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17 Feb 2011, 00:20
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
45% (02:45) correct
55%
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wrong
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For what range of values of 'x' will the inequality 15x - (2/x) > 1?
A. x > 0.4
B. x < 1/3
C. -1/3 < x < 0.4, x > 15/2
D. -1/3 < x < 0, x > 2/5
E. x < -1/3 and x > 2/5
A. x > 0.4
B. x < 1/3
C. -1/3 < x < 0.4, x > 15/2
D. -1/3 < x < 0, x > 2/5
E. x < -1/3 and x > 2/5
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17 Feb 2011, 03:38
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fluke
It's not a good question because it is not formed correctly or because it is subpar or both?
What if the question read, "Which of the following represents the complete range of x over which 15x - (2/x) > 1"?
Then; D would be the only choice, right?
What if the question read, "Which of the following represents the complete range of x over which 15x - (2/x) > 1"?
Then; D would be the only choice, right?
Right.
fluke
More important question is:
How x<-1/3, -1/3<x<0, 0<x<2/5 and x>2/5; get reduced to -1/3<x<0, x>2/5.
Is there any comprehensive material available for these types of inequalities and ranges apart from the shortcut method suggested by gurpreetsingh?
How x<-1/3, -1/3<x<0, 0<x<2/5 and x>2/5; get reduced to -1/3<x<0, x>2/5.
Is there any comprehensive material available for these types of inequalities and ranges apart from the shortcut method suggested by gurpreetsingh?
This is actually quite simple. We want to solve \(\frac{(3x+1)(5x-2)}{x}>0\):
1. Find zeros of the multiples (the values of x for which x, 3x+1 and 5x-2 equal to zero): -1/3, 2/5 and 0;
2. Arrange them in ascending order to get 4 ranges: \(x<-\frac{1}{3}\), \(-\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\);
3. Check extreme value: if \(x\) is some very large number (so some number from the 4th range) then \(\frac{(3x+1)(5x-2)}{x}\) will obviously be positive;
4. Trick: as in the 4th range expression is positive then in 3rd it'll be negative, in 2nd it'l be positive again and finally in 1st it'll be negative: - + - +, so expression is positive for 2nd range: \(-\frac{1}{3}<x<0\), and for fourth range: \(x>\frac{2}{5}\).
This works for all such kind of inequalities.
Links in from previous post:
everything-is-less-than-zero-108884.html?hilit=extreme#p868863, here: inequalities-trick-91482.html
xy-plane-71492.html?hilit=solving%20quadratic#p841486
One more: data-suff-inequalities-109078.html
17 Feb 2011, 02:58
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fluke
For what range of values of 'x' will the inequality 15x - (2/x) > 1?
A. x > 0.4
B. x < 1/3
C. -1/3 < x < 0.4, x > 15/2
D. -1/3 < x < 0, x > 2/5
E. x < -1/3 and x > 2/5
A. x > 0.4
B. x < 1/3
C. -1/3 < x < 0.4, x > 15/2
D. -1/3 < x < 0, x > 2/5
E. x < -1/3 and x > 2/5
\(15x-\frac{2}{x}>1\) ---> \(\frac{15x^2-x-2}{x}>0\) --> \(\frac{(3x+1)(5x-2)}{x}>0\) -->3 expressions change their signs at: -1/3, 0, 2/5, so 4 ranges: \(x<-\frac{1}{3}\), \(-\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\) --> in rightmost range expression is positive: so -+-+ --> \(-\frac{1}{3}<x<0\), and \(x>\frac{2}{5}\).
Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve \(15x-\frac{2}{x}>1\) for the true ranges then the answer is D but if the question asks for which of the following ranges inequality \(15x-\frac{2}{x}>1\) holds true than the answer could be D as well as A.
P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero.
Check for more about the approach used here: everything-is-less-than-zero-108884.html?hilit=extreme#p868863, here: inequalities-trick-91482.html and here: xy-plane-71492.html?hilit=solving%20quadratic#p841486
Not a good question.
