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JEE Main 30 June 2022 Shift 1 Solved Paper
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© examsnet.com
Question : 23
Total: 90
Suppose
lim
x
→
0
F
(
x
)
x
3
exists and is equal to
L
, where
F
(
x
)
=
|
a
+
sin
x
2
−
b
cos
x
0
−
b
cos
x
0
a
+
sin
x
2
0
a
+
sin
x
2
−
b
cos
x
|
Then,
−
112
L
is equal to___
[30-Jun-2022-Shift-1]
Your Answer:
Validate
Solution:
👈: Video Solution
Given,
F
(
x
)
=
|
a
+
s
i
n
x
2
−
b
cos
x
0
−
b
cos
x
0
a
+
s
i
n
x
2
0
a
+
s
i
n
x
2
−
b
cos
x
|
=
(
a
+
s
i
n
x
2
)
(
−
(
a
+
s
i
n
x
2
)
2
)
+
b
cos
x
×
b
2
cos
2
x
=
−
(
a
+
s
i
n
x
2
)
3
−
b
3
cos
3
x
Now,
lim
x
→
0
F
(
x
)
x
3
=
lim
x
→
0
−
(
a
+
s
i
n
x
2
)
3
−
b
3
cos
3
x
x
3
Given limit exists, it only possible when
a
=
0
and
b
=
0
.
=
lim
x
→
0
−
(
s
i
n
x
2
)
3
x
3
=
lim
x
→
0
−
(
1
2
×
(
s
i
n
x
2
π
2
)
)
3
=
lim
x
→
0
−
(
1
2
)
3
×
(
s
i
n
x
2
x
2
)
3
=
−
1
8
×
1
=
L
∴
−
112
L
=
−
112
×
−
1
8
=
14
© examsnet.com
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