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TS EAMCET · Maths · Functions

If \(f:[1, \infty) \rightarrow[5, \infty)\) is given by \(f(x)=3 x+\frac{2}{x}\), then \(f^{-1}(x)=\)

  1. A \(\frac{1}{6}\left[x+\sqrt{x^2-24}\right]\)
  2. B \(\frac{x}{3 x^2+2}\)
  3. C \(\frac{1}{6}\left[x-\sqrt{x^2-24}\right]\)
  4. D \(\frac{1}{2}\left[1+\sqrt{x^2-4}\right]\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{6}\left[x+\sqrt{x^2-24}\right]\)

Step-by-step Solution

Detailed explanation

We have, \(f:[1, \infty) \rightarrow[5, \infty)\) given by \(f(x)=3 x+\frac{2}{x}\) Let \(\quad y=f(x)=3 x+\frac{2}{x}\). Then, we get…