Let g(x)= (f(x)) where f(x) is a twice differentiable positive fu… - Vidyadip

MCQ

Let $g(x)=\log (f(x))$ where $f(x)$ is a twice differentiable positive function on $(0, \infty)$ such that $f(x+1)=x f(x)$. Then, for $\mathrm{N}=1,2,3, \ldots$,

$\mathrm{g}^{\prime \prime}\left(\mathrm{N}+\frac{1}{2}\right)-\mathrm{g}^{\prime \prime}\left(\frac{1}{2}\right)=$

✓
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 \mathrm{~N}-1)^2}\right\}$
B
$4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 \mathrm{~N}-1)^2}\right\}$
C
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 \mathrm{~N}+1)^2}\right\}$
D
$4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 \mathrm{~N}+1)^2}\right\}$

✓

Answer

Correct option: A.

$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 \mathrm{~N}-1)^2}\right\}$

a
$g(x+1)=\log (f(x+1))=\log x+\log (f(x))$

$\Rightarrow \mathrm{g}(\mathrm{x}+1)-\mathrm{g}(\mathrm{x})=\log \mathrm{x}$

$\Rightarrow \mathrm{g}^{\prime \prime}(\mathrm{x}+1)-\mathrm{g}^{\prime \prime}(\mathrm{x})=-\frac{1}{\mathrm{x}^2}$

$\mathrm{g}^{\prime \prime}\left(1+\frac{1}{2}\right)-\mathrm{g}^{\prime \prime}\left(\frac{1}{2}\right)=-4$

$\mathrm{g}^{\prime \prime}\left(2+\frac{1}{2}\right)-\mathrm{g}^{\prime \prime}\left(1+\frac{1}{2}\right)=-\frac{4}{9}$

$\mathrm{g}^*\left(\mathrm{~N}+\frac{1}{2}\right)-\mathrm{g}^*\left(\mathrm{~N}-\frac{1}{2}\right)=-\frac{4}{(2 \mathrm{~N}-1)^2}$

Summing up all terms

Hence, $\mathrm{g}^{\prime \prime}\left(\mathrm{N}+\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{1}{2}\right)=-4\left(1+\frac{1}{9}+\cdots+\frac{1}{(2 \mathrm{~N}-1)^2}\right)$.

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