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Question

The function f(θ)=ddθθ0dx1cosθcosx satisfies the differential equation

A
dfdθ+2f(θ)cotθ=0
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B
dfdθ2f(θ)cotθ=0
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C
dfdθ+2f(θ)=0
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D
dfdθ2f(θ)=0
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Solution

The correct option is B dfdθ+2f(θ)cotθ=0

f(θ)=ddθθ0dx1cosxcosθ
f(θ)=dθdθ[11cos2θ]=cosec2θ
dfdθ=2cosecθ(cosecθcotθ)
dfdθ=2cosec2θcotθ
dfdθ=2f(θ)cotθ
df(θ)dθ+2f(θ)cotθ=0


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