If x is a positive real number, find the range of values for which $\frac{1}{3x}$ + $\frac{1}{2}$ > $\frac{1}{4x}$

0 > - $\frac{1}{6}$

x > 0

0 < x < 4

0 < x < $\frac{1}{6}$

Correct answer is D

$\frac{1}{3x}$ + $\frac{1}{2}$ > $\frac{1}{4x}$

= $\frac{2 + 3x}{6x}$ > $\frac{1}{4x}$

= 4(2 + 3x) > 6x = 12x² - 2x = 0

= 2x(6x - 1) > 0 = x(6x - 1) > 0

Case 1 (-, -) = x < 0, 6x -1 < 0

= x < 0, x < $\frac{1}{6}$ = x < $\frac{1}{6}$ (solution)

Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > $\frac{1}{6}$

Combining solutions in cases(1) and (2)

= x > 0, x < $\frac{1}{6}$ = 0 < x < $\frac{1}{6}$

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