Question

At what point the tangent line to the curve $y=\cos (x+y),(-2\pi \le x\le 2\pi )$ is parallel to $x+2y=0$

• A. $(\frac{\pi }{2},0)$
• B. $(-\frac{\pi }{2},0)$
• C. $(\frac{3\pi }{2},0)$
• D. $(-\frac{3\pi }{2},\frac{\pi }{2})$

Answer 1

The correct option is A $(\frac{\pi }{2},0)$

$y=\cos (x+y)$ $(-2\pi \le x\le 2\pi )$
$\Rightarrow \frac{dy}{dx}=-\sin (x+y)(1+\frac{dy}{dx})$
$\Rightarrow \frac{dy}{dx}(1+\sin (x+y))=-\sin (x+y)$
$\Rightarrow \frac{dy}{dx}=-\frac{\sin (x+y)}{1+\sin (x+y)}$
Given tangent is parallel to the line, $x+2y=0$
$\Rightarrow \frac{dy}{dx}=-\frac{\sin (x+y)}{1+\sin (x+y)}=-\frac{1}{2}$
$2\sin (x+y)=1+\sin (x+y)$
$\sin (x+y)=1\Rightarrow \cos (x+y)=0\Rightarrow x+y=\frac{\pi }{2}$
Also the point lies on the given curve,
$y=0\Rightarrow x=\frac{\pi }{2}$
Thus the point is, $(\frac{\pi }{2},0)$
Hence, option 'A' is correct.