rahulvsd wrote:Is x negative?
(1) 2x > x^2
(2) x < 1
[spoiler]OA: A. I have a doubt in how we arrive at the answer from statement 1.
Simplifying we get,
x^2 - 2x <0
x(x-2)<0
x<0 x-2<0
x<0 x<2. What am I doing wrong here?[/spoiler]
I received a PM requesting that I show how statement 1 can be handled algebraically.
Statement 1: 2x > x^2.
x² - 2x < 0.
x(x-2) < 0.
The critical points are x=0 and x=2.
These are the only values where the lefthand side is equal to 0.
When x is any other value, either 2x > x² or 2x < x².
To determine the range of x, test one value to the left and right of each critical point.
Plug x = -1 into 2x > x²:
-2 > 1.
Doesn't work.
x<0 is NOT part of the range.
We can stop here.
Since x<0 does not satisfy 2x > x², we know that x is not negative.
SUFFICIENT.
If we needed to determine the full range of x, we would proceed as follows:
Plug x=1 into 2x > x²:
2 > 1.
This works.
0<x<2 is part of the range.
Plug x=3 into 2x > x²:
6 > 9.
Doesn't work.
x>2 is NOT part of the range..
The only range that satisfies 2x > x² is 0<x<2.
Statement 2: x<1.
Since could be negative, 0, or a positive fraction, INSUFFICIENT.
The correct answer is
A.
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