Let [x] denote the greatest integer less than or equal to x and g(x) be given by
$g(x)=\left\{\begin{array}{cl} [f(x)], & x \in(0, \pi / 2) \cup(\pi / 2, \pi) \\ 3 \quad, & x=\frac{\pi}{2} \end{array}\right.$
where, $f(x)=\frac{2\left(\sin x-\sin ^n x\right)+\left|\sin x-\sin ^n x\right|}{2\left(\sin x-\sin ^n x\right)-\left|\sin x-\sin ^n x\right|}, n \in R^{+},$
then at $x=\frac{\pi}{2}, g(x)$ is
Answer & explanation
Correct answer: option 1
Clearly,
$0<\sin x<1$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$.
CASE I: When n > 1
In this case, we have
$\sin x>\sin ^n x$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$
$\Rightarrow \sin x-\sin ^n x>0$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$
$\Rightarrow \left|\sin x-\sin ^n x\right|=\sin x-\sin ^n x$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$
∴ $f(x)=\frac{2\left(\sin x-\sin ^n x\right)+\left(\sin x-\sin ^n x\right)}{2\left(\sin x-\sin ^n x\right)-\left(\sin x-\sin ^n x\right)}=3 $
$\Rightarrow [f(x)]=3$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$
Thus, we have
$g(x)=3$ for all $x \in(0, \pi)$
Clearly, it is continuous and differentiable at $x=\pi / 2$
CASE II: When 0 < n < 1
In this case, we have
$\sin x<\sin ^n x$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$
$\Rightarrow \sin x-\sin ^n x<0$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$
$\Rightarrow \left|\sin x-\sin ^n x\right|=-\left(\sin x-\sin ^n x\right)$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$
∴ $f(x)=\frac{2\left(\sin x-\sin ^n x\right)-\left(\sin x-\sin ^n x\right)}{2\left(\sin x-\sin ^n x\right)+\left(\sin x-\sin ^n x\right)}=\frac{1}{3}$
$\Rightarrow [f(x)]=0$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$
Thus, we have
$g(x)= \begin{cases}0, & \text { for all } x \in(0, \pi / 2) \cup(\pi / 2, \pi) \\ 3, & \text { for } x=\pi / 2\end{cases}$
Clearly, it is discontinuous and hence non-differentiable also at $x=\pi / 2$.