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Correct answer is C
|kk4k|+|23−1k|=6
|kk4k|=(k2−4k)
|23−1k|=(2k+3)
\therefore (k^{2} - 4k) + (2k + 3) = k^{2} - 2k + 3 = 6
k^{2} - 2k - 3 = 0, factorising, we have k + 1 = 0 or k - 3 = 0
Since k > 0, k = 3.
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