Discrete Math Quiz: Injections Surjections And Bijections

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Injections Surjections And BijectionsQuestion 1 of 18

Let $\mathcal{F} = \{f: \{1,2,3\} \to \{1,2,3\} : f \text{ is injective}\}$ and $\mathcal{G} = \{g: \{1,2,3\} \to \{1,2,3\} : g \text{ is surjective}\}$ . What are $|\mathcal{F}|$ , $|\mathcal{G}|$ , and $|\mathcal{F} \cap \mathcal{G}|$ ?

|\mathcal{F}| = 6

|\mathcal{G}| = 6

, and

|\mathcal{F} \cap \mathcal{G}| = 6

, since injectivity and surjectivity are equivalent for finite sets of equal size

|\mathcal{F}| = 9

|\mathcal{G}| = 6

, and

|\mathcal{F} \cap \mathcal{G}| = 3

, since some injective functions are not surjective

|\mathcal{F}| = 6

|\mathcal{G}| = 9

, and

|\mathcal{F} \cap \mathcal{G}| = 3

, since some surjective functions are not injective

|\mathcal{F}| = 3

|\mathcal{G}| = 3

, and

|\mathcal{F} \cap \mathcal{G}| = 6

, based on the number of ways to assign elements

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Discrete Math Quiz

Discrete Math Quiz: Injections Surjections And Bijections

Practice Injections Surjections And Bijections in Discrete Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Injections Surjections And Bijections, giving you a quick way to practice the rules, question types, and explanations that matter most for Discrete Math.

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Question 1

$|\mathcal{F}| = 6$ , $|\mathcal{G}| = 6$ , and $|\mathcal{F} \cap \mathcal{G}| = 6$ , since injectivity and surjectivity are equivalent for finite sets of equal size (correct answer)
$|\mathcal{F}| = 9$ , $|\mathcal{G}| = 6$ , and $|\mathcal{F} \cap \mathcal{G}| = 3$ , since some injective functions are not surjective
$|\mathcal{F}| = 6$ , $|\mathcal{G}| = 9$ , and $|\mathcal{F} \cap \mathcal{G}| = 3$ , since some surjective functions are not injective
$|\mathcal{F}| = 3$ , $|\mathcal{G}| = 3$ , and $|\mathcal{F} \cap \mathcal{G}| = 6$ , based on the number of ways to assign elements

Explanation: For functions between finite sets of the same size, injectivity and surjectivity are equivalent to bijectivity. Since

|\{1,2,3\}| = |\{1,2,3\}| = 3

, an injective function from

\{1,2,3\}

\{1,2,3\}

must also be surjective (if three distinct inputs map to three distinct outputs, all outputs must be used). Similarly, a surjective function must be injective (if all three outputs are achieved by three inputs, no two inputs can map to the same output). Therefore,

\mathcal{F} = \mathcal{G} = \mathcal{F} \cap \mathcal{G}

= the set of all bijections from

\{1,2,3\}

\{1,2,3\}

. The number of bijections is the number of permutations of 3 elements, which is

3! = 6

. So

|\mathcal{F}| = |\mathcal{G}| = |\mathcal{F} \cap \mathcal{G}| = 6

. Choice B incorrectly calculates

|\mathcal{F}| = 9

(this would be

3^2

, but not all functions are injective). Choice C incorrectly calculates

|\mathcal{G}| = 9

. Choice D gives completely wrong values, including an impossible intersection size larger than either individual set.

Question 2

Consider the function $\Phi: \mathcal{P}(\mathbb{N}) \to \{0,1\}^\mathbb{N}$ that maps each subset $S \subseteq \mathbb{N}$ to its characteristic function $\chi_S$ , where $\chi_S(n) = 1$ if $n \in S$ and $\chi_S(n) = 0$ if $n \notin S$ . Which property does $\Phi$ possess?

