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Equation 1+x2+2xsin(cos1y)=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 y
1+x2+2xsin(cos1y)=0
For the above equation to have real solution, determinant has to be 0.
Determinant, D=4sin2(cos1y)40
sin2(cos1y)1
sin2(cos1y)=1( Range of sinx is [1,1])
cos1y=(2n+1)π2, nI
y=0
For y=0, equation becomes
1+x2+2x=0, which give x=1
1+x2+2xsin(cos1y)=0, is satisfied by one value of x;x=1 and one value of y;y=0.

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