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Mathematics Question on Functions

Let RR be the set of all real numbers and f:R→Rf : R \rightarrow R be given by f(x)=3x2+1f(x) = 3x^2 + 1.Then the set f−1([1,6])f ^{-1}\left(\left[1, 6\right]\right) is

A

\left\\{-\sqrt{\frac{5}{3}},0, \sqrt{\frac{5}{3}}\right\\}

B

[−53,53]\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]

C

[−13,13]\left[-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}\right]

D

(−53,53)\left(-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right)

Answer

[−53,53]\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]

Explanation

Solution

The correct answer is B:[−53,53][-\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}}]
Given, y=3x2+1y=3 x^{2}+1
⇒3x2=y−1\Rightarrow 3 x^{2} =y-1
⇒x2=y−13\Rightarrow x^{2}=\frac{y-1}{3}
⇒x=±y−13\Rightarrow x=\pm \sqrt{\frac{y-1}{3}}
∴f−1(x)=±x−13\therefore f^{-1}(x)=\pm \sqrt{\frac{x-1}{3}}
When x∈[1,6]x \in[1,6]
Then, f−1(x)∈[−53,53]f^{-1}(x) \in\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]
root