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19 Mar 2015, 06:26
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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:
75% (01:36) correct
25%
(01:49)
wrong
based on 665
sessions
History
Date
Time
Result
Not Attempted Yet
If \(f(x) = 12 - \frac{x^2}{2}\) and \(f(2k) = 2k\), what is one possible value for k?
A. 2
B. 3
C. 4
D. 6
E. 8
A. 2
B. 3
C. 4
D. 6
E. 8
19 Mar 2015, 09:16
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12 - ( 2k)^2/2 = 2k
4k = 24 - 4k^2
(k-2)(k+3) = 0
k = 2
k = -3
Answer : A
4k = 24 - 4k^2
(k-2)(k+3) = 0
k = 2
k = -3
Answer : A
19 Mar 2015, 21:27
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Bunuel
Once you form the equation f(2k) = \(12 - (2k)^2/2 = 2k\)
\(12 - 2k^2 = 2k\), you don't even need to solve it. Note that 12 is a small number and when you subtract twice of a square from it, you still get a positive number 2k (all values of k given in the options are positive). So k must be very small --> k^2 must be less than 6 so k can only be 2 from the given options.
Verify: \(12 - 2*2^2 = 2*2\) - Correct
Answer (A)
