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: Write the given function.
Given, \[ f(x)=\frac{x-2}{2x+1} \] We need to solve \[ f(f(x))=-x \] Let \[ y=f(x)=\frac{x-2}{2x+1} \]

Step 2: Find \(f(f(x))\).
Now, \[ f(f(x))=f(y)=\frac{y-2}{2y+1} \] Substitute \[ y=\frac{x-2}{2x+1} \] So, \[ f(f(x))= \frac{\frac{x-2}{2x+1}-2}{2\left(\frac{x-2}{2x+1}\right)+1} \]

Step 3: Simplify numerator and denominator.
Numerator: \[ \frac{x-2}{2x+1}-2 = \frac{x-2-2(2x+1)}{2x+1} \] \[ = \frac{x-2-4x-2}{2x+1} = \frac{-3x-4}{2x+1} \] Denominator: \[ 2\left(\frac{x-2}{2x+1}\right)+1 = \frac{2x-4}{2x+1}+1 \] \[ = \frac{2x-4+2x+1}{2x+1} = \frac{4x-3}{2x+1} \] Therefore, \[ f(f(x))= \frac{\frac{-3x-4}{2x+1}}{\frac{4x-3}{2x+1}} \] \[ f(f(x))=\frac{-3x-4}{4x-3} \]

Step 4: Use the given equation.
Given, \[ f(f(x))=-x \] So, \[ \frac{-3x-4}{4x-3}=-x \] Cross-multiplying, \[ -3x-4=-x(4x-3) \] \[ -3x-4=-4x^2+3x \] \[ 4x^2-6x-4=0 \] Dividing by \(2\), \[ 2x^2-3x-2=0 \]

Step 5: Find \(\alpha+\beta\) and \(\alpha\beta\).
If \(\alpha,\beta\) are roots of \[ 2x^2-3x-2=0, \] then \[ \alpha+\beta=\frac{3}{2} \] and \[ \alpha\beta=\frac{-2}{2}=-1 \] Now, \[ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta \] \[ \alpha^2+\beta^2=\left(\frac{3}{2}\right)^2-2(-1) \] \[ =\frac{9}{4}+2 \] \[ =\frac{9}{4}+\frac{8}{4} \] \[ =\frac{17}{4} \]

Step 6: Find the required value.
We need \[ 4(\alpha^2+\beta^2) \] Thus, \[ 4(\alpha^2+\beta^2)=4\cdot \frac{17}{4} \] \[ =17 \] Therefore, \[ \boxed{17} \]