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Manager
Joined: 27 Jan 2016
Posts: 150
Schools: ISB '18
GMAT 1: 700 Q50 V34

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14 Aug 2017, 23:39
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Thanks a lot for the collection Bunuel, I've been struggling with Inequalities, Probability & Combinatorics from so long. This post is very helpful to strengthen my base in Inequalities.

Do we have any such comprehensive thread for Probability and Combinatorics?
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Joined: 02 Sep 2009
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14 Aug 2017, 23:52
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Manager
Joined: 27 Jan 2016
Posts: 150
Schools: ISB '18
GMAT 1: 700 Q50 V34

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15 Aug 2017, 23:02
nikhilpoddar wrote:
What is the range for a ??

a + a^(-1) >2

a+1/a>2

(a-1)^2 / a >0

So a > 1

What is the OA?
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15 Aug 2017, 23:06
srikanth9502 wrote:
nikhilpoddar wrote:
What is the range for a ??

a + a^(-1) >2

a+1/a>2

(a-1)^2 / a >0

So a > 1

What is the OA?

a + 1/a > 2 holds true for 0 < a < 1 and a > 1.
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Manager
Joined: 27 Jan 2016
Posts: 150
Schools: ISB '18
GMAT 1: 700 Q50 V34

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15 Aug 2017, 23:38
Bunuel wrote:
srikanth9502 wrote:
nikhilpoddar wrote:
What is the range for a ??

a + a^(-1) >2

a+1/a>2

(a-1)^2 / a >0

So a > 1

What is the OA?

a + 1/a > 2 holds true for 0 < a < 1 and a > 1.

Is there any specified approach to solve these kind of questions?
Manager
Joined: 12 Aug 2017
Posts: 56
Location: United Kingdom
GMAT 1: 740 Q50 V40
GPA: 4

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14 Oct 2017, 13:48
WilDThiNg wrote:
Bunuel wrote:
Inequalities with Complications - Part I

BY KARISHMA, VERITAS PREP

Above we learned how to handle inequalities with many factors i.e. inequalities of the form $$(x – a)(x – b)(x – c)(x – d) > 0$$. This week, let’s see what happens in cases where the inequality is not of this form but can be manipulated and converted to this form. We will look at how to handle various complications.

Complication No. 1: $$(a – x)(x – b)(x – c)(x – d) > 0$$

We want our inequality to be of the form $$(x – a)$$, not $$(a – x)$$ because according to the logic we discussed last week, when x is greater than a, we want this factor to be positive. The manipulation involved is pretty simple: $$(a – x) = -(x – a)$$

So we get: $$– (x – a)(x – b)(x – c)(x – d) > 0$$

But how do we handle the negative sign in the beginning of the expression? We want the values of x for which the negative of this expression should be positive. Therefore, we basically want the value of x for which this expression itself (without the negative sign in the beginning) is negative.

We can manipulate the inequality to $$(x – a)(x – b)(x – c)(x – d) < 0$$

Or simply, multiply $$– (x – a)(x – b)(x – c)(x – d) > 0$$ by -1 on both sides. The inequality sign flips and you get $$(x – a)(x – b)(x – c)(x – d) < 0$$

e.g. Given: $$(4 – x)(2 – x)(-9 – x) < 0$$

We can re-write this as $$–(x – 4)(2 – x)(-9 – x) < 0$$

$$(x – 4)(x – 2)(-9 – x) < 0$$

$$-(x – 4)(x – 2)(x + 9) < 0$$

$$(x – 4)(x – 2)(x – (-9)) > 0$$ (multiplying both sides by -1)

Now the inequality is in the desired form.

Complication No 2: $$(mx – a)(x – b)(x – c)(x – d) > 0$$ (where m is a positive constant)

How do we bring $$(mx – a)$$ to the form $$(x – k)$$? By taking m common!

$$(mx – a) = m(x – a/m)$$

The constant does not affect the sign of the expression so we don’t have to worry about it.

e.g. Given: $$(2x – 3)(x – 4) < 0$$

We can re-write this as $$2(x – \frac{3}{2})(x – 4) < 0$$

When considering the values of x for which the expression is negative, 2 has no role to play since it is just a positive constant.

Now let’s look at a question involving both these complications.

Question 1: Find the range of x for which the given inequality holds.

$$-2x^3 + 17x^2 – 30x > 0$$

Solution:

Given: $$-2x^3 + 17x^2 – 30x > 0$$

$$x(-2x^2 + 17x – 30) > 0$$ (taking x common)

$$x(2x – 5)(6 – x) > 0$$ (factoring the quadratic)

$$2x(x – \frac{5}{2})(-1)(x – 6) > 0$$ (take 2 common)

$$2(x – 0)(x – \frac{5}{2})(x – 6) < 0$$ (multiply both sides by -1)

This inequality is in the required form. Let’s draw it on the number line.

We are looking for negative value of the expression. Look at the ranges where we have the negative sign.

