Limit Continuity and Differentiability 7 Question 13
13. Let $f(x)=x _0^{2}\left|\cos \frac{\pi}{x}\right| \begin{gathered}x \neq 0, x \in R \text {, then } f \text { is } \ x=0\end{gathered}$
(2012)
(a) differentiable both at $x=0$ and at $x=2$
(b) differentiable at $x=0$ but not differentiable at $x=2$
(c) not differentiable at $x=0$ but differentiable at $x=2$
(d) differentiable neither at $x=0$ nor at $x=2$
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Answer:
Correct Answer: 13. $(b, c)$
Solution:
- Since, $f(x)=e^{g(x)} \Rightarrow e^{g(x+1)}=f(x+1)=x f(x)=x e^{g(x)}$
and
$$ g(x+1)=\log x+g(x) $$
i.e.
$$ g(x+1)-g(x)=\log x $$
Replacing $x$ by $x-\frac{1}{2}$, we get
$$ \begin{array}{rl} \quad g & x+\frac{1}{2}-g x-\frac{1}{2}=\log x-\frac{1}{2}=\log (2 x-1)-\log 2 \\ \therefore \quad g^{\prime \prime} \quad & x+\frac{1}{2}-g^{\prime \prime} \quad x-\frac{1}{2}=\frac{-4}{(2 x-1)^{2}} \end{array} $$
On substituting, $x=1,2,3, \ldots, N$ in Eq. (ii) and adding, we get
$$ g^{\prime \prime} \quad N+\frac{1}{2}-g^{\prime \prime} \frac{1}{2}=-4 \quad 1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^{2}} \text {. } $$
