Let f(x) = sin x, g(x) = [x + 1] and h(x) = gof(x), where [.] is the greatest integer function. Then, $h' \left(\frac{\pi}{2}\right)$ is

non-existent

We have,

f(x) = sin x and g(x) = [x + 1]

∴ h(x) = gof(x) = g(f(x)) = g(sin x) = [sin x + 1]

Now,

(LHD at x = $\frac{\pi}{2}$) = $\lim\limits_{\alpha \rightarrow 0} \frac{h(\pi / 2-\alpha)-h(\pi / 2)}{-\alpha}$

⇒ (LHD at x = $\frac{\pi}{2}$) = $\lim\limits_{\alpha \rightarrow 0} \frac{\left[\sin \left(\frac{\pi}{2}-\alpha\right)+1\right]-\left[\sin \frac{\pi}{2}+1\right]}{-\alpha}$

⇒ (LHD at x = $\frac{\pi}{2}$) = $\lim\limits_{\alpha \rightarrow 0} \frac{1-2}{-\alpha}=\infty$

Similarly, we have

(RHD at x = $\frac{\pi}{2}$) = $-\infty$

Hence, $h'\left(\frac{\pi}{2}\right)$ does not exist.