The equation of the normal to the curve y=x+sin⁡xcos⁡x at x=π2, is

A
$x = 2$
B
$x = π$
C
$x + π = 0$
D
$2 x = π$

Solution:

When $x = \frac{π}{2}$ , we have

$y = \frac{π}{2} + \sin \frac{π}{2} \cos \frac{π}{2} = \frac{π}{2}$

So, the coordinates of the point are $(π / 2, π / 2)$

Now,

$\begin{array}{l} y = x + \sin x \cos x \\ \Rightarrow \frac{d y}{d x} = 1 + \cos^{2} x - \sin^{2} x \Rightarrow {(\frac{d y}{d x})}_{x = \frac{π}{2}} = 1 + 0 - 1 = 0 \end{array}$

This means that the tangent at $(π / 2, π / 2)$ is parallel to $x$ -axis.

So, the normal is parallel to $y$ -axis and hence its equation is

$x = \frac{π}{2}$ .

The equation of the normal to the curve $y = x + \sin x \cos x$ at $x = \frac{π}{2},$ is

Solution:

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