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Question

If x=a(cost+logtant2),y=asint, then evaluate d2ydx2 at t=π3.

A
43a
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B
3a
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C
83a
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D
835a
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Solution

The correct option is C 83a
x=a(cost+logtant2)
1=a(sint+sec2t22tant2)dtdx
dtdx=1a(sint+sec2t22tant2)
dydt=acost

dydx=cost(sint+sec2t22tant2)

dydx=cost(sint+12sint2cost2)
dydx=cost(sint+1sint)
dydx=costsint1sin2t
dydx=tant
d2ydx2=sec2tdtdx
At t=π3
d2ydx2=4a(sint+sec2t22tant2)=83a


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