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Re: Is |x/x^1/2|>1 [#permalink]
08 May 2013, 06:16

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

the function is not defined for x<0 because in \sqrt{x} x cannot be <0

so the question is : Is \frac{x}{\sqrt{x}}> 1 or \sqrt{x}>1, x>1?

(1) x^2 -5x + 6 > 0 x>3 or x<2. Not sufficient to say that x>1

(2) |x|^2 - 5|x| + 6 > 0 Because x cannot be negative this becomes x^2 -5x + 6 > 0, exactly the same as (1). Bad question IMO Of course is not sufficient

Because 1 and 2 are the same, together they add no new info. E What is the source? _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Is |x/x^1/2|>1 [#permalink]
08 May 2013, 06:34

Great question.. Solve the the stem...the equation boils down to Mod(Sqr root (x))>1 Using the statement 1 u can get x<2 or x>3 ......So substitute values and evaluate a yes no condition...u can get both the answers...

Using statement 2 ...u end up getting values for x that are +ve values less than 2 and + values greater than 3.... So here is a catch...u have X = 1 hence square root of 1 is 1 itself.....combining both the situation remains the same as in 2nd statement and hence the answer is E.

Re: Is |x/x^1/2|>1 [#permalink]
08 May 2013, 06:50

Zarrolou wrote:

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

the function is not defined for x<0 because in \sqrt{x} x cannot be <0

so the question is : Is \frac{x}{\sqrt{x}}> 1 or \sqrt{x}>1, x>1?

(1) x^2 -5x + 6 > 0 x>3 or x<2. Not sufficient to say that x>1

(2) |x|^2 - 5|x| + 6 > 0 Because x cannot be negative this becomes x^2 -5x + 6 > 0, exactly the same as (1). Bad question IMO Of course is not sufficient

Because 1 and 2 are the same, together they add no new info. E What is the source?

sorry Zarrolou,

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

SKM _________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

Re: Is |x/x^1/2|>1 [#permalink]
10 May 2013, 07:20

Expert's post

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.

Re: Is |x/x^1/2|>1 [#permalink]
26 Jun 2013, 07:35

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)

gmatclubot

Re: Is |x/x^1/2|>1
[#permalink]
26 Jun 2013, 07:35