For all non-zero real numbers u and v, the custom operation \Delta is defined by u \Delta v = \frac{u}{v} - \frac{v}{u} The function f is defined for all non-zero real numbers x by f(x) = x \Delta 2. Which of the following statements must be true for all non-zero real numbers a and b? Select all suc

$a \Delta b = -(b \Delta a)$; $f(-a) = -f(a)$

For all non-zero real numbers u and v, the custom operation \Delta is defined by u \Delta … — GRE General Test

For all non-zero real numbers

u

and

v

, the custom operation

\Delta

is defined by

u \Delta v = \frac{u}{v} - \frac{v}{u}

The function

f

is defined for all non-zero real numbers

x

f(x) = x \Delta 2

. Which of the following statements must be true for all non-zero real numbers

a

and

b

? Select all such statements.

$a \Delta b = -(b \Delta a)$ Answer
$f(-a) = -f(a)$ Answer
C
$f(a^2) = (f(a))^2$
D
$f(a + b) = f(a) + f(b)$

Answer

The statements

a \Delta b = -(b \Delta a)

and

f(-a) = -f(a)

must be true.

The custom operation is anti-commutative because swapping variables negates the difference of the fractions, making the statement

a \Delta b = -(b \Delta a)

true. Additionally, substituting

-a

into

f(x) = \frac{x}{2} - \frac{2}{x}

yields

-f(a)

, showing

f(-a) = -f(a)

is also true.

Step-by-Step Solution

Analyze the first identity

a \Delta b = -(b \Delta a)

a \Delta b = \frac{a}{b} - \frac{b}{a}

and

b \Delta a = \frac{b}{a} - \frac{a}{b} = -\left(\frac{a}{b} - \frac{b}{a}\right) = -(a \Delta b)

Factoring

-1

out of

b \Delta a

proves anti-commutativity.

Evaluate

f(x)

explicitly and test

f(-a)

f(x) = \frac{x}{2} - \frac{2}{x}

. Substituting

-a

gives

f(-a) = \frac{-a}{2} - \frac{2}{-a} = -\left(\frac{a}{2} - \frac{2}{a}\right) = -f(a)

This confirms that

f

is an odd function.

Compare

f(a^2)

with

(f(a))^2

f(a^2) = \frac{a^2}{2} - \frac{2}{a^2}

, but

(f(a))^2 = \frac{a^2}{4} - 2 + \frac{4}{a^2}

. Testing

a = 1

yields

f(1) = -1.5 \implies (f(1))^2 = 2.25

, whereas

f(1^2) = -1.5 \neq 2.25

Counterexample shows the equality does not hold in general.

Compare

f(a+b)

with

f(a) + f(b)

For

a=1, b=1

f(1+1) = f(2) = \frac{2}{2} - \frac{2}{2} = 0

, but

f(1) + f(1) = -1.5 + (-1.5) = -3 \neq 0

Functions do not in general preserve additivity

f(a+b) = f(a) + f(b)

unless they are linear through the origin.

Key Concept

Properties of algebraic custom operations and odd functions

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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