The equation of the normal to the curveĀ
A. $x=2$
B. $x=\pi$
C. $x+\pi=0$
D. $2 x=\pi$
Given that the curve $y=x+\sin x \cos x$
Differentiating both the sides w.r.t. $x$,
$\frac{d y}{d x}=1+\cos ^{2} x-\sin ^{2} x$
Now,
Slope of the tangent $\frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}=\frac{\pi}{2}\right)=1+\cos ^{2} \frac{\pi}{2}-\sin ^{2} \frac{\pi}{2}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1-1+0=0$
When $\mathrm{x}=\frac{\pi}{2}, \mathrm{y}=\frac{\pi}{2}$
Equation of the normal:
$\left(y-y_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(x-x_{1}\right)$
$\Rightarrow\left(y-\frac{\pi}{2}\right)=\frac{-1}{0}\left(x-\frac{\pi}{2}\right)$
$\Rightarrow 2 x=\pi$
Hence option D is correct.
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