Inequalities trick

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.

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.

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.


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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!

Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12?

Cheers,
Danny

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.

e.g.
(x + 2)(x + 3) < 2
x^2 + 5x + 6 - 2 < 0
x^2 + 5x + 4 < 0
(x+4)(x+1) < 0

Now use the concept.

I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. a<b<c<d

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Hi Gurpreet,
Thanks for the wonderful method.
I am trying to understand it so that i can apply it in tests.
Can you help me in applying this method to the below expression to find range of x.
x^3 – 4x^5 < 0?

I am getting the roots as -1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.

Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0
Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2.

Now let me add another factor: (x+8)(x+2)(x-1)(x-7)
Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes.

So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region.

Note: Make sure that the factors are of the form (ax - b), not (b - ax)...

e.g. (x+2)(x-1)(7 - x)<0

Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')
Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'
Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.

i have a doubt in the highlighted region. U have said that it will be always +ve but in a bunel post he has asked to substitute the extreme values and if the f(x) is -ve then the right most of the inequality will be -ve.

Please clarify me on this.

Can you please explain the above mentioned concept in relation to the following question?
Is a > O?
(1) a^3 - 0 < 0
(2) 1- a^2 > 0

Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0?
Sorry, but finding it difficult to understand.