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# INEQUATIONS(Inequalities) Part 2

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Updated on: 26 Aug 2013, 09:45
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This is the second part of this Article. If you have not gone thru the first part, it is highly recommended to study the same at first and then study this. Here is the link to the first Part of the Article inequations-inequalities-part-154664.html#p1238004

Modulus or Absolute Values |x|
Many times students scared seeing the modules sign in an inequality question. Indeed, dealing with inequality question that involves expressions in modules can be tricky. However once you studied the theory and mastered the results pertaining to modules, any such question will be appeared like an ordinary Inequality question.

Some Important Results

In all there are 6 results out of which 4 are basic and remaining 2 are derived from the first two basic results.

4 basic inequality with absolute values to remember:

(A) |x| ≤ a -------> -a ≤ x ≤ a
(B) |x| ≥ a -------> x ≤ -a or x ≥ a
(C) |x| < 0 -------> No solution (Remember :- |x|≤ 0 can have a solution for x=0)
(D) |x| ≥ 0 -------> -∞ < x < ∞ (all real numbers)

Result 1 :- If a is a positive real number, then |x| < a ----> -a < x < a

Let’s Prove this result

We have |x| < a
We know that |x| = (x if x ≥ 0 ) and (-x if x < 0)
So, we consider the following cases :

CASE I :- When x ≥ 0
In this case |x| = x -----------> Therefore x < a
Thus in this case the solution set of the given Inequation is given by
X ≥ 0 and x < a -----------> 0 ≤ x < a

CASE II :- When x < 0
In this case |x| = - x -----------> Therefore - x < a --------> x > -a
Thus in this case the solution set of the given Inequation is given by
X < 0 and x > - a -----------> -a < x < 0

Combining two cases together, we get (-a < x < 0) or ( 0 ≤ x < a)

Once we take extreme points, we get –a < x < a

Result 2 :- If a is a positive real number, then |x| > a -----> x > a or x < -a

We have |x| > 0

CASE I :- When x ≥ 0
In this case |x| = x -----------> Therefore x > a
Thus in this case the solution set of the given Inequation is given by
x ≥ 0 and x > a -----------> x > a

CASE I :- When x < 0
In this case |x| = -x --------------> Therefore - x > a -----------> x < -a
Thus in this case the solution set of the given Inequation is given by
x < 0 and x < -a ---------------> x < -a

Combining two cases, we get x > a or x < -a

Result 3 :- For |x| < 0 -----> No Solution.

Result 4 :- For |x| ≥ 0 ------> -∞ < x < ∞ (all real numbers)

Here are another two results which are derived from above basic results

Result 5 :- Let r be positive real and a be a fixed number, then |x - a|< r ------> (a – r) < x < (a + r)
We know that (From Result 1) |x| < a = -a < x < a
We have |x - a|< r ------------> -r < x-a < r ----------> a – r < x < a + r

Result 6 :- Let r be positive real and a be a fixed number, then |x - a|> r ------> x < a – r or x > a + r

We know that (From Result 2) |x| > a = x > a or x < -a
We have |x - a|> r ------------> x – a > r -------> x > a + r ----------or ---------> x –a < - r -------> x < a – r

Example 1 :- |3x-2| ≤ $$\frac{1}{2}$$

We know that |x-a| ≤ r = a - r ≤ x ≤ a + r

|3x-2| ≤ $$\frac{1}{2}$$ ------> 2 - $$\frac{1}{2}$$ ≤ 3x ≤ 2 + $$\frac{1}{2}$$ ------> $$\frac{3}{2}$$ ≤ 3x ≤ $$\frac{5}{2}$$ --------> $$\frac{1}{2}$$ ≤ x ≤ $$\frac{5}{6}$$ -----> [ $$\frac{1}{2}$$, $$\frac{5}{6}$$ ]

Example 2 :- |x - 2| ≥ 5
|x - a| ≥ r = x ≥ a + r or x ≤ a – r
|x – 2| ≥ 5 -------> x ≥ 2 + 5 or x ≤ 2 – 5 -----------> x ≥ 7 or x ≤ -3 --------> (-∞, -3] or [7, ∞)

