Critical points method for inequality Is x^2 > 1/x?

[Mo2men]

Is x^2 > 1/x?

(1) x^2>x

(2) 1>1/x

Source: Math rev
OA: D


How can we critical point method in Statement 2???

[Brent@GMATPrepNow]

Is x^2 > 1/x?

(1) x^2 > x

(2) 1 > 1/x

Target question: Is x² > 1/x ?

Statement 1: x² > x
First off, this inequality tells us that x ≠ 0
Second, we can conclude that x² is POSITIVE.
So, we can safely divide both sides of the inequality by x² to get: 1 > 1/x
If 1 > 1/x, then there are two possible cases:
Case a: x > 1. If x is a positive number greater than 1, then 1/x will definitely be less than 1.
Case b: x is negative. If x is negative, then 1/x will definitely be less than 1.

IMPORTANT: So how do these two cases affect the answer to the target question? Let's find out.
Case a: If x > 1, then x² is greater than 1, AND 1/x is less than 1. This means x² > 1/x
Case b: If x is negative, then x² is positive, AND 1/x is negative. This means x² > 1/x
Perfect - in both cases, we get the SAME answer to the target question
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: 1 > 1/x
Notice that this inequality is the SAME as the inequality derived from statement 1 (we got 1 > 1/x)
Since we already saw that statement 1 is sufficient, it must be the case that statement 2 is also SUFFICIENT

Answer: D

Cheers,
Brent

[GMATGuruNY]

Is x^2 > 1/x?

(1) x^2>x

(2) 1>1/x

CRITICAL POINTS occur where an inequality is UNDEFINED and where the two sides are EQUAL.
To determine the valid range for an inequality, test one value to the left and right of each critical point.

Question stem: x² > 1/x ?
Here, the critical points are x=0 (in which case 1/x is undefined) and x=1 (in which case the two sides are equal).
If we test x=-1 (a value to the left of 0), x=1/2 (a value between 0 and 1) and x=2 (a value to the right of 1), only x=-1 and x=2 satisfy x² > 1/x.
Implication:
The valid ranges for x² > 1/x are x<0 and x>1.
Question stem, rephrased:
Is x<0 or x>1?

Statement 1: x² > x
Here, the critical points are x=0 and x=1, since each of these values makes the two sides of the inequality equal.
If we test x=-1 (a value to the left of 0), x=1/2 (a value between 0 and 1) and x=2 (a value to the right of 1), only x=-1 and x=2 satisfy x² > x.
Implication:
The valid ranges for x² > x are x<0 and x>1.
Thus, the answer to the question stem is YES.
SUFFICIENT.

Statement 2: 1 > 1/x
Here, the critical points are x=0 (in which case 1/x is undefined) and x=1 (in which case the two sides are equal).
If we test x=-1 (a value to the left of 0), x=1/2 (a value between 0 and 1) and x=2 (a value to the right of 1), only x=-1 and x=2 satisfy 1 > 1/x.
Implication:
The valid ranges for 1 > 1/x are x<0 and x>1.
Thus, the answer to the question stem is YES.
SUFFICIENT.

The correct answer is D.