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Question

# If f′(x)=1−2sin2xf(x), f(x)>0, ∀x∈R and f(0)=1. Then f(x) is a periodic function with the period

A
2π
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B
π
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C
π2
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D
not periodic
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Solution

## The correct option is B πLetf′(x)=1−2sin2xf(x)or, f(x)f′(x)=cos2xIntegrating both sides we get,(f(x))22=sin2x2+c [c being integrating constant]Using f(0)=1 we get c=12.So, we have (f(x))2=1+sin2x.As given f(x)>0,∀x∈R we have f(x)=√1+sin2x,∀x∈R.It is clear that f(π+x)=f(x), this means that f(x) is a periodic function of period π.So, option (B) is correct.

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