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Question

Let f be a non-negative continuous and twice differentiable function defined on R such that f(x)+f(x+32)=6, xR. 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
3000f(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
A f(x) is a periodic function
C 3000f(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, xR (1)
Replacing x by x+32, we get
f(x+32)+f(x+3)=6 (2)
6f(x)+f(x+3)=6 [Using (1)]
f(x+3)=f(x)
f(x) is periodic.

I=3000f(x) dx
I=10030f(x) dx
I=100⎜ ⎜3/20f(x) dx+33/2f(x) dx⎟ ⎟
I=1003/20f(x) dx+3/20f(x+32) dx
I=1003/206 dx=100×6×32=900


From Rolle's theorem,
f′′(x)=0 has at least 3 roots.


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