Question

If $f(x + 2y,x - 2y) = xy$ then find $f(x,y)$ is equal to
A. $\dfrac{1}{4}xy$
B. $\dfrac{1}{4}({x^2} - {y^2})$
C. $\dfrac{1}{8}({x^2} - {y^2})$
D. $\dfrac{1}{2}({x^2} + {y^2})$

Answer 1

Hint: Here, we will find the value of $f(x,y)$ by converting the given function into another variable and simplify it, then re-convert into your desired variable.

Complete step-by-step answer:
Given;
$f(x + 2y, x - 2y) = xy \to (1)$
Let $p = x + 2y$ and $q = x - 2y$
Add and subtract p and q respectively,

\Rightarrow p + q = 2x{\text{ & }}p - q = 4y \\\ \Rightarrow \dfrac{{p + q}}{2} = x{\text{ & }}\dfrac{{p - q}}{4} = y \\\

From equation 1
$\Rightarrow f(p,q) = xy = \left( {\dfrac{{p + q}}{2}} \right)\left( {\dfrac{{p - q}}{4}} \right) = \dfrac{{{p^2} - {q^2}}}{8} \\\ \therefore f(p,q) = \dfrac{{{p^2} - {q^2}}}{8} \\\$

Now in the place of p and q, put x and y respectively
$\therefore f(x,y) = \dfrac{1}{8}({x^2} - {y^2})$
Therefore, option c is correct.

Note: These types of problems might confuse you but the approach to follow here is simplification in order to achieve the answer and then treat the new variables as dummy variables and replace them with the original variables.