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If $g(x)=x^2+x-2$ and $\frac{1}{2} g o f(x)=2 x^2-5 x+2$ then $f(x)$ is equal to
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The correct answer is:
$2 x-3$
$2 x-3$
We have, $\frac{1}{2} go f(x)=2 x^2-5 x+2$
$\begin{aligned}
& \Rightarrow g(f(x))=4 x^2-10 x+4 \\
& \Rightarrow(f(x))^2+f(x)-2=4 x^2-10 x+4 \\
& \Rightarrow(f(x))^2+f(x)-\left(4 x^2-10 x+6\right)=0 \\
& \Rightarrow f(x)=\frac{-1 \pm \sqrt{1+4\left(4 x^2-10 x+6\right)}}{2} \\
& \Rightarrow f(x)=\frac{-1 \pm \sqrt{16 x^2-40 x+25}}{2} \\
& \Rightarrow f(x)=\frac{-1 \pm(4 x-5)}{2}=2 x-3,-2 x+2
\end{aligned}$
Hence, $f(x)=2 x-3$
$\begin{aligned}
& \Rightarrow g(f(x))=4 x^2-10 x+4 \\
& \Rightarrow(f(x))^2+f(x)-2=4 x^2-10 x+4 \\
& \Rightarrow(f(x))^2+f(x)-\left(4 x^2-10 x+6\right)=0 \\
& \Rightarrow f(x)=\frac{-1 \pm \sqrt{1+4\left(4 x^2-10 x+6\right)}}{2} \\
& \Rightarrow f(x)=\frac{-1 \pm \sqrt{16 x^2-40 x+25}}{2} \\
& \Rightarrow f(x)=\frac{-1 \pm(4 x-5)}{2}=2 x-3,-2 x+2
\end{aligned}$
Hence, $f(x)=2 x-3$
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