A) 1
B) 2
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{2}\]
Correct Answer: D
Solution :
Let\[y=\cos (x+y)\] \[\Rightarrow \]\[\frac{dy}{dx}=-\sin (x+y)\left( 1+\frac{dy}{dx} \right)\] ?.(1) Now, given equation of tangent is\[x+2y=k\] \[\Rightarrow \]Slope\[=\frac{-1}{2}\] So, \[\frac{dy}{dx}=\frac{-1}{2}\]put this value in (1), we get \[\frac{-1}{2}=-\sin (x+y)\left( 1-\frac{1}{2} \right)\] \[\Rightarrow \]\[\sin (x+y)=1\]\[\Rightarrow \]\[x+y=\frac{\pi }{2}\]\[\Rightarrow \]\[y=\frac{\pi }{2}-x\] Now, \[\frac{\pi }{2}-x=\cos (x+y)\]\[\Rightarrow \]\[x=\frac{\pi }{2}\]and\[y=0\] Thus\[x+2y=k\Rightarrow \frac{\pi }{2}=k\]You need to login to perform this action.
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