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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

माना \(f: S \rightarrow S\), जहाँ \(S =(0, \infty)\) है, दो बार अवकलनीय फलन है जिसके लिए \(f( x +1)= x f( x )\) है। यदि \(g : S \rightarrow R , g ( x )=\log _{ e } f( x )\) द्वारा परिभाषित है, तो \(\left|g^{\prime \prime}(5)-g^{\prime \prime}(1)\right|\) का मान बराबर है

  1. A \(\frac{205}{144}\)
  2. B \(\frac{197}{144}\)
  3. C \(\frac{187}{144}\)
  4. D \(1\)
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Answer & Solution

Correct Answer

(A) \(\frac{205}{144}\)

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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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