If -1 < a < 0, then which of the following statements must be true?

Let a =-1/2

Negative can't be higher than positive number......Eliminate B

1/4 > 1/16.......................................................Eliminate A &C

-1/8 > -1/2.......................................................Eliminate E

Answer: D

The numbers that can be found in the given range are negative proper fractions like -1/2, -1/3 and so on. So, the quickest and the easiest method to solve this question is to take a couple of values from this range and quickly falsify options, since this is a ‘Must be’ kind of a question.

To start with, let’s take a = -\(\frac{1}{2}\). \(a^4\) = \(\frac{1}{16}\), \(a^2\) = ¼, \(a^3\) = -\(\frac{1}{8}\). Clearly, \(a^2\) is the biggest and a is the smallest.
So, options A, B, C and E can be eliminated. The only option left is D. This HAS to be the right answer.

You could maybe try one more value for a, in case you want to be surer. Let a = -\(\frac{1}{3}\). Then, \(a^4\) = \(\frac{1}{81}\), \(a^2\) = \(\frac{1}{9}\) and \(a^3\) = -\(\frac{1}{27}\). Here again, the order is \(a^2\) > \(a^4\) > \(a^3\) >a.
Therefore, the correct answer option is D.

This is a fairly simple question on Inequalities, so you should be able to solve this in less than a minute.

Hope this helps!

We see that a is a negative number between 0 and -1; thus, a^2, will be the largest value, followed by a^4 (since these values will both be positive).

We also may recall that when a negative number between 0 and -1 is raised to an odd exponent, the larger the odd exponent, the larger the result. Thus, a^3 > a.

So we see that D is correct.

Answer: D