Question

Let \( A = \mathbb{R} - \{ 3 \} \) and \( B = \mathbb{R} - \{ 1 \} \). Let \( f : A \to B \) be defined by \( f(x) = \frac{x-2}{x-3} \). What is the value of \( f^{-1} \left( \frac{1}{2} \right) \)?

• A. \( \frac{3}{5} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{2}{3} \)
• D. \( 1 \)

Hint:

To find the inverse of a function, set \( f(x) \) equal to the given value and solve for \( x \).

Answer 1

The Correct Option is A

We are given the function \( f(x) = \frac{x-2}{x-3} \). We need to find \( f^{-1} \left( \frac{1}{2} \right) \).
First, set \( f(x) = \frac{1}{2} \): \[ \frac{x-2}{x-3} = \frac{1}{2} \] Cross-multiply: \[ 2(x - 2) = x - 3 \] Simplify: \[ 2x - 4 = x - 3 \quad \Rightarrow \quad x = \frac{3}{5} \] Therefore, \( f^{-1} \left( \frac{1}{2} \right) = \frac{3}{5} \).