If x < x^3 < x^2, then x^4 must be:

We have: x < x^3 < x^2

Observations: x^2 (even power) is greater than x^3 (a higher power, odd) and x^1 (a lower power, odd)
This is possible ONLY if x were a negative number.

Again. considering the odd powers: x^3 > x^1 (a higher power is greater than a lower power where x is negative)
Thus, x must be a fractional value => -1 < x < 0

Thus, x^4 is a fraction between 0 and 1 (a positive quantity)

Looking at the options:
(A) less than x => Incorrect since x^4 is positive and x is negative
(B) between x and x^3 => Incorrect since x^4 is positive and x and x^3 are negative
(C) between x^3 and x^2 => Correct; since x is a negative value between 0 and -1, x^4 must be smaller than x^2; hence is lies between x^3 (negative) and x^2 (a larger positive value)
(D) greater than x^2 => Incorrect; since x is a negative value between 0 and -1, x^4 must be smaller than x^2 - higher the power, smaller the value
(E) undefined => Incorrect, since x is clearly defined

Answer C


Alternative:

We need to consider 4 regions on the number line and verify x < x^3 < x^2:
# x>1: Say, x=2: 2 < 8 < 4 - false
# 0<x<1: Say, x = 1/2: 1/2 < 1/8 < 1/4 - false
# -1<x<0: Say, x = -1/2: -1/2 < -1/8 < 1/4 - true
# x <-1: Say, x = -2: -2 < -8 < 4 - false

So, x must be between 0 and -1. Taking x = -1/2, we have x^4 = 1/16
Checking the options:
(A) less than x => 1/16 < -1/2 - Incorrect
(B) between x and x^3 => -1/2 < 1/16 < -1/8 - Incorrect
(C) between x^3 and x^2 => -1/8 < 1/16 < 1/4 - Correct
(D) greater than x^2 => 1/16 > 1/4 - Incorrect
(E) undefined => Incorrect, since x is clearly defined

Answer C