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B
dydx∣∣t=π4=12
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C
d2ydx2∣∣∣t=π4=−14
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D
d2ydx2∣∣∣t=π6=−2
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Solution
The correct option is Cd2ydx2∣∣∣t=π4=−14 dydt=2cos2t,dxdt=−4sin2t dydx=−12cot2t⇒dydx∣∣t=3π8=−12(−1)=12 d2ydx2=csc2(2t)×1−4sin2t d2ydx2=−14csc32t, d2ydx2∣∣∣t=π4=−14×1=−14