Question

If $f(2x^2+\frac{y^2}{8},2x^2-\frac{y^2}{8})=xy$ then

• A. $f(x,y)=x^2-y^2$
• B. $f(x,y)=\sqrt{x^2-y^2}$
• C. $f(5,4)+f(10,6)=11$
• D. $f(0,0)+f(2,2)=0$

Answer 1

The correct option is A $f(x,y)=x^2-y^2$

$let2x^2+\left(\frac{y^2}{8}\right)=u2x^2-\left(\frac{y^2}{8}\right)=vthenu+v=4x^2\therefore x=\sqrt{\left(\frac{u+v}{4}\right)}u-v=2\times \left(\frac{y^2}{8}\right)=\left(\frac{y^2}{4}\right)\therefore y=2\sqrt{u-v}\therefore f(u,v)=\sqrt{\left(\frac{u+v}{4}\right)}\times 2\sqrt{u-v}=\sqrt{u^2-v^2}f(x,y)=\sqrt{x^2-y^2}$