Question

Let $f:S\to S$ where $S=(0,\infty )$ be a twice differentiable function such that $f(x+1)=xf\left(x\right).$ If $g:S\to R$ be defined as $g\left(x\right)={\log }_{e}f\left(x\right),$ then the value of $|g^{\prime \prime }\left(5\right)-g^{\prime \prime }\left(1\right)|$ is equal to :

• A. $\frac{187}{144}$
• B. $\frac{197}{144}$
• C. $1$
• D. $\frac{205}{144}$

Answer 1

The correct option is D $\frac{205}{144}$

$f(x+1)=xf\left(x\right)$
$g(x+1)={\log }_e\left(f\right(x+1\left)\right)$
$\Rightarrow g(x+1)={\log }_{e}x+{\log }_{e}f\left(x\right)$
$\Rightarrow g(x+1)-g\left(x\right)={\log }_{e}x$
$g^{\prime \prime }(x+1)-g^{\prime \prime }\left(x\right)=-\frac{1}{x^2}$
$g^{\prime \prime }\left(2\right)-g^{\prime \prime }\left(1\right)=-1$
$g^{\prime \prime }\left(3\right)-g^{\prime \prime }\left(2\right)=-\frac{1}{4}$
$g^{\prime \prime }\left(4\right)-g^{\prime \prime }\left(3\right)=-\frac{1}{9}$
$g^{\prime \prime }\left(5\right)-g^{\prime \prime }\left(4\right)=-\frac{1}{16}$
$g^{\prime \prime }\left(5\right)-g^{\prime \prime }\left(1\right)=-(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16})$
$\therefore |g^{\prime \prime }\left(5\right)-g^{\prime \prime }\left(1\right)|=\left(\frac{144+36+16+9}{16\times 9}\right)=\frac{205}{144}$