A) 1
B) 2
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{2}\]
Correct Answer: D
Solution :
[d] 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 \[x+2y=k\Rightarrow \frac{\pi }{2}=k\]
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