NCERT Solutions class 12 Maths Exercise 6.4

NCERT Solutions class 12 Maths Exercise 6.4 Class 12 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 Maths chapter wise NCERT solution for Maths part 1 and Maths part 2 for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

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NCERT Solutions for Class 12 Maths Application of Derivatives

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),

= =  ……….(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.

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2. 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), 

   = 28.21

Therefore, approximate value of  is 28.21.

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NCERT Solutions class 12 Maths Exercise 6.4

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% =

     ……….(iii)

Since approximate change in volume V of cube =  =

=    cubic meters

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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)

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NCERT Solutions class 12 Maths Exercise 6.4

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.

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NCERT Solutions class 12 Maths Exercise 6.4

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) =

 

 

 

·

=  square meters

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NCERT Solutions class 12 Maths Exercise 6.4

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.

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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% =

    ……….(iii)

Since approximate change in volume V of cube =  =

=    cubic meters

Therefore, option (C) is correct.

NCERT Solutions class 12 Maths Exercise 6.4

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