Mathematics - Functions Question with Solution | TestHub

$$\begin{array}{rl}f(x,y)& =f(2x+2y,2y-2x)\\ & =f(2(2x+2y)+2(2y-2x),2(2y-2x)-2(2x+2y))\\ & =f(8y,-8x)\\ & =f(8(-8x),-8(8y))\\ & =f(-64x,-64y)\\ & =f(-64(-64x),-64(-64y))\\ & =f(2^{12}x,2^{12}y)\end{array}$$

Thus, we have $f(x,y)=f(2^{12}x,2^{12}y)$.

Consider $f(x,0)$. Then $f(x,0)=f(2^{12}x,0)$.

Let $g(x)=f(2^x,0)$.

$$\begin{array}{rl}g(x)& =f(2^x,0)\\ & =f(2^{12}{2}^x,0)\\ & =f(2^{x+12},0)\\ & =g(x+12)\end{array}$$

This implies that $g(x)=g(x+12)$, so the period of $g(x)$ is $T=12$.

Therefore, $\frac{T}{4}=\frac{12}{4}=3$.