Question

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

• A. $1$
• B. $0$
• C. $2$
• D. $3$

Answer 1

The correct option is B

$0$

Explanation for the correct option:

Step 1. Finding the value of $\mathit{f}\mathbf{}\mathbf{(}\mathit{x}\mathbf{,}\mathbf{}\mathit{y}\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathit{f}\mathbf{}\mathbf{(}\mathit{y}\mathbf{,}\mathbf{}\mathit{x}\mathbf{)}$:

Given that $f(2x+(\frac{y}{8}),2x-(\frac{y}{8}\left)\right)=xy,$

put $h=2x+\left(\frac{y}{8}\right)...\left(i\right)$

and $k=2x–\left(\frac{y}{8}\right)...\left(ii\right)$

Step 2. Add equation $\left(i\right)\&\left(ii\right)$, we get

$x=\frac{(h+k)}{4}$, $y=4(h-k)$

$f(h,k)=\frac{(h+k)}{4}\times 4(h-k)$

$\Rightarrow$$f(h,k)=(h^2-k^2)$

Similarly,

$f(x,y)=(x^2-y^2)$

$\Rightarrow$ $f(y,x)=(y^2-x^2)$

$\begin{array}{rcl}\mathbf{\therefore }\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{+}\mathbf{f}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{x}\mathbf{)}& =& (x^2–y^2)+(y^2–x^2)\\ & =& \mathbf{0}\end{array}$

Hence, Option ‘B’ is Correct.