The number of choices for $\Delta \in \{ \wedge , \vee , \Rightarrow , \Leftrightarrow \}$, such that $( p \Delta q ) \Rightarrow ( ( p \Delta \sim q ) \vee ( ( \sim p ) \Delta q ) )$ is a tautology, is
(1) 1
(2) 2
(3) 3
(4) 4

Let a set $A = A _ { 1 } \cup A _ { 2 } \cup \ldots \cup A _ { k }$, where $A _ { i } \cap A _ { j } = \phi$ for $i \neq j ; 1 \leq i , j \leq k$. Define the relation $R$ from $A$ to $A$ by $R = \left\{ ( x , y ) : y \in A _ { i } \right.$ if and only if $\left. x \in A _ { i } , 1 \leq i \leq k \right\}$. Then, $R$ is:
(1) reflexive, symmetric but not transitive
(2) reflexive, transitive but not symmetric
(3) reflexive but not symmetric and transitive
(4) an equivalence relation

Consider the following statements: $P$: Ramu is intelligent. $Q$: Ramu is rich. $R$: Ramu is not honest. The negation of the statement ``Ramu is intelligent and honest if and only if Ramu is not rich'' can be expressed as:
(1) $((P \wedge (\sim R)) \wedge Q) \wedge ((\sim Q) \wedge ((\sim P) \vee R))$
(2) $((P \wedge R) \wedge Q) \vee ((\sim Q) \wedge ((\sim P) \vee (\sim R)))$
(3) $((P \wedge R) \wedge Q) \wedge ((\sim Q) \wedge ((\sim P) \vee (\sim R)))$
(4) $((P \wedge (\sim R)) \wedge Q) \vee ((\sim Q) \wedge ((\sim P) \wedge R))$

Negation of the Boolean expression $p \leftrightarrow (q \rightarrow p)$ is
(1) $\sim p \wedge q$
(2) $p \wedge \sim q$
(3) $\sim p \vee \sim q$
(4) $\sim p \wedge \sim q$

Let the operations $*, \odot \in \{\wedge, \vee\}$. If $p * q \odot p \odot {\sim}q$ is a tautology, then the ordered pair $(*, \odot)$ is
(1) $(\vee, \wedge)$
(2) $(\vee, \vee)$
(3) $(\wedge, \wedge)$
(4) $(\wedge, \vee)$

The relation $R = a , b : \operatorname { gcd} a , b = 1 , \quad 2 a \neq b , \quad a , \quad b \in \mathbb { Z }$ is:
(1) transitive but not reflexive
(2) symmetric but not transitive
(3) reflexive but not symmetric
(4) neither symmetric nor transitive

The number of values of $r \in \{p, q, \sim p, \sim q\}$ for which $((p \wedge q) \Rightarrow (r \vee q)) \wedge ((p \wedge r) \Rightarrow q)$ is a tautology, is:
(1) 1
(2) 2
(3) 4
(4) 3

Among the relations $S = \left\{(a,b) : a, b \in R - \{0\},\ 2 + \frac{a}{b} > 0\right\}$ and $T = \left\{(a,b) : a, b \in R,\ a^2 - b^2 \in Z\right\}$,
(1) $S$ is transitive but $T$ is not
(2) both $S$ and $T$ are symmetric
(3) neither $S$ nor $T$ is transitive
(4) $T$ is symmetric but $S$ is not

Among the statements: $(S1): 2023^{2022} - 1999^{2022}$ is divisible by 8. $(S2): 13(13)^{n} - 11n - 13$ is divisible by 144 for infinitely many $n \in \mathbb{N}$
(1) Only $(S2)$ is correct
(2) Only $(S1)$ is correct
(3) Both $(S1)$ and $(S2)$ are correct
(4) Both $(S1)$ and $(S2)$ are incorrect

