Check whether $f'(x)$ is greater than zero or not for all $x\in \Bbb R$. Consider the function $f(x) =
\begin{cases}
x+x^2\cos(\frac{\pi}{x}),  & \text{if $x \neq 0$ } \\[2ex]
0, & \text{if $x=0$}
\end{cases}$
I have calculated that $f'(x)=1+2x\cos(\frac{\pi}{x})+\pi \sin(\frac{\pi}{x})$ and $f'(0)=1$.
Check whether $f'(x)$ is greater than zero or not for all $x\in \Bbb R$.! Any idea?
Edit: After seeing the comment I have realised that the claim is false. Let me explain what leads me to make this claim. Actually, I am solving the previous years' questions of an exam(See the picture attached) and the answer key is given in this way. Now a) is true which leads that b) and c) are not true so d) has to be true but let me change the heading of the question a bit now, Sorry for the confusion.
 A: I don't see how such a claim can be correct, since as you calculated, $$f'(x) = 1 + 2x \cos \frac{\pi}{x} + \pi \sin \frac{\pi}{x},$$ which implies when $x = 1$, $$f'(1) = 1 + 2 \cos \pi + \pi \sin \pi = 1 - 2 + 0 = -1.$$

Now that you have edited the question, it is clear that (iv) is false.  We can also show that (i) is true using the definition of derivative:
$$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x-0} = \lim_{x \to 0} 1 + x \cos \frac{\pi}{x}.$$  Since $\cos \frac{\pi}{x}$ is well-defined for all $x \ne 0$ and bounded between $-1$ and $1$, it follows that $f'(0) = 1$.  This also means (iii) is false.
We now consider whether (ii) is true or false.  We should observe that as $x \to 0$, the term $2x \cos \frac{\pi}{x} \to 0$, so if we are to show that there always exists a choice of $x$ in any neighborhood of $0$ such that $f'(x) < 0$, our hopes should lie in finding $x$ such that $\sin \frac{\pi}{x} = -1$.  This of course happens when $$\frac{\pi}{x} \in \left\{ -\frac{\pi}{2} + 2\pi k : k \in \mathbb Z \right\}$$ or equivalently, $$x \in \left\{ \frac{2}{4k-1} : k \in \mathbb Z \right\}.$$  And clearly, as $k \to \pm \infty$, $x \to 0$.  So for any neighborhood $|x| < \epsilon < 1/2$, we can always choose $$k > \left\lceil \frac{2+\epsilon}{4\epsilon} \right\rceil$$ that makes $f'(x) < 0$, since then in this neighborhood, $$1 + 2x \cos \frac{\pi}{x} \le 1 + 2(1/2) < \pi.$$  So (ii) is true:  any neighborhood of $0$ will always contain some point for which the derivative is negative, hence the function will not be increasing.  All told, (i) and (ii) are true; (iii) and (iv) are false, and the answer is (c).
A: Are you sure it is right? Take a negative $x$ with a large absolute value, say $x=-10000$, and then you have $\cos(\pi/x)\approx1,\ \sin(\pi/x)\approx0\implies f(-10000)\approx 1+2\times(-10000)\times1+\pi\times0\approx -19999<0$.
