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CUET · MATHS · PYQ PAPER 2023

Let \(R\) be the set of all real numbers and \(f: R \rightarrow\) Range \(f\) be given by \(f(x)=3 x^2+1\). Then \(f^{-1}\{1,2\}\) is :

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

Answer & Solution

Correct Answer

(B) \(\left\{-\frac{1}{\sqrt{3}}, 0, \frac{1}{\sqrt{3}}\right\}\)

Step-by-step Solution

Detailed explanation

\(3x^2+1 = 1 \implies 3x^2=0 \implies x=0\) \(3x^2+1 = 2 \implies 3x^2=1 \implies x^2=\frac{1}{3} \implies x=\pm\frac{1}{\sqrt{3}}\) \(f^{-1}\{1,2\} = \left\{-\frac{1}{\sqrt{3}}, 0, \frac{1}{\sqrt{3}}\right\}\)
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