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 pExpand 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

_________________

Brent Hanneson – Founder of gmatprepnow.com