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Difficulty:
65%
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Question Stats:
60% (01:48) correct
40%
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wrong
based on 35
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The function \(max(a, b)\) returns the largest, and \(min(a, b)\) returns the smallest value among the given numbers \(a\) and \(b\). If \(a\) and \(b\) are distinct numbers, which of the following must be true?
I. \(max(a, b) + min(a, b) > 0\)
II. \(max(a, b) = \frac{a + b + |a - b|}{2}\)
III. \(min(a, b) < \frac{a + b - |a - b|}{2}\)
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
I. \(max(a, b) + min(a, b) > 0\)
II. \(max(a, b) = \frac{a + b + |a - b|}{2}\)
III. \(min(a, b) < \frac{a + b - |a - b|}{2}\)
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
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Official Solution:
The function \(max(a, b)\) returns the largest, and \(min(a, b)\) returns the smallest value among the given numbers \(a\) and \(b\). If \(a\) and \(b\) are distinct numbers, which of the following must be true?
I. \(max(a, b) + min(a, b) > 0\)
II. \(max(a, b) = \frac{a + b + |a - b|}{2}\)
III. \(min(a, b) < \frac{a + b - |a - b|}{2}\)
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
Let's evaluate each option
I. \(max(a, b) + min(a, b) > 0\)
If we assume \(a > b\), then the above expression becomes \(a + b > 0\). However, this is not necessarily true, as the fact that one number is greater than another does not guarantee a positive sum. The same result occurs if we assume \(a < b\). Therefore, this option is not always true.
II. \(max(a, b) = \frac{a + b + |a - b|}{2}\)
If \(a > b\), then the left-hand side, \(max(a, b)\), equals \(a\). Since in this case, \(|a - b| = a-b\), the right-hand side, \(\frac{a + b + |a - b|}{2}\), equals \(\frac{a + b + a - b}{2}=a\). Thus, for this case, LHS = RHS.
If \(a < b\), then the left-hand side, \(max(a, b)\), equals \(b\). Since in this case, \(|a - b| = -(a-b)\), the right-hand side, \(\frac{a + b + |a - b|}{2}\), equals \(\frac{a + b - (a - b)}{2}=b\). Again, for this case, LHS = RHS.
Hence, this option must be true.
III. \(min(a, b) < \frac{a + b - |a - b|}{2}\)
If \(a > b\), then the left-hand side, \(min(a, b)\), equals \(b\). In this case, \(|a - b| = a-b\), and the right-hand side, \(\frac{a + b - |a - b|}{2}\), equals \(\frac{a + b - (a - b)}{2}=b\). Therefore, for this case, LHS = RHS.
If \(a < b\), then the left-hand side, \(min(a, b)\), equals \(a\). Here, \(|a - b| = -(a-b)\), and the right-hand side, \(\frac{a + b - |a - b|}{2}\), equals \(\frac{a + b + (a - b)}{2}=a\). So, once again, for this case, LHS = RHS.
Thus, this option must NOT be true.
Therefore, only option II is always true!
Answer: B
The function \(max(a, b)\) returns the largest, and \(min(a, b)\) returns the smallest value among the given numbers \(a\) and \(b\). If \(a\) and \(b\) are distinct numbers, which of the following must be true?
I. \(max(a, b) + min(a, b) > 0\)
II. \(max(a, b) = \frac{a + b + |a - b|}{2}\)
III. \(min(a, b) < \frac{a + b - |a - b|}{2}\)
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
Let's evaluate each option
I. \(max(a, b) + min(a, b) > 0\)
If we assume \(a > b\), then the above expression becomes \(a + b > 0\). However, this is not necessarily true, as the fact that one number is greater than another does not guarantee a positive sum. The same result occurs if we assume \(a < b\). Therefore, this option is not always true.
II. \(max(a, b) = \frac{a + b + |a - b|}{2}\)
If \(a > b\), then the left-hand side, \(max(a, b)\), equals \(a\). Since in this case, \(|a - b| = a-b\), the right-hand side, \(\frac{a + b + |a - b|}{2}\), equals \(\frac{a + b + a - b}{2}=a\). Thus, for this case, LHS = RHS.
If \(a < b\), then the left-hand side, \(max(a, b)\), equals \(b\). Since in this case, \(|a - b| = -(a-b)\), the right-hand side, \(\frac{a + b + |a - b|}{2}\), equals \(\frac{a + b - (a - b)}{2}=b\). Again, for this case, LHS = RHS.
Hence, this option must be true.
III. \(min(a, b) < \frac{a + b - |a - b|}{2}\)
If \(a > b\), then the left-hand side, \(min(a, b)\), equals \(b\). In this case, \(|a - b| = a-b\), and the right-hand side, \(\frac{a + b - |a - b|}{2}\), equals \(\frac{a + b - (a - b)}{2}=b\). Therefore, for this case, LHS = RHS.
If \(a < b\), then the left-hand side, \(min(a, b)\), equals \(a\). Here, \(|a - b| = -(a-b)\), and the right-hand side, \(\frac{a + b - |a - b|}{2}\), equals \(\frac{a + b + (a - b)}{2}=a\). So, once again, for this case, LHS = RHS.
Thus, this option must NOT be true.
Therefore, only option II is always true!
Answer: B
