If $\mathrm{g}(x)=[\mathrm{f}(2 \mathrm{f}(x)+2)]^2$ and $\mathrm{f}(0)=-1, \mathrm{f}^{\prime}(0)=1$, then $g^{\prime}(0)$ is

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Aicharya · Accepted Answer

1. (A). Std. $12 \mid$ Part-2 $|\mathrm{Ch}-1|$ Exercise-1.1 $\begin{aligned} \mathrm{g}(x) & =\{\mathrm{f}[2 \mathrm{f}(x)+2]\}^2 \ 	herefore \quad \mathrm{~g}^{\prime}(x) & =2 \mathrm{f}[2 \mathrm{f}(x)+2] \cdot \mathrm{f}^{\prime}[2 \mathrm{f}(x)+2] \cdot 2 \mathrm{f}^{\prime}(x) \ 	herefore \quad \mathrm{g}^{\prime}(0) & =2 \mathrm{f}[2 \mathrm{f}(0)+2] \cdot \mathrm{f}^{\prime}[2 \mathrm{f}(0)+2] \cdot 2 \mathrm{f}^{\prime}(0) \ & =2 \mathrm{f}[2(-1)+2] \cdot \mathrm{f}^{\prime}(2(-1)+2) \cdot 2(1) \ \Rightarrow \mathrm{g}^{\prime}(0) & =4 \mathrm{f}(0) \cdot \mathrm{f}^{\prime}(0) \ & =4(-1)(1) \ & =-4 \end{aligned}$

If $\mathrm{g}(x)=[\mathrm{f}(2 \mathrm{f}(x)+2)]^2$ and $\mathrm{f}(0)=-1, \mathrm{f}^{\prime}(0)=1$, then…

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