A) \[\sqrt{2}+1\]
B) \[\sqrt{2+1}\]
C) \[\frac{\sqrt{2+1}}{2}\]
D) None of these
Correct Answer: A
Solution :
\[\frac{dx}{dt}=-2\sin t+2\sin 2t\] and \[\frac{dy}{dt}=2\cos t-2\cos 2t\] Þ \[\frac{dy}{dx}=\frac{\cos t-\cos 2t}{\sin 2t-\sin t}\] Put \[t=\frac{\pi }{4},\] we have \[{{\left[ \frac{dy}{dx} \right]}_{t=\pi /4}}\] \[=\frac{\cos \pi /4-\cos \pi /2}{\sin \pi /2-\sin \pi /4}=\sqrt{2}+1\].
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