$\Phi$ is injective but not surjective because not every sequence in $\{0,1\}^\mathbb{N}$ corresponds to a finite subset of $\mathbb{N}$
$\Phi$ is surjective but not injective because multiple subsets can have the same characteristic function pattern
$\Phi$ is neither injective nor surjective due to the fundamental differences between discrete sets and function spaces
$\Phi$ is bijective, establishing a one-to-one correspondence between subsets of $\mathbb{N}$ and binary sequences (correct answer)

Explanation: When analyzing functions between sets, you need to determine two key properties: injectivity (one-to-one) and surjectivity (onto). For characteristic functions, this connection between subsets and binary sequences is fundamental in discrete mathematics. The function

\Phi

is bijective because it establishes a perfect one-to-one correspondence. To see why, consider that each subset

S \subseteq \mathbb{N}

produces a unique binary sequence: for any natural number

n

, either

n \in S

(giving

\chi_S(n) = 1

) or

n \notin S

(giving

\chi_S(n) = 0

). This makes

\Phi

injective—different subsets must produce different characteristic functions. Conversely, every possible binary sequence

f: \mathbb{N} \to \{0,1\}

corresponds to exactly one subset:

S = \{n \in \mathbb{N} : f(n) = 1\}

. This makes

\Phi

surjective. Option A incorrectly assumes the mapping only works for finite subsets, but characteristic functions handle infinite subsets perfectly well—like the subset of even numbers producing the sequence

(0,1,0,1,0,1,\ldots)

. Option B falsely claims multiple subsets can share the same characteristic function, which is impossible since the characteristic function completely determines which elements belong to the subset. Option C incorrectly suggests fundamental incompatibility between sets and functions, missing that this is precisely what characteristic functions elegantly resolve. Remember: characteristic functions create natural bijections between subsets and binary sequences. This bijection is a cornerstone result showing these seemingly different mathematical objects are essentially equivalent.

Question 3

Let $A$ be a finite set with $|A| = n$ , and let $\text{Inj}(A,A)$ denote the set of all injective functions from $A$ to $A$ . If we randomly select a function $f \in \text{Inj}(A,A)$ and then randomly select a distinct function $g \in \text{Inj}(A,A)$ , what is the probability that $g \circ f$ is also injective?

$1$ , because the composition of any two injective functions is always injective, regardless of the domain and codomain (correct answer)
$\frac{n!-1}{n!}$ , because there are $n!-1$ choices for $g$ after fixing $f$ , and $n!-1$ of them yield injective compositions
$\frac{1}{n}$ , because only $\frac{1}{n}$ of the possible compositions result in injective functions due to the constraint structure
$\frac{(n-1)!}{n!-1}$ , because we must account for the reduced number of valid choices after excluding the original function

Explanation: Since

A

is finite with

|A| = n

, every injective function

f: A \to A

is automatically bijective (an injective function from a finite set to itself must be surjective). Similarly, every

g \in \text{Inj}(A,A)

is bijective. The composition of two bijective functions is always bijective, hence injective. This is true regardless of which specific bijections

f

and

g

we choose, as long as they're both in

\text{Inj}(A,A)

. Therefore, for any choice of distinct

f, g \in \text{Inj}(A,A)

, the composition

g \circ f

is guaranteed to be injective (in fact, bijective). The probability is 1. Choice B incorrectly suggests some compositions might not be injective. Choice C gives an arbitrary small probability. Choice D provides a complex expression that doesn't apply to this situation where all compositions are guaranteed to be injective.

Question 4

Consider the function $g: \{1, 2, 3, 4, 5\} \to \{a, b, c, d\}$ defined by $g(1) = a$ , $g(2) = b$ , $g(3) = c$ , $g(4) = d$ , and $g(5) = b$ . How many elements must be removed from the domain to make $g$ injective while maintaining surjectivity?

Exactly 1 element, and there are 2 different ways to choose which element (correct answer)
Exactly 1 element, and there are 3 different ways to choose which element
Exactly 2 elements, and there are 4 different ways to choose which elements
It is impossible to maintain surjectivity while making $g$ injective by removing elements

Explanation: Currently

g

is surjective (every element in codomain is hit) but not injective because

g(2) = g(5) = b

. To make it injective, we must eliminate this collision by removing either 2 or 5 from the domain. If we remove 2, we get

g: \{1,3,4,5\} \to \{a,b,c,d\}

with

g(1)=a, g(3)=c, g(4)=d, g(5)=b

, which is bijective. If we remove 5, we get

g: \{1,2,3,4\} \to \{a,b,c,d\}

with

g(1)=a, g(2)=b, g(3)=c, g(4)=d

, which is also bijective. Removing any other single element (1, 3, or 4) would eliminate a unique image and destroy surjectivity. So exactly 1 element must be removed, with exactly 2 choices. Choice B gives wrong count of ways. Choice C suggests removing 2 elements, but this would make

|\text{domain}| < |\text{codomain}|

, preventing surjectivity. Choice D is incorrect as we showed two valid solutions exist.

