Question

If \( f(x) = \frac{1}{x^2 + 1} \), \( 0<x<1 \), then \( f^{-1} \left( \frac{1}{4} \right) + f^{-1} \left( \frac{3}{4} \right) \)= : ?

• A. 1
• B. 3
• C. \( 4\sqrt{3} \)
• D. \( \sqrt{3} \)

Hint:

When solving for the inverse of a function, ensure that you express the function in terms of \( y \), solve for \( x \), and then substitute the values carefully.

Answer 1

The Correct Option is C

We know that \( f(x) = \frac{1}{x^2 + 1} \). To find \( f^{-1} \), we first solve for \( x \) in terms of \( y \): \[ y = \frac{1}{x^2 + 1} \quad \Rightarrow \quad x^2 = \frac{1}{y} - 1 \quad \Rightarrow \quad x = \sqrt{\frac{1}{y} - 1} \] Now, substitute \( \frac{1}{4} \) and \( \frac{3}{4} \) into the inverse: \[ f^{-1} \left( \frac{1}{4} \right) = \sqrt{\frac{1}{\frac{1}{4}} - 1} = \sqrt{4 - 1} = \sqrt{3} \] \[ f^{-1} \left( \frac{3}{4} \right) = \sqrt{\frac{1}{\frac{3}{4}} - 1} = \sqrt{\frac{4}{3} - 1} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \] Adding both results: \[ f^{-1} \left( \frac{1}{4} \right) + f^{-1} \left( \frac{3}{4} \right) = \sqrt{3} + \frac{1}{\sqrt{3}} = 4\sqrt{3} \]