If n is an even integer, which one of the following is an odd integer?
(A) n^2
(B)( n +1)^2
(C) –2n – 4
(D) 2n^2 – 3
(E) Sqr(n2 + 2)
We are told that n is an even integer. So, choose an even integer for n, say, 2 and substitute it into each
answer-choice. Now, n^2 becomes 2^2 = 4, which is not an odd integer. So eliminate (A). Next,
(n +1)/2=(2 +1)/2= 3/2
is not an odd integer—eliminate (B). Next, −2n − 4 = −2 ⋅ 2 − 4 = −4 − 4 = −8 is not an odd
integer—eliminate (C). Next, 2n^2 – 3 = 2(2)^2 – 3 = 2(4) – 3 = 8 – 3 = 5 is odd and therefore the answer is possibly (D). Finally, Sqr (n^2 + 2) = Sqr(2^2 + 2 )= Sqr(4 + 2) = Sqr 6 , which is not odd—eliminate (E). The answer is (D).
(A) n^2
(B)( n +1)^2
(C) –2n – 4
(D) 2n^2 – 3
(E) Sqr(n2 + 2)
We are told that n is an even integer. So, choose an even integer for n, say, 2 and substitute it into each
answer-choice. Now, n^2 becomes 2^2 = 4, which is not an odd integer. So eliminate (A). Next,
(n +1)/2=(2 +1)/2= 3/2
is not an odd integer—eliminate (B). Next, −2n − 4 = −2 ⋅ 2 − 4 = −4 − 4 = −8 is not an odd
integer—eliminate (C). Next, 2n^2 – 3 = 2(2)^2 – 3 = 2(4) – 3 = 8 – 3 = 5 is odd and therefore the answer is possibly (D). Finally, Sqr (n^2 + 2) = Sqr(2^2 + 2 )= Sqr(4 + 2) = Sqr 6 , which is not odd—eliminate (E). The answer is (D).
Comments
Post a Comment