If the equation of a tangent drawn to the curve $y=\cos (x+y),-1\le x\le 1+\pi$ is $x+2y=\mathrm{k},$ then k=

Slope of given line $=-1 / 2$

$\Rightarrow \text{ Slope of tangent }=\left(\frac{dy}{dx}\right)=-1/2\cdots \cdots \cdots \cdots \cdots \left(i\right)$

The equation of given curve is $y=\cos (x+y)$

Differentiating the curve w.r.t. x we get,

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

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

= slope of tangent $\cdots \cdots \cdots \cdots \cdots \left(ii\right)$

From (i) and (ii) ,$\frac{-\sin (x+y)}{1+\sin (x+y)}=-\frac{1}{2}$

$\begin{array}{ll}\Rightarrow & \sin (x+y)=1\\ \Rightarrow & \cos (x+y)=0\end{array}\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(iii\right)$

From the given curve $y=\cos (x+y)\Rightarrow y=0$ ……….(iv)

and $\sin \left(x\right)=1$ 

$\Rightarrow x=\frac{\pi }{2},-\frac{3\pi }{2}$

$\therefore$ The points are $P(\pi /2,0),Q(-3\pi /2,0)$

Tangent at $P$ is $x+2y=\frac{\pi }{2}$and tangent at $Q$ is $x+2y=-\frac{3\pi }{2}$