Question

Let $R$ be the set of all real numbers and $f=R\to R$ be given by $f\left(x\right)=3x^2+1$. Then the set $f^{-1}\left(\right[1,6\left]\right)$ is

• A. $\{-\sqrt{\frac{5}{3}},0,\sqrt{\frac{5}{3}}\}$
• B. $[-\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}}]$
• C. $\{-\sqrt{\frac{1}{3}},\sqrt{\frac{1}{3}}\}$
• D. $(-\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}})$

Answer 1

The correct option is B $[-\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}}]$

$y=3x^2+1$

$\therefore x=\sqrt{\frac{y-1}{3}}$

To find $f^{-1}\left(x\right)$ replace $y$ by $x$ in the above equation

$\therefore f^{-1}\left(x\right)=\sqrt{\frac{x-1}{3}}$

For $1\le x\le 6$

$-\sqrt{\frac{5}{3}}\le f^{-1}\left(x\right)\le \sqrt{\frac{5}{3}}$

So, the range of $f^{-1}\left(x\right)$ is $[-\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}}]$