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JEE Main Math Class 12 Continuity and Differentiability Part 2 Questions
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© examsnet.com
Question : 58
Total: 92
Let
f
:
R
→
R
be a function defined as
f
(
x
)
=
{
sin
(
a
+
1
)
x
+
sin
2
x
2
x
if
x
<
0
b
√
x
+
b
x
3
−
√
x
if
x
<
0
b
x
5
∕
2
if
x
>
0.
If
f
is continuous at
x
=
0
, then the value of
a
+
b
is equal to
[2021, 18 March Shift-11]
−
5
2
−
2
−
3
−
3
2
Validate
Solution:
Given,
f
(
x
)
=
{
sin
(
a
+
1
)
x
+
sin
2
x
2
x
x
<
0
b
x
=
0
√
x
+
b
x
3
−
√
x
b
x
5
∕
2
x
>
0.
∵
f
(
x
)
is continuous at
x
=
0
.
∴
lim
x
→
0
−
f
(
x
)
=
lim
x
→
0
+
f
(
x
)
=
f
(
0
)
...(i)
∵
f
(
0
)
=
b
.
.
.
...(ii)
lim
x
→
0
−
f
(
x
)
=
lim
x
→
0
−
(
sin
(
a
+
1
)
x
+
sin
2
x
2
x
)
⇒
lim
x
→
0
−
f
(
x
)
=
lim
x
→
0
−
(
sin
(
a
+
1
)
x
2
x
+
sin
2
x
2
x
)
=
lim
x
→
0
−
(
sin
(
a
+
1
)
x
(
a
+
1
)
x
×
(
a
+
1
2
)
+
sin
2
x
2
x
)
=
a
+
1
2
+
1
. . . (iii)
Again,
lim
x
→
0
+
f
(
x
)
=
lim
x
→
0
+
(
√
x
+
b
x
3
−
√
x
b
x
5
∕
2
)
=
lim
x
→
0
+
(
√
x
+
b
x
3
−
√
x
)
(
√
x
+
b
x
3
+
√
x
)
b
x
5
∕
2
(
√
x
+
b
x
3
+
√
x
)
=
lim
x
→
0
+
(
x
+
b
x
3
−
x
)
b
x
5
∕
2
(
√
x
+
b
x
3
+
√
x
)
=
lim
x
→
0
+
√
x
√
x
(
√
1
+
b
x
2
+
1
)
⇒
lim
x
→
0
+
f
(
x
)
=
1
2
. . . (iv)
From Eq. (i), (ii), (iii) and (iv)
1
2
=
b
=
a
+
1
2
+
1
⇒
b
=
1
2
,
a
=
−
2
∴
a
+
b
=
−
3
2
© examsnet.com
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