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 = 12x2 – 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