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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 π4∫0g(t)dt≤π232 C π∫0g(t)dt=π D The roots of the equation 1∫0f(t)x2 dt+π∫−πg(t)x dt−2=0 are ±2√sinααf(t)=1(t+cosα)2+sin2α I=π∫−πg(t) dt is an odd function ∴I=0 J=1∫0f(t) dt =[1sinαtan−1t+cosαsinα]10⇒J=α2sinα ⇒Jx2−2=0⇒x=±2√sinαα g(t)=t|sin2nt|≤t⇒π4∫0g(t) dt≤π4∫0t dt⇒π4∫0g(t) dt≤π232 K=π∫0g(t) dt⇒K=π∫0t|sin(2nt)| dt⇒K=π∫0(π−t)|sin2nt| dt⇒2K=π∫0(π−t)|sin2nt| dt Put 2nt=x⇒dx=2n dt ⇒2K=2nπ∫0π2n|sint|dt⇒2K=∫π0π|sint|dt⇒K=π

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