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Is \(2\mathrm{x} - 3\mathrm{y} < \mathrm{x}^2\) ?
We need to determine: Is \(2\mathrm{x} - 3\mathrm{y} < \mathrm{x}^2\)?
This is a yes/no question. To have sufficiency, we need to be able to answer definitively YES (the inequality is always true) or NO (the inequality is always false).
We're comparing a linear expression (\(2\mathrm{x} - 3\mathrm{y}\)) with a quadratic expression (\(\mathrm{x}^2\)). The key insight: quadratic terms grow much faster than linear terms, especially as x increases.
Remember that \(\mathrm{x}^2\) is always non-negative (\(\geq 0\)) for any real value of x. This simple fact alone might be enough to answer our question in certain scenarios. Additionally, when x takes on larger values, \(\mathrm{x}^2\) will increasingly dominate the linear term \(2\mathrm{x}\).
Statement 1: \(2\mathrm{x} - 3\mathrm{y} = -2\)
This directly gives us the value of the left side of our inequality. So our question becomes:
Is \(-2 < \mathrm{x}^2\)?
Since \(\mathrm{x}^2\) is always non-negative for any real value of x:
- If \(\mathrm{x} \neq 0\), then \(\mathrm{x}^2 > 0\), which means \(-2 < \mathrm{x}^2\) ✓
- Even if \(\mathrm{x} = 0\), then \(\mathrm{x}^2 = 0\), and we still have \(-2 < 0\) ✓
The answer is always YES because a negative number (-2) is always less than a non-negative number (\(\mathrm{x}^2\)).
Statement 1 is sufficient to answer YES.
[STOP - Sufficient!]
This eliminates choices B, C, and E.
Now let's forget Statement 1 completely and analyze Statement 2 independently.
Statement 2: \(\mathrm{x} > 2\) and \(\mathrm{y} > 0\)
We have constraints on both variables:
- x must be greater than 2
- y must be positive
When \(\mathrm{x} > 2\), we enter the region where quadratic dominance becomes crystal clear:
At the boundary (\(\mathrm{x} = 2\)):
- \(\mathrm{x}^2 = 4\)
- \(2\mathrm{x} = 4\)
- They're exactly equal
For any \(\mathrm{x} > 2\):
- \(\mathrm{x}^2\) grows faster than \(2\mathrm{x}\)
- The gap widens as x increases
Now consider our expression \(2\mathrm{x} - 3\mathrm{y}\):
- We start with \(2\mathrm{x}\) (which already loses to \(\mathrm{x}^2\) when \(\mathrm{x} > 2\))
- We subtract a positive quantity \(3\mathrm{y}\) (since \(\mathrm{y} > 0\))
- This makes the left side even smaller
Example to visualize: If \(\mathrm{x} = 3\), then:
- \(\mathrm{x}^2 = 9\)
- \(2\mathrm{x} = 6\)
- Even before subtracting \(3\mathrm{y}\), we have \(6 < 9\)
- After subtracting any positive \(3\mathrm{y}\), the gap only increases
The critical question: Could \(2\mathrm{x} - 3\mathrm{y}\) ever catch up to \(\mathrm{x}^2\)?
- This would require \(2\mathrm{x} - 3\mathrm{y} \geq \mathrm{x}^2\)
- In the best case for the left side (\(\mathrm{y} \to 0\)), we'd need \(2\mathrm{x} \geq \mathrm{x}^2\)
- This means \(\mathrm{x}^2 - 2\mathrm{x} \leq 0\), or \(\mathrm{x}(\mathrm{x} - 2) \leq 0\)
- This only happens when \(0 \leq \mathrm{x} \leq 2\)
- But we're told \(\mathrm{x} > 2\)!
Statement 2 is sufficient to answer YES - the quadratic dominance guarantees the inequality.
[STOP - Sufficient!]
Both statements independently guarantee that \(2\mathrm{x} - 3\mathrm{y} < \mathrm{x}^2\), making each statement alone sufficient.
Answer Choice D: "Each statement alone is sufficient."