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If x>0 Is (x)^1/2>x
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Updated on: 02 Sep 2013, 07:55
Updated on: 02 Sep 2013, 07:55
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Difficulty:
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Question Stats:
73% (01:36) correct
27%
(01:46)
wrong
based on 192
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If \(x>0\), Is \(\sqrt{x} > x\) ?
(1) x is not equal to 1
(2) \(x\sqrt{x} >x^2\)
My Reasoning:
(1) x is not equal to 1
(2) \(x\sqrt{x} >x^2\)
My Reasoning:
Request you to please post your reasoning for this. I know, putting values can be a good option, however I want to know where I am getting wrong.
1) Ofcourse, statement 1 is not sufficient.
2) \(x\sqrt{(x)}> x^2\)
\(x\sqrt{(x)} -x^2 > 0\)
Taking x common
\(x(\sqrt{(x)}-x)>0\)
it implies
eitherx>0 or\(\sqrt{(x)} > x\)
My question is how can we infer that \(\sqrt{(x)} > x\) is true.
Please explain
1) Ofcourse, statement 1 is not sufficient.
2) \(x\sqrt{(x)}> x^2\)
\(x\sqrt{(x)} -x^2 > 0\)
Taking x common
\(x(\sqrt{(x)}-x)>0\)
it implies
eitherx>0 or\(\sqrt{(x)} > x\)
My question is how can we infer that \(\sqrt{(x)} > x\) is true.
Please explain
Originally posted by imhimanshu on 02 Sep 2013, 07:54.
Last edited by Bunuel on 02 Sep 2013, 07:55, edited 1 time in total.
Last edited by Bunuel on 02 Sep 2013, 07:55, edited 1 time in total.
Edited the question.
If x>0 Is (x)^1/2>x
[#permalink]
02 Sep 2013, 08:03
02 Sep 2013, 08:03
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Expert Reply
imhimanshu
If \(x>0\), Is \(\sqrt{x} > x\) ?
(1) x is not equal to 1
(2) \(x\sqrt{x} >x^2\)
My Reasoning:
(1) x is not equal to 1
(2) \(x\sqrt{x} >x^2\)
My Reasoning:
Request you to please post your reasoning for this. I know, putting values can be a good option, however I want to know where I am getting wrong.
1) Ofcourse, statement 1 is not sufficient.
2) \(x\sqrt{(x)}> x^2\)
\(x\sqrt{(x)} -x^2 > 0\)
Taking x common
\(x(\sqrt{(x)}-x)>0\)
it implies
eitherx>0 or\(\sqrt{(x)} > x\)
My question is how can we infer that \(\sqrt{(x)} > x\) is true.
Please explain
1) Ofcourse, statement 1 is not sufficient.
2) \(x\sqrt{(x)}> x^2\)
\(x\sqrt{(x)} -x^2 > 0\)
Taking x common
\(x(\sqrt{(x)}-x)>0\)
it implies
eitherx>0 or\(\sqrt{(x)} > x\)
My question is how can we infer that \(\sqrt{(x)} > x\) is true.
Please explain
If \(x>0\), Is \(\sqrt{x} > x\) ?
(1) x is not equal to 1. Not sufficient.
(2) \(x\sqrt{x} >x^2\). Since x is positive we can safely reduce by it: \(\sqrt{x} >x\). Sufficient.
Answer: B.
P.S. The question asks: is \(\sqrt{x} > x\)? Since both parts are positive, then we can sagely square the whole inequality: is \(x>x^2\)? --> reduce by positive x: is \(1>x\)?
Re: If x>0 Is (x)^1/2>x
[#permalink]
02 Sep 2013, 08:17
02 Sep 2013, 08:17
Thanks Bunuel for your reply.
However, it would be great if you could point out problem in my solution. Basically, I need to know is that if there are two conditions, then how would I ignore one condition. such as x>0 is ignored and we came to conclusion that \sqrt{x}>x
Please comment.
Thanks
However, it would be great if you could point out problem in my solution. Basically, I need to know is that if there are two conditions, then how would I ignore one condition. such as x>0 is ignored and we came to conclusion that \sqrt{x}>x
Please comment.
Thanks
