If x is a positive real number find the range o... - JAMB Mathematics 1998 Question
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 < x < 4
D
0 < x < \(\frac{1}{6}\)
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Correct Option: D
(\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 < 0, 6x -1 < 0
= x < 0, x < (\frac{1}{6}) = 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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