Let f be a non-negative continuous and twice differentiable function defined on R such that f(x)+f(x+32)=6,∀x∈R. Then
A
f(x) is a periodic function
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B
f(x) is a non-periodic function
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C
300∫0f(x)dx=900
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D
If f(0)=3, then f′′(x) has at least 3 roots in (0,6)
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Solution
The correct options are Af(x) is a periodic function C300∫0f(x)dx=900 D If f(0)=3, then f′′(x) has at least 3 roots in (0,6) f(x)+f(x+32)=6,∀x∈R⋯(1) Replacing x by x+32, we get f(x+32)+f(x+3)=6⋯(2) ⇒6−f(x)+f(x+3)=6[Using(1)] ⇒f(x+3)=f(x) ∴f(x) is periodic.