A) \[1-\sqrt{2}\]
B) \[-(1+\sqrt{2})\]
C) \[\sqrt{2}\]
D) \[1/\sqrt{2}\]
Correct Answer: A
Solution :
\[x=2\,\cos t+\cos 2t,y=2\sin t-\sin 2t\] \[\frac{dx}{dt}=-2\sin t-2\sin 2t,\] \[\frac{dy}{dt}=2\cos t-2\cos 2t\] \[\frac{dy}{dx}=\frac{2\cos t-2\cos 2t}{-2\sin t-2\sin 2t}=-\frac{\cos t-\cos 2t}{\sin t+\sin 2t}\] \[\frac{dy}{dx}=-\left\{ \frac{\cos \frac{\pi }{4}-\cos \frac{\pi }{4}}{\sin \frac{\pi }{4}+\sin \frac{\pi }{2}} \right\}\Rightarrow \frac{dy}{x}=-\left\{ \frac{\frac{1}{2}-0}{\frac{1}{\sqrt{2}}+1} \right\}\]\[\frac{dy}{dx}=-\left\{ \frac{\sqrt{2}}{\sqrt{2}(1+\sqrt{2})} \right\}=-\frac{\sqrt{2}-1}{1}=1-\sqrt{2}\]
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