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Questions  

The range of the function y=sin1x21+x2 is

a
0,π2
b
0,π2
c
0,π2
d
None of these

detailed solution

Correct option is B

Clearly, for y to be defined, x21+x2≤1 which is true for all x ∈ R. So, the domain = (– ∞, ∞). Now, y=sin−1⁡x21+x2⇒x21+x2=sin⁡y ⇒x=x21+x2For x to be real, sin y ≥ 0 and 1 – sin y > 0⇒0≤sin⁡y<1⇒y∈0,π2∴ Range =0,π2

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