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Tardigrade
Question
Mathematics
The value of √24 . 99 is
Q. The value of
24.99
is
3976
169
KCET
KCET 2019
Application of Derivatives
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A
5.001
8%
B
4.999
63%
C
4.897
17%
D
4.899
15%
Solution:
24.99
Δ
y
=
f
(
x
+
Δ
x
)
−
f
(
x
)
Take
f
(
x
)
=
x
L
e
t
x
=
25
Δ
x
=
−
0.01
∴
f
(
24.99
)
=
25
+
Δ
y
≈
5
−
0.001
≈
4.999
Δ
y
≈
d
x
d
y
×
Δ
x
≈
2
x
1
×
−
0.01
≈
−
10
0.01