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WBJEE · Maths · Functions

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

  1. A \(\left\{-\sqrt{\frac{5}{3}}, 0, \sqrt{\frac{5}{3}}\right\}\)
  2. B \(\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]\)
  3. C \(\left[-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}\right]\)
  4. D \(\left(-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]\)

Step-by-step Solution

Detailed explanation

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,…