GMAT Club World Cup 2022 (DAY 9): If m^4/|m|<(m^2)^(1/2), then which

\(\frac{m^4 }{ |m|} < \sqrt{m^2}\)

\(\frac{m^4 }{ |m|} < |m| \)

As |m| is non negative, we can multiply |m| on both sides without change of sign

\(m^4 < m^2\)

\(m^4 - m^2 < 0\)

\(m^2(m^2 - 1) < 0\)

\(m^2(m + 1)(m-1) < 0\)

The valid values of m for this range

-1 < m < 1

Let's look at the statements -

I. m < \(\pi\) - TRUE

II. \(m^2 < 1\)

\(m^2 - 1 < 0\)

TRUE

III. \(m^3 > -8\)

TRUE again !

IMO E

Official Solution:


If \(m \neq 0\) and \(\frac{m^4}{|m|} < \sqrt{m^2}\), then which of the following must be true?

I. \(m < \pi\)

II. \(m^2-8\)


A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III


Since \(\sqrt{m^2} = |m|\), we have \(\frac{m^4}{|m|} < |m|\).

Multiplying both sides by \(|m|\), we get \(m^4 < m^2\).

Dividing both sides by \(m^2\), we have \(m^2 < 1\).

This inequality implies that \(-1 < m < 1\).

So, essentially, the question asks: Given that \(-1 < m < 1\), which of the following statements must be true?

I. \(m < \pi\). Since \(-1 < m < 1\), it is correct to say that for any \(m\) from this range, \(m < \pi\).

II. \(m^2 < 1\). This statement is already established to be true from our earlier steps. So, this statement is always true.

III. \(m^3 > -8\). This inequality implies that \(m > -2\). Since \(-1 < m < 1\), it is true for all values of \(m\) in this range that \(m > -2\).


Answer: E

Hi can someone show me the wavy method for this question - i got -1....(-)...0....(-).....1. So basically the value is -1 <x< 1. Is this correct?

m^2 - 1 < 0
(m + 1)(m - 1) < 0

Transition points are -1 and 1. So, we get three ranges: m < -1, -1 < m < 1, and m > 1. When m > 1, the expression is positive, so we get that in the third range the expression is positive, in the second range it's negative, and in the first it's positive: + - +. Therefore, (m + 1)(m - 1) < 0, when -1 < m < 1. However, m^2 < 1 can be solved easier. Taking the square root gives |m| < 1, which implies that -1 < m < 1.

Hope it helps.