Equation `1+x^2+2x"sin"(cos^(-1)y)=0` is satisfied by exactly one value of `x` exactly two value of `x` exactly one value of `y` exactly two value of `y`

Options

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

Answer

A, C

Explanation

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