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Equation 1+x^2+2x"sin"(cos^(-1)y)=0 is s...

Equation `1+x^2+2x"sin"(cos^(-1)y)=0` is satisfied by

A

exactly one value of x

B

exactly two values of x

C

exactly one value of y

D

exactly two values of y

Text Solution

Verified by Experts

The correct Answer is:
A, C
Given equation is `x^(2) + 2x sin(cos^(-1) y) + 1 =0`
Since x is real, `D ge 0`. Therefore,
`4(sin(cos^(-1)y))^(2) -4 ge 0`
or `(sin(cos^(-1)y))^(2) ge 1`
or `sin(cos^(-1)y) = +- 1`
or `cos^(-1) y = (pi)/(2) rArr y = 0`
Putting value of y in the original equation, we have
`x^(2) + 2x + 1 = 0 rArr x -1`
Hence, the equation has only one solution
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