Example 3 :- 1 ≤ |x – 2| ≤ 3
1 ≤ |x – 2| -------> x ≥ 3 or x ≤ 1
|x – 2| ≤ 3 -------> -1 ≤ x ≤ 5

Solution of inequality -1 ≤ x ≤ 1 and 3 ≤ x ≤ 5 -------> [-1, 1] U [3, 5]

Example 4 :- |$$\frac{2}{(x-4)}$$| > 1 and x ≠ 4

$$\frac{2}{(|x-4|)}$$ > 1 ----------------------------------------------------------[|$$\frac{a}{b}$$| = $$\frac{(|a|)}{(|b|)}$$ ]

2 > |x-4|----------------- (We can multiply both sides of inequality by |x-4|, since we know |x-4|>0 always, for all x ≠ 4)

-2 + 4 < x < 4 + 2 -------> 2 < x < 6

We know x ≠ 4. Hence the solution set of inequality will be (2<x<4) U (4<x<6)

Example 5 :- If $$\frac{(|x|)}{(|3|)}$$ > 1, which of the following must be true?

A. x > 3
B. x < 3
C. x = 3
D. x ≠ 3
E. x < -3

|x| > |3| ----------------------------[Since |3| will always positive, we can multiply both sides by |3|]

|x| - |3| > 0 -------> |x|- 3 > 0 ---------> |x| > 3

x > 3 or x < -3

A. x > 3 Incorrect. This could be true. (x can be greater than 3 or less than -3)
B. X < 3 Incorrect. This could be true. (x can be greater than 3 or less than -3)
C. X = 3 Incorrect. This can never be true. (x can be greater than 3 or less than -3)
D. X ≠ 3 Correct. This must be true
E. X <-3 Incorrect. This could be true. (x can be greater than 3 or less than -3)

EXAMPLE 6 :- |1 – x| > 3

|x – 1| > 3 ----------------------------[|1 – x| = |x – 1|]

x > 4 or x < -2

if $$\frac{x}{(|x|)}$$ < x, Which of the following must be true about x ?

A) x > 1
B) x > -1
C) |x| < 1
D) |x| = 1
E) $$|x|^2$$ > 1

$$\frac{x}{(|x|)}$$ < x ------> Since |x| has to be positive, we can multiply both sides of an inequality by |x|

x < x |x| -----> x|x| - x > 0 -------> x (|x| - 1) > 0

Case I :- x and (|x|-1) are positive

In this case x > 0 and |x| - 1 > 0 -------> x > 1 or x < -1 -----------{ Applying |x| - a > r ----> x > a + r or x < a – r}
Therefore Range of x is x > 1

Case II :- x and (|x|-1) are negative

In this case x < 0 and |x| - 1 < 0 --------> -1 < x < 1 -------------{ Applying |x| - a < r ----> a – r < x < a + r}
Therefore range of x is -1 < x < 0

So we have that -1 < x < 0 or x > 1
x > 1 Incorrect
x > -1 Correct
|x| < 1 Incorrect
|x| = 1 Incorrect
|x|2 > 1 Incorrect

Here is the way to tackle Quadratic Inequalities questions.

Example 1 :- $$3x^2$$ – 7x + 4 ≤ 0

$$3x^2$$ – 7x + 4 ≤ 0 ---> $$3x^2$$ – 3x – 4x + 4 ≤ 0 -------> 3x(x-1) - 4(x-1) ≤ 0 -------> (3x-4)(x-1) ≤ 0

we get 1 and $$\frac{4}{3}$$ as critical points. We will place them on number line.

