Question

Let $x,y$ be real numbers. Let $f(x,y)=|x+y|;F\left[f\right(x,y\left)\right]=-f(x,y)$ and $G\left[f\right(x,y\left)\right]=-F\left[f\right(x,y\left)\right]$ then which of the following is true ?

• A. $F\left[f\right(x,y\left)\right]\cdot G\left[f\right(x,y\left)\right]=-F\left[f\right(x,y\left)\right]\cdot G\left[f\right(x,y\left)\right]$
• B. $F\left[f\right(x,y\left)\right]\cdot G\left[f\right(x,y\left)\right]>-F\left[f\right(x,y\left)\right]\cdot G\left[f\right(x,y\left)\right]$
• C. $F\left[f\right(x,y\left)\right]\cdot G\left[f\right(x,y\left)\right]\ne G\left[f\right(x,y\left)\right]\cdot F\left[f\right(x,y\left)\right]$
• D. $F\left[f\right(x,y\left)\right]+G\left[f\right(x,y\left)\right]+f(x,y)=f(-x,-y)$

Answer 1

Answer: (D)

$F\left[f\right(x,y\left)\right]+G\left[f\right(x,y\left)\right]+f(x,y)=f(-x,-y)$

Given that,

$f(x,y)=|x+y|$

$F\left[f(x,y)\right]=-f(x,y)\dots \left(1\right)$

$G\left[f(x,y)\right]=-F\left[f(x,y)\right]\dots \left(2\right)$

Considering $\left(1\right)$

$F\left[f(x,y)\right]-f(x,y)=0\dots \left(3\right)$

Considering $\left(2\right)$ and $\left(1\right)$

$G\left[f(x,y)\right]=f(x,y)=f(-x,-y)(\because f(x,y)=|x+y|)$

$\therefore G\left[f(x,y)\right]=f(-x,-y)\dots \left(4\right)$

Adding $\left(3\right)$ and $\left(4\right)$. We get,

$F\left[f\right(x,y\left)\right]+G\left[f\right(x,y\left)\right]+f(x,y)=f(-x,-y)$

$\therefore Option\left(D\right)$ is correct