QJEEM24Ma6544: JEE Main 2024 Mathematics - Class 12 JEEM24Ma6544 Solution: Let , and be a function such that for all . Then is equal to :

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

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