Since the number line is divided into three regions, now we can get 3 ranges of x i) x < 1 ii) 1 ≤ x ≤ $$\frac{4}{3}$$ iii) x > $$\frac{4}{3}$$

At this point we should understand that for the inequality (3x-4)(x-1) ≤ 0 to hold true, exactly one of (3x-4)and (x-1) should be negative and other one be positive.
Let’s examine 3 possible ranges one by one.

i) If x > $$\frac{4}{3}$$, obviously both the factors i.e. (3x-4)and (x-1) will be positive and in that case inequality would not hold true. So this cannot be the range of x

ii) If x is between 1 and $$\frac{4}{3}$$ both inclusive, (3x-4)will be negative or equal to zero and (x-1) will be positive or equal to zero. Hence with this range inequality holds true. Correct.

iii) If x < 1, both (3x-4)and (x-1) will be negative hence inequality will not hold true.

So the range of x that satisfies the inequality $$3x^2$$ – 7x + 4 ≤ 0 is (1 ≤ x ≤ $$\frac{4}{3}$$)

Let’s look at another question

Example 2 :- $$x^2$$ – 14x – 15 > 0

$$x^2$$ – 14x – 15 > 0 ------>
$$x^2$$ – 15x + x – 15 > 0 ------>
$$x^2$$ + x - 15x – 15 > 0 -------> (x+1)(x – 15) > 0

We have -1 and 15 as critical points. For the inequality above be true we would need both the factors either positive or negative

We can see here that whenever x takes the value greater than 15 or less than -1 both the factors becomes positive or negative respectively thereby satisfying the inequality. However whenever x takes the value between -1 and 15 one factor becomes positive and other one becomes negative. In that case inequality does not hold true. Hence the solution of inequality is x > 15 or x < -1

Algorithm to solve Quadratic Inequations ax2 + bx + c > or < 0

1) Obtain the Inequation

2) Obtain the factors of Inequation

3) Place them on number line. The number line will get divided into the three regions

4) Mark the leftmost region with + sign and rest two regions with – and + signs respectively.

5) If the Inequation is of the form $$ax^2$$ + bx + c < 0, the region having minus sign will be the solution of inequality

6) If the Inequation is of the form $$ax^2$$ + bx + c > 0, the region having plus sign will be the solutions of inequality

Most useful methods for solving quadratic inequalities –Proposed by legendary Members

BUNUEL :- if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476
VeritasPrepKARISHMA :- inequalities-trick-91482.html#p804990
Zarrolou
:- tips-and-tricks-inequalities-150873.html#p1211920

Example 3 :- if-9-x-2-0-which-of-the-following-describes-all-153160.html#p1227496

If 9 - $$x^2$$ ≥ 0, which of the following describes all possible values of x ?

A. 3 ≥ x ≥ -3
B. x ≥ 3 or x ≤ -3
C. 3 ≥ x ≥ 0
D. -3 ≥ x
E. 3 ≤ x

$$x^2$$ - 9 ≤ 0 ----> (x-3)(x+3) ≤ 0 -------> -3 ≤ x ≤ 3 ---------> Option A

Example 4 :- if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p728428

If x is an integer, what is the value of x?
(1) $$x^2$$ - 4x + 3 < 0
(2) $$x^2$$ + 4x +3 > 0

S1) $$x^2$$ - 4x + 3 < 0 ----> $$x^2$$ –x – 3x + 3 < 0 -------> x(x-1)-3(x-1)<0 -------> (x-1)(x-3)<0 -------> 1<x<3 ------------------> since x is an integer, x has to be 2. Sufficient.

S2) $$x^2$$ + 4x +3 > 0 -------> $$x^2$$ +x + 3x + 3 > 0 -------> x(x+1)+3(x+1)>0 -------> (x+1)(x+3)>0 -------> x < -3 or x > -1 -------------> x can take multiple values such as 0, 5, -4 etc… Insufficient.

Choice A.

Example 5 :- is-k-2-k-147133.html#p1181488

Is $$k^2$$ + k - 2 > 0 ?

(1) k < 1
(2) k < -2

We have $$k^2$$ + k - 2 > 0 -------> k^2 + k – 2 > 0 -------> $$k^2$$ –k + 2k – 2 > 0 -------> k(k-1)+2(k-1) > 0 -------> (k-1)(k+2)>0 -------> K>1 or k<-2

So the question can be rephrased as ‘Is K greater than 1 or less than -2’?