Question 5

Let $S = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1\}$ and $T = \{(x,y) \in \mathbb{R}^2 : 0 \leq x \leq 1, 0 \leq y \leq 1\}$ . Consider the function $f: S \to T$ defined by $f(x,y) = \left(\frac{x+1}{2}, \frac{y+1}{2}\right)$ . What can be concluded about $f$ ?

$f$ is injective but not surjective, since it preserves distinctness but maps the unit disk to only part of the unit square (correct answer)
$f$ is surjective but not injective, since the unit disk has smaller area than the unit square
$f$ is bijective, since it's a linear transformation that maps the unit disk onto the unit square
$f$ is neither injective nor surjective, since the transformation is not well-defined on the boundary of the disk

Explanation: The function

f(x,y) = \left(\frac{x+1}{2}, \frac{y+1}{2}\right)

is a linear transformation (translation and scaling). Linear transformations are injective, so if

(x_1,y_1) \neq (x_2,y_2)

, then

f(x_1,y_1) \neq f(x_2,y_2)

. For surjectivity, note that

S

is the closed unit disk (

x^2 + y^2 \leq 1

), so

-1 \leq x,y \leq 1

. Thus

f

maps

S

\{(u,v) : 0 \leq u,v \leq 1, (2u-1)^2 + (2v-1)^2 \leq 1\}

, which is a disk inside

T

, not all of

T

. So

f

is not surjective. Choice B reverses the properties. Choice C incorrectly claims bijection. Choice D is wrong - the function is well-defined everywhere.

Question 6

Let $A = \{1, 2, 3, 4, 5\}$ and consider functions $f: A \to A$ . If $f$ satisfies the condition that $f(f(x)) = x$ for all $x \in A$ , which statement about $f$ must be true?

$f$ may be neither injective nor surjective, since the condition only constrains the composition $f \circ f$ , not $f$ itself
$f$ must be injective but may not be surjective, since $f(f(x)) = x$ prevents any element from mapping to itself
$f$ must be surjective but may not be injective, since every element must be in the range of $f \circ f$
$f$ must be bijective, since the condition $f \circ f = \text{id}_A$ implies $f$ is its own inverse function (correct answer)

Explanation: When you encounter a function satisfying

f(f(x)) = x

for all elements in its domain, you're looking at a function that is its own inverse. This condition is called an involution, and it has powerful implications for the function's properties. The key insight is that if

f(f(x)) = x

holds for every

x

in the domain, then

f \circ f = \text{id}_A

(the identity function). This means

f

has a left inverse (itself) and a right inverse (also itself). For functions between finite sets of the same size, having both a left and right inverse guarantees the function is bijective. To see why

f

must be injective: if

f(a) = f(b)

for some

a \neq b

, then applying

f

again gives

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

, which means

a = b

by our condition—a contradiction. To see why

f

must be surjective: for any

y \in A

, we have

y = f(f(y))

, so

y

is in the range of

f

(specifically,

y = f(f(y))

). Answer choice A incorrectly suggests the condition is too weak to determine

f

's properties. Answer choice B wrongly claims

f

prevents self-mapping (elements can map to themselves in involutions). Answer choice C has the injectivity and surjectivity backwards in terms of what "may not" hold. Answer choice D correctly identifies that

f

must be bijective since it serves as its own inverse. Remember: when

f \circ f = \text{id}

, the function

f

is automatically bijective on finite domains of equal size.

Question 7

Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = x|x|$ . Consider the restriction $g: \mathbb{R} \to [0,\infty)$ where $g(x) = |f(x)|$ . Which statement correctly describes the relationship between the injectivity and surjectivity of $f$ and $g$ ?

