Bunuel wrote:
If \(\sqrt{a} > b^2 > c^4\), which of the following could be true?
I. a > b > c
II. c > b > a
III. a > c > b
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III only
Are You Up For the Challenge: 700 Level QuestionsSince √a > b^2 > c^4, squaring all 3 quantities, we have a > b^4 > c^8. Since both b^4 and c^8 are nonnegative and a is greater than both of them, then a must be positive. Furthermore, if both b and c are at least 1, then a > b^4 > c^8 means a > b > c. For example, we can let a = 25, b = 2 and c = 1. We see that √a = 5, b^2 = 4 and c^4 = 1 satisfies √a > b^2 > c^4. Thus, statement I could be true.
In order for c > b > a, c and b must be positive since a is positive. However, as mentioned earlier, if both b and c are at least 1, then a > b > c. So the only way that it is possible for c > b > a is for both c and b to be between 0 and 1; in that case, a must also be between 0 and 1. For example, if a = ¼, b = ⅓, and c = ½, we have √a = 1/2, b^2 = 1/9 and c^4 = 1/16 satisfies √a > b^2 > c^4. Thus, II could be true.
If both b and c are negative, obviously, a is greater than either of them since a is positive. However, since b^2 > c^4, square rooting both sides, we have |b| > c^2. Therefore, it’s possible for c > b and still have |b| > c^2. For example, if we let b = -2 and c = -1 and we can still have a = 25, we have √a = 5, b^2 = 4 and c^4 = 1 satisfies √a > b^2 > c^4. But in this case, we have a > c > b. Thus, III could be true.
Answer: E
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