If an equation of a tangent to the curve, $$y = \cos \left( {x + y} \right),\, - 1 \leqslant x \leqslant 1 + \pi ,$$       is $$x + 2y = k$$   then $$k$$ is equal to :

Solution :
Let $$y = \cos \left( {x + y} \right)$$
$$ \Rightarrow \frac{{dy}}{{dx}} = - \sin \left( {x + y} \right)\left( {1 + \frac{{dy}}{{dx}}} \right)......\left( 1 \right)$$
Now, given equation of tangent is $$x + 2y = k$$
$$\eqalign{ & \Rightarrow {\text{Slope}} = \frac{{ - 1}}{2} \cr & {\text{So, }}\frac{{dy}}{{dx}} = \frac{{ - 1}}{2}\,{\text{put this value in equation}}\left( 1 \right),{\text{ we get}} \cr & \frac{{ - 1}}{2} = - \sin \left( {x + y} \right)\left( {1 - \frac{1}{2}} \right) \cr & \Rightarrow \sin \left( {x + y} \right) = 1 \cr & \Rightarrow x + y = \frac{\pi }{2} \cr & \Rightarrow y = \frac{\pi }{2} - x \cr & {\text{Now, }}\frac{\pi }{2} - x = \cos \left( {x + y} \right) \cr & \Rightarrow x = \frac{\pi }{2}{\text{ and }}y = 0 \cr & {\text{Thus }}x + 2y = k \Rightarrow \frac{\pi }{2} = k \cr} $$