Slope of the tangent to the curve \(y=2 \mathrm{e}^x \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)\) where \(0 \leq x \leq 2 \pi\) is minimum at \(x=\)

\(y=2 \mathrm{e}^x \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)\)
\(\begin{aligned} & =\mathrm{e}^x \sin \left(\frac{\pi}{2}-x\right) \ldots[2 \sin \theta \cos \theta=\sin 2 \theta] \\ & =\mathrm{e}^x \cos x\end{aligned}\)
\(\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{e}^x(\cos x-\sin x)\)
Let \(\mathrm{T}=\mathrm{e}^x(\cos x-\sin x)\)
\(\therefore \quad \frac{\mathrm{dT}}{\mathrm{d} x}=\mathrm{e}^x(\cos x-\sin x)+\mathrm{e}^x(-\sin x-\cos x)\)
\(=-2 \mathrm{e}^x \sin x\)
Now, \(\frac{\mathrm{dT}}{\mathrm{d} x}=0\)
\(\begin{aligned} & \Rightarrow-2 \mathrm{e}^x \sin x=0 \\ & \Rightarrow \sin x=0\end{aligned}\)
\(\Rightarrow x=0, \pi, 2 \pi \quad \ldots[\because 0 \leq x \leq 2 \pi]\)
At \(x=0, \mathrm{~T}=\mathrm{e}^0(\cos 0-\sin 0)=1\)
At \(x=\pi, \mathrm{T}=\mathrm{e}^\pi(\cos \pi-\sin \pi)=-\mathrm{e}^\pi\)
At \(x=2 \pi, \mathrm{T}=\mathrm{e}^{2 \pi}(\cos 2 \pi-\sin 2 \pi)=\mathrm{e}^{2 \pi}\)
\(\therefore \quad\) Slope of the tangent is minimum at \(x=\pi\)