Is |x/x^1/2|>1

sorry Zarrolou,

as you dint liked it..
this question was framed by me.

SKM

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.

(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)

Is (x/√x)>1

First things first, x cannot be negative or zero, as you cannot take the square root of a negative number, nor can x be = ) as that would leave ) in the denominator which isn't valid.

(1) x^2 -5x + 6 > 0
(x - 2)*(x - 3) > 0
x>3 OR x<2
If x>3 then (x-2)*(x-3) is (+)*(+) which is greater than zero.
if x<2 then (x-2)*(x-3) is (-)*(-) which is positive thus greater than zero.

So
x>3
(x/√x)>1
(3/√3)>1 TRUE

x<1
(1/√1)>1 FALSE (1/√1 =1)
INSUFFICIENT

(2) |x|^2 - 5|x| + 6 > 0
As we determined in the stem, x cannot be negative nor can it be zero. Therefore, x must be positive.
|x|^2 - 5|x| + 6 > 0
(x)^2 - 5(x) + 6 > 0
x^2-5x +6 > 0
(x-2)*(x-3) > 0
x>3 OR x<2
If x>3 then (x-2)*(x-3) is (+)*(+) which is greater than zero.
if x<2 then (x-2)*(x-3) is (-)*(-) which is positive thus greater than zero.

We're left with the same information we determined in #1)
INSUFFICIENT

(E)

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