CUET Preparation Today
Let $f: R-\left\{\frac{3}{5}\right\} → R$ be defined by $f(x)=\frac{3 x+2}{5 x-3}$, then |
$f^{-1}(x)=f(x)$ $f^{-1}(x)=-f(x)$ $(fof)(x)=x^2$ $f^{-1}|x|=\frac{1}{19} f(x)$ |
$f^{-1}(x)=f(x)$ |
The correct answer is Option (1) → $f^{-1}(x)=f(x)$ $y=\frac{3x+2}{5x-3}⇒5xy-3y=3x+2$ So $5xy-3x=3y+2$ $x=\frac{3y+2}{5y-3}$ So $f^{-1}(x)=\frac{3x+2}{5x-3}$ $⇒f^{-1}(x)=f(x)$ |
