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Let $f(x)=x^5+2 x^3+3 x+1, x \in \mathbf{R}$, and $g(x)$ be a function such that $g(f(x))=x$ for all $x \in \mathbf{R}$. Then $\frac{g(7)}{g^{\prime}(7)}$ is equal to :
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The correct answer is:
14
$\begin{aligned} & f(x)=x^5+2 x^3+3 x+1 \\ & f^{\prime}(x)=5 x^4+6 x^2+3 \\ & f^{\prime}(1)=5+6+3=14 \\ & g(f(x))=x \\ & g^{\prime}(f(x)) f^{\prime}(x)=1 \\ & \text { for } f(x)=7 \\ & \Rightarrow x^5+2 x^3+3 x+1=7 \\ & \Rightarrow x=1 \\ & g^{\prime}(7) f^{\prime}(1)=1 \Rightarrow g^{\prime}(7)=\frac{1}{f^{\prime}(1)}=\frac{1}{14}\end{aligned}$
$\begin{aligned} & x=1, f(x)=7 \Rightarrow g(7)=1 \\ & \frac{g(7)}{g^{\prime}(7)}=\frac{1}{1 / 14}=14\end{aligned}$
$\begin{aligned} & x=1, f(x)=7 \Rightarrow g(7)=1 \\ & \frac{g(7)}{g^{\prime}(7)}=\frac{1}{1 / 14}=14\end{aligned}$
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