Exercise 6.4
Download as PDF.
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.
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.
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
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)
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.
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
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.
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.