# Which of the folowing is true (NBHM-2014)

1. If $f$ is twice continously differentiable in (a,b) and if for all x $\in (a,b)$ $f''(x) + 2f'(x) + 3f(x) = 0$, the $f$ is infinitely differentiable on (a,b)

2. Let $f \in C[a,b]$ be a differentiable in (a,b). If $f(a) = f(b) = 0$, then for any real number $\alpha$, there exist x $\in$ (a,b) such that $f'(x) + \alpha f(x) = 0$

3. The function defined below is not differentiable at x = 0 $$\begin{equation} f(x)=\begin{cases} x^2|cos\frac{\pi}{x}|, & \text{ x \neq 0}\\ 0, & \text{x = 0 }. \end{cases} \end{equation}$$

For (1) and (2) are true , but i am not sure

For (3) is false

Thank you for sparing your valuable time in checking my solutions

(3) is true. We have to use the Squeeze theorem.

$|\cos(\frac{\pi}{x})| \leq 1$. So, $-x^2 \leq |\cos(\frac{\pi}{x})|\leq x^2$, and now take limits as $x \rightarrow 0$.

Unless I am missing something obvious, (1) and (2) are true. For (1) we can solve the given differential equation,we get that the resulting solution is infinitely differentiable. So by , and by existence uniqueness of solution, we are done. (2) should be true by some mean value theorem trick- haven't completely checked it though.

Since you got the answers for 1) and 3)

Hint for 2) Consider $$g(x)=e^{\alpha x}f(x)$$

For (3) $f'(0) = lim_{h\rightarrow 0} \frac{f(h) - f(0)}{h} =\lim_{h\rightarrow 0} \frac{h^2|cos\frac{\pi}{h}|}{h}$ = $0$.

HINT: For (b), you can see that the solution is: $f(x)=A e^{-\alpha x}$, where $A$ being arbitrary constant. Now $f(a)=f(b)=0$ gives $A=0$ which implies $f(x)=0$ is the solution for any $\alpha\in (a,b)$.