$f$ is bijective and $g$ is injective but not surjective, since both preserve order but $g$ has restricted codomain
$f$ is bijective and $g$ is surjective but not injective, since $f(x) = x|x|$ is strictly increasing while $g(±x) = x^2$ (correct answer)
$f$ is bijective and $g$ is neither injective nor surjective, since $g$ maps both positive and negative inputs to the same outputs
$f$ is bijective and $g$ is bijective, since $|f(x)| = x^2$ establishes a bijection from $\mathbb{R}$ to $[0,\infty)$

Explanation: When analyzing functions for injectivity and surjectivity, you need to examine how inputs map to outputs and whether the mapping covers the entire codomain without repetition. First, let's analyze

f(x) = x|x|

. For

x \geq 0

, we have

f(x) = x \cdot x = x^2

. For

x < 0

, we have

f(x) = x \cdot (-x) = -x^2

. This means

f(x) = x^2

when

x \geq 0

and

f(x) = -x^2

when

x < 0

. The function

f

is strictly increasing: as

x

increases,

f(x)

increases without bound in both directions, mapping

\mathbb{R}

onto

\mathbb{R}

with no repeated outputs. Therefore,

f

is bijective. Now for

g(x) = |f(x)|

. Since

f(x) = x^2

for

x \geq 0

and

f(x) = -x^2

for

x < 0

, we get

g(x) = |x^2| = x^2

for

x \geq 0

and

g(x) = |-x^2| = x^2

for

x < 0

. Thus

g(x) = x^2

for all

x \in \mathbb{R}

. Since every non-negative real number has a square root,

g

is surjective onto

[0,\infty)

. However,

g

is not injective because

g(2) = g(-2) = 4

. Answer choice A incorrectly claims

g

is injective. Answer choice C wrongly states

g

is not surjective. Answer choice D incorrectly claims

g

is bijective when it's clearly not injective. Strategy tip: When dealing with absolute value compositions, always check both positive and negative inputs separately, and remember that absolute values typically destroy injectivity by making

f(x) = f(-x)

Question 8

Define $g: \mathbb{R} \setminus \{1\} \to \mathbb{R} \setminus \{2\}$ by $g(x) = \frac{2x}{x-1}$ . To show that $g$ is bijective, which of the following correctly identifies both the inverse function and the key step in proving surjectivity?

The inverse is $g^{-1}(y) = \frac{y+1}{y-2}$ , and surjectivity follows from verifying $g$ maps $\mathbb{R} \setminus \{1\}$ onto $\mathbb{R} \setminus \{2\}$
The inverse is $g^{-1}(y) = \frac{y+2}{y-2}$ , and surjectivity follows from showing the range equals $\mathbb{R} \setminus \{2\}$
The inverse is $g^{-1}(y) = \frac{y}{y-2}$ , and surjectivity follows from showing $\lim_{x \to \pm\infty} g(x) = 2$
The inverse is $g^{-1}(y) = \frac{y}{y-2}$ , and surjectivity follows from solving $y = \frac{2x}{x-1}$ for any $y \neq 2$ (correct answer)

Explanation: When proving a function is bijective, you need to establish both injectivity (one-to-one) and surjectivity (onto). For rational functions like this one, finding the inverse function is often the most efficient approach since it simultaneously proves both properties. To find the inverse of

g(x) = \frac{2x}{x-1}

, start with

y = \frac{2x}{x-1}

and solve for

x

. Cross-multiplying gives

y(x-1) = 2x

, which expands to

yx - y = 2x

. Rearranging:

yx - 2x = y

, so

x(y-2) = y

, yielding

x = \frac{y}{y-2}

. Therefore,

g^{-1}(y) = \frac{y}{y-2}

. For surjectivity, you must show that for every

y \in \mathbb{R} \setminus \{2\}

, there exists some

x \in \mathbb{R} \setminus \{1\}

such that

g(x) = y

. The key step is demonstrating that the equation

y = \frac{2x}{x-1}

has a solution for any

y \neq 2

, which we just showed:

x = \frac{y}{y-2}

. Option A has the wrong inverse function (

\frac{y+1}{y-2}

instead of

\frac{y}{y-2}

). Option B also gives an incorrect inverse (

\frac{y+2}{y-2}

). Option C has the correct inverse but incorrectly identifies the surjectivity approach—limits don't prove surjectivity directly. Study tip: When proving bijection for rational functions, always find the inverse algebraically by solving

y = f(x)

for

x

. This inverse both proves the function is bijective and gives you the explicit formula you need.

Question 9

Let $h: [0,1] \to [0,1]$ be defined by $h(x) = \begin{cases} 2x & \text{if } 0 \leq x \leq \frac{1}{2} \\ 2-2x & \text{if } \frac{1}{2} < x \leq 1 \end{cases}$ . Analyzing the properties of $h$ , which conclusion is correct?

