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Another Tough Absolute & Inequality

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Another Tough Absolute & Inequality

by Mo2men » Sun Nov 27, 2016 2:36 am
If |(x-3)^2+2|<|x-7|, which of the following expresses the allowable range for x?

(A) 1 < x < 4

(B) 1 < x < 7

(C) - 1 < x < 4 and 7 < x

(D) x < - 1 and 4 < x < 7

(E) - 7 < x < 4 and 7 < x

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by GMATGuruNY » Sun Nov 27, 2016 4:33 am
Mo2men wrote:If |(x-3)² + 2|<|x-7|, which of the following expresses the allowable range for x?

(A) 1 < x < 4

(B) 1 < x < 7

(C) - 1 < x < 4 and 7 < x

(D) x < - 1 and 4 < x < 7

(E) - 7 < x < 4 and 7 < x
Test x=0 in |(x-3)² + 2|<|x-7|:
|(0-3)² + 2|<|0-7|
11 < 7.
Does not work.
Eliminate any answer choice with a range that includes x=0.
Eliminate C and E.

Test x=5 in |(x-3)² + 2|<|x-7|:
|(5-3)² + 2|<|5-7|
6 < 2.
Does not work.
Eliminate any remaining answer choice with a range that includes x=5.
Eliminate B and D.

The correct answer is A.
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by flprg » Sun Nov 27, 2016 8:38 am
If you are comfortable enough with absolute values there's another way, which requires more calculus though.

First you can notice that (x-3)^2+2 is always positive, thus the absolute value is useless : you get x^2-6x+11<|x-7|

Then, you split your equation in 2 parts : when x>7 and x<7

1st case : x^2-6x+11<x-7 which gives x^2-7x+18<0 ==> No solution since x^2-7x+18 is always positive.

2nd case : x^2-6x+11<7-x which gives x^2-5x+4<0, that is (x-1)(x-4)<0, finally gives 1<x<4 hence A.
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by Mo2men » Sun Nov 27, 2016 10:52 am
flprg wrote: Then, you split your equation in 2 parts : when x>7 and x<7

1st case : x^2-6x+11<x-7 which gives x^2-7x+18<0 ==> No solution since x^2-7x+18 is always positive.
Can you elaborate more on first case? how do you find no solution?
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by Mo2men » Sun Nov 27, 2016 11:04 am
GMATGuruNY wrote:
Mo2men wrote:If |(x-3)² + 2|<|x-7|, which of the following expresses the allowable range for x?

(A) 1 < x < 4

(B) 1 < x < 7

(C) - 1 < x < 4 and 7 < x

(D) x < - 1 and 4 < x < 7

(E) - 7 < x < 4 and 7 < x
Test x=0 in |(x-3)² + 2|<|x-7|:
|(0-3)² + 2|<|0-7|
11 < 7.
Does not work.
Eliminate any answer choice with a range that includes x=0.
Eliminate C and E.

Test x=5 in |(x-3)² + 2|<|x-7|:
|(5-3)² + 2|<|5-7|
6 < 2.
Does not work.
Eliminate any remaining answer choice with a range that includes x=5.
Eliminate B and D.

The correct answer is A.
Dear Mitch,

Does the critical point method work here> if yes, what are those points??
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by flprg » Sun Nov 27, 2016 11:21 am
Mo2men wrote:
flprg wrote: Then, you split your equation in 2 parts : when x>7 and x<7

1st case : x^2-6x+11<x-7 which gives x^2-7x+18<0 ==> No solution since x^2-7x+18 is always positive.
Can you elaborate more on first case? how do you find no solution?
That's from your high school maths classes. You calculate the discriminant D = b^2-4ac (here it is (-7)^2-4*1*18). When D is negative, your quadratic expression has the sign of a (the coefficient of x^2). Here a = 1 so x^2-6x+11 is always positive, thus no solution to the equation above. Yes, I admit it is more brutal if your not familiar with these concepts.
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by coolhabhi » Sun Nov 27, 2016 12:21 pm
Mo2men wrote:If |(x-3)^2+2|<|x-7|, which of the following expresses the allowable range for x?

(A) 1 < x < 4

(B) 1 < x < 7

(C) - 1 < x < 4 and 7 < x

(D) x < - 1 and 4 < x < 7

(E) - 7 < x < 4 and 7 < x

Source: Magoosh
I tried it by plugging the answer choices
(A) 1 < x < 4 which means x is 2 or 3
When I substituted x for 2 the equation holds true. Same is the case with 3. So I knew the answer was A

But I tried the other options too.
(B) 1 < x < 7 which means x is 2 or 3 or 4 or 5 or 6
When I substitute x for 6 the equation does not hold true. So ruled out B and D.

(C) - 1 < x < 4 and 7 < x which means x is 0 or 1 or 2 or 3 or 8,9,10...
When I substitute x for 0 the equation does not hold true. So ruled out C and E.

Time taken 15 to 20 seconds.
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by GMATGuruNY » Sun Nov 27, 2016 4:02 pm
Mo2men wrote:
(A) 1 < x < 4

(B) 1 < x < 7

(C) - 1 < x < 4 and 7 < x

(D) x < - 1 and 4 < x < 7

(E) - 7 < x < 4 and 7 < x

Dear Mitch,

Does the critical point method work here> if yes, what are those points??
The CRITICAL POINTS are the values of x such that the two sides of the inequality are EQUAL.
Here, the fastest way to determine the critical points is to test the endpoints given in the answer choices:
-7, -1, 1, 4, 7.
Of these options for x, only 1 and 4 make the two sides of the inequality equal.
Since the critical points are 1 and 4, the correct answer choice must include 1 and 4 as endpoints.

The correct answer is A.
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