If x 0 and x^2

Question

If x ≠ 0 and x^2 < |x| , which of the following must be true? (A) x > 1 (B) x < -1 (C) -1 < x < 1 (D) |x| > 1 (E) x = -1

Bunuel · Accepted Answer

Divide x^2 < |x| by |x| to get |x| < 1. This translates to -1 < x < 1 (note that dividing by |x| is valid because |x| is always positive).

Answer: C.

MalachiKeti · Answer

X^2<|x|

This can only be true if X must be a positive or negative fraction because for an integer except 0 or 1 or -1 this equation wont hold true as square will always be positive and greater than the Mod. For 0,-1 and 1 it will be equal to sign than the inequality.

X = -1/2
X^2 = 1/4
|x|=1/2
 X^2<|x|

Same 
 X = +1/2
X^2 = 1/4
|x|=1/2 
 X^2<|x|

C

BrushMyQuant · Answer

x^2 < |x| and x ≠ 0 As we have |x| in the equation so we will have two cases - Case 1: x > 0 => |x| = x => x^2 < x => x^2 - x < 0 => x * (x - 1) < 0 => 0 < x < 1 But condition was x > 0 => 0 < x < 1 is a SOLUTION - Case 2: x < 0 => |x| = -x => x^2 < -x => x^2 + x < 0 => x * (x + 1) < 0 => x * (x - (-1)) < 0 => -1 < x < 0 But condition was x < 0 => -1 < x < 0 is a SOLUTION => -1 < x < 1 So, Answer will be C Hope it helps! Playlist on Solved problems on Absolute Values here Watch the following video to MASTER Absolute Values https://www.youtube.com/watch?v=huFiBk8PaGY

If x 0 and x^2 < |x| , which of the following must be true?

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As we have \|x\| in the equation so we will have two cases
-Case 1: x > 0 => \|x\| = x => \(x^2 < x\) => \(x^2 - x < 0\) => x * (x - 1) < 0 => 0 < x < 1 But condition was x > 0 => 0 < x < 1 is a SOLUTION	-Case 2: x < 0 => \|x\| = -x => \(x^2 < -x\) => \(x^2 + x < 0\) => x * (x + 1) < 0 => x * (x - (-1)) < 0 => -1 < x < 0 But condition was x < 0 => -1 < x < 0 is a SOLUTION

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