Please submit your concern

Explanation:

y = f(x) = $\frac{1-x}{1+x}$

⇒ y(1 + x) = 1 – x

⇒ y + xy = 1 – x

⇒ x + xy = 1 – y

⇒ x(1 + y) = (1 – y)

⇒ x = $\frac{1-y}{1+y}$ = f(y)

Since here, y = f(x) and x = f(y), the given function refleccts on itself.

Hence, option (d).