Which of the following COULD be correct?

Take x=-2
x^2=4
x^3=-8
x^4=16

This condition satisfies III and IV.
Now if we look at the options with both III and IV only D and E survives.
Lets analyze II and we can conlude
For II to be true, x<x^3<x^2<x^4
but x^3<x^2<x^4 only when x<0 and when x<0 , x^3<x
Therefore, II cannot be true

Hence D

Whenever we see such problems we should try four types of numbers: \(+ integer\), \(+ fraction\), \(-integer\), and \(-fraction\).

If none of them make the inequality true, then it can NOT be true. If any of them makes the inequality true, then there is NO need to try others, since it already CAN be true.

I. \(x<x^3<x^2\)

If \(x = \frac{-1}{3}\), then \(\frac{-1}{3}<\frac{-1}{27}<\frac{1}{9}\) is true

II. \(x<x^3<x^2<x^4\)

If \(x = 3\), then \(3<27<9<81\) is not true

If \(x = \frac{1}{3}\), then \(\frac{1}{3}<\frac{1}{27}<\frac{1}{9}<\frac{1}{81}\) is not true

If \(x = -3\), then \(-3<-27<9<81\) is not true

If \(x = \frac{-1}{3}\), then \(\frac{-1}{3}<\frac{-1}{27}<\frac{1}{9}<\frac{1}{81}\) is not true

III. \(x^3<x^2<x^4\)

If \(x = -3\), then \(-27<9<81\) is true

IV. \(x^3<x<x^2<x^4\)

If \(x = -3\), then \(-27<-3<9<81\) is true

Hence D

I. \(x<x^3<x^2\)
All choices contain (I), then (I) definitely could be TRUE.

II. \(x<x^3<x^2<x^4\)
\(x=-2\) --> \(-2 < -8 < 4 <16\) (NO)
\(x=-\frac{1}{2}\) --> \(-\frac{1}{2} < -\frac{1}{8} < \frac{1}{4} < \frac{1}{16}\) (NO)
\(x=\frac{1}{2}\) --> \(\frac{1}{2} < \frac{1}{8} < \frac{1}{4} < \frac{1}{16}\) (NO)
\(x=2\) --> \(2 < 8 < 4 <16\) (NO)
This statement must be FALSE for any x.

III. \(x^3<x^2<x^4\)
\(x=-2\) --> \(-8 < 4 <16\) (YES!)
This statement could be TRUE

IV. \(x^3<x<x^2<x^4\)
\(x=-2\) --> \(-8 < -2 < 4 <16\) (YES!)
This statement could be TRUE


Answer is (D) I, III, and IV only

We just need to establish that statements I, II, III, and IV is valid for a value of x in order to conclude that it could be correct. If the condition does not hold for any possible value, then we can conclude that it cannot be correct.

I. x<x^3<x^2. This is true for -1<x<0. For example x=-1/2. x^2=1/4, x^3=-1/8.
Since x(-1/2)<x^3(-1/8)<1/4, I could be correct.

II. x<x^3<x^2<x^4
Not true for x>1, not true for x<-1, not true for 0<x<=1, not true for -1<=x<=0. Hence this statement could never be correct.

III. x^3<x^2<x^4
III. is true for x<-1, for example x=-2, x^2=4, x^3=-8, x^4=16. Since x^3(-8)<x^2(4)<x^3(8), III could be correct.

IV. x^3<x<x^2<x^4
This is true for x<-1. For example x=-2,
x^3(-8)<x(-2)<x^2(4)<x^4(16)
Hence IV. could be correct.

We know that I, III, and IV could be correct. The answer is therefore D.

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Which of the following COULD be correct?

I. x < x3 < x2
II. x < x3 < x2 < x4
III. x3 < x2 < x4
IV. x3 < x < x2 < x4

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

Though each option has an equal chance as an answer but evaluating the options here we can say that options A and B have the highest chances of being the answer. ‘E’ deals with all so ignoring it for the time being.

Options ‘C’ and ‘D’ mentions only and have ‘I’ and ‘IV’ common with option ‘E’. Thus, check for the ‘COULD be correct’ condition for ‘II’ and ‘III’ first.

II. x < x3 < x2 < x4
x < x3 when x > 1 OR -1 < x < 0
x3 < x2 when x < 0 OR 0 < x < 1
x2 < x4 when x > 1 OR x < -1

As we have three conditions i.e. when
i) x < -1
ii) -1 < x < 0
iii) 0 < x < 1
iv) x > 1

We have total four conditions. Let’s refer table snapshot:

In totality, x < x3 < x2 < x4 is never true ever. Since ‘II’ is present in options ‘C’ and ’E’ both, option ‘D’ is the answer.

But for completeness check ‘III’:

III. x3 < x2 < x4
Since this is a part of ‘II’ table for ‘II’ suffices to illustrate that this could be true(x < -1).

IMO Answer (D)

PS: Also, ‘I’ can be checked through above table that is true for -1 < x < 0 AND ‘IV’ is true for x < -1 (reverse of x < x3 for x < -1 is true).

Which of the following COULD be correct?

I. x<x^3<x^2
II. x<x^3<x^2<x^4
III. x^3<x^2<x^4
IV. x^3<x<x^2<x^4

Let's take random numbers and see what happens:


When we have negative integers there is a possibility that I, II, and IV could be true.

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

D is the answer.

Archit3110

No, we don't have to test them all for the same value of x cause they may be true for different ranges of x


I. x<x3<x2 this could be true for x = -1/2
II.x<x3<x2<x4 . x<x3 is true for x between 0 and -1 or for values greater than 1 but x<x2<x4 will be true only for x < -1 or x > 1. Since there is no common raneg of values of x therefore it can not be true
III. x3<x2<x4 . This could be true for x = -2
IV. x3<x<x2<x4 . This could be true for x = -2

Answer: Option D

we need to check between option II and III


clearly x^2 <x^4 => x>1 or x<-1


now x^3>x^2=> x is negative


thus x^3 cannot be less greater than x


two is not possible

If you try with x=-1/2
only 1 is right
so ans is 1