A) ?2
B) 2
C) \[\frac{1}{2}\]
D) \[-\frac{1}{2}\]
Correct Answer: C
Solution :
Let \[x=\sin t\cos 2t\] .....(i) and \[y=\cos t\sin 2t\] .....(ii) Differentiate (i) w.r.t. t, we get \[\frac{dx}{dt}=\cos t.\cos 2t-2\sin t\sin 2t\] .....(iii) Again, differentiate (ii), we get \[\frac{dy}{dt}=2\cos t\cos 2t-\sin t\sin 2t\] .....(iv) \ Dividing equation (iv) by (iii), we get \[\frac{dy}{dx}=\frac{2\cos t\cos 2t-\sin t\sin 2t}{\cos t\cos 2t-2\sin t\sin 2t}\] At \[t=\frac{\pi }{4},\,\,\,\frac{dy}{dx}=\frac{1}{2}\].
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