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 :
- A \(\left\{-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\}\)
- B \(\left\{-\frac{1}{\sqrt{3}}, 0, \frac{1}{\sqrt{3}}\right\}\)
- C \(\left[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right]\)
- D \(\left[0, \frac{1}{\sqrt{3}}\right]\)
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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