If `f''(x)=-f(x) and g(x)=f^(prime)(x)` and `F(x)=(f(x/2))^2+(g(x/2))^2` and given that `F(5)=5,` then `F(10)` is (a)5 (b) 10 (c) 0 (d) 15

Options

A. 5
B. 10
C. 0
D. 15

Explanation

`F'(x)=[f((x)/(2))f'((x)/(2))+g((x)/(2))g'((x)/(2))]`
`"Here, "g(x) =f'(x)`
`"and "g'(x)=f'''(x)=-f(x)`
`"So, F'(x)=f((x)/(2))g((x)/(2))-f((x)/(2))g((x)/(2))=0`
Hence, F(x) is a constant function.
Therefore, F(10) =5.