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

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Re: For any real number x, the operator & is defined as: &(x) = x(1 − x) [#permalink]
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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

First write &(p+1) by plugging in (p+1) for x:

> &(x) = x(1 − x)

> &(p+1) = (p+1)(1 − (p+1))

Distribute the minus sign:

> (p+1)(1 – (p + 1)) = (p+1)(1 – p – 1) = (p+1)(-p).

For p + 1 = &(p + 1) to be true, this has to equal p+1:

> p+1 = (p+1)(-p)

which means

> (p+1)(1) = (p+1)(-p)

For this to be true, -p must equal 1. That will only happen if p = -1.
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Re: For any real number x, the operator & is defined as: &(x) = x(1 − x) [#permalink]
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Re: For any real number x, the operator & is defined as: &(x) = x(1 − x) [#permalink]
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