A function `f: RvecR` satisfies `xcosy(f(2x+2y)-f(2x-2y)=cosxsiny(f(2x+2y)+f(2x-2y))dot` If `f^(prime)(0)=1/2,t h e n` `f^(prime)(x)=f(x)=0` `4f^(x)+f(x)=0` `f^(x)+f(x)=0` `4f^(x)-f(x)=0`

Options

A. `f''(x)=f(x)=0`
B. `4f''(x)+f(x)=0`
C. `f''(x)+f(x)=0`
D. `4f''(x)-f(x)=0`

Explanation

`"We have "(f(2x+2y))/(f(2x-2y))=(sin(x+y))/(sin (x-y))`
`"or "(f(alpha))/(sin""(alpha)/(2))=(f(beta))/(sin""(beta)/(2))=K`
`"or "f(x)=K sin""(x)/(2)`
`therefore" "f'(x)=(K)/(2)cos""(x)/(2)`
`"and "f''(x)=(-K)/(4)sin""(x)/(2)`
`"or "4f''(x)+f(x)=0`