JEE Mains · Maths · STD 12 - 5. continuity and differentiation
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 :
- A \(7\)
- B \(42\)
- C \(1\)
- D \(14\)
Answer & Solution
Correct Answer
(D) \(14\)
Step-by-step Solution
Detailed explanation
\( f(x)=x^5+2 x^3+3 x+1 \) \( f^{\prime}(x)=5 x^4+6 x^2+3 \) \( f^{\prime}(1)=5+6+3=14 \) \( g(f(x))=x \) \( g^{\prime}(f(x)) f^{\prime}(x)=1 \) \( \text { for } f(x)=7 \) \( \Rightarrow x^5+2 x^3+3 x+1=7 \) \( \Rightarrow x=1 \)…
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