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 $x=$

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Aicharya · Accepted Answer

$y=2 \mathrm{e}^x \sin \left(\frac{\pi}{4}-\frac{x}{2}ight) \cos \left(\frac{\pi}{4}-\frac{x}{2}ight)$ $\begin{aligned} & =\mathrm{e}^x \sin \left(\frac{\pi}{2}-xight) \ldots[2 \sin 	heta \cos 	heta=\sin 2 	heta] \ & =\mathrm{e}^x \cos x\end{aligned}$ $	herefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{e}^x(\cos x-\sin x)$ Let $\mathrm{T}=\mathrm{e}^x(\cos x-\sin x)$ $	herefore \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 $x=2 \pi, \mathrm{T}=\mathrm{e}^{2 \pi}(\cos 2 \pi-\sin 2 \pi)=\mathrm{e}^{2 \pi}$ $	herefore \quad$ Slope of the tangent is minimum at $x=\pi$

Slope of the tangent to the curve $y=2 \mathrm{e}^x \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos…

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