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If \(0 < \mathrm{r} < 1 < \mathrm{s} < 2\). Which of the following must be less than 1.
Let's understand what we have here. We're told that \(\mathrm{r}\) is a number between 0 and 1, so \(\mathrm{r}\) could be something like 0.3, 0.7, or 0.9. We're also told that \(\mathrm{s}\) is a number between 1 and 2, so \(\mathrm{s}\) could be something like 1.2, 1.5, or 1.8.
We need to figure out which of these three expressions will always be less than 1, no matter what specific values we pick for \(\mathrm{r}\) and \(\mathrm{s}\):
The key word here is "must" - we're looking for expressions that are guaranteed to be less than 1.
Process Skill: TRANSLATE - Converting the constraint language into concrete understanding
To really test these expressions, let's be smart about our choices. We want to see if any of these expressions can get close to 1 or even exceed 1. So let's pick values that might push each expression to its limits.
For trying to make expressions as large as possible:
Let's test with \(\mathrm{r = 0.9}\) and \(\mathrm{s = 1.1}\) first:
Now let's try \(\mathrm{r = 0.9}\) and \(\mathrm{s = 1.9}\):
Interesting! We're already seeing that expression II can exceed 1, and expression III can equal 1.
Now let's think about each expression more carefully:
Expression I: \(\mathrm{r/s}\)
Since \(\mathrm{r}\) is less than 1 and \(\mathrm{s}\) is greater than 1, we're dividing a number smaller than 1 by a number larger than 1. This will always give us something smaller than the original \(\mathrm{r}\), which was already less than 1. So \(\mathrm{r/s}\) must always be less than 1.
Expression II: \(\mathrm{rs}\)
Here we're multiplying \(\mathrm{r}\) (less than 1) by \(\mathrm{s}\) (greater than 1). As we saw in our example, when \(\mathrm{r = 0.9}\) and \(\mathrm{s = 1.9}\), we get \(\mathrm{rs = 1.71}\), which is definitely greater than 1. So expression II doesn't always stay below 1.
Expression III: \(\mathrm{s - r}\)
We're subtracting \(\mathrm{r}\) (less than 1) from \(\mathrm{s}\) (greater than 1). In our example with \(\mathrm{r = 0.9}\) and \(\mathrm{s = 1.9}\), we got exactly 1.0. Since \(\mathrm{s}\) can be as large as 1.99... and \(\mathrm{r}\) can be as small as 0.01..., this difference can definitely be 1 or larger. So expression III doesn't always stay below 1.
Process Skill: CONSIDER ALL CASES - Testing boundary conditions to find counterexamples
Let's confirm our counterexamples:
For Expression II: Choose \(\mathrm{r = 0.8}\) and \(\mathrm{s = 1.5}\)
\(\mathrm{rs = 0.8 × 1.5 = 1.2 > 1}\) ✗
For Expression III: Choose \(\mathrm{r = 0.1}\) and \(\mathrm{s = 1.9}\)
\(\mathrm{s - r = 1.9 - 0.1 = 1.8 > 1}\) ✗
For Expression I: Let's verify it's always less than 1
Since \(0 < \mathrm{r} < 1\) and \(1 < \mathrm{s} < 2\), we have \(\mathrm{r < 1 < s}\), so \(\mathrm{r/s < 1/s < 1/1 = 1}\) ✓
Therefore, only Expression I must always be less than 1.
Only expression I (\(\mathrm{r/s}\)) must always be less than 1, while expressions II and III can equal or exceed 1 under certain conditions within the given constraints.
The answer is A. I only.
1. Misunderstanding what "must be less than 1" means: Students often confuse "must be less than 1" with "can be less than 1." They may think they need to find expressions that are sometimes less than 1, rather than expressions that are ALWAYS less than 1 for all valid values of \(\mathrm{r}\) and \(\mathrm{s}\). This leads them to incorrectly include expressions II and III.
2. Not recognizing the need to test boundary cases: Students may only test "nice" middle values like \(\mathrm{r = 0.5}\) and \(\mathrm{s = 1.5}\), missing the critical insight that they need to test extreme cases within the constraints (like \(\mathrm{r}\) close to 1 and \(\mathrm{s}\) close to 2) to see if expressions can approach or exceed 1.
3. Misinterpreting the constraint notation: Students may misread \(0 < \mathrm{r} < 1 < \mathrm{s} < 2\) and think that \(\mathrm{r}\) and \(\mathrm{s}\) can actually equal the boundary values (\(\mathrm{r = 1}\) or \(\mathrm{s = 1}\)), rather than understanding these are strict inequalities where \(\mathrm{r}\) must be less than 1 and \(\mathrm{s}\) must be greater than 1.
1. Arithmetic errors in testing examples: When calculating expressions like \(\mathrm{rs = 0.9 × 1.9 = 1.71}\) or \(\mathrm{s - r = 1.9 - 0.9 = 1.0}\), students may make basic multiplication or subtraction errors that lead them to incorrect conclusions about whether expressions exceed 1.
2. Not testing sufficient examples: Students may test only one set of values and conclude incorrectly. For instance, testing only \(\mathrm{r = 0.9, s = 1.1}\) gives \(\mathrm{rs = 0.99 < 1}\), leading them to think expression II is always less than 1, without testing larger values of \(\mathrm{s}\).
3. Incorrect algebraic reasoning: When trying to prove that \(\mathrm{r/s < 1}\), students may make logical errors in their inequality manipulations or fail to properly use the given constraints that \(\mathrm{r < 1}\) and \(\mathrm{s > 1}\).
1. Confusing which expressions were proven vs. disproven: After finding that expressions II and III can exceed 1 in some cases, students may accidentally select answer choices that include these expressions, forgetting that they need expressions that are ALWAYS less than 1.
2. Misreading the answer choices: Students may correctly identify that only expression I works, but then accidentally select "E. I and III" instead of "A. I only" due to careless reading of the multiple choice options.
Step 1: Choose strategic smart numbers within the given constraints
Since \(0 < \mathrm{r} < 1 < \mathrm{s} < 2\), let's choose values that will help us test each expression effectively:
These values are chosen strategically because they push the expressions toward their maximum possible values within the given constraints.
Step 2: Test each expression with our smart numbers
Expression I: \(\mathrm{r/s}\)
\(\mathrm{r/s = 0.8/1.9 ≈ 0.42}\)
Since \(\mathrm{r < 1}\) and \(\mathrm{s > 1}\), we have \(\mathrm{r/s}\) = (something less than 1)/(something greater than 1), which must be less than 1.
Expression II: \(\mathrm{rs}\)
\(\mathrm{rs = 0.8 × 1.9 = 1.52}\)
This is greater than 1! Let's verify this makes sense: when \(\mathrm{r}\) is close to 1 and \(\mathrm{s}\) is close to 2, their product approaches \(1 × 2 = 2\).
Expression III: \(\mathrm{s - r}\)
\(\mathrm{s - r = 1.9 - 0.8 = 1.1}\)
This is also greater than 1! This makes sense because \(\mathrm{s}\) can be close to 2 while \(\mathrm{r}\) can be close to 0, making the difference approach 2.
Step 3: Verify our findings with boundary analysis
To confirm Expression I always stays below 1, note that no matter what values we choose:
Step 4: Conclusion
Using our smart numbers, we've shown that:
Therefore, only I must be less than 1.
Answer: A