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imhimanshu
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imhimanshu
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

Notice that the stem says that x>0. So, we are not ignoring anything: \(x(\sqrt{(x)}-x)>0\) --> since x>0, then the other multiple must also be greater than 0 --> \(\sqrt{(x)}-x>0\).

Hope it's clear.
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Hi,

The best way to solve inequality questions is to breakdown the question stem and rephrase the question.

The question here gives us x > 0 and asks, Is \sqrt{x} > x.

Since x is positive, we can square both sides, so the question now is, Is x > x^2 -----> Is x^2 < x ----> Is 0 < x < 1

Statement 1 : x is not equal to 1

This clearly is not sufficient, since we get a YES and a NO.

Statement 2 : x√x > x^2

Since x is positive, we can divide both sides by x. This gives us √x > x. Sufficient.

Answer: B
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imhimanshu
If \(x>0\), Is \(\sqrt{x} > x\) ?

(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
From (a), an we derive that √x>x will only be less or equal (x = 1)?
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