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B
1, -1
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C
0,12
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D
None
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Solution
The correct option is C0,12 We have, sin1(1−x)=(π2−sin−1x)−sin−1x=π2−2sin−1x
Taking sine of both sides, 1−x=sin(π2−2sin−1x)=cos(2sin−1x)=cos2θ=1−2sin2θ, Where sin−1x=θ
Or 1−x=1−2x2 or x (1-2x)=0 ∴x=0,12