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Is x2 greater than x ?

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by Reshu » Fri Nov 06, 2015 4:01 pm
late4thing wrote:Is x2 greater than x ?

(1) x2 is greaer than 2x.
(2) 2x2 is greater than x.
My approach on this would be:

1. x2 is greater than 2x

Using plug in lets say x=-1/2, then x2 = 1/4 and 2x =-1

Here, x2 > 2x (1/4>-1) and is x2>x? (Yes).

Test another situation, x=3, x2=9 and 2x=6, so, x2>2x and x2 > x (Yes)
To test any odd scenario, lets say x= root of 2, then, x2=2, 2x=2root2, Is x2>x? (Yes!)

Note, We can't use x=0, 1, 1/2, 1/3, 3/2 etc cause it will not satisfy condition 1.

Therefore, answer could be either A or D. Lets test condition 2 now.

2. 2x2 is greater than x

Lets say x=-1/2, then, x2 = 1/4 and 2x2 =1/2. since, 1/2 > -1/2 so it satisfies 1 and Is x2> x? (Yes)

Now, if x=1, then 2x2 =2, and 2>1 which satisfies condition 2, but is x2 > x? No! (cause x2=x=1).

So , the answer is A

Hope it was helpful! :)[spoiler][/spoiler]
Reshu

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by theCEO » Fri Nov 06, 2015 8:40 pm
late4thing wrote:Is x2 greater than x ?

(1) x2 is greaer than 2x.
(2) 2x2 is greater than x.

Please explain using the Critical Points method.
1) x^2 > 2x
x^2 - 2x > 0
x(x - 2) > 0

a x b = positive when a and b are either negative or positive
therefore x and x-2 are either positive or negative

when x and x-2 are positive: x>2
when x and x-2 are negative: x<0

when x>2: is x^2>x? if x = 3, 3^2>3 is true,
when x<0: is x^2>x? if x = -1/2, 1/4>-1/2 is true,
both choices give the same answer, so statement is sufficent

2) 2x^2 > x
2x^2 - x > 0
x (2x - 1) > 0

a x b = positive when a and b are either negative or positive
therefore x and 2x-1 are either positive or negative
when x and 2x-1 are positive: x>1/2
when x and 2x-1 are negative: x<0

when x>1/2 is x^2>2x? if x = 1; is 1>2? no
when x<0 is x^2>2x? is x = -1, is 1>-2? yes
both choices give different answer, so statement is not sufficent

ans = a
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