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25 Jun 2023, 02:06
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
42% (02:07) correct
58%
(02:03)
wrong
based on 237
sessions
History
Date
Time
Result
Not Attempted Yet
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
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19 Mar 2024, 07:26
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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
General Discussion
25 Jun 2023, 03:12
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Bunuel
Two approaches to solving this question
Approach 1: Using Number line
Let's plot x on a number line -
------------------------- x -------------------------
x - 1 will be a point one unit left of x on the number line
-------------- x-1 --------- x -------------------------
- |x| → denotes the distance of x from 0.
- |x-1| → denotes the distance of (x-1) from 0.
- |x| - |x-1| → denotes the difference of distance of x from 0 and the distance of x-1 from 0. This distance will always be equal to 1 (as x-1 lies one unit left of x).
Given:
|x| - |x-1| = |x-2|!
Therefore we can conclude that
|x-2|! = 1
a! = 1 for the below values of a
- a = 0
- a = 1
Case 1: |x-2| = 0
x = 2
Case 2: |x-2| = 1
a) x-2 = 1 → x = 3
b) x-2 = -1 → x = 1
Therefore possible values of x = {1,2,3}
Approach 2: Using even odd rule
x and x-1 represent consecutive integers. Hence the difference \(|x| - |x - 1|\) will always be odd.
Therefore |x-2|! = 1
Case 1: |x-2| = 0
x = 2
Case 2: |x-2| = 1
a) x-2 = 1 → x = 3
b) x-2 = -1 → x = 1
Therefore possible values of x = {1,2,3}
Option D
