Let $\mathrm{f}: \mathrm{S} \rightarrow \mathrm{S}$ where $\mathrm{S}=(0, \infty)$ be a twice differentiable function such that $\mathrm{f}(\mathrm{x}+1)=\mathrm{xf}(\mathrm{x})$ If $g: S \rightarrow R$ be defined as $g(x)=\log _{e} f(x)$, then the value of $\left|g^{\prime \prime}(5)-\mathrm{g}^{\prime \prime}(1)\right|$ is equal to :
Correct Option: 1
$\operatorname{lnf}(x+1)=\ln (x f(x))$
$\operatorname{lnf}(x+1)=\ln x+\operatorname{lnf}(x)$
$\Rightarrow g(x+1)=\ln x+g(x)$
$\Rightarrow g(x+1)-g(x)=\ln x$
$\Rightarrow \quad g^{\prime \prime}(x+1)-g^{\prime \prime}(x)=-\frac{1}{x^{2}}$
Put $x=1,2,3,4$
$\mathrm{g}^{\prime \prime}(2)-\mathrm{g}^{\prime \prime}(1)=-\frac{1}{1^{2}} \quad \ldots(1)$
$\mathrm{g}^{\prime \prime}(3)-\mathrm{g}^{\prime \prime}(2)=-\frac{1}{2^{2}} \quad \ldots(2)$
$\mathrm{g}^{\prime \prime}(4)-\mathrm{g}^{\prime \prime}(3)=-\frac{1}{3^{2}}$
$g^{\prime \prime}(5)-g^{\prime \prime}(4)=-\frac{1}{4^{2}} \quad \ldots(4)$
Add all the equation we get
$\mathrm{g}^{\prime \prime}(5)-\mathrm{g}^{\prime \prime}(1)=-\frac{1}{1^{2}}-\frac{1}{2^{2}}-\frac{1}{3^{2}}-\frac{1}{4^{2}}$
$\left|g^{\prime \prime}(5)-g^{\prime \prime}(1)\right|=\frac{205}{144}$
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