ordering

[divya23]

if x is +ve which of the following could be the correct ordering of 1/x,2x and x^2.....
x^2<2x<1/x
x^2<1/x<2x
2x<x^2<1/x

ans is 1 and 2

i dont know how gmat uses values ..anyway i got 1 luckily but 2 is also they say is coorect...can somebody please tell the value of x here

[Anurag@Gurome]

if x is +ve which of the following could be the correct ordering of 1/x,2x and x^2...

x^2<2x<1/x
x^2<1/x<2x
2x<x^2<1/x

For 0 < x < 1/√2 : x² < 2x < 1/x
For 1/√2 < x < 1 : x² < 1/x < 2x
For 1 < x < 2 : 1/x < x² < 2x
For 2 < x : 1/x < 2x < x²

Hence, 1 and 2 are possible.

[vikram4689]

@Anurag: Can you please explain what was your thought process while solving this ques. Actually i could do this in 1 minute but the prob. is that sometime pluging numbers takes lot more time. So better to have solid strategy that works everytime.

[Anurag@Gurome]

@Anurag: Can you please explain what was your thought process while solving this ques...

As soon as I saw that I have to determine the ordering of 1/x, 2x, and x², I concluded there are 3 (as the three terms can be paired in three ways) critical points where the ordering will change. Equating each pair I will get the critical points.

Pair 1: 1/x and 2x --> 1/x = 2x --> x = 1/√2
Pair 2: 1/x and x² --> 1/x = x² --> x = 1
Pair 1: x² and 2x --> x² = 2x --> x = 2

Hence, I have four regions and in each of them the ordering will be different. Now I will just pick some easy number in each region (For example, 1/2, 3/4, 3/2, and 4) and determine the order.

[winniethepooh]

For 0 < x < 1/√2 : x² < 2x < 1/x
For 1/√2 < x < 1 : x² < 1/x < 2x
For 1 < x < 2 : 1/x < x² < 2x
For 2 < x : 1/x < 2x < x²

Hence, 1 and 2 are possible.



Sir, I really appreciate your responses.
But, I didn't understand this explanation!
Can it be any clearer?

Also, the Values for x can be 1 and 2 or is it that the Inequalities 1 and 2 are possible?

[Anurag@Gurome]

Sir, I really appreciate your responses.
But, I didn't understand this explanation!
Can it be any clearer?

Also, the Values for x can be 1 and 2 or is it that the Inequalities 1 and 2 are possible?

Yes, the values of x can be 1 and 2 and then some of the quantities will be equal. And that is the key point of the solution.

For example, when x = 1, 1/x = x².
Hence, we can expect that on two sides of 1, the ordering of 1/x and x² will be different. In fact, for x < 1, x² < 1/x and for x > 1, 1/x < x².

Similarly, when x = 2, 2x = x²
Hence, we can expect that on two sides of 2, the ordering of 2x and x² will be different. In fact, for x < 2, x² < 2x and for x > 2, 2x < x².

Similarly, when x = 1/√2, 1/x = 2x
Hence, we can expect that on two sides of 1/√2, the ordering of 2x and 1/x will be different. In fact, for x < 1√2, 2x < 1/x and for x > 1/√2, 1/x < 2x

Now, we have to combine these results to determine the combined ordering.

At each of these three values two of the three quantities will have same value. But the options do not have any equalities. Hence, we don't have to consider them. But they are possible and they are the key point to determine the ordering.

[goalevan]

This problem is possible to complete in under 30 seconds by sketching a simple graph of these functions, y=x^2, y=2x, y=1/x

It will look like this:

Statements are very simple to verify once you have this.