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JEE Main Math Class 12 Integrals Part 3 Questions
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© examsnet.com
Question : 10
Total: 100
Let
f
(
x
)
=
∫
dx
(
3
+
4
x
2
)
√
4
−
3
x
2
,
|
x
|
<
2
√
3
. If
f
(
0
)
=
0
and
f
(
1
)
=
1
α
β
tan
−
1
(
α
β
)
α
,
β
>
0
, then
α
2
+
β
2
is equal to
[15-Apr-2023 shift 1]
Your Answer:
Validate
Solution:
f
(
x
)
=
∫
d
x
(
3
+
4
x
2
)
√
4
−
3
x
2
x
=
1
t
=
∫
−
1
t
2
dt
(
3
t
2
+
4
)
t
2
√
4
t
2
−
3
t
=
∫
−
d
t
⋅
t
(
3
t
2
+
4
)
√
4
t
2
−
3
:
Put
4
t
2
−
3
=
z
2
=
−
1
4
∫
z
dx
(
3
(
z
2
+
3
4
)
+
4
)
z
=
∫
−
d
z
3
z
2
+
25
=
−
1
3
∫
d
z
z
2
+
(
5
√
3
)
2
=
−
1
3
√
3
5
tan
−
1
(
√
3
z
5
)
+
C
=
−
1
5
√
3
tan
−
1
(
√
3
5
√
4
t
2
−
3
)
+
C
f
(
x
)
=
−
1
5
√
3
tan
−
1
(
√
3
5
√
4
−
3
x
2
x
2
)
+
C
∵
f
(
0
)
=
0
∵
c
=
π
10
√
3
f
(
1
)
=
−
1
5
√
3
tan
−
1
(
√
3
5
)
+
π
10
√
3
f
(
1
)
=
1
5
√
3
cot
−
1
(
√
3
5
)
=
1
5
√
3
tan
−
1
(
5
√
3
)
α
=
5
:
β
=
√
3
∴
α
2
+
β
2
=
28
© examsnet.com
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