Question

If \( x \), \( y \), and \( z \) are positive integers and \( p = \left( \left( (x - 1)^2 / |x| \right) + 2 \right) + \left( \left( (y - 1)^2 / |y| \right) + 2 \right) + \left( \left( (z - 1)^2 / |z| \right) + 2 \right), \) then \( p<6 \).

• A. Always
• B. Sometimes
• C. Never
• D.
• E.

Hint:

For problems involving inequalities, test specific values and simplify to determine the range of outcomes.

Answer 1

The Correct Option is B

Let’s evaluate the expression for \( p \) for different positive integer values of \( x \), \( y \), and \( z \).
Step 1:
For each of \( x \), \( y \), and \( z \), we evaluate \( \left( (x - 1)^2 / |x| \right) + 2 \).
- For \( x = 1 \): \( \left( (1 - 1)^2 / |1| \right) + 2 = 0 + 2 = 2 \)
- For \( x = 2 \): \( \left( (2 - 1)^2 / |2| \right) + 2 = 1 + 2 = 3 \)
- For \( x = 3 \): \( \left( (3 - 1)^2 / |3| \right) + 2 = 4 / 3 + 2 \approx 3.33 \)
Step 2:
Now calculate for different values of \( p \). Let’s assume \( x = 2 \), \( y = 2 \), and \( z = 2 \):
- For \( x = y = z = 2 \): \( p = (3) + (3) + (3) = 9 \), which is greater than 6.
Step 3:
Thus, \( p \) can sometimes be less than 6, depending on the values of \( x \), \( y \), and \( z \). Therefore, the correct answer is (B) Sometimes.