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.

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

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Properties of algebraic custom operations and odd functions

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru