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

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Is |x/x^1/2|>1 [#permalink] New post 08 May 2013, 04:22
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Question Stats:

50% (02:51) correct 50% (01:26) wrong based on 95 sessions
Is |\frac{x}{\sqrt{x}}| > 1

(1) x^2 -5x + 6 > 0
(2) |x|^2 - 5|x| + 6 > 0
[Reveal] Spoiler: OA

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Re: Is |x/x^1/2|>1 [#permalink] New post 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?
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Re: Is |x/x^1/2|>1 [#permalink] New post 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.

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Re: Is |x/x^1/2|>1 [#permalink] New post 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
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Re: Is |x/x^1/2|>1 [#permalink] New post 10 May 2013, 07:20
Expert's post
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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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Re: Is |x/x^1/2|>1 [#permalink] New post 26 Jun 2013, 07:35
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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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Re: Is |x/x^1/2|>1 [#permalink] New post 28 Aug 2014, 21:20
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Re: Is |x/x^1/2|>1   [#permalink] 28 Aug 2014, 21:20
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