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(ii)
√49.5
(iii)
√0.6
(v)
(0.999)
1/10
(vi)
(15)
1/4
(vii)
(26)
1/3
(viii)
(255)1/4
(ix)
(82)
1/4
(x)
(401)1/2
(xi)
(0.0037)1/2
(xii)
(26.57)
1/3
(xiii)
(81.5)
1/4
(xiv)
(3.968)
3/2
(xv) (32.15)
1/5
Hence, the approximate value of
f
(2.01) is 28.21.
Hence, the approximate value of
f
(5.001) is −34.995.
Hence, the approximate change in the volume of the cube is 0.03
x
3
m
3
.
Question
5. Find the approximate change in the surface area of a cube of side x meters caused by decreasing the side by 1%.
Solution :
The surface area of a cube (
S)
of side
x
is given by
S
= 6
x
2
.
Hence, the approximate change in the surface area of the cube is 0.12
x
2
m
2
.
Hence, the approximate error in calculating the volume is 3.92 π m
3
.
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 :
Let
r
be the radius of the sphere and Δ
r
be the error in measuring the radius.
Then,
r
= 9 m and Δ
r
= 0.03 m
Now, the surface area of the sphere (
S)
is given by,
S
= 4πr
2
Hence, the approximate error in calculating the surface area is 2.16π m
2
.
Question
8. Iff (x) = 3x
2
+ 15x + 5, then the approximate value of f (3.02) is
A. 47.66
B. 57.66
C. 67.66
D.77.66
Solution :
Therefore, option (D) is correct.
Question
9. The approximate change in the volume of a cube of sidex metres caused by increasing the side by 3% is
A. 0.06x
3
m
3
B. 0.6x
3
m
3
C. 0.09x
3
m
3
D.0.9x
3
m
3
Solution :
The volume of a cube (
V
) of side
x
is given by
V
=
x
3
.
Therefore, option (C) is correct.
