Question

Let $f : S \rightarrow S$ where $S =(0, \infty)$ be a twice differentiable function such that $f ( x +1)= xf ( x )$ If $g: S \rightarrow R$ 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)

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