (1) w.r.t. x, we get 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
  =  
  ……….(2) 
Now, given expression can be written as, 
 
Here,  and , then  
=  
  
Since,  and  is approximately equal to  and  respectively. 
 From equation (2), 
 
 = 0.03 
Hence, approximately value of  is 5 + 0.03 = 5.03. 
(ii)  
Consider,  ……….(1) 
On differentiating equation (1) w.r.t. x, we get 
  =  
  ……….(2) 
Now, from equation (1),  
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
 
=  
Here,  and , then 
 
=  
  
Since,  and  is approximately equal to  and  respectively. 
 From equation (2),  
= 0.0357 
So, approximately value of  is 7 + 0.0357 = 7.0357. 
(iii)  
Consider,  ……….(1) 
On differentiating equation (1) w.r.t. x, we get 
  =  
  ……….(2) 
Now, from equation (1),  
=  
Here,  and , then 
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
=  
  
Since,  and  is approximately equal to  and  respectively. 
 From equation (2),  =  
Therefore, approximately value of  is 0.8 – 0.025 = 0.775. 
(iv)  
Consider,  ……….(1) 
On differentiating equation (1) w.r.t. x, we get 
  =  
  ……….(2) 
Now, from equation (1),  
=  
Here,  and , 
then  
=  
  
Since,  and  is approximately equal to  and  respectively. 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 6 Application of 
Derivatives 
 From equation 
(2),  
=  = 0.0083 
Therefore, approximately value of  is 0.2 + 0.0083 = 0.2083. 
(v)  
Consider,  ……….(1) 
On differentiating equation (1) w.r.t. x, we get 
  =  
  ……….(2) 
Now, from equation (1),  
=  =  ……….(3) 
Here  and  
Then  
=  
  =  
Since,  and  is approximately equal to  and  respectively.