Functions and Custom Symbol Operations Practice Questions — GRE General Test — Page 2

Question 21Question →

For all positive real numbers $x$ and $y$ , the custom operation $\star$ is defined by $x \star y = \frac{x^2 + y^2}{xy}$ . The function $g$ is defined by $g(t) = 4(t \star 3) - 7$ . If $g(k) = 3$ , where $k > 3$ , what is the value of $k$ ?

Show answer & explanation

Answer: 6

Answer

The value of

k

is 6.

Solving

g(k) = 3

yields

4(k \star 3) - 7 = 3

, so

k \star 3 = \frac{5}{2}

. Replacing

k \star 3

with

\frac{k^2+9}{3k}

results in the quadratic equation

2k^2 - 15k + 18 = 0

. The roots are

1.5

and

6

. Because

k

must be strictly greater than

3

, the only valid solution is

6

.

Step-by-Step Solution

1

Isolate the custom operation term in the function equation

k \star 3 = \frac{5}{2}

Setting

g(k) = 4(k \star 3) - 7 = 3

allows us to find the required value of

k \star 3

by adding 7 to both sides and dividing by 4.

2

Express the custom operation algebraically using its definition

\frac{k^2 + 9}{3k} = \frac{5}{2}

By definition,

x \star y = \frac{x^2 + y^2}{xy}

, so

k \star 3 = \frac{k^2 + 3^2}{3(k)} = \frac{k^2 + 9}{3k}

.

3

Form and solve the quadratic equation

k = 1.5 \text{ or } k = 6

Cross-multiplying gives

2(k^2 + 9) = 5(3k)

, which simplifies to

2k^2 - 15k + 18 = 0

. Factoring as

(2k - 3)(k - 6) = 0

yields two roots:

k = 1.5

and

k = 6

.

4

Apply the domain constraint

k = 6

The problem specifies that

k > 3

. Since

1.5 \le 3

,

k = 1.5

is rejected, leaving

k = 6

as the unique correct answer.

Key Concept

Custom Binary Symbol Operations and Composite Functions

Question 22Question →

For all real numbers $x$ , the function $g$ is defined by $g(x) = cx + 5$ , where $c$ is a constant. The custom operation $\diamond$ is defined for all real numbers $a$ and $b$ by $a \diamond b = g(a+b) - g(a-b)$ . If $3 \diamond 1 = 12$ , what is the value of $g(4)$ ?

Show answer & explanation

Answer: 29

Answer

29

Applying the function definition

g(x) = cx + 5

to the custom operation yields

a \diamond b = [c(a+b)+5] - [c(a-b)+5] = 2bc

. Substituting

a=3

and

b=1

into

3 \diamond 1 = 12

gives

2(1)c = 12

, so

c = 6

. Consequently,

g(x) = 6x + 5

, and evaluating at

x = 4

yields

g(4) = 6(4) + 5 = 29

.

Step-by-Step Solution

1

Substitute

(a+b)

and

(a-b)

into the function definition

g(x) = cx + 5

to simplify

a \diamond b

.

a \diamond b = [c(a+b) + 5] - [c(a-b) + 5] = 2bc

.

Applying the definition of the custom binary operation in terms of function

g

eliminates the constant term

5

.

2

Use the given equality

3 \diamond 1 = 12

to determine the constant

c

.

2(1)c = 12 \implies 2c = 12 \implies c = 6

.

Plugging

a=3

and

b=1

into

2bc = 12

yields an equation in terms of

c

.

3

Evaluate

g(4)

using

c = 6

.

g(4) = 6(4) + 5 = 29

.

Substituting

x = 4

into

g(x) = 6x + 5

calculates the required numerical value.

Key Concept

Functions and Custom Symbol Operations

Estimated Time:1m 30s

Question 23Question →

For all real numbers

x

, the function

f

is defined by

f(x) = x^2 - k

, where

k

is a constant. The custom operation

\star

is defined for all real numbers

a

and

b

by

a \star b = f(a + b) - f(a) - f(b)

If

5 \star (-3) = -12

, what is the value of

k

?

18

12

-12

0

-42

Show answer & explanation

Answer: 18

Answer

The value of

k

is 18.

Evaluating

f(2) = 4 - k

,

f(5) = 25 - k

, and

f(-3) = 9 - k

yields

5 \star (-3) = (4 - k) - (34 - 2k) = -30 + k

. Setting

-30 + k = -12

correctly gives

k = 18

.

Step-by-Step Solution

1

Evaluate

f(a+b)

for

a=5

and

b=-3

a + b = 5 + (-3) = 2

, so

f(2) = 2^2 - k = 4 - k

.

Substitute the input

a+b = 2

into the function definition

f(x) = x^2 - k

.

