If `f(x)=x+1,g(x)=x^2+1`, then `(f+g)/(fg) (0)` =?

Explanation

`f(x,y)=f(2x+2y,2y-2x)`
` " " `(Replacing x by `2x+2y` and y by `2y-2x`)
`=f(2(2x+2y)+2(2y-2x),2(2y-2x)-2(2x+2y))`
` :. f(x,y)=f(8y,-8x)=f(8(-8x),-8(8y))`
`=f(-64x,-64y)`
`=f(-64(-64x),-64y(-64y))`
`=f(2^(12)x,2^(12)y)`
` :. f(x,0)=f(2^(12)x,0)`
` :. f(2^(y),0)=f(2^(12)*2^(y),0)=f(2^(12+y),0)`
`or g(y)=g(y+12)`
Hence, `g(x)` is periodic and its period is 12.