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What is the scope of x such that (x-1)(x+4)

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What is the scope of x such that (x-1)(x+4)  [#permalink]

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14 Aug 2018, 19:26
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45% (medium)

Question Stats:

63% (02:07) correct 37% (02:27) wrong based on 55 sessions

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What is the scope of x such that $$\frac{(x-1)(x+4)}{(x+2)} <x$$?

A. $$-2<x<4$$
B. $$-4<x<4$$
C. $$-2<x<2$$
D. $$-2<x$$
E. $$0<x<3$$
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Joined: 02 Aug 2009
Posts: 7764
Re: What is the scope of x such that (x-1)(x+4)  [#permalink]

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14 Aug 2018, 19:53
1
ammuseeru wrote:
What is the scope of x such that $$\frac{(x-1)(x+4)}{(x+2)} <x$$?

A. $$-2<x<4$$
B. $$-4<x<4$$
C. $$-2<x<2$$
D. $$-2<x$$
E. $$0<x<3$$

$$\frac{(x-1)(x+4)}{(x+2)} <x.........\frac{x^2+3x-4}{x+2}$$-x<0
$$\frac{x^2+3x-4-x^2-2x}{x+2}<0.............\frac{x-4}{x+2}<0$$..
So two cases...
1) x-4>0, then x+2<0........x<-2 and x>4.... No overlap in two regions
2) x-4<0, then x+2>0.......-2<x<4.....
So A

Also use choices..
Check for X at -3, 3 and you will be able to get our answer..
1) x=3
$$\frac{(x-1)(x+4)}{(x+2)} <x........ \frac{(3-1)(3+4)}{3+2}<3........2*7/5<3.......14<15$$ yes
Eliminate all those which do not contain 3, so C and E are out.

2) x=-3
$$\frac{(x-1)(x+4)}{(x+2)} <x.... ..\frac{(-3-1)(-3+4)}{-3+2}<-3........(-4)*1/(-1)<-3......4<-3$$.. No
So eliminate those containing -3, so B out

3) remaining A and D
Look for x = 5..
(5-1)(5+4)/(5+2)<5......4*20/7<5.....80<35.....no
So eliminate D

A
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Re: What is the scope of x such that (x-1)(x+4)  [#permalink]

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14 Aug 2018, 20:05
ammuseeru wrote:
What is the scope of x such that $$\frac{(x-1)(x+4)}{(x+2)} <x$$?

A. $$-2<x<4$$
B. $$-4<x<4$$
C. $$-2<x<2$$
D. $$-2<x$$
E. $$0<x<3$$

Another easy way is to consider the oprions.
Put x=0, LHS<RHS TRUE
Option E is out.

Put x=3 LHS<RHS TRUE
Option C is out.

Put x=-3 LHS>RHS FALSE
Option B is out.

Put x=5 LHS>RHS FALSE
Option D is out.

Hence Option A is correct.

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Re: What is the scope of x such that (x-1)(x+4)  [#permalink]

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14 Aug 2018, 22:47
ammuseeru wrote:
What is the scope of x such that $$\frac{(x-1)(x+4)}{(x+2)} <x$$?

A. $$-2<x<4$$
B. $$-4<x<4$$
C. $$-2<x<2$$
D. $$-2<x$$
E. $$0<x<3$$

$$\frac{(x-1)(x+4)}{(x+2)} <x$$

Or, $$x^2+3x−4 < x^2 + 2x$$

Or, $$x < 4$$

Now check using options given, Answer must be (A), as shown in the above posts....
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Posts: 95
Re: What is the scope of x such that (x-1)(x+4)  [#permalink]

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14 Aug 2018, 22:58
ammuseeru wrote:
What is the scope of x such that $$\frac{(x-1)(x+4)}{(x+2)} <x$$?

A. $$-2<x<4$$
B. $$-4<x<4$$
C. $$-2<x<2$$
D. $$-2<x$$
E. $$0<x<3$$

$$\frac{(x-1)(x+4)}{(x+2)} <x$$

Solving for x =>

$$(x^2 - x + 4x - 4) < (x^2 + 2x)$$

(3x - 2x) < 4

Therefore, x < 4

Now if we look at the options, we find two options namely A and B having,

x < 4

On comparing its range start parameters, we get 3 as the eliminator between both

Substituting x = 3 in $$\frac{(x-1)(x+4)}{(x+2)} <x$$

$$\frac{(-4) * (1)}{(-1)} < (-3)$$

On solving we get
LHS = 4
RHS = -3

as LHS is not less than RHS.

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Re: What is the scope of x such that (x-1)(x+4)  [#permalink]

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15 Aug 2018, 09:34
1
ammuseeru wrote:
What is the scope of x such that $$\frac{(x-1)(x+4)}{(x+2)} <x$$?

A. $$-2<x<4$$
B. $$-4<x<4$$
C. $$-2<x<2$$
D. $$-2<x$$
E. $$0<x<3$$

Given, $$\frac{(x-1)(x+4)}{(x+2)} <x$$
Or, $$\frac{\left(x-1\right)\left(x+4\right)}{x+2}-x<0$$
Or, $$\frac{\left(x-1\right)\left(x+4\right)-x\left(x+2\right)}{x+2}<0$$
Or, $$\frac{x-4}{x+2}<0$$
Or, $$-2<x<4$$

Ans. (A)
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Re: What is the scope of x such that (x-1)(x+4)   [#permalink] 15 Aug 2018, 09:34
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