COMEDK · Maths · 24. Functions
If \(f:[0, \infty) \rightarrow[2, \infty)\) is given by \(f(x)=x+\frac{1}{x}\), then \(f^{-1}(x)\) equals
- A \(\frac{x+\sqrt{x^{2}-4}}{2}\)
- B \(\frac{x}{1+x^{2}}\)
- C \(\frac{2 x-\sqrt{x^{2}-4}}{2}\)
- D \(1+\sqrt{x^{2}-4}\)
Answer & Solution
Correct Answer
(A) \(\frac{x+\sqrt{x^{2}-4}}{2}\)
Step-by-step Solution
Detailed explanation
Let \(f(x)=y=x+\frac{1}{x}\) or \(x y=x^{2}+1\) or \(x^{2}-x y+1=0\) Since, \(x \in[0, \infty)\) \(\therefore \quad D \geq 0 \Rightarrow y^{2}-4 \geq 0 \Rightarrow y \in[2, \infty)\) \(\therefore \quad x=\frac{y \pm \sqrt{y^{2}-4}}{2}\) \(\quad x=\frac{y-\sqrt{y^{2}-4}}{2}\) or…
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