For all real numbers x and y, the custom operation \diamondsuit is defined by x \diamondsuit y = x^2 - y^2 - 2x + 2y The function f is defined for all real numbers t by f(t) = t + 1. Which of the following statements must be true for all real numbers a and b? Select all such statements.

If $a \neq b$, then $a \diamondsuit b = 0$ if and only if $a + b = 2$.; $f(a) \diamondsuit f(b) = a^2 - b^2$

For all real numbers x and y, the custom operation \diamondsuit is defined by x \diamondsu… — GRE General Test

For all real numbers

x

and

y

, the custom operation

\diamondsuit

is defined by

x \diamondsuit y = x^2 - y^2 - 2x + 2y

The function

f

is defined for all real numbers

t

f(t) = t + 1

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

a

and

b

? Select all such statements.

A
$f(a + b) = f(a) + f(b)$
If $a \neq b$ , then $a \diamondsuit b = 0$ if and only if $a + b = 2$ .Cevap
$f(a) \diamondsuit f(b) = a^2 - b^2$ Cevap
D
$a \diamondsuit (-a) = 0$
E
$(a + b) \diamondsuit 0 = a^2 + b^2 - 2a - 2b$

Cevap

The statements 'If

a \neq b

, then

a \diamondsuit b = 0

if and only if

a + b = 2

' and '

f(a) \diamondsuit f(b) = a^2 - b^2

' must be true.

The statement claiming that if

a \neq b

a \diamondsuit b = 0

if and only if

a + b = 2

is correct because factoring

a \diamondsuit b

gives

(a - b)(a + b - 2) = 0

, which requires

a + b = 2

when

a - b \neq 0

. The statement claiming

f(a) \diamondsuit f(b) = a^2 - b^2

is also correct because substituting

f(a) = a + 1

and

f(b) = b + 1

into the operation yields

[(a + 1) - (b + 1)][(a + 1) + (b + 1) - 2] = (a - b)(a + b) = a^2 - b^2

Adım Adım Çözüm

Factor the custom operation definition

x \diamondsuit y = (x^2 - y^2) - 2(x - y) = (x - y)(x + y) - 2(x - y) = (x - y)(x + y - 2)

Factoring groups terms into products that simplify statement testing.

Analyze the condition for

a \diamondsuit b = 0

when

a \neq b

(a - b)(a + b - 2) = 0 \implies a + b - 2 = 0 \implies a + b = 2

Since

a \neq b

a - b \neq 0

, allowing division by

(a - b)

to isolate

a + b = 2

Evaluate

f(a) \diamondsuit f(b)

using

f(t) = t + 1

f(a) = a + 1, f(b) = b + 1 \implies f(a) \diamondsuit f(b) = [(a + 1) - (b + 1)][(a + 1) + (b + 1) - 2] = (a - b)(a + b) = a^2 - b^2

Direct substitution into the factored form simplifies the evaluation.

Evaluate remaining statements to identify counterexamples

f(a + b) = a + b + 1 \neq a + b + 2 = f(a) + f(b)

;

a \diamondsuit (-a) = -4a \neq 0

;

(a + b) \diamondsuit 0 = a^2 + 2ab + b^2 - 2a - 2b \neq a^2 + b^2 - 2a - 2b

Confirming false statements ensures accurate multi-selection.

Anahtar Kavram

Algebraic manipulation of custom binary operations and function evaluation

Soru

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