# Let f : S → S where S = (0,∞) be a twice

Question:

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 $\left|g^{\prime \prime}(5)-g^{\prime \prime}(1)\right|$ is equal to :

1. $\frac{205}{144}$

2. $\frac{197}{144}$

3. $\frac{187}{144}$

4. 1

Correct Option: 1

Solution:

$\operatorname{lnf}(x+1)=\ln (x f(x))$

$\operatorname{lnf}(x+1)=\ln x+\operatorname{lnf}(x)$

$\Rightarrow \mathrm{g}(\mathrm{x}+1)=\ln \mathrm{x}+\mathrm{g}(\mathrm{x})$

$\Rightarrow \mathrm{g}(\mathrm{x}+1)-\mathrm{g}(\mathrm{x})=\ln \mathrm{x}$

$\Rightarrow \mathrm{g}^{\prime \prime}(\mathrm{x}+1)-\mathrm{g}^{\prime \prime}(\mathrm{x})=-\frac{1}{\mathrm{x}^{2}}$

Put $x=1,2,3,4$

$g^{\prime \prime}(2)-g^{\prime \prime}(1)=-\frac{1}{1^{2}}$.......(1)

$g^{\prime \prime}(3)-g^{\prime \prime}(2)=-\frac{1}{2^{2}}$......(2)

$g^{\prime \prime}(4)-g^{\prime \prime}(3)=-\frac{1}{3^{2}}$........(3)

$g^{\prime \prime}(5)-g^{\prime \prime}(4)=-\frac{1}{4^{2}}$.......(4)

Add all the equation we get

$g^{\prime \prime}(5)-g^{\prime \prime}(1)=-\frac{1}{1^{2}}-\frac{1}{2^{2}}-\frac{1}{3^{2}}-\frac{1}{4^{2}}$

$\left|g^{\prime \prime}(5)-g^{\prime \prime}(1)\right|=\frac{205}{144}$