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21 Jan 2024, 11:00
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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) = (a + b + |a - b|)/2
III. min(a, b) < (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) = (a + b + |a - b|)/2
III. min(a, b) < (a + b - |a - b|)/2
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
22 Jan 2024, 09:04
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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
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
General Discussion
21 Jan 2024, 11:26
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Bunuel
This is a "MUST be true" type of question.
Let's evaluate the options -
I. max(a, b) + min(a, b) > 0
If 'a' and 'b' are negative, both max(a,b) and min(a,b) will be negative. In that case, max(a, b) + min(a, b) will also be negative. Hence, the stated inequality is not always true. We can eliminate all options that state option I must be true. Eliminate A, C and D.
II. max(a, b) = (a + b + |a - b|)/2
There are two ways to approach this statement
1) Test out numbers
Case 1: a = -5; b = -10
max(a,b) = -5
min (a,b) = -10
(a + b + |a - b|)/2 = (-5 - 10 + 5)/2 = -5
Hence, max(a, b) = (a + b + |a - b|)/2
Case 2: a = -5; b = 0
max(a,b) = 0
min (a,b) = -5
(a + b + |a - b|)/2 = (-5 - 0 + 5)/2 = 0
Hence, max(a, b) = (a + b + |a - b|)/2
Case 3: a = 0 ; b = 5
max(a,b) = 5
min (a,b) = 0
(a + b + |a - b|)/2 = (0 + 5 + 5)/2 = 5
Hence, max(a, b) = (a + b + |a - b|)/2
Case 4: a = 5 ; b = 6
max(a,b) = 6
min (a,b) = 5
(a + b + |a - b|)/2 = (5 + 6 + 1)/2 = 6
Hence, max(a, b) = (a + b + |a - b|)/2
We can conclude II holds.
2) Logical Approach
|a - b| → Represents the distance between the numbers 'a' and 'b'.
Assume between 'a' and 'b', 'a' is the smaller number. Therefore, max(a,b) = b
In this case, if we add |a - b|, i.e. the distance between 'a' and 'b', to 'a', the resultant number is 'b'.
Therefore, a + |a-b| = b
(a + b + |a - b|)/2 = (b + b)/2 = b
max(a, b) is always equals to (a + b + |a - b|)/2
We can conclude II holds.
Between B and C ⇒ We can eliminate C.
Option B
