Question

Which of the following expressions yields a positive value for every pair of non-zero real numbers \((x, y)\)?

• 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. Both \(x\) and \(y\) are negative
• B. Both \(x\) and \(y\) are positive
• C. Both \(x\) and \(y\) are less than \(-1\)
• D. \(y>x\)

Answer 1

The Correct Option is A

Step 1: Rewrite in terms of \(s=x+y\).
\((x+y)^{0.5}\) is real exactly when \(s \ge 0\). So
if \(s \ge 0\): \(f=\sqrt{s}\), \(g=s^2\);
if \(s<0\): \(f=s^2\), \(g=-s=|s|\).

Step 2: Test option (4) \(f-g\).
For \(s=4\): \(2-16<0\). Not always positive.

Step 3: Test option (2) \(f-g^2\).
For \(s=4\): \(2-256<0\). Not always positive.

Step 4: Test option (3) \(g-f^2\).
For \(s=\tfrac14\): \(\tfrac1{16}-\tfrac14<0\). Not always positive.

Step 5: Option (1) \(f+g\).
If \(s>0\): \(\sqrt{s}+s^2>0\). If \(s<0\): \(s^2+|s|>0\). So \(f+g\) is positive for every pair with \(x+y\neq 0\) (it is 0 only in the degenerate case \(x+y=0\), where \(f=g=0\)). It is the only option that never goes negative.

Final Answer:
The correct option is (1): \(f(x,y)+g(x,y)\)

Answer 2

Step 1: Express with \(s=x+y\).
If \(s\ge0\): \(f=\sqrt{s}\), \(g=s^2\). If \(s<0\): \(f=s^2\), \(g=-s=|s|\).

Step 2: Option (3): both less than \(-1\).
Then \(s<-2<0\), so \(f=s^2\) and \(g=|s|\). Since \(|s|>2>1\), \(s^2>|s|\), i.e. \(f>g\). This always holds.

Step 3: Option (2): both positive.
\(s>0\): \(f=\sqrt{s}\), \(g=s^2\). For \(x=y=2\), \(s=4\): \(f=2<g=16\). Fails.

Step 4: Option (1): both negative.
\(s<0\): \(f=s^2\), \(g=|s|\). For \(x=y=-0.1\), \(s=-0.2\): \(f=0.04<g=0.2\). Fails.

Step 5: Option (4): \(y>x\).
Take \(x=1,\ y=3\): \(f=2<g=16\). Fails.

Final Answer:
The correct option is (3): Both \(x\) and \(y\) are less than \(-1\)