Question

If x²y³ < 0, which of the following must be true?
Indicate all such answers.
[Note: Select one or more answer choices]

• A. x\textgreater0
• B. xy\textless0
• C. x\^{}2y\textless0
• D. xy\^{}2\textgreater0
• E. x\^{}2\textgreater0

Hint:

For any inequality involving variables raised to powers, pay close attention to whether the exponent is even or odd. An even power (like \(x\^{}2\)) hides the sign of the base, but guarantees a non-negative result. An odd power (like \(y\^{}3\)) preserves the sign of the base.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question tests the properties of inequalities involving powers of variables. The sign (positive or negative) of a product depends on the signs of its factors. The key is to analyze the signs of x² and y³.
Step 2: Key Formula or Approach:
1. The term x², being a square of a real number, is always non-negative (≥ 0).
2. The term y³ has the same sign as y. If y is positive, y³ is positive. If y is negative, y³ is negative.
3. The product of two terms is negative if and only if one term is positive and the other is negative.
4. From the given inequality x²y³ < 0, we can deduce the signs of x and y, and then check each statement.
Step 3: Detailed Explanation:
We are given the inequality x²y³ < 0.
First, notice that the product is not zero. This implies that neither x² nor y³ can be zero.
If x² ≠ 0, then x ≠ 0.
If y³ ≠ 0, then y ≠ 0.
Since x ≠ 0, x² must be strictly positive (x² > 0).
The inequality can be seen as (a positive number) × y³ < 0.
For this product to be negative, the second factor, y³, must be negative.
So, y³ < 0, which implies that y must be negative (y < 0).
So, what we know for sure is: x ≠ 0 and y < 0.
Now let's evaluate each statement:
(A) x > 0: This is not necessarily true. x could be negative (e.g., x = -1). Then x² = 1 > 0. So, this is not a "must be true".
(B) xy < 0: We know y < 0. The sign of xy depends on the sign of x. If x > 0, then xy < 0. If x < 0, then xy > 0. Since we don't know the sign of x, this is not a "must be true".
(C) x²y < 0: We established that x² > 0 and y < 0. The product of a positive number and a negative number is always negative. So, this must be true.
(D) xy² > 0: We know y < 0, which means y ≠ 0, so y² > 0. The sign of xy² depends on the sign of x, which is unknown. This is not a "must be true".
(E) x² > 0: As shown above, since the overall product is not zero, x cannot be zero. The square of any non-zero number is positive. So, this must be true.
(F) y² > 0: As shown above, since the overall product is not zero, y cannot be zero. The square of any non-zero number is positive. So, this must be true.
(G) xy ≠ 0: Since we deduced that x ≠ 0 and y ≠ 0, their product cannot be zero. So, this must be true.
Step 4: Final Answer:
The statements that must be true are (C), (E), (F), and (G).