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Hi Experts, In solving x^2<2x=>x^2-2x<0=>x(x-2)<0=>x<0 or x<2

why did we ignore x<0 is it bcz we are told in question that x is positive number...it is so then why can't it be ignored in statement 3...btw I understood the number picking methodology, just little bit curious...Thanks in advance.

Re: If x is positive, which of the following could be correct ordering of [#permalink]

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12 Oct 2016, 10:06

Bunuel wrote:

ykaiim wrote:

IMO B. for 0<x<1, only statement I holds. Brunuel, if u put x=1/2: II. II. x^2<1/x<2x >>>>> will not hold true. x^2 = 1/4, 1/x=2 and 2x=1 then this expression will not hold. 1/4<2<1 [Incorrect]

If x=1/9 then: x^2=1/81, 1/x=9 and 2x=2/9 1/81<9<2/9 [Incorrect]

Let's check the III option for above values: III. 2x<x^2<1/x For x=1/2: 1<1/4<2 [Incorrect] For x=1/9: 2/9<1/81<9 [Incorrect]

So, B should be the correct answer. Please check.

OA IS D.

Algebraic approach is given in my solution. Here is number picking:

I. \(x^2<2x<\frac{1}{x}\) --> \(x=\frac{1}{2}\) --> \(x^2=\frac{1}{4}\), \(2x=1\), \(\frac{1}{x}=2\) --> \(\frac{1}{4}<1<2\). Hence this COULD be the correct ordering.

II. \(x^2<\frac{1}{x}<2x\) --> \(x=0.9\) --> \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) --> \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.

III. \(2x<x^2<\frac{1}{x}\) --> \(x^2\) to be more than \(2x\), \(x\) must be more than 2 (for positive \(x-es\)). But if \(x>2\), then \(\frac{1}{x}\) is the least value from these three and can not be more than \(2x\) and \(x^2\). So III can not be true.

Thus I and II could be correct ordering and III can not.

Answer: D.

how did u choose the numbers for plugging in. i guess thats the trick in inequalities

Re: If x is positive, which of the following could be correct ordering of [#permalink]

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27 Mar 2017, 20:53

There are three equations. All three equations are fairly simple to test, so the quickest way to get the answer is to guess and check. But what points should we use to test? Well since there are three equations, we can set them to each other to find three critical points. Once we get the critical points, test above and below those critical points and you can find all the different ways the equations relate to each other.

Set \(\frac{1}{x} = 2x\) so a critical point is: \(\frac{1}{\sqrt{2}}=x\)

Set \(\frac{1}{x} = x^2\) so a critical point is: \(1 = x\)

Set \(2x = x^2\) so a critical point is: \(2 = x\)

We have to test below and above each critical point. So the minimum tests are four:

At \(x=\frac{1}{10}\) the order is \(x^2\) < 2x < \(\frac{1}{x}\)

At \(x=\frac{9}{10}\) the order is \(x^2\) < \(\frac{1}{x}\) < 2x

At \(x=\frac{3}{2}\) the order is \(\frac{1}{x}\) < \(x^2\) < 2x

At \(x=\frac{5}{2}\) the order is \(\frac{1}{x}\) < 2x < \(x^2\)

Out of the results, only the first two are provided as answers. So the correct answer is D, or "I and II only."

Re: If x is positive, which of the following could be correct ordering of [#permalink]

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28 Mar 2017, 04:35

I see everyone used numbers ranging 0 to 1. I just used fractions and got the right answer. Is it luck or using fractions is a viable strategy since fractions are just a representation of the numbers of 0-1

I had a question- @veritaskarishma also gave a solution to this question using inequalities, in which she solved each answer choice by segmenting the inequalities and checking out if the ranges for the inequalities overlap. However, my question is that, for each option, won t we have to check for three inequalities i.e 1<2, and 2<3 and 1<3. right now the official solution just considers 1<2 and 2<3. am i missing something here , is there a need for such a consideration- please explain why we don't need to compare 3 inequalities within each system. i believe that a third comparison might even further narrow down our range. please correct me if i am wrong. thanks

thanks
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If x is positive, which of the following could be correct ordering of [#permalink]

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19 Jul 2017, 04:37

1

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In all the three statements X^2 < 1/X. Which means, X^2 < X^(-1). So, greater power having lesser value and lesser power having greater value is possible only if 0 < x < 1. So, if x = 0.1, Statement-I is true. If x = 0.9, Statement-II is true. And for any value of x statement -III if not true.

If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?

I. \(x^2 < 2x < \frac{1}{x}\)

II. \(x^2 < \frac{1}{x} < 2x\)

III. \(2x < x^2 < \frac{1}{x}\)

(A) none (B) I only (C) III only (D) I and II (E) I, II and III

We need to equate these expressions first. Of course, we can only equate two of them at a time. So we have three equations to solve, 1/x = 2x, 1/x = x^2, and 2x = x^2.

1) 1/x = 2x

2x^2 = 1

x^2 = 1/2

x = √(1/2) = (√2)/2 ≈ 1.4/2 = 0.7

2) 1/x = x^2

x^3 = 1

x = ∛1 = 1

3) 2x = x^2

Dividing both sides by x (since we know x > 0), we have:

2 = x

From the three equations above, we see that x = (√2)/2, 1, and 2. These numbers are critical since they make two of the three expressions equal to one another. Thus, we need to consider all the values that are not exactly these numbers in order to determine the order of these expressions. That is, we need to consider the following intervals:

i) 0 < x < (√2)/2 ii) (√2)/2 < x < 1 iii) 1 < x < 2 iv) x > 2

However, for each of these intervals, we can just pick a representative number (for example, in 1 < x < 2, we can pick 1.5) to determine the order of these expressions.

i) 0 < x < (√2)/2

Since (√2)/2 ≈ 0.7, we can let x = ½. Then 1/x = 2, 2x = 1, and x^2 = ¼. Thus, we have x^2 < 2x < 1/x, and hence Roman numeral I could be true.

ii) (√2)/2 < x < 1

We can let x = ¾. Then 1/x = 4/3, 2x = 3/2, and x^2 = 9/16. Thus, we have x^2 < 1/x < 2x, and hence Roman numeral II could be true.

iii) 1 < x < 2

We can let x = 3/2. Then 1/x = ⅔, 2x = 3, and x^2 = 9/4. Thus, we have 1/x < x^2 < 2x. (However, this is not one of the given Roman numerals.

iv) x > 2

We can let x = 3. Then 1/x = 1/3, 2x = 6, and x^2 = 9. Thus, we have 1/x < 2x < x^2. However, this is not one of the given Roman numerals.

We see that only Roman numerals I and II could be true.

Answer: D
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(a) none (b) I only (c) III only (d) I and II (e) I, II and III

could be correct ordering

So if we can find any example that satisfy the inequation, that statement will be correct

(I) x = 0.1 => 0.01 < 0.2 < 10 (II) x= 1/2 => 1/4 < 1/2 < 1

(III) 2x < x^2 <=> x ( 2 -x) < 0, x > 0 then x > 2

with x > 2 ==> x^2 < 1/x <=> x^3 < 1 <=> x < 1

So (III) can't happen

The answer is D

Would you please give an exmaple to illustrate (II) In your example: x= 1/2 => 1/4 < 1/2 < 1 How did you prove by taking x=1/2?? 1/4 < 2 < 1 is not correct. Did you miss something here?

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