If `f: R->R` is an invertible function such that `f(x) and f^-1(x)` are symmetric about the line `y=-x,` then (a) `f(x)` is odd (b) `f(x) and f^-1(x)` may be symmetric (c) `f(c)` may not be odd (d) none of these

Options

A. `f(x)` is odd
B. `f(x) and f^(-1)(x)` may not be symmetric about the line `y=x`
C. `f(x)` may not be odd
D. None of these

Answer

A

Explanation

Since `f(x)` and `f^(-1)(x)` are symmetric about the line `y= -x`,
if `(alpha, beta)` lies on `y= f(x)`, then `(-beta, -alpha)` lies on `y= f^(-1) (x).`
Therefore, `(alpha, beta)` lies on `y= f(x).`
Hence, `y=f(x)` is odd.