Among the statements $(S1): (p \Rightarrow q) \vee ((\sim p) \wedge q)$ is a tautology $(S2): (q \Rightarrow p) \Rightarrow ((\sim p) \wedge q)$ is a contradiction
(1) Neither $(S1)$ and $(S2)$ is True
(2) Both $(S1)$ and $(S2)$ are True
(3) Only $(S2)$ is True
(4) Only $(S1)$ is True

Negation of $p \wedge ( q \wedge \sim ( p \wedge q ) )$ is
(1) $( \sim ( p \wedge q ) ) \vee p$
(2) $p \vee q$
(3) $\sim ( p \vee q )$
(4) $( \sim ( p \wedge q ) ) \wedge q$

Let $x = (8\sqrt{3} + 13)^{13}$ and $y = (7\sqrt{2} + 9)^{9}$. If $[t]$ denotes the greatest integer $\leq t$, then
(1) $[x] + [y]$ is even
(2) $[x]$ is odd but $[y]$ is even
(3) $[x]$ is even but $[y]$ is odd
(4) $[x]$ and $[y]$ are both odd

Consider the relations $R_1$ and $R_2$ defined as $a R_1 b \Leftrightarrow a^2 + b^2 = 1$ for all $a, b \in \mathbb{R}$ and $(a,b) R_2 (c,d) \Leftrightarrow a + d = b + c$ for all $a, b, c, d \in \mathbb{N} \times \mathbb{N}$. Then
(1) Only $R_1$ is an equivalence relation
(2) Only $R_2$ is an equivalence relation
(3) $R_1$ and $R_2$ both are equivalence relations
(4) Neither $R_1$ nor $R_2$ is an equivalence relation

Let $X = \mathbf { R } \times \mathbf { R }$. Define a relation $R$ on $X$ as : $\left( a _ { 1 } , b _ { 1 } \right) R \left( a _ { 2 } , b _ { 2 } \right) \Leftrightarrow b _ { 1 } = b _ { 2 }$
Statement I : $\quad \mathrm { R }$ is an equivalence relation.
Statement II : For some $( a , b ) \in X$, the set $S = \{ ( x , y ) \in X : ( x , y ) R ( a , b ) \}$ represents a line parallel to $y = x$.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both Statement I and Statement II are false
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are true
(4) Statement I is false but Statement II is true

The relation $R = \{ ( x , y ) : x , y \in \mathbb { Z }$ and $x + y$ is even $\}$ is:
(1) reflexive and symmetric but not transitive
(2) an equivalence relation
(3) symmetric and transitive but not reflexive
(4) reflexive and transitive but not symmetric

A student made an error while proving the following claim that he thought was true.
Claim: For any sets $A$, $B$, $C$, we have $A \backslash ( B \cap C ) \subseteq ( A \backslash B ) \cap ( A \backslash C )$.
The student's proof:
If I show that every element of the set $A \backslash ( B \cap C )$ is in the set $( A \backslash B ) \cap ( A \backslash C )$, the proof is complete.
Now, let $x \in A \backslash ( B \cap C )$. (I) From this, $x \in A$ and $x \notin ( B \cap C )$. (II) From this, $x \in A$ and $( x \notin B$ and $x \notin C )$. (III) From this, $( x \in A$ and $x \notin B )$ and $( x \in A$ and $x \notin C )$. (IV) From this, $x \in A \backslash B$ and $x \in A \backslash C$. (V) From this, $x \in [ ( A \backslash B ) \cap ( A \backslash C ) ]$.
In which of the numbered steps did this student make an error?
A) I
B) II
C) III
D) IV
E) V

Let A, B and C be sets. I. If $A \cup B = A \cup C$ then $B = C ^ { \prime }$. II. If $\mathrm { A } \cap \mathrm { B } = \varnothing$ then $\mathrm { A } \backslash \mathrm { B } = \mathrm { A } ^ { \prime }$. III. If $A \cup B = A$ then $B \backslash A = \varnothing$. Which of these propositions are always true?
A) Only I
B) Only II
C) Only III
D) I and II
E) II and III

