If `f(2-x)=f(2+x)` and `f(7-x)=f(7+x)` and `f(0)=0`. If the minimum number of roots of `f(x)=0` where `0lexle100` is `lamda` then `lamda//3` equals

Answer

7

Explanation

`f(2-x)=f(2+x)` ..(1)
& `f(7-x)=f(7+x)` .(2)
replace x by `x+5` we get `|x|le100`
`f(2-(x+5))=f(2+x+5)`
`impliesx in [-100,100]`
`impliesf(-3-x)=f(x+7)`
`impliesf(-3-x)=f(7-x)` [from 2]
but `x=-3-x` we get `f(-3-(-3-x))=f(7-(-3-x))`
`f(x)=f(x+10)` impliesf(x)` has a period 10
`impliesf(0)=f(10)=f(20)=...=f(100)=0`
also from (1) to `f(2-x)=f(2+x)` but `x=2`
we have `f(2-2)=f(2+2)impliesf(4)=f(0)=0`
`impliesf(4)=f(14)=f(24)=..f(94)=0`
10 roots
minimum number of roots `=21`