Is |x| ≤ 4 ? |x + 2| + |x − 4| 2|x − 1| |(x + 4)(x − 3)| 8

Question

Is |x| ≤ 4 ?

(1) |x + 2| + |x − 4| > 2|x − 1|
(2) |(x + 4)(x − 3)| > 8

Bunuel · Accepted Answer

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chetan2u · Answer

ans A.. 1. |x + 2| + |x − 4| > 2|x − 1| statement I tells us .. a) the value of left side will remain SAME when x is between -2 and 4 and that is |-2|+|4|=6... here RHS is 2|x-1| is always less than 6.. b) Any value above 4 and below -2 will have similar effect on LHS and RHS as the increase / decrease will have SAME effect on both sides since there are 2xs on each side.. so -2 8 when x = 0, |(x + 4)(x − 3)| = 4*3=12 when x = 10, |(x + 4)(x − 3)| = 14*3 so |x| ≤ 4 is NOT necessary insuff A

rahul16singh28 · Answer

Hi  chetan2u ,

Please correct me if I am wrong.

For Statement I, the Range should be  -2≤x<4  but not  -2≤x≤4  as you have mentioned above. When  x = 4,   L.H.S = R.H.S . So, Can we say that  -2≤x<4  is a sub-set of  x≤4 .?? Please advise.

chetan2u · Answer

Hi..

The actual range is -2<x<4, as at -2 and 4 both sides are equal to 6..
But still it is a subset of |x|<=4..
|x|<=4 means -4<=x<=4...
-2<x<4 is a subset of -4<=x<=4..
And will remain A

gmatexam439 · Answer

Hello chetan2u , I understand that by solving statement 1, we get 1<=x<4; sufficient However, please point out the mistake in my statement 2 analysis: |x^2+x-12| > 8 if x^2+x-12 > 0; then --> (x+5)(x-4)>0 Thus range of x is --> x>-5 and x>4 if both are +ve x<-5 and x<4 if both are -ve We get, x<-5 and x>4 combining above statements -------------------------B if x^2+x-12 < 0; then --> x^2+x-4<0 (x+2.5)(x-1.5)<0 Thus range of x is --> x>-2.5 and x<1.5 if (x+2.5)>0 x<-2.5 and x>1.5 if (x+2.5)<0 We get, x<-2.5 and x>1.5 combining above statements ----------------------A Combining A and B we get x<-5 and x>4 --- which is sufficient to answer the question. Please point out where am I going wrong? Regards

akshata19 · Answer

I find this question tough. 
It would be very helpful if some can share a trick, how to select numbers to plug in and check range fast

Posted from my mobile device

zamiraolortegui · Answer

Please can you explain me the reasoning to analyze this type of questions? How did you know that for (1) the value of left side will remain SAME when x is between -2 and 4 and that is |-2|+|4|=6...?

Thank you

Mo2men · Answer

I will use other visual approach (1) |x + 2| + |x − 4| > 2|x − 1| Critical points -2, 1, 4 Representing on number line -----Invalid---- -2 -- +++ ---- 1 ---- +++ ------- 4 ----Invalid-------- Now let's examine each region x\leq{-2} : Let x = -10..........|-8| + |-14| > 2|-11|..........22 > 22.......Invalid.............Hence NO number satisfy the region. -2 2|-1|..........6 > 4.........Hence every Number satisfies the region...........Answer Is Yes 1 2|2|..........6 > 4.........Hence every Number satisfies the region............Answer Is Yes x\geq{4} Let x = 10..........|12| + |6| > 2|-9|..........18 > 18.......Invalid.............Hence NO number satisfy the region. From above: Every x in region -2 8 critical points: -4 & 3 ------ ++++ ------ -4 -- ++++ ------- 3 ---- +++ -------- x\leq{-4} : Let x = -10..........|6 * -13 | > 8....................Answer is NO -4 8........................Answer is Yes We can stop here as two different answers but I will continue the last region for clarification X >3 Let x = 10..........|14* 7 | > 8......................Answer is NO. Insufficient Answer: A

analytica233 · Answer

Here is my approach: |x| ≤ 4 ? <=> -4 ≤ x ≤ 4 ? (1) |x + 2| + |x − 4| > 2|x − 1| <=> |x + 2| + |x - 4| > |2x - 2|. Notice that x + 2 + x - 4 = 2x - 2 --> |x + 2| + |x - 4| > |(x + 2) + (x - 4)|. This means that (x + 2) and (x - 4) must have opposite signs. So (x + 2)(x - 4) < 0 --> -2 < x < 4. SUFFICIENT. (2) |(x + 4)(x − 3)| > 8 (2) holds true when x = 0 or x = 10. So x can be in the range -4 ≤ x ≤ 4 or not. INSUFFICIENT. Answer: (A)

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Is |x| ≤ 4 ? |x + 2| + |x − 4| > 2|x − 1| |(x + 4)(x − 3)| > 8

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