$h$ is injective but not surjective, since $h$ is strictly increasing on each piece but has range $[0,1] \setminus \{1\}$
$h$ is surjective but not injective, since $h$ achieves every value in $[0,1]$ but $h(x) = h(1-x)$ for all $x$ (correct answer)
$h$ is bijective, since $h$ is continuous and maps the compact interval $[0,1]$ onto itself
$h$ is neither injective nor surjective, since $h$ fails the horizontal line test and has maximum value $1$

Explanation: When analyzing whether a function is injective (one-to-one) or surjective (onto), you need to examine how inputs map to outputs and whether every possible output is achieved. Let's check if

h

is surjective by finding its range. On

[0, \frac{1}{2}]

h(x) = 2x

ranges from

h(0) = 0

h(\frac{1}{2}) = 1

. On

(\frac{1}{2}, 1]

h(x) = 2-2x

ranges from values approaching

h(\frac{1}{2}) = 1

down to

h(1) = 0

. So

h

achieves every value in

[0,1]

, making it surjective. For injectivity, notice that

h

is not one-to-one because different inputs can produce the same output. For example,

h(\frac{1}{4}) = 2 \cdot \frac{1}{4} = \frac{1}{2}

and

h(\frac{3}{4}) = 2 - 2 \cdot \frac{3}{4} = \frac{1}{2}

. The pattern

h(x) = h(1-x)

holds generally, confirming the function is not injective. Choice A is wrong because

h

is surjective—it does achieve the value 1 at

x = \frac{1}{2}

. Choice C is wrong because continuity and mapping a compact set to itself doesn't guarantee bijectivity; the function must also be injective. Choice D is wrong because

h

is surjective—it does achieve its maximum value of 1 and every other value in

[0,1]

. Study tip: For piecewise functions, check injectivity by looking for symmetries or repeated values, and check surjectivity by carefully finding the range of each piece. The horizontal line test is your friend for visualizing injectivity.

Question 10

Consider the function $\phi: \mathcal{P}(\{1,2,3\}) \to \{0,1\}^3$ defined by $\phi(S) = (x_1, x_2, x_3)$ where $x_i = 1$ if $i \in S$ and $x_i = 0$ if $i \notin S$ . Here $\mathcal{P}(\{1,2,3\})$ denotes the power set. Which property does $\phi$ have?

$\phi$ is injective but not surjective, since distinct subsets map to distinct bit strings but not all bit strings correspond to valid subsets
$\phi$ is surjective but not injective, since every bit string corresponds to some subset but multiple subsets can yield the same bit string
$\phi$ is bijective, since it establishes a one-to-one correspondence between subsets of $\{1,2,3\}$ and binary strings of length 3 (correct answer)
$\phi$ is neither injective nor surjective, since the power set has more elements than the set of binary strings of length 3

Explanation: The function

\phi

maps each subset

S \subseteq \{1,2,3\}

to its characteristic vector. For injectivity: if

S \neq T

, then there exists

i

such that

i \in S

and

i \notin T

(or vice versa), so

\phi(S)

and

\phi(T)

differ in the

i

-th coordinate. For surjectivity: given any

(x_1,x_2,x_3) \in \{0,1\}^3

, take

S = \{i : x_i = 1\}

; then

\phi(S) = (x_1,x_2,x_3)

. Since

|\mathcal{P}(\{1,2,3\})| = 2^3 = 8 = |\{0,1\}^3|

, the bijection can also be seen by counting. Choice A incorrectly claims not all bit strings correspond to subsets. Choice B incorrectly claims multiple subsets can yield the same bit string. Choice D incorrectly states the cardinalities are different.

Question 11

A function $\phi: \mathbb{N} \to \mathbb{N}$ satisfies the property that for any $m, n \in \mathbb{N}$ , if $\gcd(m,n) = 1$ , then $\phi(mn) = \phi(m)\phi(n)$ . Additionally, $\phi(p^k) = p^{k-1}(p-1)$ for any prime $p$ and $k \geq 1$ . Which statement about $\phi$ is correct?

