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For any real number x, the operator & is defined as: &(x) = x(1 − x)

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Math Expert
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V
Joined: 02 Sep 2009
Posts: 50585
For any real number x, the operator & is defined as: &(x) = x(1 − x)  [#permalink]

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New post 16 Feb 2016, 00:05
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

72% (01:47) correct 28% (01:48) wrong based on 111 sessions

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Senior Manager
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Joined: 20 Aug 2015
Posts: 389
Location: India
GMAT 1: 760 Q50 V44
Re: For any real number x, the operator & is defined as: &(x) = x(1 − x)  [#permalink]

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New post 23 Feb 2016, 03:15
Bunuel wrote:
For any real number x, the operator & is defined as:

&(x) = x(1 − x)
If p + 1 = &(p + 1), then p =

A. −1
B. 0
C. 1
D. 2
E. 3


&(x) = x(1 − x)
&(p + 1) = (p + 1)(1 - p - 1) = -p(p+1)

We are given that p + 1 = &(p + 1)
Therefore -p(p+1) = (p + 1)

Or (p + 1) + p(p+1) = 0
(p + 1)^2 = 0
p = -1
Option A
CEO
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Re: For any real number x, the operator & is defined as: &(x) = x(1 − x)  [#permalink]

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New post 15 Apr 2018, 06:19
Top Contributor
Bunuel wrote:
For any real number x, the operator & is defined as:

&(x) = x(1 − x)
If p + 1 = &(p + 1), then p =

A. −1
B. 0
C. 1
D. 2
E. 3


Let's look at a few examples of this operator (&) in action.
If &(x) = x(1 − x), then...
&(3) = 3(1 − 3) = 3(-2) = -6
&(7) = 7(1 − 7) = 7(-6) = -42
&(-5) = -5(1 − -5) = (-5)(6) = -30
And now.....
&(p+1) = (p+1)[1 − (p+1)] = (p + 1)(-p)

Now onto the question.....
We're told that p + 1 = &(p + 1)
We can now rewrite the right side of the equation as: p + 1 = (p + 1)(-p)
We need to solve this equation for p
Expand the right side to get: p + 1 = -p² - p
Add p² to both sides: p² + p + 1 = -p
Add p to both sides: p² + 2p + 1 = 0
Factor: (p +1)(p +1) = 0
So, p = -1

Answer: A

Cheers,
Brent
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Brent Hanneson – GMATPrepNow.com
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Re: For any real number x, the operator & is defined as: &(x) = x(1 − x) &nbs [#permalink] 15 Apr 2018, 06:19
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