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For any real number x, the operator & is defined as: &(x) = x(1 − x)
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15 Mar 2021, 10:28
15 Mar 2021, 10:28
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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
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
&(x) = x(1 − x)
If p + 1 = &(p + 1), then p =
A. −1
B. 0
C. 1
D. 2
E. 3
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Re: For any real number x, the operator & is defined as: &(x) = x(1 − x)
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15 Mar 2021, 10:33
15 Mar 2021, 10:33
1
Carcass 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
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
&(x) = x(1 − x)
If p + 1 = &(p + 1), then p =
A. −1
B. 0
C. 1
D. 2
E. 3
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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
For any real number x, the operator & is defined as: &(x) = x(1 − x)
[#permalink]
15 Mar 2021, 10:36
15 Mar 2021, 10:36
Solution:
We can start by simplifying the equation
p+1=p+1(1-(p+1)
=p+1(1-p-1)
=p+1(-p)
=-\((p^2)\)-p
p+1=-\((p^2)\)-p
2p+1=-\((p^2)\)
Now we can either solve it as a quadratic equation or substitute answer option. I used the second method
When we substitute p=-1
LHS=RHS
IMO A
Hope this helps!
We can start by simplifying the equation
p+1=p+1(1-(p+1)
=p+1(1-p-1)
=p+1(-p)
=-\((p^2)\)-p
p+1=-\((p^2)\)-p
2p+1=-\((p^2)\)
Now we can either solve it as a quadratic equation or substitute answer option. I used the second method
When we substitute p=-1
LHS=RHS
IMO A
Hope this helps!
