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Byju's Answer
Standard XII
Mathematics
Theorems for Differentiability
Find the numb...
Question
Find the number of points where the function
f
(
x
)
=
(
x
2
−
1
)
+
sin
|
x
|
is not differentiable.
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Solution
Given function is
f
(
x
)
=
x
2
−
1
+
sin
|
x
|
,
consider two cases,
case 1 :
x
>
0
,
⟹
f
(
x
)
=
x
2
−
1
+
sin
x
⟹
f
′
(
x
)
=
2
x
+
cos
x
⟹
f
′
(
0
)
=
1
case 2 :
x
<
0
,
⟹
f
(
x
)
=
x
2
−
1
−
sin
x
⟹
f
′
(
x
)
=
2
x
−
cos
x
⟹
f
′
(
0
)
=
−
1
So here at point zero the left derivatives and the right derivaties are not equal.
So, the function is not differentiable at
0
, this means the graph may contain a kink, as the graph is continuous everywhere.
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Q.
Number of points where
f
(
x
)
=
(
1
−
x
)
∣
∣
x
−
x
2
∣
∣
+
x
is not differentiable is
Q.
Let
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)
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√
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|
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e
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2
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. The points where
f
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)
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Q.
Assertion :The number of points of discontinuity of
f
(
x
)
are
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, where
f
(
x
)
=
∫
x
0
t
sin
(
1
t
)
d
t
Reason: The function
g
(
x
)
=
m
a
x
{
−
x
,
1
,
x
2
}
∀
x
∈
R
is not differentiable at two values of
x
.
Q.
If
f
(
x
)
=
2
x
+
1
;
x
≤
1
=
x
2
+
2
;
1
<
x
≤
2
=
4
x
2
+
2
;
x
>
2
then number of points where
f
(
x
)
is not differentiable is
Q.
Number of point(s) where the function
f
(
x
)
=
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,
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)
is non-differentiable, is/are
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