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Joint Entrance Examination
Mathematics
Functions
Question 58
Question 58 of 162
Medium
MCQ
The range of
f
(
x
)
=
4
sin
−
1
(
x
2
x
2
+
1
)
f(x)=4{\sin }^{-1}(\frac{{x}^{2}}{{x}^{2}+1})
f
(
x
)
=
4
sin
−
1
(
x
2
+
1
x
2
)
is
A
[
0
,
2
π
]
[0,2\pi ]
[
0
,
2
π
]
B
[
0
,
2
π
)
[0,2\pi )
[
0
,
2
π
)
C
[
0
,
π
)
[0,\pi )
[
0
,
π
)
D
[
0
,
π
]
[0,\pi ]
[
0
,
π
]
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Solutions · 1
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Official solution
x
2
1
+
x
2
=
1
−
1
1
+
x
2
<
1
∴
0
≤
x
2
1
+
x
2
<
1
⇒
0
≤
sin
−
1
(
x
2
1
+
x
2
)
<
π
2
⇒
0
≤
4
sin
−
1
(
x
2
1
+
x
2
)
<
2
π
\begin{gathered} \frac{{x}^{2}}{1+{x}^{2}}=1-\frac{1}{1+{x}^{2}}<1 \\ \\ \therefore 0\le \frac{{x}^{2}}{1+{x}^{2}}<1 \\ \\ \Rightarrow 0\le {\sin }^{-1}(\frac{{x}^{2}}{1+{x}^{2}})<\frac{\pi }{2} \\ \\ \Rightarrow 0\le 4{\sin }^{-1}(\frac{{x}^{2}}{1+{x}^{2}})<2\pi \end{gathered}
1
+
x
2
x
2
=
1
−
1
+
x
2
1
<
1
∴
0
≤
1
+
x
2
x
2
<
1
⇒
0
≤
sin
−
1
(
1
+
x
2
x
2
)
<
2
π
⇒
0
≤
4
sin
−
1
(
1
+
x
2
x
2
)
<
2
π
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