1f(2-x)=f(x) symmetric about x=1
f(5-x)=f(x)symmetric about x=5
this is only possible for a periodic function
with period 4
so f'(-3)=0
f'(1+4k)=0
f'(3+4k)=0
3. must be 2k ?
1..if f:R→R be a differential finction such that that f(2-x)=f(x) and f(5-x)=f(5+x) then
1.f'(-3)=0
2..f'(9)=0
.3.if \int_{3}^{5}{f(x)dx=k},then \int_{11}^{15}{f(x) dx} wil l be k
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5 Answers
1
341f(x) = f(2-x) → f'(x) = -f'(2-x)
Putting x=1, we get f'(1) = 0
similarly f'(5+x) = - f'(5-x).
Now putting x = 4, we obtain f'(9) = -f'(1) = 0
Again putting x=0, we get f'(5) = 0
Put x = -3 in the first equation to obtain f'(-3) = -f'(5) = 0
