Question

Let $f:S\to S$ where $S=\left(0,\infty \right)$ 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 $\left|g\text{'}\text{'}\left(5\right)-g\text{'}\text{'}\left(1\right)\right|$ is equal to:

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

Answer 1

The correct option is C

$\frac{205}{144}$

Determine the value of $\left|g\text{'}\text{'}\left(5\right)-g\text{'}\text{'}\left(1\right)\right|$

Given that:

$f(x+1)=xf\left(x\right)$

$g\left(x\right)={\log }_{e}f\left(x\right)$

Now,

$$\begin{gathered} g(x+1)={\log }_{e}f\left(x+1\right)\left[\text{Given}\right] \\ \Rightarrow g(x+1)={\log }_{e}xf\left(x\right) \\ \Rightarrow g(x+1)={\log }_{e}x+{\log }_{e}f\left(x\right) \\ \Rightarrow g(x+1)-{\log }_{e}f\left(x\right)={\log }_{e}x \\ \Rightarrow g(x+1)-g\left(x\right)={\log }_{e}x\left[{\log }_{e}f\left(x\right)=g\left(x\right)\right] \\ \mathrm{Double}\mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}\mathrm{x}: \\ g\text{'}\text{'}(x+1)-g\text{'}\text{'}\left(x\right)=-\frac{1}{x^2} \\ Letx=1, \\ g\text{'}\text{'}\left(2\right)-g\text{'}\text{'}\left(1\right)=-1.....\left(1\right) \\ Letx=2, \\ g\text{'}\text{'}\left(3\right)-g\text{'}\text{'}\left(2\right)=-\frac{1}{4}.....\left(2\right) \\ Letx=3, \\ g\text{'}\text{'}\left(4\right)-g\text{'}\text{'}\left(3\right)=-\frac{1}{9}......\left(3\right) \\ Letx=4, \\ g\text{'}\text{'}\left(5\right)-g\text{'}\text{'}\left(4\right)=-\frac{1}{16}.....\left(4\right) \\ \left(1\right)\left(2\right)+\left(3\right)+\left(4\right): \\ g\text{'}\text{'}\left(5\right)-g\text{'}\text{'}\left(1\right)=-1-\frac{1}{4}-\frac{1}{9}-\frac{1}{16} \\ \left|g\text{'}\text{'}\left(5\right)-g\text{'}\text{'}\left(1\right)\right|=\frac{205}{144} \end{gathered}$$

Therefore, the correct answer is Option (C).