Absolute value inequality: |x - 4| + |x - 3| < a, where a > 0, then...

[speedway_joe]

Hello,

I could use some help with the following multiple choice problem;

|x - 4| + |x - 3| < a, where a > 0, then ________.

A) .01 < a < 1
B) 0 < a < 1
C) 0 < a <= 1
D) a > 1

Note: I didn't know how to type a less than or equal to sign in choice C.

The key says the answer is D, but I'm not sure how they got this. The problem is from a set of
practice problems for the Indiana Academic Superbowl.

Thanks

 

[Deleted member 4993]

Hello,

I could use some help with the following multiple choice problem;

|x - 4| + |x - 3| < a, where a > 0, then ________.

A) .01 < a < 1
B) 0 < a < 1
C) 0 < a <= 1
D) a > 1

Note: I didn't know how to type a less than or equal to sign in choice C.

The key says the answer is D, but I'm not sure how they got this. The problem is from a set of
practice problems for the Indiana Academic Superbowl.

Thanks

What are your thoughts?

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[speedway_joe]

Hello,

I could use some help with the following multiple choice problem;

|x - 4| + |x - 3| < a, where a > 0, then ________.

A) .01 < a < 1
B) 0 < a < 1
C) 0 < a <= 1
D) a > 1

Note: I didn't know how to type a less than or equal to sign in choice C.

The key says the answer is D, but I'm not sure how they got this. The problem is from a set of
practice problems for the Indiana Academic Superbowl.

Thanks

I've been asked to provide my thoughts. Well, we know that
|a + b| <= |a| + |b|
So,
|(x - 4) + (x - 3)| < |x - 4| + |x - 3| < a
So,
|2x - 7| < a
So,
-a < 2x - 7 < a
(7 - a)/2 < x < (7 + a)/2
I can see by inspection that this inequality would not be true if a = 0
I can plug in various values for a, like a = 1
(7 - 1)/2 < x < (7 + 1)/2
3 < x < 4, which is true
That does not agree with the answer key.
I have found some errors in the answer key, but I'm not confident (yet) in saying that this problem has an error.

 

[Ishuda]

I've been asked to provide my thoughts. Well, we know that
|a + b| <= |a| + |b|
So,
|(x - 4) + (x - 3)| < |x - 4| + |x - 3| < a
So,
|2x - 7| < a
So,
-a < 2x - 7 < a
(7 - a)/2 < x < (7 + a)/2
I can see by inspection that this inequality would not be true if a = 0
I can plug in various values for a, like a = 1
(7 - 1)/2 < x < (7 + 1)/2
3 < x < 4, which is true
That does not agree with the answer key.
I have found some errors in the answer key, but I'm not confident (yet) in saying that this problem has an error.

Think it through: If x were more that 4, then a would have to be greater than 1 since x is more than 1 away from 3. Similarly, if x were less than three it would be more than 1 away from 4, so again a must be greater than 1. If x were between 3 and 4 the the sum of the distances between x and 3 and between x and 4 would be one. Now put that in a proof

Highlight below for more of a hint:
>>>>>>>>>
Look at three cases:
(1) x \(\displaystyle \ge\) 4: Then 1 \(\displaystyle \le\) 2x-7 = (x-3) + (x-4) = |x-3| + |x-4| < a. So, a>1
(2) x \(\displaystyle \le\) 3: Then ...
(3) 3 \(\displaystyle \le\) x \(\displaystyle \le\) 4: Then ...
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