Q. If $f(x + 2y,x,x - 2y) = xy$, then $f(x,\ y)$ equals
A) $\frac{x^{2} - y^{2}}{8}$
B) $\frac{x^{2} - y^{2}}{4}$
C) $\frac{x^{2} + y^{2}}{4}$
D) $\frac{x^{2} - y^{2}}{2}$
Explanation
(a)
We have, $f(x + 2y,\ x - 2y) = xy$ ...(i)
Let $x + 2y = u$ and $x - 2y = v$
Then, $x = \frac{u + v}{2}$ and $y = \frac{u - v}{4}$
Subtracting the values of $x$ and $y$ in Eq. (i), we obtain
$f(u,\ v) = \frac{u^{2} - v^{2}}{8} \Rightarrow f(x,\ y) = \frac{x^{2} - y^{2}}{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.