verify rolles theorem for this function f(x) sin 2x in [0,pi/2]

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$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\sin 2\mathrm{x} \mathrm{where} \mathrm{x}\in \left[0, \frac{\mathrm{\pi }}{2}\right] \\ \mathrm{f}\left(0\right)=\sin 0=0 \\ \mathrm{f}\left(\frac{\mathrm{\pi }}{2}\right)=\sin \left(2.\frac{\mathrm{\pi }}{2}\right) \\ =\mathrm{sin\pi } \\ \Rightarrow \mathrm{f}\left(\frac{\mathrm{\pi }}{2}\right)=0 \\ \Rightarrow \mathrm{f}\left(0\right)=\mathrm{f}\left(\frac{\mathrm{\pi }}{2}\right)=0 \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{sin\theta } \mathrm{is} \mathrm{a} \mathrm{continuous} \mathrm{and} \mathrm{differentiable} \mathrm{function} \mathrm{for} \mathrm{all} \mathrm{x}\in \mathrm{R}. \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\sin 2\mathrm{x} \mathrm{is} \mathrm{continuous} \mathrm{and} \mathrm{differentiable} \mathrm{for} \mathrm{x}\in \left[0, \frac{\mathrm{\pi }}{2}\right]. \\ \mathrm{Also} \mathrm{f}\left(0\right)=\mathrm{f}\left(\frac{\mathrm{\pi }}{2}\right) \\ \mathbf{\Rightarrow }\mathbf{Rolle}\mathbf{\text{'}}\mathbf{s}\mathbf{ }\mathbf{theorem}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{applicable}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{x}\mathbf{\in }\left[\mathbf{0}\mathbf{,}\mathbf{ }\frac{\mathbf{\pi }}{\mathbf{2}}\right]\mathbf{.} \\ \mathrm{It} \mathrm{means} \mathrm{there} \mathrm{exists} \mathrm{at} \mathrm{least} \mathrm{one} \mathrm{c}\in \left[0, \frac{\mathrm{\pi }}{2}\right] \mathrm{for} \mathrm{which} \mathrm{f}\text{'}\left(\mathrm{c}\right)=0 \\ \mathrm{f}\left(\mathrm{x}\right)=\sin 2\mathrm{x} \\ \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)=2.\cos 2\mathrm{x} \\ \mathrm{Let} \mathrm{f}\text{'}\left(\mathrm{x}\right)=0 \mathrm{for} \mathrm{x}=\mathrm{c} \\ \Rightarrow 2.\cos 2\mathrm{c}=0 \\ \Rightarrow \cos 2\mathrm{c}=0 \\ \Rightarrow 2\mathrm{c}=\frac{\mathrm{\pi }}{2} \\ \mathbf{\Rightarrow }\mathbf{c}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{4}} \\ \mathbf{\Rightarrow }\mathbf{Rolle}\mathbf{\text{'}}\mathbf{s}\mathbf{ }\mathbf{theorem}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{verified}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{x}\mathbf{\in }\left[\mathbf{0}\mathbf{,}\mathbf{ }\frac{\mathbf{\pi }}{\mathbf{2}}\right]\mathbf{.} \end{gathered}$$

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