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Absolute Value question

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Absolute Value question

by Nupur.nk » Fri Feb 14, 2014 3:55 pm
If x is a # such that -2 ≤ x ≤ 2, which of the
following has the largest possible absolute value?

A. 3x-1
B. X2-x
C. 3-x
D. x-3
E. x2+1

Kaplan's solution states that (C) and (D) can be eliminated because they are negatives of each other
and have the same absolute value.

What does the above statement mean? How does being negative of each other guarantee the same absolute value?
Can someone give an intuitive understanding of the above.
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by [email protected] » Fri Feb 14, 2014 4:15 pm
Hi Nupur.nk,

This question gives us a range of values to work with (-2 to +2, inclusive) and asks for the answer that would provide the largest possible ABSOLUTE value. This type of set-up certainly makes me think that plugging -2 into one or more of the answers will be required (as a way to prove which option has the highest possible absolute value.

The largest possible absolute value for each answer would be:

A: when x = -2, |3(-2) -1| = 7
B: when x = -2, |(-2)^2 -(-2)| = 6
C: when x = -2, |3 - (-2)| = 5
D: when x = -2, |-2 -3| = 5
E: when x = 2 or -2, |2^2 + 1| = 5

Final Answer: A

The logic behind eliminating C and D is that when you plug ANY value of x into those options and then take the absolute value of the result, you will get the same number. Thus, neither could possibly be "the largest result."

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by GMATGuruNY » Sat Feb 15, 2014 4:01 am
Nupur.nk wrote:If x is a # such that -2 ≤ x ≤ 2, which of the
following has the largest possible absolute value?

A. 3x-1
B. X2-x
C. 3-x
D. x-3
E. x2+1
|a-b| = the DISTANCE between a and b on the number line.

C: |3-x| = the distance between 3 and x.
D: |x-3| = the distance between x and 3.
Since the distance is the same in each case -- and both answer choices can't be correct -- eliminate C and D.

A: |3x-1| = the distance between 3x and 1.
Since -2≤x≤2, the greatest possible distance between 3x and 1 occurs when x=-2, implying that 3x=-6:
|-6-1| = 7.

B: |x²-x| = the distance between x² and x
Since -2≤x≤2, and the square of a value cannot be negative, 0≤x²≤4.
Thus, the greatest possible distance between x² and x occurs when x=-2, implying that x²=4:
|4 - (-2)| = 6.
Since the maximum possible distance in A is greater than the maximum possible distance in B, eliminate B.

E: |x²+1| = |x² - (-1)| = the distance between x² and -1.
Since -2≤x≤2, and the square of a value cannot be negative, 0≤x²≤4.
Thus, the greatest possible distance between x² and -1 occurs when x=±2, implying that x²=4:
|4 - (-1)| = 5.
Since the maximum possible distance in A is greater than the maximum possible distance in E, eliminate E.

The correct answer is E.
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