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Que: A set is such that if m is in the set.........

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by Max@Math Revolution » Sat Feb 13, 2021 9:20 pm

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E

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Que: A set is such that if m is in the set, \(m^2+3\) is also in the set. If −1 is in the set, which of the following is also in the set?

I. −2
II. 4
III. 19

(A) Only I
(B) Only II
(C) Only I and II
(D) Only II and III
(E) I, II, and III
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Re: Que: A set is such that if m is in the set.........

by Max@Math Revolution » Tue Feb 16, 2021 9:05 pm
Solution: According to the problem: If m is in the set, \(m^2+3\) is also in the set.

However, it does NOT imply that if \(m^2+3\) is in the set, then m must be in the set.

What it does imply is that: If \(m^2+3\) is NOT in the set, m is NOT in the set.

Thus, if we have m = −1 as a member of the set, \(m^2+3\) = \(\left(-1\right)^2+3\) = 4 is also a member of the set. Thus, statement II is correct.

Proceeding in the same way: Since m = 4 is a member of the set, then \(m^2+3\) = \(4^2+3\) = 19 is a member of the set. Thus, statement III is also correct.

Therefore, D is the correct answer.

Answer D
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