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Question

For a positive integer n, let fn(θ)=(2cosθ+1)(2cosθ1)(2cos2θ1)(2cos(22)θ1)......(2cos(2n1)θ1). Which one of the following hold(s) not good?

A
f2(π/6)=0
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B
f3(π/8)=1
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C
f4(π/32)=1
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D
f5(π/128)=1+2
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Solution

The correct option is D f5(π/128)=1+2
f(θ)=(2cosθ+1)(2cosθ1)(2cos2θ1)(2cos(22)θ1)......(2cos(2n1)θ1).=(2cosθ+1)+(2cosθ1)....(2cos2n1θ+1)Now,fn(θ)=2cos2nθ+1f2(π6)=2cos(4×π6)+1=2cos2π3+1=0f3(π8)=2cos(8×π8)+1=1f4(π32)=2cos(16×π32)+1=1f5(π128)=2cos(32×π128)+1=1+2Hence,theoptionDisthecorrectanswer.

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