GMAT Club World Cup 2022 (DAY 9): If m^4/|m|

Question

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

I.  m < \pi 
II.  m^2<1 
III.  m^3>-8

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

Bunuel · Accepted Answer

Official Solution: If m eq 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

ThatDudeKnows · Answer

We can plot  \frac{m^4}{|m|}  and  \sqrt{m^2}  and see where the first is lower than the second.

m is between -1 and 1, exclusive.

I, II, and III all work.

Answer choice E.

Kinshook · Answer

Asked: If \frac{m^4}{|m|}<\sqrt{m^2} , then which of the following must be true? \frac{m^4}{|m|}<\sqrt{m^2} Since \sqrt{m^2}= |m| \frac{m^4}{|m|}<|m| \frac{m^4}{|m|}-|m|<0 \frac{(m^4-m^2)}{|m|}<0 \frac{m^2(m^2-1)}{|m|}<0 (m+1)(m-1)<0 -1-8 : m^3>-1>-8; MUST BE TRUE IMO E

gmatophobia · Answer

\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

hloonker · Answer

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?

Bunuel · Answer

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.

bumpbot · Answer

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

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