Question

A function $f:R\to R$ is defined as $f\left(x\right)=3x^2+1$. Then $f^{-1}\left(x\right)$ is :

• A. $\frac{\sqrt{x-1}}{3}$
• B. $\frac{1}{3}\sqrt{x}-1$
• C. $\frac{1}{4}\sqrt{x}-1$
• D. $\frac{\sqrt{x+1}}{3}$

Answer 1

The correct option is A $\frac{\sqrt{x-1}}{3}$

$f\left(x\right)=3x^2+1$

Let$f\left(x\right)=y$

$y=3x^2+1$

Putting $xforyx=3y^2+13y^2=x-1y^2=\frac{x-1}{3}y=sqrt\frac{x-1}{3}{f}^{-1}\left(x\right)=\sqrt{\frac{x-1}{3}}$