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Hi Whoisdmx15,

Thank you for the solution. However, option C is not true for X=0.5 or values in between 0 to 1. And these are positive terms as well. Am I missing something here?

Best,
JohnnyMe
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If x is positive, which of the following may be true?

Let's try for numbers greater than 1, (If several options remain, then we will try numbers between 0 and 1)

As x has been given to be only a number (hence can be a fraction), Assume x = 1.1



A. \(\frac{1}{1.1}<\frac{1}{1.21}<1.1\)
Not true, as \(\frac{1}{1.1}>\frac{1}{1.21}\)


B. \(1+1.1<\frac{1}{1.1}<2*1.1\)
Not true, as \(2.1>\frac{1}{1.1}\)


C.\(\frac{1}{1.1}<1+1.1<2*1.1\)
Can be true as: \(\frac{1}{1.1}<2.1<2.2\)


D.\(2+1.1<\frac{1}{1.1}<4*1.1\)
Not true, as \(3.1>\frac{1}{1.1}\)


E.\(1+1.1<\frac{1}{1.1}<1.1\)
Not true, as \(2.1>\frac{1}{1.1}\)



Hence IMO Ans C
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The question is may be true and not must be true.
JhonnyMe
Hi Whoisdmx15,

Thank you for the solution. However, option C is not true for X=0.5 or values in between 0 to 1. And these are positive terms as well. Am I missing something here?

Best,
JohnnyMe
Whoisdmx15
If x is positive, which of the following may be true?

Let's try for numbers greater than 1, (If several options remain, then we will try numbers between 0 and 1)

As x has been given to be only a number (hence can be a fraction), Assume x = 1.1



A. \(\frac{1}{1.1}<\frac{1}{1.21}<1.1\)
Not true, as \(\frac{1}{1.1}>\frac{1}{1.21}\)


B. \(1+1.1<\frac{1}{1.1}<2*1.1\)
Not true, as \(2.1>\frac{1}{1.1}\)


C.\(\frac{1}{1.1}<1+1.1<2*1.1\)
Can be true as: \(\frac{1}{1.1}<2.1<2.2\)


D.\(2+1.1<\frac{1}{1.1}<4*1.1\)
Not true, as \(3.1>\frac{1}{1.1}\)


E.\(1+1.1<\frac{1}{1.1}<1.1\)
Not true, as \(2.1>\frac{1}{1.1}\)



Hence IMO Ans C
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How would we know which number to choose to start the testing?
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Shan0110
How would we know which number to choose to start the testing?
You can test one number from each of these intervals, 0 and 1, 0 and -1, 0, 1, greater than 1, and less than 1

Check out: The 4 Math Strategies Everyone Must Master, Part 1 (1. Test Cases and 2. Choose Smart Numbers.)

Hope it helps.
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logically option 3 has to be true, even though its a may be true question, you don't have to do any calculations
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To do such questions:

Put x to be the possbile smallest integer value. Now we cannot put x = 1 since we would get the same numbers like (1/x and 1/x^2 would be same).

Hence lets put x = 2.

Now, lets go through options:

A.) 1/x < 1/x^2 < x

1/2 < 1/4 < 2

Clearly 1/2 > 1/4. Hence option A eliminated.

B.) 1+x < 1/x < 2x

3 < 1/2 < 4

Again 3 > 1/2. So option B eliminated

C.) 1/x < 1+x < 2x

1/2 < 3 < 4

This could be option. Lets go through others as well

D.) 2 + x < 1/x < 4x

3 < 1/2 < 8. Again D eliminated

E.) 1+x < 1/x < x

3 < 1/2 < 2

Again eliminated. So clearly C is the answer.


Remember if we put x between 0 and 1, we might not get the answer. The question says may be true and not must be true. So we get our answer as C
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