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a is the maximum integer and b is the minimum integer satisfying

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a is the maximum integer and b is the minimum integer satisfying

by Max@Math Revolution » Wed Mar 25, 2020 5:49 am

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[GMAT math practice question]

a is the maximum integer and b is the minimum integer satisfying \(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}<5\) What is the value of \(a^2+b^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6
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Re: a is the maximum integer and b is the minimum integer satisfying

by Max@Math Revolution » Fri Mar 27, 2020 12:52 am
=>
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\) = |x - 1| + |x - 2|.

Case 1: x ≥ 1
|x - 1| + |x + 2| < 5
=> (x - 1) + (x + 2) < 5, since x – 1 ≥ 0 and x + 2 ≥ 0
=> 2x - 1 < 5
=> 2x < 4
=> x < 2
Thus, we have 1 ≤ x < 2.

Case 2: -2 ≤ x < 1
|x - 1| + |x + 2| < 5
=> -(x - 1) + (x + 2) < 5, since x – 1 < 0 and x + 2 ≥ 0
=> -x + 1 + x + 2 < 5
=> 3 < 5, which is always valid.
Thus, we have -2 ≤ x < 1.

Case 3: x < -2
|x - 1| + |x + 2| < 5
=> -(x - 1) - (x + 2) < 5, since x – 1 < 0 and x + 2 < 0
=> -2x - 1 < 5
=> -2x < -6
=> x > -3
Thus, we have -3 < x < -2.

When we combine those solutions, we have -3 < x < 2.
However, x is an integer. a, the maximum possible integer of x is 1, and b, the minimum possible integer is -2.
Then a^2 + b^2 = 1^2 + (-2)^2 = 1 + 4 = 5.

Therefore, the answer is D.
Answer: D
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