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JEE Main Math Class 12 Limits and Derivatives Part 1 Questions
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© examsnet.com
Question : 5
Total: 100
If the function
f
(
x
)
=
{
log
e
(
1
−
x
+
x
2
)
+
log
e
(
1
+
x
+
x
2
)
sec
x
−
cos
x
,
x
∈
(
−
π
2
,
π
2
)
−
{
0
}
k
,
x
=
0.
is continuous at
x
=
0
, then
k
is equal to:
[26-Jul-2022-Shift-1]
1
−
1
e
0
Validate
Solution:
👈: Video Solution
f
(
x
)
=
{
log
e
(
1
−
x
+
x
2
)
+
log
e
(
1
+
x
+
x
2
)
sec
x
−
cos
x
x
∈
(
−
π
2
,
π
2
)
−
{
0
}
k
x
=
0
.
for continuity at
x
=
0
lim
x
→
0
f
(
x
)
=
k
∴
k
=
lim
x
→
0
log
e
(
x
4
+
x
2
+
1
)
sec
x
−
cos
x
(
0
0
form
)
=
lim
x
→
0
cos
x
log
e
(
x
4
+
x
2
+
1
)
sin
2
x
=
lim
x
→
0
log
e
(
x
4
+
x
2
+
1
)
x
2
=
lim
x
→
0
ln
(
1
+
x
2
+
x
4
)
x
2
+
x
4
⋅
x
2
+
x
4
x
2
=
1
© examsnet.com
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