Plug Using Transition Points

Quarter Wit, Quarter WisdomLet’s take a break from Geometry today and discuss the concept of transition points. This is especially useful in questions where you are tempted to plug in values. A question often asked is: how do I know which values to plug and how do I know that I have covered the entire range in the 3-4 values I have tried? What transition points do is that they give you the ranges in which the relationships differ. All you have to do is try one value from each range. If you do, you would have figured out all the different relationships that can hold. We will discuss this concept using a GMAT Prep question. You can solve it using our discussion on inequalities too. But if number plugging is what comes first to your mind in this question, then it will be a good idea to get the transition points.

Let’s begin:

If x is positive, which of the following could be correct ordering of 1/x, 2x and x^2?

(I) x^2 < 2x < 1/x

(II) x^2 < 1/x < 2x

(III) 2x < x^2 < 1/x

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

Solution:

Notice that the question says “could be correct ordering”. This means that for different values of x, different orderings could hold. We need to find the one (or two or three) which will not hold in any case. So what do we do? We cannot try every value that x can take so how do we know for sure that one or more of these relations cannot hold? What if we try 4-5 values and only one relation holds for all of those values? Can we say for sure that the other two relations will not hold for any value of x? No, we cannot since we haven’t tried all values of x. So there are two options you have in this case:

1. Use logic to figure out which relations can hold and which cannot. This you can do using inequalities (but we will not discuss that today).

2. You can figure out the ranges in which the relationships are different and then try one value from each range. This is our transition points concept which we will discuss today.

Let’s discuss the second option in more detail.

First of all, we are just dealing with positives so there is less to worry about. That’s good.

To picture the relationship between two functions, we first need to figure out the points where they are equal.

x^2 = 2x

x^2 – 2x = 0

x = 0 or 2

x cannot be 0 since x must be positive so this equation holds when x = 2

So x =2 is the transition point of their relation. x^2 is less than 2x when x is less than 2 and it will be greater than 2x when x is greater than 2.

Let’s try to figure out the relation between 1/x and 2x now.

1/x = 2x

x = 1/√2

Since 1/x is less than 2x when x is greater than 1/√2, it will be more than 2x when x is less than 1/√2.

Move on to the relation between 1/x and x^2.

1/x = x^2

x^3 = 1 (notice that since x must be positive, we can easily multiply/divide by x without any complications)

x = 1

So you have got three transition points: 1/√2, 1 and 2.

Now all you need to do is try a number from each of these ranges:

(i) x < 1/√2

(ii) 1/√2 < x < 1

(iii) 1 < x < 2

(iv) x > 2

If a relation doesn’t hold in any of these ranges, it will not hold for any value of x.

(i) For x < 1/√2, put x = a little more than 0 (e.g. 0.01)

1/x = 100, 2x = 0.02, x^2 = 0.0001

We get x^2 < 2x < 1/x is possible. So (I) is possible

(ii) For 1/√2 < x < 1, put x = a little less than 1 (e.g. 0.99)

1/x = slightly more than 1, 2x = slightly less than 2, x^2 = slightly less than 1

We get x^2 < 1/x < 2x is possible. So (II) is also possible.

(iii) For 1 < x < 2, put x = 3/2

1/x = 2/3, 2x = 3, x^2 = 9/4 = 2.25

We get 1/x < x^2 < 2x is possible.

(iv) For x > 2, put x = 3

1/x = 1/3, 2x = 6, x^2 = 9

We get 1/x < 2x < x^2 is possible.

We see that for no positive value of x is the third relation possible. We have covered all different ranges of values of x.

Answer (D)

Try using inequalities instead of number plugging to see if solving the question becomes easier in that case.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

4 Responses

  1. Sri G says:

    Karishma,

    Thanks for the valuable posts. I like this method more than inequalities method.

    • Karishma says:

      Yes, I instinctively used this method too when I did this question. Algebra is usually an afterthought (though it is good to know both methods)

  2. Mike says:

    x^2 is less than 2x when x is less than 2 and it will be greater than 2x when x is greater than 2.

    Since 1/x is less than 2x when x is greater than 1/√2, it will be more than 2x when x is less than 1/√2.

    Hi Karishma,

    Can you please elaborate on the above statements. How did you know that “Since 1/x is less than 2x when x is greater than 1/√2″? Did you make that conclusion by just looking at it or did you try to plug in some extreme numbers to test the result? Thank you.

    • Karishma says:

      “Since 1/x is less than 2x when x is greater than 1/√2″

      The transition point of 1/x and 2x is 1/√2.
      Put x = 1 (which is greater than 1/√2). In this case 1/x < 2x
      So when x>1/√2, 1/x<2x

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