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FIT1058 Foundations of computing - S1 2025

FIT1058 Foundations of computing - S1 2025

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Which of the following Boolean expressions is equivalent to $A \wedge \neg (B \wedge C)$ A \wedge \neg (B \wedge C)?

$(A \vee \neg B) \wedge (\neg B \vee \neg C)$ (A \vee \neg B) \wedge (\neg B \vee \neg C)

$(A \vee \neg B) \wedge (A \vee \neg C)$ (A \vee \neg B) \wedge (A \vee \neg C)

$(A \wedge \neg B) \vee (\neg B \wedge \neg C)$ (A \wedge \neg B) \vee (\neg B \wedge \neg C)

$(A \wedge \neg B) \vee (A \wedge \neg C)$ (A \wedge \neg B) \vee (A \wedge \neg C)

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Let $A = \{1, 2\}, B = \{x, y\}$ A = \{1, 2\}, B = \{x, y\}, and $C = \{2, 3\}$ C = \{2, 3\}. Which of the following is equal to $(A \triangle C) \times B$ (A \triangle C) \times B?

$\{(1, x), (1, y), (3, x), (3, y)\}$ \{(1, x), (1, y), (3, x), (3, y)\}

$\{1, 3, x, y\}$ \{1, 3, x, y\}

$\{(2, x), (2, y)\}$ \{(2, x), (2, y)\}

$\{1, 2, 3, x, y\}$ \{1, 2, 3, x, y\}

The empty set.

$\{(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)\}$ \{(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)\}

$\{2, x, y\}$ \{2, x, y\}

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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ f: \mathbb{R} \rightarrow \mathbb{R} and $g: \mathbb{R} \rightarrow \mathbb{R}$ g: \mathbb{R} \rightarrow \mathbb{R} be defined as:

f(x) = 2x + 3

g(x) = x^2 - 1

Which of the following statements is true?

$(f\circ g)(x) = 2x^2 + 1$ (f\circ g)(x) = 2x^2 + 1

$(g\circ f)(x) = 2x^2 + 1$ (g\circ f)(x) = 2x^2 + 1

$f^{-1}(x)$ f^{-1}(x) and $g^{-1}(x)$ g^{-1}(x) exist for all $x \in \mathbb{R}$ x \in \mathbb{R}.

Neither $f^{-1}(x)$ f^{-1}(x) nor $g^{-1}(x)$ g^{-1}(x) exist for all $x \in \mathbb{R}$ x \in \mathbb{R}.

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Let R be a binary relation on the set $\mathbb{Z}$ \mathbb{Z}. Which of the following definitions for R would make it an equivalence relation?

For all $x, y \in \mathbb{Z}, xRy$ x, y \in \mathbb{Z}, xRy if and only if x = -y.

For all $x, y \in \mathbb{Z}, xRy$ x, y \in \mathbb{Z}, xRy if and only if $xy \ge 0$ xy \ge 0.

For all $x, y \in \mathbb{Z}, xRy$ x, y \in \mathbb{Z}, xRy if and only if x < y.

For all $x, y \in \mathbb{Z}, xRy$ x, y \in \mathbb{Z}, xRy if and only if x - y is even.

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Which of the following is a geometric sequence?

1, 2, 6, 24, 120, …

1, 2, 4, 8, 16, 32, …

100, 81, 64, 49, 36, …, 9, 4, 1

100, 98, 96, 94, 92, … , 4, 2, 0

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Which of the following is an arithmetic sequence?

100, 98, 96, 94, 92, … , 4, 2, 0

1, 2, 4, 8, 16, 32, …

100, 81, 64, 49, 36, …, 9, 4, 1

1, 2, 6, 24, 120, …

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Given the following recursively defined sequence:

$f_1 = 1, f_n = (\sqrt{f_{n-1}} + 1)^2$ f_1 = 1, f_n = (\sqrt{f_{n-1}} + 1)^2

Which of the following is a correct non-recursive definition for the same sequence?

$f_n = \sqrt{n+1}$ f_n = \sqrt{n+1}

$f_n = \sqrt{n}$ f_n = \sqrt{n}

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Consider the sequence defined by $f_n = 2^{n}$ f_n = 2^{n}, which begins 1, 2, 4, 8, 16, 32...

Which of the following is a correct recursive definition of this sequence?

$f_1 = 1, f_n = f_{n-1} \times 2$ f_1 = 1, f_n = f_{n-1} \times 2

$f_1 = 1, f_n = (n-1) \times 2$ f_1 = 1, f_n = (n-1) \times 2

$f_1 = 1, f_n = 2^{f_{n-1}} \times 2$ f_1 = 1, f_n = 2^{f_{n-1}} \times 2

$f_1 = 1, f_n = 2^{n-1} \times 2$ f_1 = 1, f_n = 2^{n-1} \times 2

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Suppose you have:

a function f(A) which takes some finite set of strings A as input, and outputs the cardinality of A; and
a function g(B, C) which takes two finite sets of strings B and C as input, and outputs the Cartesian product $B \times C$ B \times C.

Which of the following statements is true?

$g \circ f$ g \circ f is defined, but $f \circ g$ f \circ g is not.

$f \circ g$ f \circ g is defined, but $g \circ f$ g \circ f is not.

Neither $g \circ f$ g \circ f nor $f \circ g$ f \circ g are defined.

Both $g \circ f$ g \circ f and $f \circ g$ f \circ g are defined.

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Let:

B be the set of all possible bijections;

S be the set of all possible surjections;

I be the set of all possible injections; and

J be the set of all possible identity functions.

Which of the following statements is true?

$B \cap (I \cup S) = \emptyset$ B \cap (I \cup S) = \emptyset

$J \cap B = \emptyset$ J \cap B = \emptyset

$J \cap (I \triangle S) = \emptyset$ J \cap (I \triangle S) = \emptyset

$B \cap (I \cap S) = \emptyset$ B \cap (I \cap S) = \emptyset

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