Question

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

• A. \(17\)
• B. \(12\)
• C. \(24\)
• D. \(34\)

Hint:

For a quadratic equation \(ax^2+bx+c=0\), if roots are \(\alpha,\beta\), then use \[ \alpha+\beta=-\frac{b}{a} \] and \[ \alpha\beta=\frac{c}{a} \] to calculate expressions like \[ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta. \]

Answer 1

The Correct Option is A

Step 1: Identify the function and the composition.
We have $f(x) = \dfrac{x-2}{2x+1}$ defined on $\mathbb{R} \setminus \{-\tfrac{1}{2}\}$. We need $f(f(x)) = -x$ and then find $4(\alpha^2+\beta^2)$ where $\alpha, \beta$ are the solutions.
Step 2: Compute $f(f(x))$.
First substitute $f(x)$ into itself. Let $y = f(x) = \dfrac{x-2}{2x+1}$. Then: \[f(y) = \frac{y - 2}{2y + 1} = \frac{\dfrac{x-2}{2x+1} - 2}{2 \cdot \dfrac{x-2}{2x+1} + 1}.\]
Step 3: Simplify numerator and denominator separately.
Numerator: $\dfrac{x-2}{2x+1} - 2 = \dfrac{x-2 - 2(2x+1)}{2x+1} = \dfrac{x-2-4x-2}{2x+1} = \dfrac{-3x-4}{2x+1}$.
Denominator: $\dfrac{2(x-2)}{2x+1} + 1 = \dfrac{2x-4+2x+1}{2x+1} = \dfrac{4x-3}{2x+1}$.
So $f(f(x)) = \dfrac{-3x-4}{4x-3}$.
Step 4: Set $f(f(x)) = -x$ and form the equation.
\[\frac{-3x-4}{4x-3} = -x \implies -3x-4 = -x(4x-3) = -4x^2 + 3x.\] Rearranging: $4x^2 - 3x - 3x - 4 = 0 \implies 4x^2 - 6x - 4 = 0 \implies 2x^2 - 3x - 2 = 0$.
Step 5: Read off $\alpha + \beta$ and $\alpha\beta$ by Vieta's formulas.
From $2x^2 - 3x - 2 = 0$: $\alpha + \beta = \dfrac{3}{2}$ and $\alpha\beta = \dfrac{-2}{2} = -1$.
Step 6: Compute $\alpha^2 + \beta^2$.
\[\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \frac{9}{4} - 2(-1) = \frac{9}{4} + 2 = \frac{17}{4}.\]
Step 7: Find the required value.
\[4(\alpha^2+\beta^2) = 4 \times \frac{17}{4} = 17.\] \[ \boxed{17} \]