$\phi$ is injective because $\phi(p^k) \neq \phi(q^j)$ whenever $p \neq q$ are primes and $k,j \geq 1$
$\phi$ is not injective because $\phi(3) = \phi(4) = 2$ , but it is surjective since every positive integer appears as $\phi(n)$ for some $n$
$\phi$ is not injective because $\phi(3) = \phi(4) = \phi(6) = 2$ , and it is not surjective since some integers like 14 are not in the range (correct answer)
$\phi$ is surjective because the multiplicative property ensures all positive integers are achieved, but not injective due to overlapping values

Explanation: This function is Euler's totient function

\phi(n)

. Let's check specific values:

\phi(3) = 3^{1-1}(3-1) = 1 \cdot 2 = 2

\phi(4) = \phi(2^2) = 2^{2-1}(2-1) = 2 \cdot 1 = 2

, and

\phi(6) = \phi(2 \cdot 3) = \phi(2)\phi(3) = 1 \cdot 2 = 2

since

\phi(2) = 2^{1-1}(2-1) = 1

. So

\phi

is not injective. For surjectivity, the totient function is known to not be surjective - for example, no positive integer

n

satisfies

\phi(n) = 14

. Choice A incorrectly claims injectivity. Choice B incorrectly claims surjectivity. Choice D incorrectly claims surjectivity.

Question 12

Define $\psi: \mathbb{R} \setminus \{1\} \to \mathbb{R} \setminus \{2\}$ by $\psi(x) = \frac{2x}{x-1}$ . Which statement about $\psi$ and its inverse is correct?

$\psi$ is bijective with inverse $\psi^{-1}(y) = \frac{y}{y-2}$ , and both functions have the same domain exclusion pattern
$\psi$ is bijective with inverse $\psi^{-1}(y) = \frac{2y}{y-2}$ , but the domain exclusions are different for $\psi$ and $\psi^{-1}$
$\psi$ is not bijective because the function has vertical and horizontal asymptotes that prevent surjectivity onto the specified codomain
$\psi$ is bijective with inverse $\psi^{-1}(y) = \frac{y}{y-2}$ , and the domain exclusions correspond to the codomain exclusions (correct answer)

Explanation: When you encounter a function between restricted domains, you need to verify three things: injectivity (one-to-one), surjectivity (onto), and find the correct inverse formula. To find the inverse of

\psi(x) = \frac{2x}{x-1}

, start with

y = \frac{2x}{x-1}

and solve for

x

. Cross-multiplying gives

y(x-1) = 2x

, so

yx - y = 2x

. Rearranging:

yx - 2x = y

, which factors as

x(y-2) = y

. Therefore

x = \frac{y}{y-2}

, making

\psi^{-1}(y) = \frac{y}{y-2}

. Notice the beautiful symmetry:

\psi

excludes

x = 1

from its domain (where the denominator

x-1 = 0

), while

\psi^{-1}

excludes

y = 2

from its domain (where

y-2 = 0

). This isn't coincidental—the domain exclusion of

\psi

corresponds exactly to the codomain exclusion

\{2\}

, and vice versa. You can verify bijectivity by checking that

\psi

maps

\mathbb{R} \setminus \{1\}

onto

\mathbb{R} \setminus \{2\}

and that the inverse relationship holds. Choice A has the wrong inverse formula (missing the factor of 2). Choice B has an incorrect inverse

\frac{2y}{y-2}

and misunderstands the domain-codomain relationship. Choice C incorrectly claims the function isn't bijective—while asymptotes exist, they don't prevent bijectivity between the specified sets. Study tip: For rational functions, the domain exclusions of a function become the codomain exclusions of its inverse, and vice versa. Always check this correspondence when verifying inverse functions.

Question 13

Let $A = \{1, 2, 3, 4\}$ and consider all functions $f: A \to A$ such that $f(f(x)) = x$ for all $x \in A$ . How many of these functions are bijective?

All such functions are bijective, and there are exactly 10 of them in total (correct answer)
All such functions are bijective, and there are exactly 9 of them in total
Not all such functions are bijective, but exactly 8 of the total are bijective involutions
Not all such functions are bijective, but exactly 6 of the total are bijective involutions

Explanation: A function

f

satisfying

f(f(x)) = x

for all

x

is called an involution. We need to prove that every involution on a finite set is bijective. For injectivity: if

f(a) = f(b)

, then

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

. For surjectivity: given any

y \in A

, we have

f(f(y)) = y

, so

y = f(f(y))

, meaning

y

is the image of

f(y)

. Thus every involution is bijective. Now we count involutions on

\{1,2,3,4\}

. An involution partitions the set into fixed points (where

f(x) = x

) and 2-cycles (where

f(x) = y, f(y) = x

with

x \neq y

). Possible structures: (1) All 4 fixed points: 1 way. (2) 2 fixed points, 1 two-cycle:

\binom{4}{2} = 6

ways to choose the 2-cycle. (3) 0 fixed points, 2 two-cycles:

\frac{1}{2!}\binom{4}{2}\binom{2}{2} = \frac{6 \cdot 1}{2} = 3

ways. Total:

1 + 6 + 3 = 10

involutions. Choice B gives the wrong count. Choices C and D incorrectly suggest some involutions aren't bijective.

