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Question

# Equation 1+x2+2xsin(cosâˆ’1y)=0 is satisfied by

A
exactly one value of x
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B
exactly two values of x
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C
exactly one value of y
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D
exactly two values of y
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Solution

## The correct options are A exactly one value of x D exactly one value of y1+x2+2xsin(cos−1y)=0For the above equation to have real solution, determinant has to be ≥0.Determinant, D=4sin2(cos−1y)−4≥0⇒sin2(cos−1y)≥1⇒sin2(cos−1y)=1(∵ Range of sinx is [−1,1])⇒cos−1y=(2n+1)π2, n∈I⇒y=0For y=0, equation becomes1+x2+2x=0, which give x=−1∴1+x2+2xsin(cos−1y)=0, is satisfied by one value of x;x=−1 and one value of y;y=0.

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