JEE MAIN - Mathematics (2024 - 5th April Morning Shift - No. 8)

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 :
42
7
1
14

Explanation

$$\begin{aligned} & f(x)=x^5+2 x^3+3 x+1 \\ & g(f(x))=x . \quad \Rightarrow g^{\prime}(f(x)) f^{\prime}(x)=1 \\ \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & \text { Now } \frac{g(7)}{g^{\prime}(7)} \\ & g(7) \Rightarrow f(x)=7 \\ & x^5+2 x^3+3 x+1=7 \\ & \Rightarrow x\left(x^4+2 x^2+3\right)=0 \\ & \Rightarrow x=1 \\ & \therefore g(7) \Rightarrow g(f(1))=1 \\ & \& ~g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \\ & g^{\prime}(7) \\ & \Rightarrow f(x)=7 \Rightarrow x=1 \\ & \therefore g^{\prime}(7)=\frac{1}{f^{\prime}(1)} \\ & =\frac{1}{5 x^4+6 x^2+3} \\ & =\frac{1}{14} \\ & \therefore \frac{g(7)}{g^{\prime}(7)}=\frac{1}{\frac{1}{14}} \quad 14 \end{aligned}\\ \end{aligned}$$

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