How many of the following inequalities are possible for at least one value of x?

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

II. \(x^2 > x^3 > x\)

III. \(x^3 > x > x^2\)


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

(1) Let X=-2 then we get a yes answer
(2) Let X=-0.5 then we get a yes
(3) none of X satisfies

Ans C

Answer is D.

For (i) - Consider x=1/3. => We can have X= 2/27 which is between 1/9 and 1/27

For (ii)- Consider x= 1/2= > We can have X=0 which is less than 1/4 and 1/8.

For (iii) - Consider x= 2 => We can have X = 5 which is between 8 and 4.

Hence, we can have each of the above statements valid for different vales of x and X.

NOTE : I have solved the above question based on the assumption that x and X are two different variables.
_________________

Correct bro. I agree to what you said.

Can anyone please help here?
_________________

Bunuel: Can you please check this question? I did not find any values for which (iii) holds. Shouldn't OA be C?

For a problem that hinges on a conceptual understanding of properties of numbers with exponents, plugging in is a natural first approach. So try values in meaningfully different ranges. Say x is 2: that makes
\(x^2\)=4
and \(x^3\)=8
\(So, x^3 > x^2 > x\)
which doesn't work for any of the inequalities.
Now say x is -2: that makes \(x^2\)
\(x^3\)=-8

\(So, x^2 > x > x^3\)

Having tried both negative and positive integers, consider what other ranges are left: fractions between 0 and 1 and fractions between 0 and 1. Inequality (ii) will be true for any fraction between 0 and -1, so it's possible, but inequality (iii) never works: the only numbers for which x > \(x^2\) are fractions between 0 and 1, but for all such numbers \(x^2\) is also greater than \(x^3\)

Answer C