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

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