S1) k < 1 k may be less than -2 or may not, Insufficient
S2) k < -2 k is less than -2, sufficient.
Choice B

Example 6 :- if-y-x-1-x-and-x-0-is-xy-152900.html#p1225963

If $$y=\frac{|x+1|}{x}$$ and $$x\neq{}0$$, is $$xy>0$$ ?

A)$$x^2+2x+1>0$$

B)$$y\neq{0}$$

First lets simplify the inequality

$$\frac{|x+1|}{x}$$ –y = 0 ----------> $$\frac{|x+1|-xy}{x}$$ = 0 -----> we know $$x\neq{}0$$, then |x+1|-xy must be zero. Hence |x+1| - xy = 0 --------> |x+1| = xy

We are asked whether xy>0 --------> Whether |x+1| > 0 ? --------> We know the expression within modules can either be zero or greater than zero. For xy to be greater than zero |x+1| has to be greater than zero.
|x+1| will be zero only when x=-1 and for any other value of x, |x+1| will always be greater than zero
So the question can be rephrased as whether $$x\neq{-1}$$

Statement 1) $$x^2$$ + 2x + 1 > 0

Rule :- For any quadratic inequation $$ax^2$$ + bx + c > 0, if $$b^2$$ – 4ac = 0 and a > 0 then the inequality holds true outside the interval of roots
In our case $$b^2$$ – 4ac = 4 – 4 = 0 and a > 0 so $$x^2$$ + 2X + 1 > 0 will hold true for all values beyond the Root(s) of equation (Towards any direction - Positive or Negative)
$$x^2$$ + 2x + 1 = 0 --------> x(x+1) +1(x + 1) = 0 ---------> (x+1)(x+1) = 0 ----------> x=Root = -1
So $$x^2$$ + 2x + 1 > 0 will hold true for any of x except for -1
This reveals that $$x\neq{-1}$$ and xy>0 ----------------> Sufficient

Statement 2) $$y\neq{0}$$

From the question stem we know $$x\neq{}0$$
As per Statement 2, $$y\neq{0}$$-------------->That means both X and Y are nonzero.

|x+1| = xy
xy can be either Positive or Negative
|x+1| can be Zero or Positive
Combining both these inferences we can conclude that XY must be Positive. Sufficient

Example 7 :- which-of-the-following-values-of-x-satisfy-both-inequalities-143295.html#p1148838

(x —1)(x + 3) > 0
(x +5)(x—4) < 0

Which of the following values of x satisfy both inequalities shown?

I. -6
II. -4
III. 2
IV. 5

A. I and II only
B. I and III only
C. II and III only
D. II and IV only
E. I, II, III, and IV

[color=#f7941d]1st Inequality :- (x —1)(x + 3) > 0 -------> x > 1 or x < -3

2nd Inequality :- (x +5)(x—4) < 0 -------> -5 < x < 4

Solution = (-5 < x < -3) U (1 < x < 4)
Choice C

Example 8 :- which-of-the-following-represents-the-complete-range-of-x-108884.html

Which of the following represents the complete range of x over which $$x^3$$ – $$4x^5$$ < 0?

A. 0 < |x| < ½
B. |x| > ½
C. –½ < x < 0 or ½ < x
D. x < –½ or 0 < x < ½
E. x < –½ or x > 0

First Factories the expression

$$x^3$$(1 - $$4x^2$$) < 0 -------> $$x^3$$(1 - 2x)(1 + 2x) < 0 -------> Critical points are $$\frac{-1}{2}$$, 0, ½

Recall the steps mentioned earlier

4) Mark the leftmost region with + sign and rest two regions with – and + signs respectively.
5) If the Inequation is of the form $$ax^2$$ + bx + c < 0, the region having minus sign will be the solution of inequality
6) If the Inequation is of the form $$ax^2$$ + bx + c > 0, the region having plus sign will be the solutions of inequality

Hence, Range of x is (-1/2 < x < 0) and (x > ½)
Choice C

Hopefully the document will help to strengthen the understanding on inequalities.

