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

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Aicharya · Accepted Answer

$f(x)=\frac{x-2}{2 x+1}$ $f(f(x))=-x$ [Given] $\Rightarrow \quad \frac{f(x)-2}{2(f(x))+1}=-x$ $\Rightarrow \quad \frac{\frac{x-2}{2 x+1}-2}{\frac{2 x-4}{2 x+1}+1}=-x$ $\Rightarrow \frac{x-2-4 x-2}{2 x-4+2 x+1}=-x$ $\Rightarrow \quad \frac{-3 x-4}{4 x-3}=-x \Rightarrow \frac{3 x+4}{4 x-3}=x$ $\Rightarrow \quad 3 x+4=4 x^2-3 x$ $\begin{aligned} & \Rightarrow \quad 4 x^2-6 x-4=0 \ & \Rightarrow \quad 2 x^2-3 x-2=0\end{aligned}$ Now, $\alpha+\beta=\frac{3}{2}, \alpha \beta=-1$ Now, $(\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta$ $\Rightarrow \quad \frac{9}{4}=\alpha^2+\beta^2-2 \quad[\because \alpha \beta=-1]$ $\begin{aligned} & \Rightarrow \alpha^2+\beta^2=\frac{17}{4} \ & \Rightarrow 4\left(\alpha^2+\beta^2ight)=17\end{aligned}$

Let $f: R-\left\{\frac{-1}{2}\right\} \rightarrow R$ be defined by $f(x)=\frac{x-2}{2 x+1}$. If $\alpha$ and…

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