NCERT Solutions class 12 Maths Exercise 6.4 (Ex 6.4) Chapter 6 Application of Derivatives

NCERT Solutions for Class 12 Maths Exercise 6.4 Chapter 6 Application of Derivatives – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4 (Ex 6.4) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks.

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

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) 

Ans. (i)  

Let  ……….(i)

   = 

   ……….(ii)

Now, from eq. (i), 

= 

Here,  and , then 

= 

 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),

 = 0.03

Therefore, approximately value of  is 5 + 0.03 = 5.03.

(ii) 

Let  ……….(i)

   = 

   ……….(ii)

Now, from eq. (i), 

= 

Here,  and , then

= 

 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),  

= 0.0357

Therefore, approximately value of  is 7 + 0.0357 = 7.0357.

(iii) 

Let  ……….(i)

   = 

   ……….(ii)

Now, from eq. (i), 

= 

Here,  and , then

= 

 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),   = 

Therefore, approximately value of  is 0.8 – 0.025 = 0.775.

(iv) 

Let  ……….(i)

   = 

     ……….(ii)

Now, from eq. (i), 

= 

Here,  and ,

then 

= 

 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),  

=  = 0.0083

Therefore, approximately value of  is 0.2 + 0.0083 = 0.2083.

(v)  

Let  ……….(i)

   = 

     ……….(ii)

Now, from eq. (i), 

=  =  ……….(iii)

Here  and 

Then 

= 

  = 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),

Therefore, approximate value of  is 1 – 0.0001 = 0.9999.

(vi) 

Let  ……….(i)

   = 

   ……….(ii)

Now, from eq. (i), 

= =  ……….(iii)

Here  and 

Then 

= 

  = 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),  

Therefore, approximate value of  is  = 1.96875.

(vii) 

Let  ……….(i)

   = 

    ……….(ii)

Now, from eq. (i), 

= 

Here,  and ,

then  = 

 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),   = 

Therefore, approximately value of  is  = 2.9629.

(viii) 

Let  ……….(i)

   = 

   ……….(ii)

Now, from eq. (i), 

= (255)1/4(255)1/4=  ……….(iii)

Here  and 

Then 

= 

 = 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),  

Therefore, approximate value of  is  = 3.9961.

(ix) 

Let  ……….(i)

 

= 

   ……….(ii)

Now, from eq. (i),  = = ……….(iii)

Here  and 

Then 

= 

 = 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),  

Therefore, approximate value of  is  = 3.0092.

(x) 

Let  ……….(i)

   = 

  ……….(ii)

Now, from eq. (i), 

= 

Here,  and , then 

= 

 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),   = 

Therefore, approximately value of  is  = 20.025.

(xi) 

Let  ……….(i)

   = 

   ……….(ii)

Now, from eq. (i), 

= 

Here,  and , then

= 

 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),  

= 

Therefore, approximately value of  is 0.06 +  = 0.060833.

(xi) 

Let  ……….(i)

   = 

     ……….(ii)

Now, from eq. (i), 

= 

Here,  and ,

then 

= 

 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),  

=  = 0.0159

Therefore, approximately value of  is 3 – 0.0159 = 2.9841.

(xii) 

Let  ……….(i)

   = 

   ……….(ii)

Now, from eq. (i), 

= =   ……….(iii)

Here  and 

Then 

= 

  = 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),  

= 0.00462

Therefore, approximate value of  is 3 + 0.00462= 3.00462.

(xiv) 

Let  ……….(i)

  

     ……….(ii)

Now, from eq. (i), 

= 

Here,  and , then 

= 

 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),

 = 

Therefore, approximately value of  is 8 – 0.096 = 7.904.

(xv) 

Let  ……….(i)

   = 

   ……….(ii)

Now, from eq. (i), 

= =   ……….(iii)

Here  and 

Then 

= 

 = 

Since,  and  is approximately equal to  and  respectively.

 From eq. (ii),

 = 0.001875

Therefore, approximate value of  is 2 + 0.001875= 2.001875.

2. Find the approximate value of  where 

 

Ans. Let f(x)=y=4x2+5x+2f(x)=y=4×2+5x+2  ……….(i) 

  

 

=   ……….(ii)

Changing  to  and  to  in eq. (i),

=   ……….(iii)

Here,  and 

 From eq. (iii),  

Since,  and  is approximately equal to  and  respectively.

 From eq. (i) and (ii),  

   = 28.21

Therefore, approximate value of  is 28.21.

3. Find the approximate value of  where 

 

Ans. Let   ……….(i) 

  

  ……….(ii)

Changing  to  and  to  in eq. (i),

=   ……….(iii)

Here,  and 

 From eq. (iii),  

Since,  and  is approximately equal to  and  respectively.

 From eq. (i) and (ii),

 

=  = 

Therefore, approximate value of  is .

4. Find the approximate change in the volume of a cube of side  meters caused by increasing the side by 1%.

 

Ans. Since Volume (V) =   ……….(i) 

  ……….(ii)

It is given that increase in side = 1% of x = 

     ……….(iii)

Since approximate change in volume V of cube =  = 

=    cubic meters

5. Find the approximate change in the surface area of a cube of side  meters caused by decreasing the side by 1%.

 

Ans. Since Surface area (S) =  

  

It is given that decrease in side =  of 

 

Since approximate change in surface area S of cube =  = 

=    square meters (decreasing)

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.

 

Ans. Let  be the radius of the sphere and  be the error in measuring the radius. 

Then, according to the question,  = 7 m and  = 0.02 m

  Volume of sphere (V) = 

  

 Approximate error in calculating the volume = Approximate value of 

=  = 

= 

= 

=  = 12.32 m3

Therefore, the approximate error in calculating volume is 12.32 m3.

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.

 

Ans. Let  be the radius of the sphere. 

  Surface area of the sphere (S) = 

  

  

  

According to question  r=9 m and =.03 m

ds=8π(9)(.03)ds=8π(9)(.03)

=  square meters

8. If  then the approximate value of  is:

 

(A) 47.66

(B) 57.66

(C) 67.66

(D) 77.66

Ans. Let   ……….(i) 

 

   ……….(ii)

Changing  to  and  to  in eq. (i),

=   ……….(iii)

Here,  and 

 From eq. (iii),  

Since,  and  is approximately equal to  and  respectively.

 From eq. (i) and (ii),

 

=  = 

Therefore, option (D) is correct.

9. The approximate change in the volume of a cube of side  meters caused by increasing the side by 3% is:

 

(A) 0.06  m3

(B) 0.6  m3

(C) 0.09  m3

(D) 0.9  m3

Ans. Since Volume (V) =    ……….(i) 

  ……….(ii)

It is given that increase in side = 3% of x= 

    ……….(iii)

Since approximate change in volume V of cube =  = 

=    cubic meters

Therefore, option (C) is correct.