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Question

If f(t)=11+t2+2tcosα and g(t)=t|sin(2nt)| , where α(0,π2] and n is odd, then

A
The roots of the equation 10f(t)x2 dt+ππg(t)x dt2=0 are ±2sinα
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B

π40g(t)dtπ232
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C
π0g(t)dt=π
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D
The roots of the equation 10f(t)x2 dt+ππg(t)x dt2=0 are ±2sinαα
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Solution

The correct options are
B
π40g(t)dtπ232
C π0g(t)dt=π
D The roots of the equation 10f(t)x2 dt+ππg(t)x dt2=0 are ±2sinαα
f(t)=1(t+cosα)2+sin2α
I=ππg(t) dt is an odd function
I=0
J=10f(t) dt =[1sinαtan1t+cosαsinα]10J=α2sinα
Jx22=0x=±2sinαα

g(t)=t|sin2nt|tπ40g(t) dtπ40t dtπ40g(t) dtπ232

K=π0g(t) dtK=π0t|sin(2nt)| dtK=π0(πt)|sin2nt| dt2K=π0(πt)|sin2nt| dt
Put 2nt=xdx=2n dt
2K=2nπ0π2n|sint|dt2K=π0π|sint|dtK=π

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