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If \(0 < \mathrm{x} < 1\), which of the following inequalities must be true ?
Let's start by understanding what we're looking for. We have a number \(\mathrm{x}\) that's between 0 and 1 - in other words, \(\mathrm{x}\) is a proper fraction like \(\frac{1}{2}\), \(\frac{3}{4}\), or 0.3. We need to figure out which of the three given inequalities must always be true for any such fraction.
The key phrase "must be true" means we're looking for inequalities that work for every single value of \(\mathrm{x}\) in this range, not just some values.
Process Skill: TRANSLATE - Converting the constraint "\(0 < \mathrm{x} < 1\)" to the meaningful concept of "proper fractions" is crucial for intuitive understanding.Here's the key insight: when you multiply a proper fraction by itself, you get a smaller number. Let's see this with a simple example using \(\mathrm{x} = \frac{1}{2}\):
Notice the pattern: \(\frac{1}{2} > \frac{1}{4} > \frac{1}{8} > \frac{1}{16} > \frac{1}{32}\)
In general terms: the higher the power, the smaller the result when \(0 < \mathrm{x} < 1\).
Mathematically: \(\mathrm{x}^1 > \mathrm{x}^2 > \mathrm{x}^3 > \mathrm{x}^4 > \mathrm{x}^5 > \ldots\)
Now let's check each inequality using our understanding:
Since higher powers are smaller, \(\mathrm{x}^5\) (5th power) must be smaller than \(\mathrm{x}^3\) (3rd power). This matches our pattern, so Inequality I is always true.
With \(\mathrm{x} = \frac{1}{2}\): \(\frac{1}{32} < \frac{1}{8}\) ✓
Let's think about this step by step. We know that \(\mathrm{x}^4 < \mathrm{x}^2\) (higher power is smaller) and \(\mathrm{x}^5 < \mathrm{x}^3\) (higher power is smaller). When we add smaller terms to smaller terms, the sum on the left (\(\mathrm{x}^4 + \mathrm{x}^5\)) will be less than the sum on the right (\(\mathrm{x}^3 + \mathrm{x}^2\)).
With \(\mathrm{x} = \frac{1}{2}\): \(\left(\frac{1}{16} + \frac{1}{32}\right) < \left(\frac{1}{8} + \frac{1}{4}\right)\)
Left side: \(\frac{3}{32}\), Right side: \(\frac{3}{8} = \frac{12}{32}\)
Indeed: \(\frac{3}{32} < \frac{12}{32}\) ✓
Inequality II is always true.
This is more subtle. We're comparing two differences. Let's think about what each difference represents:
Since both have the factor \((1 - \mathrm{x})\), we're essentially comparing \(\mathrm{x}^4\) versus \(\mathrm{x}^2\). Since \(\mathrm{x}^4 < \mathrm{x}^2\), we have \(\mathrm{x}^4(1 - \mathrm{x}) < \mathrm{x}^2(1 - \mathrm{x})\).
With \(\mathrm{x} = \frac{1}{2}\): \(\left(\frac{1}{16} - \frac{1}{32}\right) < \left(\frac{1}{4} - \frac{1}{8}\right)\)
Left side: \(\frac{1}{32}\), Right side: \(\frac{1}{8} = \frac{4}{32}\)
Indeed: \(\frac{1}{32} < \frac{4}{32}\) ✓
Inequality III is always true.
Let's double-check with another value, say \(\mathrm{x} = 0.1\):
All three inequalities hold for this example as well.
Process Skill: APPLY CONSTRAINTS - Testing multiple values within the given constraint ensures our reasoning applies universally.Since all three inequalities (I, II, and III) must be true for any value of \(\mathrm{x}\) where \(0 < \mathrm{x} < 1\), the answer is E. I, II and III.
This matches the given correct answer of E.
Faltering Point 1: Misunderstanding the constraint \(0 < \mathrm{x} < 1\)
Students often fail to grasp that when \(0 < \mathrm{x} < 1\), higher powers of \(\mathrm{x}\) become progressively smaller. They may incorrectly assume that \(\mathrm{x}^5 > \mathrm{x}^3\) (thinking higher powers are always larger) instead of recognizing that for proper fractions, \(\mathrm{x}^5 < \mathrm{x}^4 < \mathrm{x}^3 < \mathrm{x}^2 < \mathrm{x}^1\). This fundamental misunderstanding leads to incorrect evaluation of all three inequalities.
