Have u tried using fractions.
The question doesn't say they have to be positive integers

Hi,

I can use fraction as well as integers. But these conditions should hold true for all values of x. According to my understanding the relation depends upon value of x. if value of x changes.... 0.1 or 2 or 1 then these conditions are not true.

Can anyone please help me out ?

x^2<2x<1/x
If we want it to be true we need x<2 and 2*x^2<1 so x<sqrt(1/2)
so solutions are 0<x<sqrt(1/2)

x^2<1/x<2x
If we want it to be true we need x^3<1 and 1<2*x^2 so sqrt(1/2)<x
so solutions exist

2x<x^2<1/x
If we want it to be true 2<x and x^3<1 so x under 2
so it is impossible

We have 1 and 2 which can exist

Picking values for X that are less than 1 works well to illustrate the point made by the previous post.

If X=0.5 then:

1/X = 2

2X = 1

X^2 = .25

Thus, X^2< 2X <1/X

However, if X = 0.9:

1/X = 1.11

2X = 1.8

X^2 = .81

Thus, X^2 < 1/X < 2X

plugging numbers will b time consuming...

Is there a quicker method for this..??

Hi ,

Here how we can assume x is in between 0 and 1. there is no such condition mentioned in the question. And it should hold true for all values of x irrespective of 0 to 1 or any other value.

I feel either question is wrong or Answer choice.

shrikantkamble wrote:Hi ,

Here how we can assume x is in between 0 and 1. there is no such condition mentioned in the question. And it should hold true for all values of x irrespective of 0 to 1 or any other value.

I feel either question is wrong or Answer choice.

Read the question carefully, it says "what could be the correct ordering"

Therefore only way the options given could be possible is when x is between 0 and 1.

Hope this helps.