CUET Preparation Today
If A=R- {+1} and function f: A →A is defined by $f(x)=\frac{x+1}{x-1}$ then $f^{-1}(x)$ is given by : |
$\frac{1}{x-1}$ $\frac{1}{1-x}$ $\frac{x+1}{1-x}$ $\frac{x+1}{x-1}$ |
$\frac{x+1}{x-1}$ |
The correct answer is option (4) → $\frac{x+1}{x-1}$ $y=\frac{x+1}{x-1}⇒xy-y=x+1$ so $x(y-1)=y+1$ $x=\frac{y+1}{y-1}⇒f^{-1}(x)=\frac{x+1}{x-1}$ |