BUNUEL's setof Practice Questions on Inequality and Absolute Value

Courtesy for the Information
[b]

Prof. Dr. R. D. Sharma
B.Sc. (Hons) (Gold Medalist), M.Sc. (Gold Medalist), Ph.D. in Mathematics
Author of CBSE Math Books

Mr. Arun Sharma
Aluminus – IIM Bangalore

Special Thanks to

BUNUEL, doe007, Zarrolou

Regards,

Narenn

---------- End of the Article ------------
Attachments

INEQUATIONS Part 2.pdf [183.28 KiB]

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Originally posted by Narenn on 22 Jun 2013, 06:39.
Last edited by Narenn on 26 Aug 2013, 09:45, edited 3 times in total.
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13 Aug 2013, 06:47
2
Thanks Narenn.

Before jumping to the practice questions, a beginner must start inequalities with this stuff.

Good effort, well summarized basic questions and concepts at one place.
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19 Aug 2013, 12:16
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10 Mar 2014, 05:27
Narenn wrote:
[b][i][highlight]

Example 8 :- which-of-the-following-represents-the-complete-range-of-x-108884.html

Which of the following represents the complete range of x over which $$x^3$$ – $$4x^5$$ < 0?

A. 0 < |x| < ½
B. |x| > ½
C. –½ < x < 0 or ½ < x
D. x < –½ or 0 < x < ½
E. x < –½ or x > 0

First Factories the expression

$$x^3$$(1 - $$4x^2$$) < 0 -------> $$x^3$$(1 - 2x)(1 + 2x) < 0 -------> Critical points are $$\frac{-1}{2}$$, 0, ½

Recall the steps mentioned earlier

4) Mark the leftmost region with + sign and rest two regions with – and + signs respectively.
5) If the Inequation is of the form $$ax^2$$ + bx + c < 0, the region having minus sign will be the solution of inequality
6) If the Inequation is of the form $$ax^2$$ + bx + c > 0, the region having plus sign will be the solutions of inequality

Hence, Range of x is (-1/2 < x < 0) and (x > ½)
Choice C

Here it says "Mark the leftmost region with + sign and rest two regions with – and + signs respectively." but in Part 1 of this article it was saying that we should start with + from rightmost region and continue by changing the sign until leftmost region.

Which one is true ?
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17 May 2014, 03:18
For example 7, you quoted the correct answer as b , which is x>-1

Since this is a must be true question, even if we find one value for which x/|x| < x doesnt hold true, the choice cannot be correct

If x>-1, lets say x= 1/2

(1/2)/|1/2| = 1

R.H.S = 1/2

How is 1 < 1/2 ?
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17 May 2014, 04:18
gaurav1418z wrote:
For example 7, you quoted the correct answer as b , which is x>-1

Since this is a must be true question, even if we find one value for which x/|x| < x doesnt hold true, the choice cannot be correct

If x>-1, lets say x= 1/2

(1/2)/|1/2| = 1

R.H.S = 1/2

How is 1 < 1/2 ?

Hope it helps.
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16 Sep 2014, 11:21
In Part 1 of this posting, it says:

Step VII :- In the rightmost region the expression on LHS of the Inequation obtained in step IV will be Positive and in other regions it will be alternatively negative and positive. So, mark positive Sign in the right most region and then mark alternatively negative and positive signs in Other regions.

Do we start from right or left, as mentioned in this posting?

Thank you!
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19 Aug 2018, 08:49
jkim1118 wrote:
In Part 1 of this posting, it says:

Step VII :- In the rightmost region the expression on LHS of the Inequation obtained in step IV will be Positive and in other regions it will be alternatively negative and positive. So, mark positive Sign in the right most region and then mark alternatively negative and positive signs in Other regions.

Do we start from right or left, as mentioned in this posting?

Thank you!

