Question

If $g\left(x\right)=\ln \left[f\right(x\left)\right]$ and $f(x+1)=xf\left(x\right),$ then $g^{\prime \prime }\left(\frac{5}{2}\right)-g^{\prime \prime }\left(\frac{1}{2}\right)$ is equal to

• A. $-\frac{40}{9}$
• B. $\frac{40}{9}$
• C. $-\frac{20}{9}$
• D. $\frac{20}{9}$

Answer 1

The correct option is A $-\frac{40}{9}$

$g\left(x\right)=\ln \left[f\right(x\left)\right]$
$\Rightarrow f\left(x\right)=e^{g\left(x\right)}$
Also, $f(x+1)=xf\left(x\right)$
$\Rightarrow e^{g(x+1)}=x\cdot e^{g\left(x\right)}$
$\Rightarrow e^{g(x+1)-g\left(x\right)}=x$
$\Rightarrow g(x+1)-g\left(x\right)=\ln x$
Differentiating w.r.t. $x,$
$\Rightarrow g^{\prime }(x+1)-g^{\prime }\left(x\right)=\frac{1}{x}$
Again differentiating w.r.t. $x,$
$g^{\prime \prime }(x+1)-g^{\prime \prime }\left(x\right)=-\frac{1}{x^2}\cdots \left(i\right)$
Put $x=\frac{1}{2}$ in $\left(i\right)$
$\Rightarrow g^{\prime \prime }\left(\frac{3}{2}\right)-g^{\prime \prime }\left(\frac{1}{2}\right)=-4\cdots \left(ii\right)$
Put $x=\frac{3}{2}$ in $\left(i\right)$
$\Rightarrow g^{\prime \prime }\left(\frac{5}{2}\right)-g^{\prime \prime }\left(\frac{3}{2}\right)=-\frac{4}{9}\cdots \left(iii\right)$
Adding $\left(ii\right)$ and $\left(iii\right)$
$\Rightarrow g^{\prime \prime }\left(\frac{5}{2}\right)-g^{\prime \prime }\left(\frac{1}{2}\right)=-4-\frac{4}{9}=-\frac{40}{9}$