Question

Which expression yields positive values for non-zero and real values of $x$ and $y$?

• A. $f(x,y) - g(x,y)$
• B. $f(x,y) - [g(x,y)]^2$
• C. $g(x,y) - [f(x,y)]^2$
• D. $f(x,y) + g(x,y)$
• A. $y \ge x$
• B. Both $x$ and $y$ are less than $-1$
• C. Both $x$ and $y$ are greater than $0$
• D. Both $x$ and $y$ are less than $0$

Answer 1

The Correct Option is A

Given Functions:

The function \( f(x, y) \) is defined as: \[ f(x,y) = \begin{cases} \sqrt{x+y}, & \text{if } x+y \ge 0 \\ (x+y)^2, & \text{if } x+y < 0 \end{cases} \] The function \( g(x, y) \) is defined as: \[ g(x,y) = \begin{cases} (x+y)^2, & \text{if } x+y \ge 0 \\ -(x+y), & \text{if } x+y < 0 \end{cases} \]

Case 1: When \( x + y \ge 0 \)

In this case: \[ f(x,y) = \sqrt{x+y}, \quad g(x,y) = (x+y)^2 \] Therefore, \[ f(x,y) - g(x,y) = \sqrt{x+y} - (x+y)^2 \] Now analyze the sign of this difference:

• When \( x + y = 0 \), then difference = \(0 - 0 = 0\)
• When \( 0 < x + y < 1 \), say \( x + y = 0.25 \):
  \[ \sqrt{0.25} = 0.5,\quad (0.25)^2 = 0.0625 \Rightarrow \text{Difference} = 0.5 - 0.0625 = 0.4375 > 0 \] So the difference is **positive** in this range.
• When \( x + y > 1 \), the square grows faster than square root, so the difference becomes **negative**.

Therefore, for some values where \(0 < x + y < 1\), the difference \( f(x,y) - g(x,y) \) is **positive**.

 

Case 2: When \( x + y < 0 \)

In this case: \[ f(x,y) = (x+y)^2,\quad g(x,y) = -(x+y) \] So, \[ f(x,y) - g(x,y) = (x+y)^2 + (x+y) \] Let us denote \( s = x + y \), where \( s < 0 \). Then: \[ s^2 + s = s(s + 1) \] This expression depends on the value of \( s \):

• If \( -1 < s < 0 \), then \( s + 1 > 0 \) and \( s < 0 \), so \( s(s+1) < 0 \) → **Negative**
• If \( s < -1 \), then \( s + 1 < 0 \), \( s < 0 \), so \( s(s+1) > 0 \) → **Positive**

Therefore, the difference is **positive** in some negative regions and **negative** in others.

 

Conclusion:

The expression: \[ f(x,y) - g(x,y) \] is **positive** for some values of \(x\) and \(y\), particularly in the range: \[ 0 < x + y < 1 \] which is a clean and reliable interval with consistent positivity.

Final Answer:

\[ \boxed{f(x,y) - g(x,y)} \]

Answer 2

If $x>0$ and $y>0$, then $x+y>0$ and: \[ f(x,y) = \sqrt{x+y}, \quad g(x,y) = (x+y)^2 \] For $0<x+y<1$, $\sqrt{x+y}>(x+y)^2$ holds. Positive $x$ and $y$ make $x+y$ positive and allow for a range where $f>g$. This aligns with the intended selection of both positive $x$ and $y$. \[ \boxed{\text{Both } x \text{ and } y \text{ are greater than 0}} \]