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Question: Consider the following function \(f(x+2y,x-2y)=xy\), then \(f(x,y)\) equals \(\begin{aligned} ...

Consider the following function f(x+2y,x2y)=xyf(x+2y,x-2y)=xy, then f(x,y)f(x,y) equals
(A)x2y28 (B)x2y24 (C)x2+y24 (D)x2y22 \begin{aligned} & \left( A \right)\dfrac{{{x}^{2}}-{{y}^{2}}}{8} \\\ & \left( B \right)\dfrac{{{x}^{2}}-{{y}^{2}}}{4} \\\ & \left( C \right)\dfrac{{{x}^{2}}+{{y}^{2}}}{4} \\\ & \left( D \right)\dfrac{{{x}^{2}}-{{y}^{2}}}{2} \\\ \end{aligned}

Explanation

Solution

For solving this question, we consider the given variables x+2y and x-2y as some other variables X and Y. Then we get two equations in terms of x, y and X, Y. Using those equations, we can find the values of x and y in terms of X and Y. Then we will get the function with variables X and Y only in terms of X and Y, by substituting them in place of x and y. Then we change X as x and Y as y to modify them to get an answer as in the options.

Complete step by step answer:
We were given a function f such that f(x+2y,x2y)=xyf(x+2y,x-2y)=xy.
Now let us consider two new variables X, Y such that
X=x+2y
Y=x-2y
Now, let us find the value of x in terms of X and Y.
Let us consider the value of X+Y.
X+Y=(x+2y)+(x2y) X+Y=2x \begin{aligned} & \Rightarrow X+Y=\left( x+2y \right)+\left( x-2y \right) \\\ & \Rightarrow X+Y=2x \\\ \end{aligned}
So, we can write x in terms of X and Y as
X+Y=2x x=X+Y2 \begin{aligned} & \Rightarrow X+Y=2x \\\ & \Rightarrow x=\dfrac{X+Y}{2} \\\ \end{aligned}
Now, let us find the value of y in terms of X and Y.
Let us consider the value of X-Y.
XY=(x+2y)(x2y) XY=4y \begin{aligned} & \Rightarrow X-Y=\left( x+2y \right)-\left( x-2y \right) \\\ & \Rightarrow X-Y=4y \\\ \end{aligned}
So, we can write y in terms of X and Y as
XY=4y y=XY4 \begin{aligned} & \Rightarrow X-Y=4y \\\ & \Rightarrow y=\dfrac{X-Y}{4} \\\ \end{aligned}
So, the values of x and y in terms of X and Y are as below
x=X+Y2x=\dfrac{X+Y}{2}
y=XY4y=\dfrac{X-Y}{4}
So, we substitute these values in the given function value f(x+2y,x2y)=xyf(x+2y,x-2y)=xy.
Then, we get the function in the terms of X and Y.
f(x+2y,x2y)=xy f(X,Y)=(X+Y2)(XY4) f(X,Y)=(X+Y)(XY)8 \begin{aligned} & \Rightarrow f(x+2y,x-2y)=xy \\\ & \Rightarrow f(X,Y)=\left( \dfrac{X+Y}{2} \right)\left( \dfrac{X-Y}{4} \right) \\\ & \Rightarrow f(X,Y)=\dfrac{\left( X+Y \right)\left( X-Y \right)}{8} \\\ \end{aligned}
Now, let us consider the formula,
(a+b)(ab)=a2b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}
So, the value of product of (X+Y) and (X-Y) is
(X+Y)(XY)=X2Y2\left( X+Y \right)\left( X-Y \right)={{X}^{2}}-{{Y}^{2}}
Then, we can write the function as,
f(X,Y)=X2Y28\Rightarrow f(X,Y)=\dfrac{{{X}^{2}}-{{Y}^{2}}}{8}
So, for the function f on X and Y value of the function is f(X,Y)=X2Y28f(X,Y)=\dfrac{{{X}^{2}}-{{Y}^{2}}}{8}.
Now, let us replace X by x and Y by y, then we can write the function as,
f(x,y)=x2y28\Rightarrow f(x,y)=\dfrac{{{x}^{2}}-{{y}^{2}}}{8}

So, the correct answer is “Option A”.

Note: The chance of occurrence of mistake is at the ending of the solution, one might think that we should not change the variables X and Y into x and y. Here, we are not transforming the variables like we did in the starting of the solution, we are just changing the symbol from X to x and Y to y to make it look like the one in the given options.