Limit Continuity And Differentiability Ques 5
- 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 $N=1,2,3, \ldots, g^n\left(N+\frac{1}{2}\right)-g^g\left(\frac{1}{2}\right)$ is equal to
$(2008,3 M)$
(a) $-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$
(b) $4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$
(c) $-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N+1)^2}\right\}$
(d) $4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N+1)^2}\right\}$
Show Answer
Answer:
Correct Answer: 5.(a)
Solution: (a) Since, $f(x)=e^{g(x)} \Rightarrow e^{g(x+1)}=f(x+1)=x f(x)=x e^{g(x)}$
and $ \quad g(x+1)=\log x+g(x) $
i.e. $ \quad g(x+1)-g(x)=\log x \quad ……..(i) $
Replacing $x$ by $x-\frac{1}{2}$, we get
$ \begin{aligned} g\left(x+\frac{1}{2}\right)-g\left(x-\frac{1}{2}\right) & =\log \left(x-\frac{1}{2}\right)=\log (2 x-1)-\log 2 \\ \therefore\quad g^{\prime \prime}\left(x+\frac{1}{2}\right)-g^{\prime \prime}\left(x-\frac{1}{2}\right) & =\frac{-4}{(2 x-1)^2} \quad ……..(ii) \end{aligned} $
On substituting, $x=1,2,3, \ldots, N$ in Eq. (ii) and adding, we get
$ B^{\prime \prime}\left(N+\frac{1}{2}\right)-B^{\prime \prime}\left(\frac{1}{2}\right)=-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\} \text {. } $
BETA
