Question

Let $ f : \mathbb{R} - \left\{-\frac{1}{2}\right\} \rightarrow \mathbb{R} $ be defined by $$ f(x) = \frac{x - 2}{2x + 1} $$ If $ \alpha, \beta $ satisfy the equation $$ f(f(x)) = -x $$ then evaluate: $$ 4(\alpha^2 + \beta^2) $$

• A. 17
• B. 12
• C. 24
• D. 34

Hint:

To solve function composition equations like \( f(f(x)) = -x \), compute the inner function first, then substitute and simplify carefully.

Answer 1

The Correct Option is A

We are given: \[ f(x) = \frac{x - 2}{2x + 1} \] We must find: \[ f(f(x)) = -x \] First, compute \( f(f(x)) \). Let \( y = f(x) = \frac{x - 2}{2x + 1} \). Then, \[ f(f(x)) = f(y) = \frac{y - 2}{2y + 1} = \frac{\frac{x - 2}{2x + 1} - 2}{2\left(\frac{x - 2}{2x + 1}\right) + 1} \] Simplify numerator: \[ \frac{x - 2 - 2(2x + 1)}{2x + 1} = \frac{x - 2 - 4x - 2}{2x + 1} = \frac{-3x - 4}{2x + 1} \] Simplify denominator: \[ \frac{2(x - 2) + (2x + 1)}{2x + 1} = \frac{2x - 4 + 2x + 1}{2x + 1} = \frac{4x - 3}{2x + 1} \] So, \[ f(f(x)) = \frac{\frac{-3x - 4}{2x + 1}}{\frac{4x - 3}{2x + 1}} = \frac{-3x - 4}{4x - 3} \] We are told that \( f(f(x)) = -x \), so: \[ \frac{-3x - 4}{4x - 3} = -x \] Multiply both sides by \( 4x - 3 \): \[ -3x - 4 = -x(4x - 3) = -4x^2 + 3x \] Rearrange: \[ -3x - 4 + 4x^2 - 3x = 0 \Rightarrow 4x^2 - 6x - 4 = 0 \] Divide by 2: \[ 2x^2 - 3x - 2 = 0 \] Solve using quadratic formula: \[ x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4} \] So the roots are: \[ \alpha = \frac{8}{4} = 2, \quad \beta = \frac{-2}{4} = -\frac{1}{2} \] Now compute: \[ 4(\alpha^2 + \beta^2) = 4\left(2^2 + \left(-\frac{1}{2}\right)^2\right) = 4\left(4 + \frac{1}{4}\right) = 4 \cdot \frac{17}{4} = 17 \]