A) a
B) ? a
C) 4b
D) ? 4b
Correct Answer: D
Solution :
Given \[s=a\sin t+b\cos 2t\] \[\therefore \]\[\frac{ds}{dt}=a\cos t-2b\sin 2t\] \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=-a\sin t-4b\cos 2t\] At \[t=0\], \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=-a\sin 0{}^\circ -4b\cos 0{}^\circ =-4b\].You need to login to perform this action.
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