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

A.
0 > (frac{1}{6}) 
B.
x > 0 
C.
0 
D.
0
Correct Answer: Option D
Explanation
(frac{1}{3x}) + (frac{1}{2}) > (frac{1}{4x})
= (frac{2 + 3x}{6x}) > (frac{1}{4x})
= 4(2 + 3x) > 6x = 12x^{2} – 2x = 0
= 2x(6x – 1) > 0 = x(6x – 1) > 0
Case 1 (, ) = x
= x
Case 2 (+, +) = x > 0, 6x 1 > 0 = x > 0, x > (frac{1}{6})
Combining solutions in cases(1) and (2)
= x > 0, x