For all non-zero real numbers x and y, the custom operation \diamond is defined by x \diam… — GRE General Test

For all non-zero real numbers $x$ and $y$ , the custom operation $\diamond$ is defined by $x \diamond y = \frac{x^2 + y^2}{xy}$ . Which of the following statements must be true for all non-zero real numbers $a$ and $b$ ? Select all such statements.

$a \diamond b = b \diamond a$ Answer
$a \diamond (-a) = -2$ Answer
C
$(2a) \diamond (2b) = 2(a \diamond b)$
D
$a \diamond a = 1$

Answer

The correct statements are that

a \diamond b = b \diamond a

and that

a \diamond (-a) = -2

The operation is symmetric with respect to

a

and

b

, making

a \diamond b = b \diamond a

true. Substituting

-a

into the operation yields

\frac{2a^2}{-a^2} = -2

, which makes

a \diamond (-a) = -2

true as well.

Step-by-Step Solution

Evaluate the commutative statement

a \diamond b

a \diamond b = \frac{a^2 + b^2}{ab}

and

b \diamond a = \frac{b^2 + a^2}{ba} = \frac{a^2 + b^2}{ab}

Addition and multiplication of real numbers are commutative, so the expressions are identical.

Evaluate the statement

a \diamond (-a)

a \diamond (-a) = \frac{a^2 + (-a)^2}{a(-a)} = \frac{a^2 + a^2}{-a^2} = \frac{2a^2}{-a^2} = -2

Squaring a negative quantity

(-a)^2

yields positive

a^2

, while the denominator evaluates to

-a^2

Evaluate the scaling statement

(2a) \diamond (2b)

(2a) \diamond (2b) = \frac{4a^2 + 4b^2}{4ab} = \frac{a^2 + b^2}{ab} = a \diamond b

The factor of 4 in the numerator and denominator cancels out, showing

(2a) \diamond (2b) = a \diamond b

, which is not equal to

2(a \diamond b)

Evaluate the self-operation statement

a \diamond a

a \diamond a = \frac{a^2 + a^2}{a^2} = \frac{2a^2}{a^2} = 2

Summing identical squared terms in the numerator produces

2a^2

, which divides by

a^2

to give 2.

Key Concept

Evaluating custom binary operations by algebraic substitution and simplifying terms.

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