Let f(x) = `(x+1)^(2) - 1, (x ge - 1)`. Then, the set S = {x : f(x) = `f^(-1)(x)`} is

Options

A. `{0, -1, (-3+isqrt(3))/(2),(-3-isqrt(3))/(2)},i=sqrt(-1)`
B. {0, 1, -1}
C. {0, -1}
D. empty

Answer

C

Explanation

`f(x) = (x+1)^(2)-1" "[because xge1]`
`=x^(2)+1+2x-1=x^(2)+2x`
`S={x : f(x)-=f^(-1)(x)}`
S is the set of point of intersection of (y = x) and tf.
Now, solve y = x and `f(x) = x^(2) + 2x`
`x^(2) + 2x = x`
`x^(2) + x = 0`
x(x+1)=0
x = 0 or x = - 1