Hi Oneo,
In these types of Roman Numeral questions, it's usually beneficial to use whatever information you've been given to consider the possibilities...
(For this question, I'm going to assume that you meant that a5 means a^5....)
In this case a^5 <= a tells us a lot about the POSSIBLE values of a...
a could be...
1, 0, any positive fraction, any negative number
The question asks for what MUST be true, meaning what is ALWAYS TRUE?
Since each of the Roman Numerals offers a small subset of what a COULD be, NONE of them is ALWAYS TRUE.
Final Answer: A
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Try to show that statements I, II and III DON'T have to true.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
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.
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vietmoi999
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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.
so it is possible that a<-1 and a dose not need to satisfy one of 3 cases.
answer A.
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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.)
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.)
