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If \(x\) is an integer, how many solutions exist for the equation \(|x|- |x - 1| = |x-2|!\)?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
19 Mar 2024, 07:25
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Official Solution:
If \(x\) is an integer, how many solutions exist for the equation \(|x|- |x - 1| = |x-2|!\)?
A. 0
B. 1
C. 2
D. 3
E. 4
The equation may appear daunting at first glance, yet, a closer examination reveals a simpler path to solution. Notice that the left-hand side invariably results in an odd number (when \(x\) is odd, \(x - 1\) becomes even, and their difference is odd; similarly, if \(x\) is even, then \(x - 1\) is odd, resulting in an odd difference again). On the other side, the right-hand side represents the factorial of a non-negative integer, which is only odd for the cases of \(0! = 1\) and \(1! = 1\). For any other number, the factorial becomes even, as it includes a factor of 2. Consequently, we only need to consider those values of \(x\) for which \(|x-2|! = odd = 1\).
\(|x-2|! = 1\), when \(x-2 = -1\), \(x-2 = 0\), and \(x-2 = 1\), hence when \(x = 1\), \(x = 2\), and \(x = 3\).
Substituting these values into the equation shows that all of them satisfy it. Thus, the given equation has three solutions.
Answer: D
If \(x\) is an integer, how many solutions exist for the equation \(|x|- |x - 1| = |x-2|!\)?
A. 0
B. 1
C. 2
D. 3
E. 4
The equation may appear daunting at first glance, yet, a closer examination reveals a simpler path to solution. Notice that the left-hand side invariably results in an odd number (when \(x\) is odd, \(x - 1\) becomes even, and their difference is odd; similarly, if \(x\) is even, then \(x - 1\) is odd, resulting in an odd difference again). On the other side, the right-hand side represents the factorial of a non-negative integer, which is only odd for the cases of \(0! = 1\) and \(1! = 1\). For any other number, the factorial becomes even, as it includes a factor of 2. Consequently, we only need to consider those values of \(x\) for which \(|x-2|! = odd = 1\).
\(|x-2|! = 1\), when \(x-2 = -1\), \(x-2 = 0\), and \(x-2 = 1\), hence when \(x = 1\), \(x = 2\), and \(x = 3\).
Substituting these values into the equation shows that all of them satisfy it. Thus, the given equation has three solutions.
Answer: D
01 Jul 2024, 14:24
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I think the anser should be two solutions, since the factorial of of negative numbers (-1) is not defined.
