If x is positive real number, find the range of values for which (frac{1}{3})x + (frac{1}{2}) > (frac{1}{4})x
-
A.
x > -(frac{1}{6}) -
B.
x > 0 -
C.
0 -
D.
0
Correct Answer: Option A
Explanation
(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})
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