JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(f : S \rightarrow S\) where \(S =(0, \infty)\) be a twice differentiable function such that \(f ( x +1)= xf ( x )\) If \(g: S \rightarrow R\) be defined as \(g(x)=\log _{e} f(x),\) then the value of \(\mid g "(5)- g "(1) \mid\) is equal to :
- A \(\frac{205}{144}\)
- B \(\frac{197}{144}\)
- C \(\frac{187}{144}\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(\frac{205}{144}\)
Step-by-step Solution
Detailed explanation
\(\operatorname{lnf}(x+1)=\ln (x f(x))\) \(\operatorname{lnf}(x+1)=\ln x+\operatorname{lnf}(x)\) \(\Rightarrow g(x+1)=\ln x+g(x)\) \(\Rightarrow g(x+1)-g(x)=\ln x\) \(\Rightarrow \quad g^{\prime \prime}(x+1)-g^{\prime \prime}(x)=-\frac{1}{x^{2}}\) Put \(x=1,2,3,4\)…
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