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  •  Let denote the greatest integer less than or equal to <img alt="\mathrm{x}" src="https:

 Let \mathrm{[x]} denote the greatest integer less than or equal to \mathrm{x}. Now \mathrm{g(x)} is defined as below:
\begin{aligned} \mathrm{g(x)} & =\mathrm{][f(x)], x \in\left(0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right)} \\ & \mathrm{=3, x=\frac{\pi}{2} }\end{aligned}
where \mathrm{ 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

Option: 1

\mathrm{g(x)} is continuous and differentiable at  \mathrm{\mathrm{x=\frac{\pi}{2}}}when\mathrm{\mathrm{ n>1}}
 


Option: 2

\mathrm{ g(x) }is continuous and differentiable at \mathrm{x=\frac{\pi}{2}} when \mathrm{ 0<n<1}
 


Option: 3

\mathrm{g(x) }is continuous but not differentiable at \mathrm{x=\frac{\pi}{2}} when \mathrm{n>1}
 


Option: 4

\mathrm{g(x)} is continuous but differentiable at\mathrm{x=\frac{\pi}{2}} when \mathrm{0<n<1}


Answers (1)

best_answer

\begin{aligned} &\mathrm{If n>1, \sin x>\sin ^n x. If 0<n<1, \sin x<\sin ^n x.}\\ & \mathrm{\therefore \quad \text { if } n>1, 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 . }\\ & \text { If } \mathrm{0<n<1, 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} . }\\ & \therefore \text { if } \mathrm{n>1, g(x)=3, x \in(0, \pi)} \end{aligned}
\therefore \mathrm{g(x)} is continuous and differentiable at \mathrm{x=\frac{\pi}{2}.}

\text { If } \begin{gathered} \mathrm{0<n<1, g(x)=0, x \in\left(0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right)} \\ \mathrm{3, x=\frac{\pi}{2} . }\end{gathered}
Then\mathrm{ g\left(\frac{\pi}{2}+0\right)=0, g\left(\frac{\pi}{2}-0\right)=0, g\left(\frac{\pi}{2}\right)=3}. So, \mathrm{ g(x)} is not continuous at \mathrm{ x=\frac{\pi}{2}}.

Hence, \mathrm{g(x)} is also not differentiable at \mathrm{x=\frac{\pi}{2}}.

Posted by

Irshad Anwar

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