Let p and q be integers such that p < 0 < q, (-1)^p + (-1)^q = 0, and p^2 q + p is an even… — GRE General Test

Let $p$ and $q$ be integers such that $p < 0 < q$ , $(-1)^p + (-1)^q = 0$ , and $p^2 q + p$ is an even integer. Which of the following expressions must be negative?

A
$(-p)^q$
B
$(-q)^p$
C
$q^p$
$p^q$ Cevap
E
$(q - p)^p$

Cevap

p^q

From

(-1)^p + (-1)^q = 0

p

and

q

must have opposite parities. Factoring

p^2 q + p

into

p(pq + 1)

shows that if

p

were odd,

q

would be even, leading to an odd product

\text{odd} \times \text{odd} = \text{odd}

. Because

p^2 q + p

is given as even,

p

must be an even integer and

q

must be an odd integer. Given

p < 0

p

is a negative even integer, and

q

is a positive odd integer. The expression

p^q

represents a negative base raised to an odd exponent, which is guaranteed to be negative.

Adım Adım Çözüm

Determine the parities of

p

and

q

using the equation

(-1)^p + (-1)^q = 0

One of

p

q

is even and the other is odd.

For the sum

(-1)^p + (-1)^q

to equal

0

, one term must equal

1

and the other must equal

-1

, meaning

p

and

q

have opposite parities.

Analyze the parity of

p^2 q + p = p(pq + 1)

to determine which variable is even.

p

must be even, and

q

must be odd.

p

were odd,

q

would be even, making

pq

even,

pq+1

odd, and

p(pq+1)

odd. Since

p^2 q + p

is even,

p

must be even and

q

must be odd.

Evaluate the sign of

p^q

using sign rules for exponents.

p^q < 0

p

is negative (

p < 0

) and

q

is a positive odd integer. A negative number raised to an odd power is always negative.

Anahtar Kavram

Parity and sign rules for negative bases and integer exponents

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru