shaileshmishra wrote:

Is |\frac{x}{\sqrt{x}}| > 1

(1) x^2 -5x + 6 > 0

(2) |x|^2 - 5|x| + 6 > 0

First thing to note is that since root x is in the denominator, x must be positive.

When is

|\frac{x}{\sqrt{x}}| > 1?

When x is greater than

\sqrt{x}. When does that happen? When x is greater than 1.

So we basically need to answer whether x is greater than 1 or not.

(1) x^2 -5x + 6 > 0

(x - 2)(x - 3) > 0

This tells us that either x < 2 or x > 3. Not sufficient alone.

If you are not sure how we got this, check:

http://www.veritasprep.com/blog/2012/06 ... e-factors/(2) |x|^2 - 5|x| + 6 > 0

Since x must be positive, this boils down to x^2 - 5x + 6 > 0

This is the same as statement 1. Hence not sufficient alone.

Both together are essentially the same statement so they are not sufficient together.

Answer (E)

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