Question

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

Answer: (B)

$[-\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}}]$

Given, $f\left(x\right)=3x^2+1$
Let $y=3x^2+1$
$\Rightarrow 3x^2=y-1\Rightarrow x^2=\frac{y-1}{3}$
$\Rightarrow x=\pm \sqrt{\frac{y-1}{3}}$
$\therefore f^{-1}\left(x\right)=\pm \sqrt{\frac{x-1}{3}}$
When $x\in [1,6]$, then, $f^{-1}\left(x\right)\in [-\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}}]$