Question

Which of tangents to the curve $y=\cos (x+y),-2\pi \le x\le 2\pi$ is/are parallel to the line $x+2y=0$.

• A. $2x+4y+3\pi =0$
• B. $x+4y-\pi =0.$
• C. $2x+4y-\pi =0.$
• D. $x-4y-3\pi =0.$

Answer 1

Answer: (A), (C)

The correct options are
A $2x+4y+3\pi =0$
C $2x+4y-\pi =0.$

Given $y=\cos (x+y)$

$\frac{dy}{dx}=-\sin (x+y).[1+\frac{dy}{dx}]$ ...(1)

Since the tangent is parallel to $x+2y=0$

So, slope of tangent $=-\frac{1}{2}$

$\therefore \frac{dy}{dx}=slope=-\frac{1}{2}.$

$\sin (x+y)=\sin \left(\frac{\pi }{2}\right).$

$\Rightarrow x+y=\frac{\pi }{2}$

$\therefore \cos (x+y)=0$

$\therefore y=\cos (x+y)=0$

$\therefore \sin (x+y)=1$

$\Rightarrow \sin x=1$ ($\because y=0$)

$\therefore x=\frac{\pi }{2},-\frac{3\pi }{2}$ as $-2\pi <x<2\pi$

Hence the points are $[\frac{-3\pi }{2},0]$ and $[\frac{\pi }{2},0]$ where the tangents are parallel to the line x+2y=0.

$\therefore$ The equations of tangents are:

$y-0=-\frac{1}{2}(x+\frac{3\pi }{2})$

$\Rightarrow 2x+4y+3\pi =0$

and $y-0=-\frac{1}{2}(x-\frac{\pi }{2})$

$\Rightarrow 2x+4y-\pi =0.$