If x is positive real number find the range of ... - JAMB Mathematics 2020 Question
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 < x < 6
D
0 < x <\(\frac{1}{6}\)
correct option: a
\(\frac{1}{3x}\) + \(\frac{1}{2}\)x
=> \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\)
= 4(2 + 3x) > 6x = 12x\(^2\) - 2x = 0
= 2x(6x - 1) > 0 = x(6x - 1) > 0
(-, -) = x < 0, 6x - 1 > 0. - - - - - - - 1
= x < 0, x < \(\frac{1}{6}\) (solution)
(+, +) = x > 0, 6x - 1 > 0 = x > 0 - - - - -- - - - 2
x > \(\frac{1}{6}\)
Combining (1) and (2):
= x > 0, x < \(\frac{1}{6}\)
= 0 < x < \(\frac{1}{6}\)
Please share this, thanks:
Add your answer
No responses