A) \[x+2y=1\]
B) \[x+2y=\pi /2\]
C) \[x+2y=\pi /4\]
D) None of these
Correct Answer: B
Solution :
[b] \[y=\cos (x+y)....(1)\] \[\therefore \frac{dy}{dx}=-\sin (x+y)\left\{ 1+\frac{dy}{dx} \right\}\] \[=-\frac{\sin (x+y)}{1+\sin (x+y)}=-\frac{1}{2}\] \[\Rightarrow \sin (x+y)=1,\] so \[\cos (x+y)=0\] \[\therefore \] from \[(1)y=0\] and \[(x+y)=2n\pi +\frac{\pi }{2}\] Tangent at \[\left( \frac{\pi }{2},0 \right)\] is \[x+2y=\frac{\pi }{2}\]
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