The equation of tangent to the curve $y=y=\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)$ at $x=\frac{\pi}{4}$…
The equation of tangent to the curve $y=y=\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)$ at $x=\frac{\pi}{4}$, is
- $2 x+y-\frac{\pi}{2}-1=0$
- $2 x-y-\frac{\pi}{2}+1=0$
- $x+y-\frac{\pi}{2}-1=0$
- $x-y-\frac{\pi}{2}+1=0$
Solution
$\begin{aligned}
& \mathrm{y}=\sqrt{2} \sin \left(2 \mathrm{x}+\frac{\pi}{4}\right) \\
& \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{2} \cos \left(2 \mathrm{x}+\frac{\pi}{4}\right)(2) \\
& \therefore\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=\frac{\pi}{4}}=(2 \sqrt{2}) \cos \left(\frac{\pi}{2}+\frac{\pi}{4}\right)=2 \sqrt{2}\left(-\sin \frac{\pi}{4}\right)=-2
\end{aligned}$
When $\mathrm{x}=\frac{\pi}{4}, \mathrm{y}=\sqrt{2} \sin \left(\frac{\pi}{2}+\frac{\pi}{4}\right)=1$
Hence equation of required tangent is
$(y-1)=-2\left(x-\frac{\pi}{4}\right) \Rightarrow 2 x+y-\frac{\pi}{2}-1=0$
Asked in: MHT CET 2021 (24 Sep Shift 1)
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