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Let $f: R \rightarrow R$ be defined by $f(x)=2 x+3$. If $\alpha$, $\beta$ are the roots of the equation $f\left(x^2\right)-2 f\left(\frac{x}{2}\right)-1=0$, then $\alpha^2+\beta^2=$
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$5$
$\begin{array}{rlrl} & & f(x)=2 x+3 \\ & & f\left(x^2\right)-2 f\left(\frac{x}{2}\right)-1=0 \\ \Rightarrow & 2 x^2+3-2(x+3)-1=0 \\ \Rightarrow & & 2 x^2-2 x-4=0 \\ \Rightarrow & & x^2-x-2=0 \\ \Rightarrow & & x=2 \text { or } x=-1 \\ \therefore & & \alpha^2+\beta^2=4+1=5\end{array}$
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