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(A) Given $f(x) = \frac{x-2}{2x+1}$.We are given the equation $f(f(x)) = -x$.Substituting $f(x)$ into the expression:$\frac{f(x)-2}{2f(x)+1} = -x$$\fr

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) Given $f(x) = \frac{x-2}{2x+1}$.We are given the equation $f(f(x)) = -x$.Substituting $f(x)$ into the expression:$\frac{f(x)-2}{2f(x)+1} = -x$$\frac{\frac{x-2}{2x+1}-2}{2(\frac{x-2}{2x+1})+1} = -x$$\frac{x-2-2(2x+1)}{2(x-2)+1(2x+1)} = -x$$\frac{x-2-4x-2}{2x-4+2x+1} = -x$$\frac{-3x-4}{4x-3} = -x$$\frac{3x+4}{4x-3} = x$$3x+4 = x(4x-3)$$3x+4 = 4x^2-3x$$4x^2-6x-4 = 0$Dividing by $2$,we get $2x^2-3x-2 = 0$.Since $\alpha$ and $\beta$ are roots of this quadratic equation,by Vieta's formulas:$\alpha+\beta = \frac{3}{2}$ and $\alpha\beta = -1$.We need to find $4(\alpha^2+\beta^2)$.Using the identity $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$:$\alpha^2+\beta^2 = (\frac{3}{2})^2 - 2(-1) = \frac{9}{4} + 2 = \frac{17}{4}$.Therefore,$4(\alpha^2+\beta^2) = 4 	imes \frac{17}{4} = 17$.

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

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