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26 Apr 2023, 01:42
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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
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
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:
\(\frac{|x| - 2}{|x - 2|} = 1\)
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
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
02 Jun 2024, 16:20
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The answer is E because
I. |x|>2
assume x = -4 which satisfies the above equation but |x|−2| / |x−2|= 1/3
III. x^3>0
X=1 satisfies the equation but |x|−2| / |x−2|= -1
I. |x|>2
assume x = -4 which satisfies the above equation but |x|−2| / |x−2|= 1/3
III. x^3>0
X=1 satisfies the equation but |x|−2| / |x−2|= -1
