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14 Jan 2021, 11:15
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
65% (02:19) correct
35%
(02:16)
wrong
based on 188
sessions
History
Date
Time
Result
Not Attempted Yet
If there are no real numbers that satisfy the inequality \(3x – k(x + 2) ≥ 10 – 3(1 + x)\), where k is a constant, what is the value of k ?
(A) 0
(B) 1
(C) 2
(D) 4
(E) 6
(A) 0
(B) 1
(C) 2
(D) 4
(E) 6
24 Jan 2021, 02:29
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\(3x – k(x + 2) ≥ 10 – 3(1 + x)\)
\(3x-kx-2k ≥ 10-3-3x\)
\((6-k)x ≥ 7+2k\)
\(x ≥ \frac{7+2k}{(6-k)}\)
(6-k) can't be equal to 0 or k can't be equal to 6.
E
\(3x-kx-2k ≥ 10-3-3x\)
\((6-k)x ≥ 7+2k\)
\(x ≥ \frac{7+2k}{(6-k)}\)
(6-k) can't be equal to 0 or k can't be equal to 6.
E
Bunuel
24 Jan 2021, 04:32
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Bunuel
Let x=10.
Plugging x=10 into \(3x – k(x + 2) ≥ 10 – 3(1 + x)\), we get:
30 - k(12) ≥ 10 - 3(11)
-12k ≥ -53
k ≤ 53/12
Of the five answer choices, only E is not less than 53/12.
Implication:
If k=0, k=1, k=2, or k=4, then the inequality has at least one real solution (x=10).
Eliminate A, B, C and D.
