Overthinking it

We know that x^2 is always positive or 0. If 2x is greater than x^2, then x must be positive. If we try x=1/2:

2*(1/2) > (1/2)^2

1 > 1/4

It works. If we trie x=-1/2:

2*(-1/2) > (-1/2)^2

-1 > 1/4

which is not true.

Thanks for the reply Bill. But what am I doing wrong in the simplification of Statement 1? Can you help me with that.

Agree the logic you pointed out is easier, but only want to know what's wrong in my method?

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]

Hey man!

From your rearrangement:
x^2 - 2x <0
x(x-2)<0 ; this statement is correct
x<0 x-2<0 ; now this is missing a crucial construct
x<0 x<2.

Let's restart:
x(x-2) < 0; this means (x < 0) OR (x-2< 0). Think of it. if x<0 AND x-2<0 that means both of the terms are negative and a multiplication of two negative elements is positive; hence your initial inequality is invalid. Now we know that only one statement is right (i.e. : either (x<0) OR (x-2)<0)

Now let's check one by one:
Case 1: x<0 implies that x-2<0 as well.. oops.. this case is not right!
Case 2: x-2<0 --> x<2; this opens up 3 new cases .. (dammit..)
case2a: x<0 ; we know this is not true..
case2b: x = 0; if x = 0 then x*(x-2) = 0; since this contradicts the initial statement, its out.
case2c 0<x<2 ; let's take a number in this range.. x = 0.5 ; 0.5 * (0.5-2) = -0.75; since -0.75<0 this holds.. finally.. we nailed it

case2c proves that yes, x is positive. in fact we can even say that it is between 0 and 2 (exclusive)

But as Bill said it.. now this is over thinking it

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.