Please stick to the point mentioned in Part 1, there is a mistake in the solution of 8th question,
it is mentioned as "x^3(1 - 2x)(1 + 2x) < 0"
but it should be "x^3(x-1/2)(x+1/2) > 0 "
Hope this helps.
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19 Aug 2018, 08:55
lool wrote:
Narenn wrote:
[b][i][highlight]

Example 8 :- http://gmatclub.com/forum/which-of-the- ... 08884.html

Which of the following represents the complete range of x over which $$x^3$$ – $$4x^5$$ < 0?

A. 0 < |x| < ½
B. |x| > ½
C. –½ < x < 0 or ½ < x
D. x < –½ or 0 < x < ½
E. x < –½ or x > 0

First Factories the expression

$$x^3$$(1 - $$4x^2$$) < 0 -------> $$x^3$$(1 - 2x)(1 + 2x) < 0 -------> Critical points are $$\frac{-1}{2}$$, 0, ½

Recall the steps mentioned earlier

4) Mark the leftmost region with + sign and rest two regions with – and + signs respectively.
5) If the Inequation is of the form $$ax^2$$ + bx + c < 0, the region having minus sign will be the solution of inequality
6) If the Inequation is of the form $$ax^2$$ + bx + c > 0, the region having plus sign will be the solutions of inequality

Hence, Range of x is (-1/2 < x < 0) and (x > ½)
Choice C

Here it says "Mark the leftmost region with + sign and rest two regions with – and + signs respectively." but in Part 1 of this article it was saying that we should start with + from rightmost region and continue by changing the sign until leftmost region.

Which one is true ?

Stick to point mentioned in Post 1 but remember that your equation should be of the form
(x-a)(x-b)(x-c) > 0 or (x-a)(x-b)(x-c) < 0
same has been explained in above comment as well.
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15 Dec 2019, 01:05
Hi Bunuel chetan2u

would you kindly help me understand where I'm mistaken in understanding the solution of example 7 from above. I have marked my confusion in bold below:

if x(|x|)x(|x|) < x, Which of the following must be true about x ?

A) x > 1
B) x > -1
C) |x| < 1
D) |x| = 1
E) |x|2|x|2 > 1

x(|x|)x(|x|) < x ------> Since |x| has to be positive, we can multiply both sides of an inequality by |x|

x < x |x| -----> x|x| - x > 0 -------> x (|x| - 1) > 0

Case I :- x and (|x|-1) are positive

In this case x > 0 and |x| - 1 > 0 -------> x > 1 or x < -1 -----------{ Applying |x| - a > r ----> x > a + r or x < a – r}
Therefore Range of x is x > 1 - a+r gives x > 1 but how does a-r give -1 ?

Case II :- x and (|x|-1) are negative

In this case x < 0 and |x| - 1 < 0 --------> -1 < x < 1 -------------{ Applying |x| - a < r ----> a – r < x < a + r}
Therefore range of x is -1 < x < 0 - same as above : how does a-r give -1<X ??

So we have that -1 < x < 0 or x > 1
x > 1 Incorrect
x > -1 Correct
|x| < 1 Incorrect
|x| = 1 Incorrect
|x|2 > 1 Incorrect

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15 Dec 2019, 01:42
deeeuce wrote:
Hi Bunuel chetan2u

would you kindly help me understand where I'm mistaken in understanding the solution of example 7 from above. I have marked my confusion in bold below:

if x(|x|)x(|x|) < x, Which of the following must be true about x ?

A) x > 1
B) x > -1
C) |x| < 1
D) |x| = 1
E) |x|2|x|2 > 1

x(|x|)x(|x|) < x ------> Since |x| has to be positive, we can multiply both sides of an inequality by |x|

x < x |x| -----> x|x| - x > 0 -------> x (|x| - 1) > 0

Case I :- x and (|x|-1) are positive

In this case x > 0 and |x| - 1 > 0 -------> x > 1 or x < -1 -----------{ Applying |x| - a > r ----> x > a + r or x < a – r}
Therefore Range of x is x > 1 - a+r gives x > 1 but how does a-r give -1 ?