Question 14

Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = x^3 - 3x + 2$ . Which statement about $f$ is correct?

$f$ is injective because $f'(x) = 3x^2 - 3 > 0$ for all $x \in \mathbb{R}$
$f$ is not injective because $f'(x) = 3x^2 - 3 = 0$ has real solutions at $x = \pm 1$
$f$ is surjective but not injective, making it neither bijective nor one-to-one (correct answer)
$f$ is bijective because it is a cubic polynomial with positive leading coefficient

Explanation: To check injectivity, we examine if

f(x_1) = f(x_2)

implies

x_1 = x_2

. Since

f'(x) = 3x^2 - 3 = 3(x^2 - 1)

, we have

f'(x) = 0

x = \pm 1

. This creates local extrema:

f(-1) = 4

(local max) and

f(1) = 0

(local min). Since

f

has both a local maximum and minimum, there exist distinct values that map to the same output, so

f

is not injective. However, as

x \to \pm\infty

f(x) \to \pm\infty

respectively, and

f

is continuous, so by the Intermediate Value Theorem,

f

is surjective. Choice A is wrong because

f'(x)

is not always positive. Choice B correctly identifies why

f

is not injective but doesn't address surjectivity. Choice D is wrong because being cubic doesn't guarantee bijectivity.

Question 15

Let $f: \{0,1,2,\ldots,n-1\} \to \{0,1,2,\ldots,n-1\}$ be defined by $f(x) = ax \bmod n$ where $\gcd(a,n) = d > 1$ . Which statement about the range of $f$ is correct?

The range has exactly $\frac{n}{d}$ elements, and $f$ is injective on its domain but not surjective
The range has exactly $\frac{n}{d}$ elements, and $f$ is neither injective nor surjective on its full domain (correct answer)
The range has exactly $d$ elements, and $f$ is surjective but not injective on its domain
The range has exactly $n-d$ elements, and the function properties depend on the specific values of $a$ and $n$

Explanation: Since

\gcd(a,n) = d > 1

, we can write

a = da'

and

n = dn'

where

\gcd(a',n') = 1

. Then

f(x) = ax \bmod n = da'x \bmod dn'

. The range of

f

consists of all values

\{da'x \bmod dn' : x = 0,1,\ldots,n-1\}

. Since

\gcd(a',n') = 1

, as

x

runs through

\{0,1,\ldots,n'-1\}

a'x \bmod n'

takes all values

\{0,1,\ldots,n'-1\}

. Therefore,

da'x \bmod dn'

takes values

\{0,d,2d,\ldots,(n'-1)d\} = \{kd : k = 0,1,\ldots,n'-1\}

. This gives exactly

n' = \frac{n}{d}

distinct values. For injectivity:

f(x_1) = f(x_2)

means

ax_1 \equiv ax_2 \pmod{n}

, so

a(x_1-x_2) \equiv 0 \pmod{n}

. This means

n | a(x_1-x_2)

, so

dn' | da'(x_1-x_2)

, giving

n' | a'(x_1-x_2)

. Since

\gcd(a',n') = 1

, we need

n' | (x_1-x_2)

. For

x_1,x_2 \in \{0,1,\ldots,n-1\}

with

x_1 \neq x_2

, we can have

|x_1-x_2| = n' < n

, so

f

is not injective. Since

|\text{range}| = \frac{n}{d} < n

f

is not surjective. Choice A incorrectly claims injectivity.

Question 16

Consider the function $F: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $F(x,y) = (2x - y, x + y)$ . Which statement correctly describes the properties of $F$ ?

