oneo wrote:hey guys... got this question in a prep test.... please help me out... Could not understand the explanation

If a5≤a, then which of the following must be true?
i. -1≤a≤0
ii. a=0
iii. 0≤a≤1

a. None of the above
b. i only
c. ii only
d. iii only
e. i and iii

Try to show that statements I, II and III DON'T have to true.
When a problem involves EXPONENTS, consider the following types of values:
positive integers greater than 1
0
±1
negative integers less than -1
positive and negative fractions
roots

Statement I: -1≤ a≤ 0
Try a value OUTSIDE this range.
If we plug a=1 into a� ≤ a, we get:
1� ≤ 1
1 ≤ 1.
This works.
Thus, it does NOT have to be true that -1≤a≤0.
Eliminate B and E, which say that statement I must be true.

Statement II: a=0
Since it's possible that a=1, it does NOT have to be true that a=0.
Eliminate C, which says that statement II must be true.

Statement III: 0≤a≤1
Try a value OUTSIDE this range.
If we plug a=-1 into a� ≤ a, we get:
(-1)� ≤ -1
-1 ≤ -1.
This works.
Thus, it does NOT have to be true that 0≤a≤1.
Eliminate D, which says that statement III must be true.

The correct answer is A.

from a^5<a we infer a<-1 or 0<a<1

so it is possible that a<-1 and a dose not need to satisfy one of 3 cases.

answer A.

You can do this algebraically as well.

If a^5 < a and a is positive, dividing both sides by a, we get a^4 < 1. Since it's positive, we have 0 < a^4 < 1, so a^4 is anything between 0 and 1, and a itself is anything between 0 and 1.

If a^5 < a and a is negative, dividing both sides by a we get a^4 > 1. Taking the fourth root of each side, the sign will flip (because a is negative), and the right side will be negative (for obvious reasons), so a < -1.

So if a^5 < a, either 0 < a < 1 or a < -1.

(And if a^5 = a, obviously a is either 0 or 1.)