Faltering Point 2: Overlooking the "must be true" requirement
The question asks which inequalities "must be true" for all values where \(0 < \mathrm{x} < 1\). Students might test only one specific value (like \(\mathrm{x} = 0.5\)) and conclude an inequality is true without verifying it holds universally. They need to understand that "must be true" means the inequality works for every possible value in the given range, not just cherry-picked examples.
Faltering Point 3: Inadequate approach for complex inequalities
For Inequality III (\(\mathrm{x}^4 - \mathrm{x}^5 < \mathrm{x}^2 - \mathrm{x}^3\)), students may struggle to devise a systematic approach. Instead of factoring out common terms to get \(\mathrm{x}^4(1-\mathrm{x}) < \mathrm{x}^2(1-\mathrm{x})\) and then comparing \(\mathrm{x}^4\) vs \(\mathrm{x}^2\), they might attempt direct numerical substitution without understanding the underlying algebraic relationship, making verification unnecessarily complicated.
Faltering Point 1: Arithmetic errors in power calculations
When testing with specific values like \(\mathrm{x} = \frac{1}{2}\), students frequently make calculation mistakes. For example, they might incorrectly compute \(\left(\frac{1}{2}\right)^5 = \frac{1}{16}\) instead of \(\frac{1}{32}\), or miscalculate fractions when adding terms like \(\mathrm{x}^4 + \mathrm{x}^5\). These computational errors lead to wrong conclusions about whether inequalities hold.
Faltering Point 2: Incorrect algebraic manipulation
In Inequality III, when factoring \(\mathrm{x}^4 - \mathrm{x}^5 = \mathrm{x}^4(1-\mathrm{x})\) and \(\mathrm{x}^2 - \mathrm{x}^3 = \mathrm{x}^2(1-\mathrm{x})\), students may make algebraic errors such as incorrect factorization or sign errors. They might also incorrectly cancel the \((1-\mathrm{x})\) factor without recognizing that since \(0 < \mathrm{x} < 1\), we have \((1-\mathrm{x}) > 0\), making the cancellation valid.
Faltering Point 1: Partial verification leading to incomplete answers
Students might correctly verify that Inequality I is true but fail to systematically check Inequalities II and III, leading them to select "B. I only" instead of the correct answer "E. I, II and III". They may get overwhelmed by the complexity of the latter inequalities and assume they're false without proper verification.
Faltering Point 2: Misinterpreting the answer format
Even if students correctly determine that all three inequalities are true, they might misread the answer choices and select "D. I and II only" thinking it includes all their verified results, without noticing that option "E. I, II and III" is the complete and correct choice that includes all three inequalities they've proven to be true.
Step 1: Choose a strategic smart number
Since we need \(0 < \mathrm{x} < 1\), let's choose \(\mathrm{x} = \frac{1}{2}\). This is an ideal smart number because:
Step 2: Calculate the required powers
With \(\mathrm{x} = \frac{1}{2}\):
Step 3: Test each inequality
Inequality I: \(\mathrm{x}^5 < \mathrm{x}^3\)
\(0.03125 < 0.125\) ✓ TRUE
Inequality II: \(\mathrm{x}^4 + \mathrm{x}^5 < \mathrm{x}^3 + \mathrm{x}^2\)
Left side: \(0.0625 + 0.03125 = 0.09375\)
Right side: \(0.125 + 0.25 = 0.375\)
\(0.09375 < 0.375\) ✓ TRUE
Inequality III: \(\mathrm{x}^4 - \mathrm{x}^5 < \mathrm{x}^2 - \mathrm{x}^3\)
Left side: \(0.0625 - 0.03125 = 0.03125\)
Right side: \(0.25 - 0.125 = 0.125\)
\(0.03125 < 0.125\) ✓ TRUE
Step 4: Verify with a second smart number
To ensure our conclusion is robust, let's test \(\mathrm{x} = \frac{1}{3}\):
Inequality I: \(0.004 < 0.037\) ✓ TRUE
Inequality II: \(0.012 + 0.004 = 0.016 < 0.037 + 0.111 = 0.148\) ✓ TRUE
Inequality III: \(0.012 - 0.004 = 0.008 < 0.111 - 0.037 = 0.074\) ✓ TRUE
Conclusion: All three inequalities I, II, and III are true for both test values, confirming that the answer is E.