Let $f:[-1,1]\rightarrow \mathbb R$ be a function given by $f(x) = \begin{cases} x^2\cos(\frac{1}{x}), & \text{if $x\neq 0$ } \\ 0, & \text{if $x=0$ } \end{cases}$then
$(a)$ $f$ is of bounded variation on $[-1,1]$.
$(b)$ $f'$ is of bounded variation on $[-1,1]$.
$(c)$ $\vert f(x) \vert\le1 \forall x\in [-1,1]$ is of bounded variation on $[-1,1]$.
$(d)$ $\vert f(x) \vert\le3 \forall x\in [-1,1]$ is of bounded variation on $[-1,1]$.
Solution:$f'(x)=2x\cos(\frac{1}{x})-\sin(\frac{1}{x})\implies \vert f'(x)\vert\le3$ on $[1,1]\implies f$ is of bounded variation on $[-1,1]$.So,option $(a),(b)$ are correct.
Please check the following argument:
Since, the only problem creator for $f'$ being of bounded variation is $0$. Now, if I take a partition of $[-1,1]$ which contains $0$ as one of its points, then the limit of $f'$ as $x$ approaches zero does not exist, but the variation remains bounded, hence $f'$ is of bounded variation on $[-1,1]$.
Please provide some argument which can discard option $c$
The derivative $f'$ has unbounded variation on $[-1,1]$ and (b) is incorrect.
For any $n \in \mathbb{N}$, take a partition $(0,x_1,,x_2 , \ldots, x_{2n-2}, 1)$ where $x_k = [(n - (k-1)/2)\pi]^{-1}$, and we have
$$V(f',[-1,1]) > \sum_{k=2}^{2n-2} |f'(x_k) - f'(x_{k-1}| = \\ \sum_{k=2}^{2n-2} \left|\frac{\cos \left(n- \frac{k-1}{2}\right)\pi}{\left(n- \frac{k-1}{2}\right)\pi} - \frac{\cos (n- \frac{k-2}{2})\pi}{\left(n- \frac{k-2}{2}\right)\pi} -\sin \left(n- \frac{k-1}{2}\right)\pi + \sin \left(n- \frac{k-2}{2}\right)\pi\right| \\ = 1 - \frac1{2\pi} + 1 + \frac1{2\pi} +1 - \frac1{3\pi} + 1 +\frac1{3\pi} + \ldots + 1 - \frac1{n\pi} + 1 + \frac1{n\pi}= 2(n-1)$$
Note also that $\displaystyle f'(0) = \lim_{h \to 0} \frac{h^2 \cos(1/h) - 0}{h}= 0$ and your observation about the nature of $f'(x)$ as $x \to 0$ is not relevant in regards to the total variation of $f'$.