If x is positive real number, find the range of values for which \(\frac{1}{3x}\) + \(\frac{1}{2}\)x = \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\) 

A.

x > -\(\frac{1}{6}\)

B.

x > 0

C.

0 < x < 6

D.

0 < x <\(\frac{1}{6}\)

Correct answer is A

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

= 4(2 + 3x) > 6x = 12x\(^2\) - 2x = 0

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

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

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