It is given that $\lim _{x \rightarrow 0} f(x)=1, \lim _{x \rightarrow 0} g(x)=\alpha$
and $\lim _{x \rightarrow 0} \frac{2 f(x)-g(x)}{(f(x)+7)^{2 / 3}}=\frac{7}{4}$
$\Rightarrow \quad \frac{2-\alpha}{(8)^{2 / 3}}=\frac{7}{4} \Rightarrow \alpha=-5$
For $\quad h(x)= \begin{cases}\sin (\alpha x), & 0 \leq x \leq \frac{\pi}{10} \\ \cos (2 \alpha x), & \frac{\pi}{10} < x \leq \frac{\pi}{5}\end{cases}$
$\because \alpha=-5$, so $\sin (\alpha x)$ is continuous for $x \in\left[0, \frac{\pi}{10}\right)$
and $\cos (2 \alpha x)$ is also continuous for $x \in\left(\frac{\pi}{10}, \frac{\pi}{5}\right]$
Now, at $x=\pi / 10$
$\begin{aligned} \operatorname{LHL}\left(\text { at } x=\frac{\pi}{10}\right) & =\lim _{h \rightarrow 0} \sin \left(-5\left(\frac{\pi}{10}-h\right)\right) \\ & =\sin \left(-\frac{\pi}{2}\right)=-1\end{aligned}$
and $f\left(\frac{\pi}{10}\right)=\sin \left(-\frac{5 \pi}{10}\right)=\sin \left(-\frac{\pi}{2}\right)=-1$ and RHL $\left(\right.$ at $\left.x=\frac{\pi}{10}\right)=\lim _{h \rightarrow 0} \cos \left(-10\left(\frac{\pi}{10}+h\right)\right)$ $=\lim _{h \rightarrow 0} \cos (\pi+10 h)=\cos \pi=-1$
$\therefore$ The function $h(x)$ is continuous on $\left[0, \frac{\pi}{5}\right]$. Hence, option (d) is correct.