DS inequality !

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DS inequality !

by simpu » Sat Jul 24, 2010 7:20 pm
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

I could not work this out , does anyone know how to go about it?

OA C
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by Rahul@gurome » Sat Jul 24, 2010 8:38 pm
First consider (1) alone.
Square both sides of the equation.
(x+1)^2 = 4*(x-1)^2.
Or x^2+2x+1 = 4x^2 - 8x + 4.
Or 3x^2 - 10x + 3 = 0.
Or (3x - 1)(x - 3) = 0.
Or x = 3 or x = 1/3.
We do not get unique value of x.
So (1) alone is not sufficient.

Next consider (2) alone.
It just implies x is not equal to 3.
But we cannot say whether lxl < 1 or not.
So (2) alone is not sufficient.

Next combine both the statements together and check.
From (1) we have x = 3 or x = 1/3.
Since x is not equal to 3 by (2), x = 1/3 and lxl = 1/3 which is < 1.
So the answer to the question is yes.

The correct answer is hence (C).
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by simpu » Tue Jul 27, 2010 10:32 pm
wow thank you

i could not figure out that both sides of the equation had to be squared!
it seems easy now :)
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by gmatmachoman » Wed Jul 28, 2010 3:32 am
simpu wrote:wow thank you

i could not figure out that both sides of the equation had to be squared!
it seems easy now :)
https://www.beatthegmat.com/is-x-1-1-x-1 ... 58763.html