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24 Jan 2024, 09:57
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
78% (01:38) correct
22%
(01:56)
wrong
based on 615
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Time
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Which of the following represents an even integer for each integer value of n?
A. \(n^2 - 3n + 4\)
B. \(n^2 + 2n + 4\)
C. \(2n^2 + 3n - 1\)
D. \(3n^2 + 2n - 3\)
E. \(4n^2 - 3n + 1\)
2024-01-24_20-55-49.png [ 24.62 KiB | Viewed 11708 times ]
A. \(n^2 - 3n + 4\)
B. \(n^2 + 2n + 4\)
C. \(2n^2 + 3n - 1\)
D. \(3n^2 + 2n - 3\)
E. \(4n^2 - 3n + 1\)
Attachment:
2024-01-24_20-55-49.png [ 24.62 KiB | Viewed 11708 times ]
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24 Jan 2024, 10:30
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guddo
Easiest way...
Substitute n as one value of even and one value of odd.
n as 0.
A. \(n^2 - 3n + 4=4\)
B. \(n^2 + 2n + 4=4\)
C. \(2n^2 + 3n - 1=-1\)
D. \(3n^2 + 2n - 3=-3\)
E. \(4n^2 - 3n + 1=1\)
Discard C, D and E, and substitute n as 1 in A and B.
A. \(n^2 - 3n + 4=1-3+4=2\)
B. \(n^2 + 2n + 4=1+2+4=7\)
Only A left
General Discussion
24 Jan 2024, 10:34
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guddo
Answer choice elimination
A. \(n^2 - 3n + 4\)
- n ⇒ odd
\(n^2 - 3n + 4\) = odd - odd + even = even - n ⇒ even
\(n^2 - 3n + 4\) = even - even + even = even
Correct.
B. \(n^2 + 2n + 4\)
- n ⇒ odd
\(n^2 + 2n + 4\) = odd + even + even = odd - n ⇒ even
\(n^2 + 2n + 4\) = even + even + even = even
Incorrect
C. \(2n^2 + 3n - 1\)
- n ⇒ odd
\(2n^2 + 3n - 1\) = even + odd - odd = even - n ⇒ even
\(2n^2 + 3n - 1\) = even + even - odd = odd
Incorrect
D. \(3n^2 + 2n - 3\)
- n ⇒ odd
\(3n^2 + 2n - 3\) = odd + even - odd = even - n ⇒ even
\(3n^2 + 2n - 3\) = even + even - odd = odd
Incorrect
E. \(4n^2 - 3n + 1\)
- n ⇒ odd
\(4n^2 - 3n + 1\) = even - odd + even = odd - n ⇒ even
\(4n^2 - 3n + 1\) = even - even + odd = odd
Incorrect
Option A
P.S. - I wouldn't try any other options once the correct option is found.
