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JEE Main Math Class 12 Continuity and Differentiability Part 2 Questions
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© examsnet.com
Question : 42
Total: 92
If the function
f
(
x
)
=
{
1
x
log
e
(
1
+
x
a
1
−
x
b
)
x
<
0
k
x
=
0
cos
2
x
−
sin
2
x
−
1
√
x
2
+
1
−
1
x
>
0.
is continuous at
x
=
0
, then
1
a
+
1
b
+
4
k
is equal to
[2021,31 Aug. Shift-1]
-5-2
5
-4
4
Validate
Solution:
f
(
x
)
is continuous at
x
=
0
L
H
L
at
x
=
0
=
f
(
0
)
=
R
H
L
at
x
=
0
lim
x
→
0
−
ln
(
1
+
x
a
1
−
x
b
)
x
=
lim
x
→
0
−
(
1
a
)
ln
(
1
+
x
a
)
(
1
a
)
x
−
lim
x
→
0
−
(
−
1
b
)
ln
(
1
−
x
b
)
(
−
1
b
)
x
=
(
1
a
+
1
b
)
lim
x
→
0
+
cos
2
x
−
sin
2
x
−
1
√
x
2
+
1
−
1
lim
x
→
0
+
−
2
sin
2
x
√
x
2
+
1
−
1
=
lim
x
→
0
+
−
(
2
sin
2
x
x
2
)
(
√
x
2
+
1
+
1
)
=
−
4
⇒
1
a
+
1
b
=
−
4
=
k
(
1
a
+
1
b
)
+
(
4
k
)
=
−
4
−
1
=
−
5
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