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JEE Main Math Class 12 Inverse Trigonometric Functions Questions
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© examsnet.com
Question : 24
Total: 74
Let
x
*
y
=
x
2
+
y
3
and
(
x
*
1
)
*
1
=
x
*
(
1
*
1
)
. Then a value of
2
sin
−
1
(
x
4
+
x
2
−
2
x
4
+
x
2
+
2
)
is
[24-Jun-2022-Shift-2]
π
4
π
3
π
2
π
6
Validate
Solution:
👈: Video Solution
Given,
x
*
y
=
x
2
+
y
3
∴
x
*
1
=
x
2
+
1
3
=
x
2
+
1
Now,
(
x
*
1
)
*
1
=
(
x
2
+
1
)
*
1
⇒
(
x
*
1
)
*
1
=
(
x
2
+
1
)
2
+
1
3
⇒
(
x
*
1
)
*
1
=
x
4
+
1
+
2
x
2
+
1
Also,
x
*
(
1
*
1
)
=
x
*
(
1
2
+
1
3
)
=
x
*
2
=
x
2
+
2
3
=
x
2
+
8
Given that,
(
x
*
1
)
*
1
=
x
*
(
1
*
1
)
∴
x
4
+
1
+
2
x
2
+
1
=
x
2
+
8
⇒
x
4
+
x
2
−
6
=
0
⇒
x
4
+
3
x
2
−
2
x
2
−
6
=
0
⇒
x
2
(
x
2
+
3
)
−
2
(
x
3
+
3
)
=
0
⇒
(
x
2
+
3
)
(
x
2
−
2
)
=
0
⇒
x
2
=
2
,
−
3
[
x
2
=
−
3.
not possible as square of anything should be always possible]
∴
x
2
=
2
∴
Now,
2
sin
−
1
(
x
4
+
x
2
−
2
x
4
+
x
2
+
2
)
=
2
sin
−
1
(
2
2
+
2
−
2
2
2
+
2
+
2
)
=
2
sin
−
1
(
4
8
)
=
2
sin
−
1
(
1
2
)
=
2
×
π
6
=
π
3
© examsnet.com
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