A student made an error while proving the following claim that he thought was true.
Claim: Let $f : X \rightarrow Y$ be a function, and let $A$ and $B$ be subsets of $X$. Then $f ( A \cap B ) = f ( A ) \cap f ( B )$.
The student's proof: If I show that the sets $f ( A \cap B )$ and $f ( A ) \cap f ( B )$ are subsets of each other, the proof is complete.
Now let $c \in f ( A \cap B )$. I. There exists a $d \in A \cap B$ such that $c = f ( d )$. II. Since $d \in A$ and $d \in B$, we have $f ( d ) \in f ( A )$ and $f ( d ) \in f ( B )$. Thus $c = f ( d ) \in f ( A ) \cap f ( B )$.
On the other hand, let $c \in f ( A ) \cap f ( B )$. III. We have $c \in f ( A )$ and $c \in f ( B )$. From this, there exists an $a \in A$ such that $c = f ( a )$ and a $\mathrm { b } \in \mathrm { B }$ such that $c = f ( b )$. IV. Since $c = f ( a )$ and $c = f ( b )$, we have $a = b$. V. Since $a \in A , b \in B$ and $a = b$, we have $a \in A \cap B$ and thus $c = f ( a ) \in f ( A \cap B )$.
In which of the numbered steps did this student make an error?
A) I
B) II
C) III
D) IV
E) V

Ali; starting with the equality $x = y$ for non-zero, equal real numbers x and y, follows the following steps in order.
I. Let us multiply both sides of the equality by x: $$x ^ { 2 } = x \cdot y$$
II. Let us subtract $\mathrm { y } ^ { 2 }$ from both sides: $$x ^ { 2 } - y ^ { 2 } = x \cdot y - y ^ { 2 }$$
III. Let us factor both sides: $$( x + y ) ( x - y ) = y ( x - y )$$
IV. Let us divide both sides by $\mathrm { x } - \mathrm { y }$: $$x + y = y$$
V. Let us substitute y for x: $$2 y = y$$
As a result of these steps, Ali arrives at the conclusion "Every number equals twice itself."
Accordingly, in which of the numbered steps did Ali make an error?
A) I B) II C) III D) IV E) V

For propositions $p$, $q$, and $r$ $$( p \Rightarrow q ) \Rightarrow r$$ it is known that the proposition is false.
Accordingly,\ I. $p \Rightarrow q$\ II. $q \Rightarrow r$\ III. $r \Rightarrow p$\ Which of the following propositions are always true?\ A) Only I\ B) Only II\ C) Only III\ D) I and III\ E) II and III

A student made an error while proving the following claim that he believed to be true.
Claim: The number $\pi$ equals the number $e$.\ The student's proof: Let $f ( x )$ and $g ( x )$ be functions for $x > 0$ defined as $\mathrm{f} ( \mathrm{x} ) = \ln ( \pi \mathrm{x} )$ and $\mathrm{g} ( \mathrm{x} ) = \ln ( \mathrm{ex} )$.\ I. For every $x > 0$, the derivatives of functions $f ( x )$ and $g ( x )$ are equal to each other.\ II. Therefore, for every $x > 0$, functions $f ( x )$ and $g ( x )$ are equal to each other.\ III. Since $\ln ( x )$ is one-to-one and $f ( x ) = g ( x )$, we conclude that for every $x > 0$, $\pi x = ex$.\ IV. If two functions are equal for every $x > 0$, then their values at $x = 1$ are the same.\ V. Since the values of the functions $\pi \mathrm{x}$ and $ex$ at $x = 1$ are the same, we conclude that $\pi = \mathrm{e}$.\ In which of the numbered steps did this student make an error?\ A) I\ B) II\ C) III\ D) IV\ E) V

Let $a$ and $b$ be integers. The notation $\mathrm { a } \mid \mathrm { b }$ means that $a$ divides $b$ exactly.
A student wants to prove that the proposition "If integers $a$, $b$ and $c$ satisfy the conditions $a \mid c$ and $b \mid c$, then $(a + b) \mid c$ also holds." is false by using the counterexample method.
Accordingly, which of the following could be the example given by the student?