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Question

Let J=120(14x2)4dx and K=120x4(1x)4dx, then

A
JK=2
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B
JK=0
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C
J=1210x4(1x)4dx
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D
K=11260
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Solution

The correct options are
B JK=0
C J=1210x4(1x)4dx
D K=11260
J=120(14x2)4dx=120(14(12x)2)4dxK=120(xx2)4dx=120x4(1x4)dx...(i)J=K, so JK=0Put x=1yK=121(1y)2y4dyK=112x4(1x)4dx...(ii)equation(i)+(ii)2J=10x4(1x4dx)
J=1210x4(1x)4dxJ=1210x4(1x4)dxPut x=sin2θJ=π20sin9θ.cos9θdθ=(8×6×4×2)(8×6×4×2)(18×16×14×12×10×8×6×4×2)=11260

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