JAMB - Mathematics (1994 - No. 16)
Find the range of values of x for which \(\frac{1}{x}\) > 2 is true
x < \(\frac{1}{2}\)
x < 0 or x < \(\frac{1}{2}\)
0 < x < \(\frac{1}{2}\)
1 < x < 2
Explanation
\(\frac{1}{x}\) > 2 = \(\frac{x}{x^2}\) > 2
x > 2x2
= 2x2 < x
= 2x2 - x < 0
= x(2x - 10 < 0
Case 1(+, -) = x > 0, 2x - 1 < 0
x > 0, x < \(\frac{1}{2}\) (solution)
Case 2(-, 4) = x < 0, 2x - 1 > 0
x < 0, x , \(\frac{1}{2}\) = 0
x > 2x2
= 2x2 < x
= 2x2 - x < 0
= x(2x - 10 < 0
Case 1(+, -) = x > 0, 2x - 1 < 0
x > 0, x < \(\frac{1}{2}\) (solution)
Case 2(-, 4) = x < 0, 2x - 1 > 0
x < 0, x , \(\frac{1}{2}\) = 0
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