a is the maximum integer and b is the minimum integer satisfying (x-1)

Question

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

A.  2 
 
B.  3 
 
C.  4 
 
D.  5 
 
E.  6

MathRevolution · Accepted Answer

=> \sqrt{(x-1)^2}+\sqrt{(x+2)^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

ArunSharma12 · Answer

First I tried to do it the harder way, which is squaring both sides and solving for x. It is time consuming and does not guarantees that one would get the correct values for a and b.
In my second attempt, looking at options we can see that a,b can not be any integer  \ge{3} .
that means, a and b will fit this equation  -2\le b<a\le{2} 
checking for maximum value, for a =2 
 \sqrt{(2-1)^2}+\sqrt{(2+2)^2} = 5 ; this does not satisfy the condition, so a =1.
checking for minimum value, for b = -2
 \sqrt{(-2-1)^2}+\sqrt{(-2+2)^2} = 3 ; this satisfies the condition, so b =-2.

a^2+b^2 = (1)^2+(-2)^2  = 5
Ans: D

nick1816 · Answer

Case 1 x<-2 \sqrt{(x-1)^2} = |x-1| = 1-x \sqrt{(x+2)^2} = |x+2| = -x-2 1-x -x-2 <5 x >-3 Case 2 1> x ≥ -2 \sqrt{(x-1)^2} = |x-1| = 1-x \sqrt{(x+2)^2} = |x+2| = x+2 1-x+x+2 <5 3 < 5 (always true) Case 3 x≥1 \sqrt{(x-1)^2} = |x-1| = x-1 \sqrt{(x+2)^2} = |x+2| = x+2 x-1+x+2 <5 x < 2 Required range of x = -3 < x < 2 b= -2; a =1 a^2 + b^2 = 1+4 = 5

nisthagupta28 · Answer

how do we know when to put equal to sign while checking different range cases?

ThatDudeKnows · Answer

Both addends are positive. Let's just try a few numbers.

Let's look for the maximum value first. 
Does 5 work? No, the thing on the right is too big on its own.
How about 3? Still too big.
2? Now we have 1+4. That equals 5, but we need less than 5. 
1? Aha! Okay, that's the max.

From the above, we have a pretty good idea that we are looking for +/-1 under one of them and +/-4 under the other as our threshold case. 
-3? Nah, that equals 5.
-2? Okayyyy!

1^2 + -2^2 = 5

Answer choice D.

Kimberly77 · Answer

Thanks  nick1816  for explanation. So after the calculation and final Required range of x = -3 < x < 2. To clarify so we can just take a and b value from the original question as it is?

b= -2; a =1

Lastly is there another faster way, as the calculation for every possible value is difficult to achieve under 2 min and can be error prone when is under exam condition? Thanks

Kimberly77 · Answer

Thanks  ThatDudeKnows . when x=1 satisfy both square root <5. Not quite sure below parts. Could you kindly explain a bit more please thanks?

one of them and +/-4 under the other as our threshold case. 
-3? Nah, that equals 5.
-2? Okayyyy!

ThatDudeKnows · Answer

Kimberly77

I just meant that in testing numbers to try to find the maximum value, when we tried 2, we ended up with 1+4<5 and knew we needed something smaller. The odds are good (not definite, but this is how test-writers tend to think) that the minimum value will follow the same structure...that is either 1+4<5 or 4+1<5. When we plug in -3, we do, indeed, get one of those: 4+1<5. But 4+1=5, so we need something closer to zero. -2 it is!

Does that make sense?

Kimberly77 · Answer

Got it thanks  ThatDudeKnows  and thanks for your reply. 
so -2 <5 works.
But how do we decide a and b value from there?

b= -2; a =1

ThatDudeKnows · Answer

The original question stem tells us that a is the maximum and b is the minimum.

Posted from my mobile device

Kimberly77 · Answer

Thanks  ThatDudeKnows  for your prompt reply. You made a good point. 
In that case, could we say that the testing of -2 is even irrelevant here now then?

ThatDudeKnows · Answer

Definitely not irrelevant!! We need the maximum and minimum possible values. The full list of possibilities is -2, -1, 0, 1. So the maximum is 1 and the minimum is -2.

Kimberly77 · Answer

Brilliant and crystal clear now, thanks ThatDudeKnows for your great insights always

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a is the maximum integer and b is the minimum integer satisfying (x-1)

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a is the maximum integer and b is the minimum integer satisfying (x-1)

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