$F$ is injective but not surjective because the determinant of its matrix representation is positive
$F$ is surjective but not injective because the system $F(x,y) = (a,b)$ always has solutions but not unique ones
$F$ is neither injective nor surjective because its matrix representation is singular with determinant zero
$F$ is bijective because its matrix representation has determinant 3, making it invertible as a linear transformation (correct answer)

Explanation: When you encounter a linear function from

\mathbb{R}^2

\mathbb{R}^2

, the key insight is that its injectivity and surjectivity properties are completely determined by whether its matrix representation is invertible. The function

F(x,y) = (2x - y, x + y)

can be written as matrix multiplication:

F(\mathbf{v}) = A\mathbf{v}

where

A = \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix}

. The determinant is

\det(A) = 2(1) - (-1)(1) = 2 + 1 = 3

. Since the determinant is non-zero, the matrix is invertible, which means

F

is bijective (both injective and surjective). Looking at the wrong answers: Choice A incorrectly claims

F

is injective but not surjective. While it correctly notes the positive determinant, it misses that a non-zero determinant guarantees both properties, not just injectivity. Choice B states

F

is surjective but not injective with non-unique solutions. This contradicts the invertible matrix having exactly one solution for each output. Choice C claims

F

is neither injective nor surjective with determinant zero. This is factually wrong—we calculated the determinant as 3, not 0. Choice D correctly identifies that the non-zero determinant (specifically 3) makes the linear transformation invertible, ensuring bijectivity. Study tip: For linear transformations between spaces of equal dimension, always compute the determinant first. Non-zero determinant means bijective; zero determinant means neither injective nor surjective. This single calculation tells you everything about the function's properties.

Question 17

Let $h: \mathbb{Z} \to \mathbb{Z}$ be defined by $h(n) = \begin{cases} 2n & \text{if } n \geq 0 \\ 2n + 1 & \text{if } n < 0 \end{cases}$ . Which property does $h$ satisfy?

$h$ is injective but not surjective because negative integers are not in the range
$h$ is surjective but not injective because $h(0) = h(-1) = 0$
$h$ is bijective because it maps non-negative integers to even integers and negative integers to odd integers (correct answer)
$h$ is neither injective nor surjective because the piecewise definition creates overlapping ranges

Explanation: Let's analyze both properties. For injectivity: if

n \geq 0

, then

h(n) = 2n \geq 0

and even. If

n < 0

, then

h(n) = 2n + 1 < 1

and odd (since

2n

is even). The ranges don't overlap: non-negative even integers vs. negative odd integers. Within each piece, the functions

n \mapsto 2n

and

n \mapsto 2n+1

are strictly increasing, so no collisions occur. Thus

h

is injective. For surjectivity: every non-negative even integer

2k

equals

h(k)

where

k \geq 0

. Every negative odd integer

2k+1 < 0

can be written as

2k+1

where

k < 0

, so it equals

h(k)

. Every positive odd integer

2k+1 > 0

equals

h(k)

where

k = -\frac{2k+1-1}{2} < 0

. Thus

h

is surjective. Choice A is wrong about surjectivity and the range. Choice B incorrectly claims

h(0) = h(-1)

, but

h(0) = 0

and

h(-1) = -1

. Choice D incorrectly suggests the ranges overlap.

Question 18

Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = x^3 - 3x + 2$ . Consider the restriction $g: [1, \infty) \to \mathbb{R}$ where $g(x) = f(x)$ . Which statement about $g$ is correct?

$g$ is injective but not surjective, since $g$ is strictly increasing on $[1, \infty)$ but its range is $[0, \infty)$ (correct answer)
$g$ is surjective but not injective, since $g$ achieves all real values but fails the horizontal line test
$g$ is bijective, since $g$ is strictly increasing on $[1, \infty)$ and $\lim_{x \to \infty} g(x) = \infty$
$g$ is neither injective nor surjective, since $g$ has a local minimum at $x = 1$ and range $[0, \infty)$

Explanation: First, find critical points:

f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)

. So

f'(x) = 0

x = \pm 1

. For

x > 1

f'(x) > 0

, so

g

is strictly increasing on

[1, \infty)

, making it injective. Since

g(1) = 1 - 3 + 2 = 0

and

g

is increasing with

\lim_{x \to \infty} g(x) = \infty

, the range of

g

[0, \infty)

, not all of

\mathbb{R}

, so

g

is not surjective. Choice B is wrong because

g

is injective. Choice C is wrong because

g

is not surjective onto

\mathbb{R}

. Choice D is wrong because

g

is injective.