Let f(x) = x^5 + 2x^3 + 3x + 1, x ∈ R and g(x) be a function such that g(f(x)) = x

Question

Let \(f(x)=x^{5}+2 x^{3}+3 x+1, x \in R\), and \(g(x)\) be a function such that \(g(f(x))=x\) for all \(x \in R\). Then \(\frac{g(7)}{g^{\prime}(7)}\) is equal to :

(1) 7

(2) 42

(3) 1

(4) 14

Answer 1

Correct option is (4) 14

\(\mathrm {f(x)=x^{5}+2 x^{3}+3 x+1}\)

\(\mathrm {f^{\prime}(x)=5 x^{4}+6 x^{2}+3}\)

\(\mathrm{f}^{\prime}(1)=5+6+3=14\)

\(\mathrm{g}(\mathrm{f}(\mathrm{x}))=\mathrm{x}\)

\(\mathrm{g}^{\prime}\left(\mathrm{f}(\mathrm{x}) \mathrm{f}^{\prime}(\mathrm{x})=1\right.\)

for \(f(x)=7\)

\(\Rightarrow \mathrm{x}^{5}+2 \mathrm{x}^{3}+3 \mathrm{x}+1=7\)

\(\Rightarrow \mathrm{x}=1\)

\(\mathrm {g^{\prime}(7) f^{\prime}(1)=1 \Rightarrow g^{\prime}(7)=\frac{1}{f^{\prime}(1)}=\frac{1}{14}}\)

\(\mathrm{x}=1, \mathrm{f}(\mathrm{x})=7 \Rightarrow \mathrm{g}(7)=1\)

\(\frac{\mathrm{g}(7)}{\mathrm{g}^{\prime}(7)}=\frac{1}{1 / 14}=14\)