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JEE Main Math Class 12 Continuity and Differentiability Part 2 Questions
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© examsnet.com
Question : 26
Total: 92
Let
f
(
x
)
=
{
|
4
x
2
−
8
x
+
5
∣
,
if
8
x
2
−
6
x
+
1
≥
0
[
4
x
2
−
8
x
+
5
]
,
if
8
x
2
−
6
x
+
1
<
0
,
where
[
α
]
denotes the greatest integer less than or equal to
α
. Then the number of points in
R
where
f
is not differentiable is
[25-Jul-2022-Shift-1]
Your Answer:
Validate
Solution:
f
(
x
)
=
{
|
4
x
2
−
8
x
+
5
|
if
8
x
2
−
6
x
+
1
≥
0
[
4
x
2
−
8
x
+
5
]
,
if
8
x
2
−
6
x
+
1
<
0.
=
{
4
x
2
−
8
x
+
5
if
x
∈
[
−
∞
,
1
4
]
∪
[
1
2
,
∞
)
[
4
x
2
−
8
x
+
5
]
if
x
∈
(
1
4
,
1
2
)
.
f
(
x
)
=
{
4
x
2
−
8
x
+
5
if
x
∈
(
−
∞
,
1
4
]
∪
[
1
2
,
∞
)
3
x
∈
(
1
4
,
2
−
√
2
2
)
2
x
∈
[
2
−
√
2
2
,
1
2
)
.
∴
Non-diff at
x
=
1
4
,
2
−
√
2
2
,
1
2
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