Explanation

(d)
Given, $f(x) = \left\{ \begin{array}{r}
x^{p}\cos{\left( \frac{1}{x} \right),\ x \neq 0} \\
0,\ \ \ x = 0
\end{array} \right.\$
Since, $f(x)$ is differentiable at $x = 0$, therefore it is continuous at $x = 0$
$\therefore\ \lim_{x \rightarrow 0}\ f(x) = f(0) = 0$
$\Rightarrow \ \lim_{x \rightarrow 0}\ \ x^{p}\cos{\left( \frac{1}{x} \right) = 0\ \ \ \Rightarrow \ \ \ \ p > 0}$
As $f(x)$ is differentiable at $x = 0$
$\therefore\ \ \ \lim_{x \rightarrow 0}\ \ \ \frac{f(x) - f(0)}{x - 0}$ exists finitely
$\Rightarrow \ \lim_{x \rightarrow 0}\ \ \frac{x^{p}\cos{\frac{1}{x} - 0}}{x}$ exists finitely
$\Rightarrow \lim_{x \rightarrow 0}\ \ x^{p - 1}\cos\frac{1}{x} - 0$ exists finitely
$\Rightarrow \ \ \ \ \ \ \ \ p - 1 > 0\ \ \ \ \ \ \Rightarrow \ \ \ \ p > 1\ \$