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25 Aug 2022, 03:32
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If \(x \neq 2\) and \(\frac{|x| - 2}{|x - 2|}= 1\), then which of the following must be true?
I. \(|x| > 2\)
II. \(x^2 < 1\)
III. \(x^3 > 0\)
A. I only
B. II only
C. III only
D. I and III only
E. None
I. \(|x| > 2\)
II. \(x^2 < 1\)
III. \(x^3 > 0\)
A. I only
B. II only
C. III only
D. I and III only
E. None
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26 Apr 2023, 01:43
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Official Solution:
If \(x \neq2\) and \(\frac{|x| - 2}{|x - 2|} = 1\), then which of the following must be true?
I. \(|x| > 2\)
II. \(x^2 < 1\)
III. \(x^3 > 0\)
A. I only
B. II only
C. III only
D. I and III only
E. None of the above
Note that \(\frac{|x| - 2}{|x - 2|} = 1\) implies that both \(x\) and \(x - 2\) must be positive. In this case, the left-hand side becomes \(\frac{x - 2}{x - 2}\), which equals 1. For both \(x\) and \(x - 2\) to be positive, \(x\) must be greater than 2. Therefore, \(\frac{|x| - 2}{|x - 2|} = 1\) implies that \(x > 2\).
Alternatively, we can derive the same conclusion using the following algebraic manipulations:
Cross-multiply:
Square both sides (we can do this safely since both sides of the equation are positive):
The above implies that \(x - 2 > 0\) (recall that \(|a| = a\) when \(a \geq 0\)).
Now, \(x - 2 > 0\) means that \(x > 2\).
So, the question essentially asks if \(x > 2\), then which of the following must be true?
I. \(|x| > 2\). Since the possible values of \(x\) are such that \(x > 2\), \(|x| > 2\) will always be true. So, this option is always true.
II. \(x^2 < 1\). This option is never true because if \(x > 2\), then \(x^2\) will be greater than 1, not less than 1.
III. \(x^3 > 0\). Since \(x\) is positive (\(x > 2\)), the cube of \(x\) will always be greater than 0. Thus, this option is always true.
Answer: D
Check other similar questions from Trickiest Inequality Questions Type: Confusing Ranges (part of our Special Questions Directory).
Hope it helps.
If \(x \neq2\) and \(\frac{|x| - 2}{|x - 2|} = 1\), then which of the following must be true?
I. \(|x| > 2\)
II. \(x^2 < 1\)
III. \(x^3 > 0\)
A. I only
B. II only
C. III only
D. I and III only
E. None of the above
Note that \(\frac{|x| - 2}{|x - 2|} = 1\) implies that both \(x\) and \(x - 2\) must be positive. In this case, the left-hand side becomes \(\frac{x - 2}{x - 2}\), which equals 1. For both \(x\) and \(x - 2\) to be positive, \(x\) must be greater than 2. Therefore, \(\frac{|x| - 2}{|x - 2|} = 1\) implies that \(x > 2\).
Alternatively, we can derive the same conclusion using the following algebraic manipulations:
- \(\frac{|x| - 2}{|x - 2|} = 1\)
Cross-multiply:
- \(|x| - 2= |x - 2|\);
\(|x|= |x - 2|+2\)
Square both sides (we can do this safely since both sides of the equation are positive):
- \(x^2 = x^2 - 4x + 4 + 4|x-2| + 4\)
\(x - 2 = |x-2|\)
The above implies that \(x - 2 > 0\) (recall that \(|a| = a\) when \(a \geq 0\)).
Now, \(x - 2 > 0\) means that \(x > 2\).
So, the question essentially asks if \(x > 2\), then which of the following must be true?
I. \(|x| > 2\). Since the possible values of \(x\) are such that \(x > 2\), \(|x| > 2\) will always be true. So, this option is always true.
II. \(x^2 < 1\). This option is never true because if \(x > 2\), then \(x^2\) will be greater than 1, not less than 1.
III. \(x^3 > 0\). Since \(x\) is positive (\(x > 2\)), the cube of \(x\) will always be greater than 0. Thus, this option is always true.
Answer: D
Check other similar questions from Trickiest Inequality Questions Type: Confusing Ranges (part of our Special Questions Directory).
Hope it helps.
11 May 2024, 06:53
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I am confused here. It says that |x|>2, that means x can have values that are less than -2 or greater than 2. If that's the case, consider x=-5. Will the equation stand with x=-5 ? I don't think that will.
Then how come |x|>2 becomes the right answer.
Bunuel can you please check this once ?
Then how come |x|>2 becomes the right answer.
Bunuel can you please check this once ?
