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JEE Main Math Class 12 Limits and Derivatives Part 2 Questions
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© examsnet.com
Question : 33
Total: 56
Let
{
x
}
denote the fractional part of
x
and
f
(
x
)
=
cos
−
1
(
1
−
{
x
}
2
)
s
i
n
−
1
(
1
−
{
x
}
)
{
x
}
−
{
x
}
3
,
x
≠
0
. If
L
and
R
respectively denotes the left hand limit and the right hand limit of
f
(
x
)
at
x
=
0
, then
32
π
2
(
L
2
+
R
2
)
is equal to___
[1-Feb-2024 Shift 1]
Your Answer:
Validate
Solution:
👈: Video Solution
Finding right hand limit
lim
x
⟶
0
+
f
(
x
)
=
lim
h
⟶
0
f
(
0
+
h
)
=
lim
h
⟶
0
f
(
h
)
=
lim
h
⟶
0
cos
−
1
(
1
−
h
2
)
s
i
n
−
1
(
1
−
h
)
h
(
1
−
h
2
)
=
lim
h
⟶
0
cos
−
1
(
1
−
h
2
)
h
(
s
i
n
−
1
1
1
)
Let
cos
−
1
(
1
−
h
2
)
=
θ
⇒
cos
θ
=
1
−
h
2
=
π
2
lim
θ
⟶
0
θ
√
1
−
cos
θ
=
π
2
lim
θ
⟶
0
1
√
1
−
cos
θ
θ
2
=
π
2
1
√
1
∕
2
R
=
π
√
2
Now finding left hand limit
L
=
lim
x
⟶
0
−
f
(
x
)
=
lim
h
⟶
0
f
(
−
h
)
=
lim
h
⟶
0
cos
−
1
(
1
−
{
−
h
}
2
)
s
i
n
−
1
(
1
−
{
−
h
}
)
{
−
h
}
−
{
−
h
}
3
=
lim
h
⟶
0
cos
−
1
(
1
−
(
−
h
+
1
)
2
)
s
i
n
−
1
(
1
−
(
−
h
+
1
)
)
(
−
h
+
1
)
−
(
−
h
+
1
)
3
=
lim
h
⟶
0
cos
−
1
(
−
h
2
+
2
h
)
s
i
n
−
1
h
(
1
−
h
)
(
1
−
(
1
−
h
)
2
)
=
lim
h
⟶
0
(
π
2
)
s
i
n
−
1
h
(
1
−
(
1
−
h
)
2
)
=
π
2
lim
h
⟶
0
(
s
i
n
−
1
h
−
h
2
+
2
h
)
=
π
2
lim
h
⟶
0
(
s
i
n
−
1
h
h
)
(
1
−
h
+
2
)
L
=
π
4
32
π
2
(
L
2
+
R
2
)
=
32
π
2
(
π
2
2
+
π
2
16
)
=
16
+
2
=
18
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