2

Evaluate

f(a)

and

f(b)

individually

f(5) = 5^2 - k = 25 - k

and

f(-3) = (-3)^2 - k = 9 - k

.

Apply the function rule to inputs 5 and -3.

3

Substitute the evaluated expressions into the custom operation definition

5 \star (-3) = (4 - k) - [(25 - k) + (9 - k)] = (4 - k) - (34 - 2k) = -30 + k

.

Simplify the algebraic expression by combining like terms and distributing the negative sign.

4

Solve for

k

using the given equation

5 \star (-3) = -12

-30 + k = -12 \implies k = 18

.

Add 30 to both sides of the equation to isolate

k

.

Key Concept

Evaluating custom binary operations by substituting function definitions and simplifying algebraic expressions.

Estimated Time:1m 30s

Question 24Question →

For all real numbers $x$ , the function $f$ is defined by $f(x) = 3 - x^2$ . For all real numbers $x \neq -2$ , the function $g$ is defined by $g(x) = \frac{2x - 1}{x + 2}$ . What is the value of $g(f(-3))$ ?

\frac{13}{4}

\frac{23}{14}

-46

-42

-\frac{1}{2}

Show answer & explanation

Answer: $\frac{13}{4}$

Answer

The correct answer is

\frac{13}{4}

.

First, evaluate the inner expression

f(-3) = 3 - (-3)^2 = 3 - 9 = -6

. Next, evaluate the outer expression at this value:

g(-6) = \frac{2(-6) - 1}{-6 + 2} = \frac{-13}{-4} = \frac{13}{4}

.

Step-by-Step Solution

1

Evaluate the inner function

f(-3)

f(-3) = 3 - (-3)^2 = 3 - 9 = -6

Applying the definition of

f(x)

where the negative base is squared to yield positive

9

.

2

Substitute the result into the outer function

g(x)

g(-6) = \frac{2(-6) - 1}{-6 + 2} = \frac{-12 - 1}{-4} = \frac{-13}{-4} = \frac{13}{4}

Evaluating

g(x)

at

x = -6

and simplifying the fraction.

Key Concept

Nested Function Evaluation

g(f(x))

Estimated Time:1m 30s

Question 25Question →

For all non-zero real numbers $a$ and $b$ , the custom binary operation $\diamondsuit$ is defined by $a \diamondsuit b = \frac{a^2 - b^2}{ab}$ . If the function $f$ is defined for all $x \neq 0$ by $f(x) = x \diamondsuit 2$ , what is the value of $f(4) - f(1)$ ?

Show answer & explanation

Answer: 3

Answer

3

Evaluating

f(4) = 4 \diamondsuit 2 = \frac{16 - 4}{8} = 1.5

and

f(1) = 1 \diamondsuit 2 = \frac{1 - 4}{2} = -1.5

, the required difference is

f(4) - f(1) = 1.5 - (-1.5) = 3

.

Step-by-Step Solution

1

Evaluate

f(4)

using the custom operation definition

f(4) = \frac{4^2 - 2^2}{4 \cdot 2} = \frac{12}{8} = 1.5

Substitute

a = 4

and

b = 2

into

a \diamondsuit b = \frac{a^2 - b^2}{ab}

.

2

Evaluate

f(1)

using the custom operation definition

f(1) = \frac{1^2 - 2^2}{1 \cdot 2} = -1.5

Substitute

a = 1

and

b = 2

into

a \diamondsuit b = \frac{a^2 - b^2}{ab}

.

3

Compute the difference

f(4) - f(1)

3

Subtracting

-1.5

from

1.5

yields

1.5 - (-1.5) = 1.5 + 1.5 = 3

.

Key Concept

Custom Symbol Operations and Function Evaluation

Estimated Time:1m 30s

Question 26Question →

The function $f$ is defined by $f(t) = 2^t$ for all real numbers $t$ , and the binary operation $\odot$ is defined for all non-zero real numbers $m$ and $n$ by $m \odot n = \frac{m}{n} - \frac{n}{m}$ . What is the value of $f(2) \odot f(-2)$ ?

\frac{255}{16}

\frac{15}{16}

\frac{15}{4}

0

-\frac{255}{16}

Show answer & explanation

Answer: $\frac{255}{16}$

Answer

\frac{255}{16}

Evaluating

f(2)

yields

2^2 = 4

, and evaluating

f(-2)

yields

2^{-2} = \frac{1}{4}

. Substituting these values into the binary operation rule

m \odot n = \frac{m}{n} - \frac{n}{m}

gives

\frac{4}{1/4} - \frac{1/4}{4} = 16 - \frac{1}{16} = \frac{255}{16}

.

