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Further, say if the same expression was (x+2)(1-x)/(x-4)(x-7) and still the question was for what values of x is the expression positive, then ... make it x-1 and with the same roots, have the rightmost as -ve. Then we look for the +ve intervals and check for those intervals if the expression is positive. for examples, in this case, -2<x<1 and 4<x<7 both depict positive interval but only first range satisfies the condition. Please confirm
However, if for the same equation as mentioned, say the expression was (x+2)(x-1)/(x-4)(x-7) >0 and then we were asked to give the range where this is valid, then we would also multiply the -ve sign and make is <0 and then make the range after extreme right root -ve and provide all the intervals where it is negative. Please confirm
When you have (x-a)^2(x-b) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x - b) should be negative i.e. x - b < 0 or x < b. .
IMHO, it should be x<b and also x is not equal to a . so, one shouldn't totally ignore the squared term. We can ignore it, if the expression is <=0 correct me, if I am wrong
my question is - do we always have a sequence of + and - from rightmost to the left side. I mean is it possible to have + and then + again ? _________________
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Re: Inequalities trick [#permalink]
18 Oct 2012, 04:17
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GMATBaumgartner wrote:
gurpreetsingh wrote:
ulm wrote:
in addition: if we have smth like (x-a)^2(x-b) we don't need to change a sign when jump over "a".
yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a
Hi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ?
Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.
If the powers are even then the inequality won't be affected.
eg if u have to find the range of values of x satisfying (x-a)^2 *(x-b)(x-c) >0
just use (x-b)*(x-c) >0 because x-a raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero. _________________
Re: Inequalities trick [#permalink]
02 Dec 2012, 04:31
Expert's post
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GMATGURU1 wrote:
This is a great method but using this I am not getting an expected answer! Can Karishma and other please help me find where I am wrong?
The problem is (x - 2)(x + 1)/x(x+2) <= 0 The graph will be +++ -2 ------ -1 +++++ 0 ------- 2 ++++++ So the answer will be 0 < x <= 2 and -2 < x < -1
But in Arun Sharma book, the answer given is -2 < x <= 2
Can you help?
Solution set for \(\frac{(x - 2)(x + 1)}{x(x+2)}\leq{0}\) is \(-2<x\leq{-1}\) and \(0<x\leq{2}\). So, you've done everything right, except < sign for the second range, which should be <=. _________________
Re: Inequalities trick [#permalink]
02 Dec 2012, 04:45
Bunuel wrote:
GMATGURU1 wrote:
This is a great method but using this I am not getting an expected answer! Can Karishma and other please help me find where I am wrong?
The problem is (x - 2)(x + 1)/x(x+2) <= 0 The graph will be +++ -2 ------ -1 +++++ 0 ------- 2 ++++++ So the answer will be 0 < x <= 2 and -2 < x < -1
But in Arun Sharma book, the answer given is -2 < x <= 2
Can you help?
Solution set for \(\frac{(x - 2)(x + 1)}{x(x+2)}\leq{0}\) is \(-2<x\leq{-1}\) and \(0<x\leq{2}\). So, you've done everything right, except < sign for the second range, which should be <=.
Yes even I think so that I am correct but in that book (famous Arun Sharma book for CAT), he writes the following...
case 1) Numerator positive and denominator negative: This occurs only between -2 < x < -1 case 2) Numerator negative and denominator positive: Numerator is negative when (x - 2) and (x + 1) take opposite signs. This can be got for ... case a) x - 2 < 0 and x + 1 > 0 i.e. x < 2 and x >- 1 case b) x - 2 > 0 and x + 1 < 0 i.e. x > 2 and x < -1. --- Cannot happen
Hence answer is -2 < x <= 2.
-------------------------------------
The point I want to make here is, using our curve savy method we are getting a different answer. Using our answer x is not coming between 0 and -1. but the answer given in the book take the full range of -2 < x <= 2
So do we need any more tweaks in out curve method?
Re: Inequalities trick [#permalink]
02 Dec 2012, 05:01
Expert's post
GMATGURU1 wrote:
Bunuel wrote:
GMATGURU1 wrote:
This is a great method but using this I am not getting an expected answer! Can Karishma and other please help me find where I am wrong?
The problem is (x - 2)(x + 1)/x(x+2) <= 0 The graph will be +++ -2 ------ -1 +++++ 0 ------- 2 ++++++ So the answer will be 0 < x <= 2 and -2 < x < -1
But in Arun Sharma book, the answer given is -2 < x <= 2
Can you help?
Solution set for \(\frac{(x - 2)(x + 1)}{x(x+2)}\leq{0}\) is \(-2<x\leq{-1}\) and \(0<x\leq{2}\). So, you've done everything right, except < sign for the second range, which should be <=.
Yes even I think so that I am correct but in that book (famous Arun Sharma book for CAT), he writes the following...
case 1) Numerator positive and denominator negative: This occurs only between -2 < x < -1 case 2) Numerator negative and denominator positive: Numerator is negative when (x - 2) and (x + 1) take opposite signs. This can be got for ... case a) x - 2 < 0 and x + 1 > 0 i.e. x < 2 and x >- 1 case b) x - 2 > 0 and x + 1 < 0 i.e. x > 2 and x < -1. --- Cannot happen
Hence answer is -2 < x <= 2.
-------------------------------------
The point I want to make here is, using our curve savy method we are getting a different answer. Using our answer x is not coming between 0 and -1. but the answer given in the book take the full range of -2 < x <= 2
So do we need any more tweaks in out curve method?
Thanks!
I'm not familiar with that source but with this particular question it's wrong.
The numerator is positive and the denominator is negative: \(-2<x\leq{-1}\); The numerator is negative and the denominator positive: \(0<x\leq{2}\) (the source didn't consider the case when the denominator is positive).
Jut to check, try x=-1/2 to see that for this value the inequality does not hold true.
Re: Inequalities trick [#permalink]
20 Dec 2012, 20:00
Expert's post
1
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maddyboiler wrote:
Excellent post. Till now we have all seen problems in the format f(x) < 0 where f(x) is written in its factors for (x-a)(x-b)...
what if we have something like f(x) < k "k is a constant" (x-a)(x-b)(x-c) < k How do we solve these kind of questions?
The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side.
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