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If $R$ be the set of all real numbers and $f: R \rightarrow R$ is given by $f(x)=3 x^{2}+1$. Then, the set $f^{-1}([1,6])$ is
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The correct answer is:
$\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]$
Given, $f(x)=3 x^{2}+1$
Let $\quad y=3 x^{2}+1$
$\Rightarrow \quad 3 x^{2}=y-1 \Rightarrow x^{2}=\frac{y-1}{3}$
$\Rightarrow \quad x=\pm \sqrt{\frac{y-1}{3}}$
$\therefore \quad f^{-1}(x)=\pm \sqrt{\frac{x-1}{3}}$
When $x \in[1,6]$
Then, $\quad f^{-1}(x) \in\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]$
Let $\quad y=3 x^{2}+1$
$\Rightarrow \quad 3 x^{2}=y-1 \Rightarrow x^{2}=\frac{y-1}{3}$
$\Rightarrow \quad x=\pm \sqrt{\frac{y-1}{3}}$
$\therefore \quad f^{-1}(x)=\pm \sqrt{\frac{x-1}{3}}$
When $x \in[1,6]$
Then, $\quad f^{-1}(x) \in\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]$
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