Question

Which of the following functions is inverse of itself?

• A. $ f\left(t\right) = \frac{\left(1-t\right)}{\left(1+t\right)} $
• B. $ f\left(t\right) = \frac{\left(1-t^{2}\right)}{\left(1+t^{2}\right)} $
• C. $ f\left(t\right) = 4^{log\, t} $
• D. $ f(t) = 2^t $

Answer 1

Answer: (A)

Let $y=f (t)$
$\therefore t=f^{-1}(y)$
Now, $y=f (t) =\frac{1-t}{1+t}$
$\Rightarrow y+ty=1-t$
$\Rightarrow t+ty=1-y$
$\Rightarrow t=\frac{1-y}{1+y}$
i.e., $f^{-1}(y) =\frac{1-y}{1+y}$
or $f^{-1}(t)=\frac{1-t}{1+t}$
Thus, this function is inverse of itself