If $\mathrm{g}(x)=1+\sqrt{x}$ and $\mathrm{f}(\mathrm{g}(x))=3+2 \sqrt{x}+x$, then $\mathrm{f}(\mathrm{f}(x))$ is

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Aicharya · Accepted Answer

$\begin{aligned} & g(x)=1+\sqrt{x} 	ext { and } \mathrm{f}(\mathrm{g}(x))=3+2 \sqrt{x}+x \ & 	herefore \quad \mathrm{f}(\mathrm{g}(x))=\left[(\sqrt{x})^2+2 \sqrt{x}+1ight]+2 \ & =(\sqrt{x}+1)^2+2 \ & =[\mathrm{g}(x)]^2+2 \ & \Rightarrow \mathrm{f}(x)=x^2+2 \ & \Rightarrow \mathrm{f}(\mathrm{f}(x))=\left(x^2+2ight)^2+2=x^4+4 x^2+6 \ & \end{aligned}$

If $\mathrm{g}(x)=1+\sqrt{x}$ and $\mathrm{f}(\mathrm{g}(x))=3+2 \sqrt{x}+x$, then…

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