Step-by-Step Solution

1

Evaluate the function

f(t) = 2^t

at

t = 2

and

t = -2

.

f(2) = 2^2 = 4

and

f(-2) = 2^{-2} = \frac{1}{2^2} = \frac{1}{4}

.

Negative exponents follow the rule

x^{-a} = \frac{1}{x^a}

.

2

Substitute

m = f(2) = 4

and

n = f(-2) = \frac{1}{4}

into the custom operation definition

m \odot n = \frac{m}{n} - \frac{n}{m}

.

4 \odot \frac{1}{4} = \frac{4}{\frac{1}{4}} - \frac{\frac{1}{4}}{4}

.

Apply the defined binary operation rule.

3

Simplify the complex fraction terms.

\frac{4}{\frac{1}{4}} = 4 \times 4 = 16

and

\frac{\frac{1}{4}}{4} = \frac{1}{4 \times 4} = \frac{1}{16}

.

Dividing by a fraction is equivalent to multiplying by its reciprocal.

4

Subtract the two simplified terms.

16 - \frac{1}{16} = \frac{256}{16} - \frac{1}{16} = \frac{255}{16}

.

Find a common denominator to compute the final value.

Key Concept

Function evaluation with exponent rules combined with custom binary symbol operations

Question 27Question →

For all non-zero real numbers $x$ and $y$ , the custom binary operation $\star$ is defined by $x \star y = \frac{x}{y} - \frac{y}{x}$ . Which of the following statements must be true for all non-zero real numbers $a$ , $b$ , and $c$ ? Select all that apply.

Select all that apply

a \star b = -(b \star a)

a \star a = 0

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

a \star (b \star c) = (a \star b) \star c

a \star (b + c) = (a \star b) + (a \star c)

Show answer & explanation

Answer: $a \star b = -(b \star a)$ ; $a \star a = 0$ ; $(a \star b)^2 = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2$

Answer

The statements

a \star b = -(b \star a)

,

a \star a = 0

, and

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

must be true for all non-zero real numbers

a

,

b

, and

c

.

The statements showing anti-commutativity, zero self-operation, and the expanded squared expression are all mathematically valid for all non-zero real numbers upon expanding their definitions using algebra.

Step-by-Step Solution

1

Evaluate

a \star b = -(b \star a)

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

and

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

.

Verify anti-commutativity property by direct substitution.

2

Evaluate

a \star a = 0

a \star a = \frac{a}{a} - \frac{a}{a} = 1 - 1 = 0

.

Verify identity property for identical inputs.

3

Evaluate

(a \star b)^2

(a \star b)^2 = \left(\frac{a}{b} - \frac{b}{a}\right)^2 = \left(\frac{a}{b}\right)^2 - 2\left(\frac{a}{b}\right)\left(\frac{b}{a}\right) + \left(\frac{b}{a}\right)^2 = \frac{a^2}{b^2} - 2(1) + \frac{b^2}{a^2} = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2

.

Apply binomial expansion to the squared custom operation.

4

Test associativity statement

a \star (b \star c) = (a \star b) \star c

with a counterexample

Let

a = 4, b = 2, c = 1

. Then

b \star c = \frac{2}{1} - \frac{1}{2} = \frac{3}{2}

.

a \star (b \star c) = 4 \star \frac{3}{2} = \frac{4}{3/2} - \frac{3/2}{4} = \frac{8}{3} - \frac{3}{8} = \frac{55}{24}

. Meanwhile,

a \star b = 4 \star 2 = \frac{4}{2} - \frac{2}{4} = \frac{3}{2}

, and

(a \star b) \star c = \frac{3}{2} \star 1 = \frac{3/2}{1} - \frac{1}{3/2} = \frac{3}{2} - \frac{2}{3} = \frac{5}{6}

. Since

\frac{55}{24} \neq \frac{5}{6}

, associativity fails.

A single counterexample disproves a general identity statement.

5

Test distributivity statement

a \star (b + c) = (a \star b) + (a \star c)

with a counterexample

Let

a = 1, b = 1, c = 1

. Then

a \star (b + c) = 1 \star 2 = \frac{1}{2} - \frac{2}{1} = -\frac{3}{2}

. Meanwhile,

(a \star b) + (a \star c) = (1 \star 1) + (1 \star 1) = 0 + 0 = 0

. Since

-\frac{3}{2} \neq 0

, distributivity fails.

A single counterexample disproves distributivity over addition.

Key Concept

Evaluating algebraic properties and identity statements for custom defined binary operations.

Question 28Question →

For all real numbers $x$ and $y$ , the custom operation $\odot$ is defined by $x \odot y = x|y| - y|x|$ . Which of the following statements must be true for all real numbers $x$ and $y$ ? Select all such statements.

