Question

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

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

Answer 1

Answer: (A)

$\ln f(x+1)=\ln (xf(x))$
$lnf(x+1)=\ln x+lnf(x)$
$\Rightarrow g(x+1)=\ln x+g(x)$
$\Rightarrow g(x+1)-g(x)=\ln x$
$\Rightarrow g^{\prime \prime }(x+1)-g^{\prime \prime }(x)=-\frac{1}{x^2}$
Put $x=1,2,3,4$
$g^{\prime \prime }(2)-g^{\prime \prime }(1)=-\frac{1}{1^2}\dots$(1)
$g^{\prime \prime }(3)-g^n(2)=-\frac{1}{2^2}\dots$(2)
$g^{\prime \prime }(4)-g^{\prime \prime }(3)=-\frac{1}{3^2}\dots$(3)
$g^{\prime \prime }(5)-g^{\prime \prime }(4)=-\frac{1}{4^2}\dots$(4)
Add all the equation we get
$g^{\prime \prime }(5)-g^{\prime \prime }(1)=-\frac{1}{1^2}-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}$
$\left|g^{\prime \prime }(5)-g^{\prime \prime }(1)\right|-\frac{205}{144}$