Question

Equation of a tangent to the curve $y=\cos (x+y),0\le x\le 2\pi$ that is parallel to the line $x+2y=0$ is

• A. $x+2y=\frac{\pi }{2}$
• B. $x+2y=\frac{\pi }{4}$
  
• C. $x+2y=\pi$
• D. $x+y=\pi$

Answer 1

The correct option is A $x+2y=\frac{\pi }{2}$

We have,

$y=\cos (x+y)$

differentiate with respect to $x$

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

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

$\frac{dy}{dx}\left(1+\sin (x+y)\right)=-\sin (x+y)$

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

Because equation of tangent is parallel to $x+2y=0$ so,

$\frac{dy}{dx}=\frac{-1}{2}=\frac{-\sin (x+y)}{1+\sin (x+y)}$

$2\sin (x+y)=1+\sin (x+y)$

$\sin (x+y)=1$

$x+y=\frac{\pi }{2}$ and $\frac{-3\pi }{2}$

So,

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

Hence, $(\frac{\pi }{2},0)$ and $(\frac{-3\pi }{2},0)$ are two points form which two tangents are passing and parallel to $x+2y=0$

Now , equation of tangents:

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

Hence, this is the answer.