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Question

If x=2 sin t+sin 2t, y=2 cos t-cos 2t, then the value of d2ydx2 at t=π2 is

A
2
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B
12
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C
34
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D
32
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Solution

The correct option is A 12
x=2sint+sin2t
dxdt=2cost+2cos2t
y=2costcos2t
dydt=2sint+2sin2t
dydx=dydtdxdt
=2sint+2sin2t2cost+2cos2t
=sint+sin2tcost+cos2t
=sin2tsintcost+cos2t
=2sin(2tt2)cos(2t+t2)2cos(2tt2)cos(2t+t2)
=2sin(t2)cos(3t2)2cos(t2)cos(3t2)
=sin(t2)cos(t2)
dydx=tan(t2)
d2ydx2=12sec2(t2)dtdx
d2ydx2=12sec2(t2)12cost+2cos2t
d2ydx2=14sec2(t2)1cost+cos2t

[d2ydx2]t=π2=14sec2⎜ ⎜π22⎟ ⎟1cosπ2+cos2×π2
=14sec2π410+cosπ

=14×2×1

=12

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