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Hi Bunuel,
what confused me in the solution is that if -x^2 and -x^3 are -1/4 and -1/8, then how are we again treating them as x individually. The way question is written it seems that the set has 3 numbers, x, -x^2, and -x^3

Please help me out with this, thanks :)
Bunuel
5. For a certain set of numbers, if \(x\) is in the set, then both \(-x^2\) and \(-x^3\) are also in the set. If the number \(\frac{1}{2}\) is in the set, which of the following must also be in the set?

I. \(-\frac{1}{64}\)

II. \(\frac{1}{64}\)

III. \(\frac{1}{\sqrt[3]{2}}\)


A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III

Since \(\frac{1}{2}\) is in the set, the following must also be in the set:


\(-x^2 = -\frac{1}{4}\);

\(-x^3 = -\frac{1}{8}\).

Since \(-\frac{1}{4}\) is in the set, the following must also be in the set:


\(-x^3 =\frac{1}{64}\)

Since \(-\frac{1}{8}\) is in the set, the following must also be in the set:


\(-x^2 =-\frac{1}{64}\)

The only number we cannot obtain is \(\frac{1}{\sqrt[3]{2}}\).

Answer: D.
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saransh2797
Hi Bunuel,
what confused me in the solution is that if -x^2 and -x^3 are -1/4 and -1/8, then how are we again treating them as x individually. The way question is written it seems that the set has 3 numbers, x, -x^2, and -x^3

Please help me out with this, thanks :)
Bunuel
5. For a certain set of numbers, if \(x\) is in the set, then both \(-x^2\) and \(-x^3\) are also in the set. If the number \(\frac{1}{2}\) is in the set, which of the following must also be in the set?

I. \(-\frac{1}{64}\)

II. \(\frac{1}{64}\)

III. \(\frac{1}{\sqrt[3]{2}}\)


A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III

Since \(\frac{1}{2}\) is in the set, the following must also be in the set:


\(-x^2 = -\frac{1}{4}\);

\(-x^3 = -\frac{1}{8}\).

Since \(-\frac{1}{4}\) is in the set, the following must also be in the set:


\(-x^3 =\frac{1}{64}\)

Since \(-\frac{1}{8}\) is in the set, the following must also be in the set:


\(-x^2 =-\frac{1}{64}\)

The only number we cannot obtain is \(\frac{1}{\sqrt[3]{2}}\).

Answer: D.

The set's rule means each number generates new elements based on the pattern. Here, -x^2 and -x^3 are treated as elements in the set, not individual variables. They follow the same rule to generate further numbers.
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