If x is a positive real number, find the range of values for which 13x 1 3 x + 12 1 2 > 14x

Assessment: JAMB UTME - Mathematics - 1998 Subject: General Mathematics

Question 1 Report

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

0 > -

\frac{1}{6}

x > 0

0 < x < 4

0 < x <

\frac{1}{6}

Answer Details

$\frac{1}{3 x}$ + $\frac{1}{2}$ > $\frac{1}{4 x}$
= $\frac{2 + 3 x}{6 x}$ > $\frac{1}{4 x}$
= 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 < $\frac{1}{6}$

View Answer

If x is a positive real number, find the range of values for which 13x 1 3 x + 12 1 2 > 14x

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