Bunuel wrote:
If 0 < a < 1, which of the following is the greatest?
A. \((-2a)^{(-2)}\)
B. \(\frac{1}{a^{(-2)}}\)
C. \((\frac{1}{a})^{(-2)}\)
D. \(a^{(-2)}\)
E. \(a^2\)
Let a = \(\frac{1}{4}\), and evaluate each answer choice. Straight algebra -- reciprocals, fractions, and squares -- became too tangled.
A. \((-2a)^{(-2)}\) --> \(\frac{1}{(-2*a)^2}\)
\(\frac{1}{(-2*\frac{1}{4})^2}\) =
\(\frac{1}{(-\frac{1}{2})^2}\) = \(\frac{1}{(\frac{1}{4})}\) =
4B. \(\frac{1}{a^{(-2)}}\) --> \(\frac{a^2}{1^2}\)
\(a^2\) = \((\frac{1}{4})^2\) =
\(\frac{1}{16}\)C.\((\frac{1}{a})^{(-2)}\) --> \(\frac{a^2}{1^2}\)=
\(a^2\). Same as Answer B =
\(\frac{1}{16}\)D. \(a^{(-2)}\)--> \(\frac{1}{a^2}\)
\(\frac{1}{(\frac{1}{4})^2}\) = \(\frac{1}{\frac{1}{16}}\) =
16E. \(a^2\) = same as Answer B =
\(\frac{1}{16}\)ANSWER D
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