Question
Let $$\left[ x \right]$$ denote the greatest integer less than or equal to $$x.$$ Now $$g\left( x \right)$$ is defined as below :
\[g\left( x \right) = \left\{ \begin{array}{l}
\left[ {f\left( x \right)} \right],\,x\, \in \left( {0,\,\frac{\pi }{2}} \right) \cup \left( {\frac{\pi }{2},\,\pi } \right)\\
\,3,\,x = \frac{\pi }{2}
\end{array} \right.\]
where $$f\left( x \right) = \frac{{2\left( {\sin \,x - {{\sin }^n}x} \right) + \left| {\sin \,x - {{\sin }^n}x} \right|}}{{2\left( {\sin \,x - {{\sin }^n}x} \right) - \left| {\sin \,x - {{\sin }^n}x} \right|}},\,n\, \in \,R.$$ Then :
A.
$$g\left( x \right)$$ is continuous and differentiable at $$x = \frac{\pi }{2}$$ when $$n > 1$$
B.
$$g\left( x \right)$$ is continuous and differentiable at $$x = \frac{\pi }{2}$$ when $$0 < n < 1$$
C.
$$g\left( x \right)$$ is continuous but not differentiable at $$x = \frac{\pi }{2}$$ when $$n > 1$$
D.
$$g\left( x \right)$$ is continuous but differentiable at $$x = \frac{\pi }{2}$$ when $$0 < n < 1$$
Answer :
$$g\left( x \right)$$ is continuous and differentiable at $$x = \frac{\pi }{2}$$ when $$n > 1$$
Solution :
$$\eqalign{
& {\text{If }}n > 1,\,\sin \,x > {\sin ^n}x.{\text{ If }}0 < n < 1,\,\sin \,x < {\sin ^n}x\, \cr
& \therefore {\text{ if }}n > 1,\,f\left( x \right) = \frac{{2\left( {\sin \,x - {{\sin }^n}x} \right) + \left| {\sin \,x - {{\sin }^n}x} \right|}}{{2\left( {\sin \,x - {{\sin }^n}x} \right) - \left| {\sin \,x - {{\sin }^n}x} \right|}} = 3 \cr
& {\text{If }}0 < n < 1,\,f\left( x \right) = \frac{{2\left( {\sin \,x - {{\sin }^n}x} \right) - \left| {\sin \,x - {{\sin }^n}x} \right|}}{{2\left( {\sin \,x - {{\sin }^n}x} \right) + \left| {\sin \,x - {{\sin }^n}x} \right|}} = \frac{1}{3} \cr
& \therefore {\text{ if }}n > 1,\,g\left( x \right) = 3,\,x\, \in \left( {0,\,\pi } \right)\, \cr} $$
$$\therefore \,g\left( x \right)$$ is continuous and differentiable at $$x = \frac{\pi }{2}$$
$$\eqalign{
& {\text{If }}\,0 < n < 1,\,g\left( x \right) = 0\,x\, \in \left( {0,\,\frac{\pi }{2}} \right) \cup \left( {\frac{\pi }{2},\,\pi } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3,\,x = \frac{\pi }{2} \cr} $$
Then $$g\left( {\frac{\pi }{2} + 0} \right) = 0,\,\,g\left( {\frac{\pi }{2} - 0} \right) = 0,\,\,g\left( {\frac{\pi }{2}} \right) = 3$$
So, $$g\left( x \right)$$ is not continuous at $$x = \frac{\pi }{2}$$
Hence, $$g\left( x \right)$$ is also not differentiable at $$x = \frac{\pi }{2}.$$