Select all that apply

x \odot y = 0

whenever

x

and

y

have the same sign

x \odot y = -(y \odot x)

If

x > 0

and

y < 0

, then

x \odot y > 0

x \odot (-x) = 0

for all non-zero real numbers

x

(x \odot y) \odot z = x \odot (y \odot z)

Show answer & explanation

Answer: $x \odot y = 0$ whenever $x$ and $y$ have the same sign; $x \odot y = -(y \odot x)$ ; If $x > 0$ and $y < 0$ , then $x \odot y > 0$

Answer

The correct statements are the statement asserting

x \odot y = 0

when

x

and

y

have the same sign, the statement asserting anti-commutativity

x \odot y = -(y \odot x)

, and the statement asserting

x \odot y > 0

when

x > 0

and

y < 0

.

The operation

x \odot y = x|y| - y|x|

produces 0 whenever

x

and

y

share the same sign because terms evaluate to identical quantities. Swapping variables negates the expression, establishing anti-commutativity. When

x

is positive and

y

is negative,

x \odot y

simplifies to

-2xy

, which is strictly greater than 0 since

xy < 0

.

Step-by-Step Solution

1

Analyze the first statement regarding same-sign inputs

If

x > 0

and

y > 0

,

|x|=x

and

|y|=y

, so

x \odot y = xy - yx = 0

. If

x < 0

and

y < 0

,

|x|=-x

and

|y|=-y

, so

x \odot y = x(-y) - y(-x) = -xy + xy = 0

. Thus,

x \odot y = 0

when

x

and

y

have the same sign.

Verifying the definition under both positive and negative cases of identical sign.

2

Analyze the second statement regarding operand order reversal

y \odot x = y|x| - x|y| = -(x|y| - y|x|) = -(x \odot y)

, which holds universally for all real numbers.

Testing anti-commutativity by algebraic substitution into the custom operation.

3

Analyze the third statement for opposite signs (

x > 0

and

y < 0

)

Since

x > 0

,

|x|=x

. Since

y < 0

,

|y|=-y

. Substituting yields

x(-y) - y(x) = -xy - xy = -2xy

. Because

x > 0

and

y < 0

, the product

xy

is negative, making

-2xy

strictly positive.

Determining the overall algebraic sign of the expression when variables have opposite signs.

4

Counter-test the remaining statements to verify incorrectness

For

x \odot (-x)

with

x=1

:

1 \odot (-1) = 1|-1| - (-1)|1| = 1 - (-1) = 2 \neq 0

. For associativity with

x=2, y=-1, z=-1

:

(2 \odot -1) \odot -1 = 4 \odot -1 = 8

, but

2 \odot (-1 \odot -1) = 2 \odot 0 = 0

.

Demonstrating specific counterexamples for false generalizations.

Key Concept

Custom Binary Operations and Absolute Value Properties

Estimated Time:1m 30s

Question 29Question →

For all positive real numbers $a$ and $b$ , the custom operation $\diamondsuit$ is defined by $a \diamondsuit b = \frac{a^2 + b^2}{ab}$ . The function $f$ is defined for all $x > 0$ by $f(x) = x \diamondsuit 4$ . If $f(x) = 2.5$ , what is the value of $x$ that is greater than $4$ ?

Show answer & explanation

Answer: 8

Answer

The value of

x

greater than

4

is

8

.

Applying the custom operator gives

f(x) = \frac{x^2 + 16}{4x}

. Setting this equal to

2.5

yields

\frac{x^2 + 16}{4x} = \frac{5}{2}

, which simplifies to

x^2 - 10x + 16 = 0

. The roots are

x = 2

and

x = 8

. Since

x

must be greater than

4

, the only valid answer is

8

.

Step-by-Step Solution

1

Substitute

a = x

and

b = 4

into the custom operation definition

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

.

f(x) = \frac{x^2 + 16}{4x}

This establishes the explicit algebraic rule for the function

f(x)

.

2

Set

f(x)

equal to

2.5

and clear the fraction.

\frac{x^2 + 16}{4x} = 2.5 \implies x^2 + 16 = 10x

Multiplying both sides by

4x

converts the rational equation into a standard polynomial equation.

3

Rearrange into standard quadratic form and solve by factoring.

x^2 - 10x + 16 = 0 \implies (x - 2)(x - 8) = 0 \implies x = 2

or

x = 8

Factoring determines all potential positive real solutions for

x

.

4

Select the solution satisfying the constraint

x > 4

.

x = 8

The question explicitly specifies that

x

must be greater than

4

, eliminating

x = 2

.

Key Concept

Evaluating custom binary operations and solving algebraic function equations involving quadratic constraints.

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