Case II :- x and (|x|-1) are negative

In this case x < 0 and |x| - 1 < 0 --------> -1 < x < 1 -------------{ Applying |x| - a < r ----> a – r < x < a + r}
Therefore range of x is -1 < x < 0 - same as above : how does a-r give -1<X ??

So we have that -1 < x < 0 or x > 1
x > 1 Incorrect
x > -1 Correct
|x| < 1 Incorrect
|x| = 1 Incorrect
|x|2 > 1 Incorrect

Very detailed discussion of that question is here: https://gmatclub.com/forum/if-x-x-x-whi ... 68886.html

Hope it helps.
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15 Dec 2019, 02:43
Bunuel wrote:
deeeuce wrote:
Hi Bunuel chetan2u

would you kindly help me understand where I'm mistaken in understanding the solution of example 7 from above. I have marked my confusion in bold below:

if x(|x|)x(|x|) < x, Which of the following must be true about x ?

A) x > 1
B) x > -1
C) |x| < 1
D) |x| = 1
E) |x|2|x|2 > 1

x(|x|)x(|x|) < x ------> Since |x| has to be positive, we can multiply both sides of an inequality by |x|

x < x |x| -----> x|x| - x > 0 -------> x (|x| - 1) > 0

Case I :- x and (|x|-1) are positive

In this case x > 0 and |x| - 1 > 0 -------> x > 1 or x < -1 -----------{ Applying |x| - a > r ----> x > a + r or x < a – r}
Therefore Range of x is x > 1 - a+r gives x > 1 but how does a-r give -1 ?

Case II :- x and (|x|-1) are negative

In this case x < 0 and |x| - 1 < 0 --------> -1 < x < 1 -------------{ Applying |x| - a < r ----> a – r < x < a + r}
Therefore range of x is -1 < x < 0 - same as above : how does a-r give -1<X ??

So we have that -1 < x < 0 or x > 1
x > 1 Incorrect
x > -1 Correct
|x| < 1 Incorrect
|x| = 1 Incorrect
|x|2 > 1 Incorrect

Very detailed discussion of that question is here: https://gmatclub.com/forum/if-x-x-x-whi ... 68886.html

Hope it helps.

Bunuel: thank you for this link. Yes it is clear now...
Math Expert
Joined: 02 Aug 2009
Posts: 8334

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15 Dec 2019, 02:46
deeeuce wrote:
Hi Bunuel chetan2u

would you kindly help me understand where I'm mistaken in understanding the solution of example 7 from above. I have marked my confusion in bold below:

if x(|x|)x(|x|) < x, Which of the following must be true about x ?

A) x > 1
B) x > -1
C) |x| < 1
D) |x| = 1
E) |x|2|x|2 > 1

x(|x|)x(|x|) < x ------> Since |x| has to be positive, we can multiply both sides of an inequality by |x|

x < x |x| -----> x|x| - x > 0 -------> x (|x| - 1) > 0

Case I :- x and (|x|-1) are positive

In this case x > 0 and |x| - 1 > 0 -------> x > 1 or x < -1 -----------{ Applying |x| - a > r ----> x > a + r or x < a – r}
Therefore Range of x is x > 1 - a+r gives x > 1 but how does a-r give -1 ?

Case II :- x and (|x|-1) are negative

In this case x < 0 and |x| - 1 < 0 --------> -1 < x < 1 -------------{ Applying |x| - a < r ----> a – r < x < a + r}
Therefore range of x is -1 < x < 0 - same as above : how does a-r give -1<X ??

So we have that -1 < x < 0 or x > 1
x > 1 Incorrect
x > -1 Correct
|x| < 1 Incorrect
|x| = 1 Incorrect
|x|2 > 1 Incorrect

x(|x|-1)>0, so either both are positive or both are negative

Both are positive
x>0 and
|x|-1>0.....|x|>1...x>1 or -x>1 => x<-1, when you multiply-x>1 by ‘-‘ and reverse the sign.
So x>1
Similarly for other too
_________________
Re: INEQUATIONS(Inequalities) Part 2   [#permalink] 15 Dec 2019, 02:46
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