Hi,
Can anyone please help me in solving the following inequalities:-
(1). x/(x + 2) <= 1/x
My approach :- x/(x + 2) - 1/x <= 0
After calculating, (x - 2)(x + 1) / x(x + 2) <= 0
What is after this ? how can i get the inequality ? What is the right approach ?
(2). x/(x - 3) <= 1/x
Please help in the above questions.
Thanks
Sachin
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Inequality doubt
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sachin_yadav
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Anju@Gurome
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Please go through the post I made here >> https://www.beatthegmat.com/gmat-prep-ea ... tml#602154
Critical points of this inequality are : -2, -1, 0, and 2.
As we are looking for less than or equal to zero region, our solution set is -2 < x ≤ -1 and 0 < x ≤ 2
(x - 2)(x + 1)/[x(x + 2)] ≤ 0sachin_yadav wrote:x/(x + 2) <= 1/x
My approach :- x/(x + 2) - 1/x <= 0
After calculating, (x - 2)(x + 1) / x(x + 2) <= 0
Critical points of this inequality are : -2, -1, 0, and 2.
As we are looking for less than or equal to zero region, our solution set is -2 < x ≤ -1 and 0 < x ≤ 2
Anju Agarwal
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--> x/(x - 3) - 1/x ≤ 0sachin_yadav wrote:(2). x/(x - 3) ≤ 1/x
--> [x² - x + 3]/[x(x - 3)] ≤ 0
Now, x² - x + 3 is always positive as
- for |x| > 1, x² > x --> (x² - x) > 0 ---> (x² - x + 3) > 0
for |x| < 1, -1 < (x² - x) < 0 ---> (x² - x + 3) > 0
For x = 0, (x² - x + 3) = 3 > 0
--> Either {x < 0 and (x - 3) > 0} or {x > 0 and (x - 3) < 0}
--> Either {x < 0 and x > 3} or {x > 0 and x < 3}
The first one is not possible for any x.
Hence, our final solution is 0 < x < 3
Last edited by Anju@Gurome on Wed Apr 17, 2013 7:57 am, edited 1 time in total.
Anju Agarwal
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Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.
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Wavy Curve Method : Solving complex inequalities in a matter of seconds.
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GMATGuruNY
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Check my post here:sachin_yadav wrote:Hi,
Can anyone please help me in solving the following inequalities:-
(1). x/(x + 2) <= 1/x
My approach :- x/(x + 2) - 1/x <= 0
After calculating, (x - 2)(x + 1) / x(x + 2) <= 0
What is after this ?
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misterholmes
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Hey Sachin, without anything to go on for x, you have to look out for two kinds of trouble, when dividing by 0 and when multiplying (or dividing) by a negative number. When you put your equation (I) terms over a common denominator, the inequality could change directions on you, depending on the x values you're considering.sachin_yadav wrote:Hi,
Can anyone please help me in solving the following inequalities:-
(1). x/(x + 2) <= 1/x
My approach :- x/(x + 2) - 1/x <= 0
After calculating, (x - 2)(x + 1) / x(x + 2) <= 0
What is after this ? how can i get the inequality ? What is the right approach ?
(2). x/(x - 3) <= 1/x
Please help in the above questions.
Thanks
Sachin
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jainpiyushjain
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Dear AnjuAnju@Gurome wrote: Now, x² - x + 3 is always positive asSo our effective inequality is 1/[x(x - 3)] ≤ 0 ---> x(x - 3) < 0 (I replaced ≤ with ≤ because 1/[x(x - 3)] can never be equal to zero)
- for |x| > 1, x² > x --> (x² - x) > 0 ---> (x² - x + 3) > 0
for |x| < 1, -1 < (x² - x) < 0 ---> (x² - x + 3) > 0
For x = 0, (x² - x + 3) = 3 > 0
--> Either {x < 0 and (x - 3) > 0} or {x > 0 and (x - 3) < 0}
--> Either {x < 0 and x > 3} or {x > 0 and x < 3}
The first one is not possible for any x.
Hence, our final solution is 0 < x < 3
Can you please explain this question again. I was not able to understand how did you prove x² - x + 3 is always positive and why did you consider |x| > 1 and |x| < 1.
Thank you
jainpiyushjain wrote:Simply put, if we consider x^2 - x + 3 => x = negative => x^2 positive, -x will be positive, +3 (positive) --> Hence, expression positive.Anju@Gurome wrote: Now, x² - x + 3 is always positive asSo our effective inequality is 1/[x(x - 3)] ≤ 0 ---> x(x - 3) < 0 (I replaced ≤ with ≤ because 1/[x(x - 3)] can never be equal to zero)
- for |x| > 1, x² > x --> (x² - x) > 0 ---> (x² - x + 3) > 0
for |x| < 1, -1 < (x² - x) < 0 ---> (x² - x + 3) > 0
For x = 0, (x² - x + 3) = 3 > 0
--> Either {x < 0 and (x - 3) > 0} or {x > 0 and (x - 3) < 0}
--> Either {x < 0 and x > 3} or {x > 0 and x < 3}
The first one is not possible for any x.
Hence, our final solution is 0 < x < 3[/quot
Dear Anju
Can you please explain this question again. I was not able to understand how did you prove x² - x + 3 is always positive and why did you consider |x| > 1 and |x| < 1.
Thank you
x= positive => x^2 = positive, -x negative but less than x^2 [hence x^2 - x is positive] +3 which makes the expression positive, irrespective of whatever value x holds.
Both the problems will have a similar approach. You have to take care of 2 things :sachin_yadav wrote:Hi,
Can anyone please help me in solving the following inequalities:-
(1). x/(x + 2) <= 1/x
My approach :- x/(x + 2) - 1/x <= 0
After calculating, (x - 2)(x + 1) / x(x + 2) <= 0
What is after this ? how can i get the inequality ? What is the right approach ?
(2). x/(x - 3) <= 1/x
Please help in the above questions.
Thanks
Sachin
[1] Denominator cannot be zero.
[2] When Numerator is positive, denominator will be negative and vice versa [since the expression is negative]
problem 1.
(x-2)(x-3)/x(x+2) <=0
Numerator positive or zero: x E [-inf, -1]U[2, inf]; Denominator negative: x E (-2,0) => x E (-2,-1]
Numerator negative or zero: x E [-1,2]; Denominator positive: x E (-inf., -2)U(0, inf)=> x E (0,2]
problem 2.
Numerator always positive, hence Denominator should be negative for the inequality to hold true.
x(x-3) < 0 ---> x E (0,3)