The ranges where the expression gives us negative values are $$5/2 < x < 6$$ and x < 0.

Hence, the inequality is satisfied if x lies in the range $$\frac{5}{2} < x < 6$$ or in the range $$x < 0$$.

Plug in some values lying in these ranges to confirm.

In the next post, we will look at some more variations which can be brought into this form.

[Reveal] Spoiler:
Attachment:
Ques4.jpg

Hi:

Please could you help me understand how to determine which sign to assign to which range in the above graph?

Thanks

as per my understanding the extreme right one you can take +ve & then assign an alternate sign to each region.
Try yourself with a few equations. You can clarify that yourself.

cheers
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Manager
Joined: 02 Jan 2017
Posts: 90
Location: Pakistan
Concentration: Finance, Technology
GMAT 1: 650 Q47 V34
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28 Dec 2017, 04:16
Bunuel wrote:
Or Just Use Inequalities!

BY KARISHMA, VERITAS PREP

If you are wondering about the absurd title of this post, just take a look at the above post's title. It will make much more sense thereafter. This post is a continuation of last week’s post where we discussed number plugging. Today, as per students’ request, we will look at the inequalities approach to the same official question. You will need to go through our inequalities post to understand the method we will use here.

Recall that, given $$a < b$$, $$(x – a)(x – b) < 0$$ gives us the range $$a < x < b$$ and $$(x – a)(x – b) > 0$$ gives us the range $$x < a$$ or $$x > b$$.

Question: If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2?
(I) x^2 < 2x < 1/x
(II) x^2 < 1/x < 2x
(III) 2x < x^2 < 1/x

(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III

Solution: The question has three complex inequalities. We will take each in turn. Note that each inequality consists of two more inequalities. We will split the complex inequality into two simpler inequalities e.g. x^2 < 2x < 1/x gives us x^2 < 2x and 2x < 1/x. Next we will find the range of values of x which satisfy each of these two inequalities and we will see if the two ranges have an overlap i.e. whether there are any values of x which satisfy both these simpler inequalities. If there are, it means there are values of x which satisfy the entire complex inequality too. Things will become clearer once we start working on it so hold on.

Let’s look at each inequality in turn. We start with the first one:

(I) x^2 < 2x < 1/x

We split it into two inequalities:

(i) x^2 < 2x

We can rewrite x^2 < 2x as x^2 – 2x < 0 or x(x – 2) < 0.

We know the range of x for such inequalities can be easily found using the curve on the number line. This will give us 0 < x < 2.

(ii) 2x < 1/x

It can be rewritten as x^2 – 1/2 < 0 (Note that since x must be positive, we can easily multiply both sides of the inequality with x)
[ Secondly would you explain this step again i did not understand how multiplying by x gives us this equation] $$-1/?2< x < 1/?2$$

This gives us the range -1/?2 < x < 1/?2 (which is 0 < x < 1/?2 since x must be positive). [ Question mark is square root i think]

Is there a region of overlap in these two ranges i.e. can both inequalities hold simultaneously for some values of x? Yes, they can hold for 0 < x < 1/?2. Hence, x^2 < 2x < 1/x will be true for the range 0 < x < 1/?2. So this could be the correct ordering. Let’s go on to the next complex inequality.

(II) x^2 < 1/x < 2x

Again, let’s break up the inequality into two parts:

(i) x^2 < 1/x

x^1 < 1/x is rewritten as x^3 – 1 < 0 which gives us x < 1.

(ii) 1/x < 2x

1/x < 2x is rewritten as x^2 – 1/2 > 0 which gives us x < -1/?2 (not possible since x must be positive) or x > 1/?2

Can both x < 1 and x > 1/?2 hold simultaneously? Sure! For 1/?2 < x < 1, both inequalities will hold and hence x^2 < 1/x < 2x will be true. So this could be the correct ordering too.

(III) 2x < x^2 < 1/x

The inequalities here are:

(i) 2x < x^2

2x < x^2 can be rewritten as x(x – 2) > 0 which gives us x < 0 (not possible) or x > 2.

(ii) x^2 < 1/x

x^2 < 1/x gives us x^3 – 1 < 0 i.e. x < 1

Can x be less than 1 and greater than 2 simultaneously? No. Therefore, 2x < x^2 < 1/x cannot be the correct ordering.

Is this method simpler?
Hi bunuel (Or any other expert) i have highlighted a portion in green and red . Could you clarify those points. Thankyou
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Joined: 02 Aug 2009
Posts: 5539

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28 Dec 2017, 05:32
mtk10 wrote:
Hi bunuel (Or any other expert) i have highlighted a portion in green and red . Could you clarify those points. Thankyou

(ii)$$2x < \frac{1}{x}$$...
Multiply by x..

$$2x*x < \frac{1}{x}*x.........2x^2<1..........x^2<\frac{1}{2}.....x^2-1/2<0$$...
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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