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# Solve the inequality

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Intern
Joined: 13 May 2013
Posts: 12
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Kudos [?]: 14 [1] , given: 27

|x^2 - 2x| <x ? [#permalink]

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17 May 2013, 19:57
1
KUDOS
Solve the inequality.

What are possible values of x?

|x^2 - 2x| <x ?

a. 1<x<3
b. -1<x<3
c. 0<x<4
d. x>3
Intern
Joined: 13 May 2013
Posts: 12
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Kudos [?]: 14 [1] , given: 27

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17 May 2013, 20:01
1
KUDOS
Solve the inequality

What are possible values of x?

|x-6|>x^2 - 5x+9 ?

a. 1<= x < 3
b. 1<x<3
c. 2<x<5
d. -3<x<1
Intern
Joined: 13 May 2013
Posts: 12
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Kudos [?]: 14 [1] , given: 27

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17 May 2013, 20:05
1
KUDOS
Solve the inequality

What are possible values of x?

x-17>=60/x ?

a. x<-3
b. x<20
c. -3<=x<0 or x>=20
d. -3<x<=0 or x>=20
Intern
Joined: 13 May 2013
Posts: 12
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Kudos [?]: 14 [1] , given: 27

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17 May 2013, 20:08
1
KUDOS
Solve the inequality

What are possible values of x?

\sqrt{x^2} < x+1 ?

a. x>0.5
b. x>0
c. All x
d. x>-0.5
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 629
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Kudos [?]: 1192 [0], given: 136

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17 May 2013, 21:25
SathyaNIT wrote:
Solve the inequality

x-17>=60/x ?

a. x<-3
b. x<20
c. -3<=x<0 or x>=20
d. -3<x<=0 or x>=20

$$x-17\geq{60/x}$$ = $$x^2(x-17)\geq{60x}$$
or $$x[x^2-17x-60]\geq{0}$$
or $$x(x-20)(x+3)\geq{0}$$
Thus, either $$x>={20} OR -3<=x<0$$[As x cannont equal 0].
C
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Last edited by mau5 on 17 May 2013, 21:48, edited 1 time in total.
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 629
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Kudos [?]: 1192 [0], given: 136

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17 May 2013, 21:38
SathyaNIT wrote:
Solve the inequality

What are possible values of x?

|x-6|>x^2 - 5x+9 ?

a. 1<= x < 3
b. 1<x<3
c. 2<x<5
d. -3<x<1

$$|x-6|>x^2-6x+9+x =| x-6|>(x-3)^2+x$$

For x>6, we have$$x-6>(x-3)^2+x$$ -->$$-6>(x-3)^2$$ = This is not possible as for all real x, $$(x-3)^2\geq{0}$$

Thus, for x<6, we have $$6-x>(x-3)^2+x$$ --> $$(x-3)^2+2x-6<0$$--> $$(x-3)^2+2(x-3)<0$$ =$$(x-3)(x-3+2)<0$$ --> (x-3)(x-1)<0

Thus, 1<x<3.

B.
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17 May 2013, 21:39
1
KUDOS
Just solve for the case when X is +ve.
You will end up getting x>= 20 or x<= -3.
So finally search for answer that states the following
C and are the ones ..but notice that d does not mentions equal to in the inequality for -3, hence cannot be the correct answer.
So the answer must be C

Consider kudos if my post helps!!!!!!!!!!

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18 May 2013, 00:13
1
KUDOS
I used the attached graph to solve this inequality

Its clear from the graph that $$\sqrt{x^2}$$ is greater than (x+1) only when x is greater than -0.5

So ans is D
Attachments

untitled1.JPG [ 10.31 KiB | Viewed 1413 times ]

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18 May 2013, 09:10
aceacharya wrote:
I used the attached graph to solve this inequality

Its clear from the graph that $$\sqrt{x^2}$$ is greater than (x+1) only when x is greater than -0.5

So ans is D

This graph doesn't seem right. The mod function a curve that intersects the line y=x somewhere between the ranges x=1 and x=3.
Therefore the range of values for f(x) < x or y< x is 1<x < 3

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Kudos [?]: 439 [0], given: 23

Re: |x^2 - 2x| <x ? [#permalink]

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19 May 2013, 23:56
SathyaNIT wrote:
Solve the inequality.

What are possible values of x?

|x^2 - 2x| <x ?

a. 1<x<3
b. -1<x<3
c. 0<x<4
d. x>3

Hi,

The best approach is to put different value of x and validate the inquality.
For ex, x=0, LHS=RHS (thus, inequality doesn't hold true.)
So, answer choice that has got 0 in the range will be eliminated.
a. 1<x<3
b. -1<x<3
c. 0<x<4
d. x>3

Put x= 1, LHS=RHS
a. 1<x<3
b. -1<x<3
c. 0<x<4
d. x>3

Put x=4, LHS = RHS
a. 1<x<3
b. -1<x<3
c. 0<x<4
d. x>3
Thus, the only choice which remains is (A)

Check, put x=2, LHS < RHS
Re: |x^2 - 2x| <x ?   [#permalink] 19 May 2013, 23:56
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