Let f(x) be a non-constant thrice differential function defined on `(-oo, oo)` such that `f((x+13)/2) = f((3-x)/2)` and `f'(0) = f'(1/2) = f'(3) = f'(9/2) =0` then the minimum number of zeros of `h(x) =(f''(x))^2 + f'(x) f''(x)` in the interval `[0,9] is 2k then k is equal to

Answer

9

Explanation

`f((x+13)/2)=f((3-x)/2)`
`f(x)=f(8-x)`
`f^(')=f(8-x)`
`f^(')(x)=-f^(')(8-x)`
`f^(')(1/2)=-f^(')(15/2)=0`
`f^(')(2)=-f^(')(6)=0`
`f^(')(3)=-f^(')(5)=0`
`f^(')(9/2)=-f^(')(7/2)=0`
`f^(')(0)=-f^(')(8),h(x)=d/(dx)(f^(')(x)f^('')(x))`
Clearly `h(x)` has minimum `18` zeroes