LFM Pure

jee-main 2020 Q59 True/False Justification View

Let $p , q , r$ be three statements such that the truth value of $( p \wedge q ) \rightarrow ( \sim q \vee r )$ is $F$. Then the truth values of $p , q , r$ are respectively :
(1) $T , T , F$
(2) $T , T , T$
(3) $T , F , T$
(4) $F , T , F$

jee-main 2020 Q59 True/False Justification View

Given the following two statements: $\left( \mathrm { S } _ { 1 } \right) : ( \mathrm { q } \vee \mathrm { p } ) \rightarrow ( \mathrm { p } \leftrightarrow \sim \mathrm { q } )$ is a tautology $\left( \mathrm { S } _ { 2 } \right) : \sim \mathrm { q } \wedge ( \sim \mathrm { p } \leftrightarrow \mathrm { q } )$ is a fallacy. Then :
(1) both ( $S _ { 1 }$ ) and ( $S _ { 2 }$ ) are not correct.
(2) only ( $S _ { 1 }$ ) is correct.
(3) only ( $S _ { 2 }$ ) is correct.
(4) both $\left( S _ { 1 } \right)$ and $\left( S _ { 2 } \right)$ are correct.

jee-main 2020 Q59 Direct Proof of a Stated Identity or Equality View

The negation of the Boolean expression $x \leftrightarrow \sim y$ is equivalent to:
(1) $( \sim x \wedge y ) \vee ( \sim x \wedge \sim y )$
(2) $( x \wedge y ) \vee ( \sim x \wedge \sim y )$
(3) $( x \wedge \sim y ) \vee ( \sim x \wedge y )$
(4) $( x \wedge y ) \wedge ( \sim x \vee \sim y )$

jee-main 2020 Q59 Direct Proof of a Stated Identity or Equality View

The negation of the Boolean expression $p \vee ( \sim p \wedge q )$ is equivalent to :
(1) $p \wedge \sim q$
(2) $\sim p \wedge \sim q$
(3) $\sim p \vee \sim \mathrm { q }$
(4) $\sim p \vee q$

jee-main 2020 Q60 Direct Proof of a Stated Identity or Equality View

Let $A , B , C$ and $D$ be four non-empty sets. The contrapositive statement of "If $A \subseteq B$ and $B \subseteq D$, then $A \subseteq C$" is
(1) If $A \nsubseteq C$, then $A \subseteq B$ and $B \subseteq D$
(2) If $A \subseteq C$, then $B \subset A$ and $D \subset B$
(3) If $A \nsubseteq C$, then $A \nsubseteq B$ and $B \subseteq D$
(4) If $A \nsubseteq C$, then $A \nsubseteq B$ or $B \nsubseteq D$

jee-main 2020 Q60 True/False Justification View

The statement $(p \rightarrow (q \rightarrow p)) \rightarrow (p \rightarrow (p \vee q))$ is:
(1) equivalent to $(p \wedge q) \vee (\sim q)$
(2) a contradiction
(3) equivalent to $(p \vee q) \wedge (\sim p)$
(4) a tautology

jee-main 2020 Q61 True/False Justification View

Which of the following is a tautology?
(1) $( \sim p ) \wedge ( p \vee q ) \rightarrow q$
(2) $( q \rightarrow p ) \vee \sim ( p \rightarrow q )$
(3) $( \sim q ) \vee ( p \wedge q ) \rightarrow q$
(4) $( p \rightarrow q ) \wedge ( q \rightarrow p )$

jee-main 2021 Q64 True/False Justification View

The negation of the statement $\sim p \wedge ( p \vee q )$ is:
(1) $\sim p \vee q$
(2) $\sim p \wedge q$
(3) $p \vee \sim q$
(4) $p \wedge \sim q$

jee-main 2021 Q65 True/False Justification View

For the statements $p$ and $q$, consider the following compound statements: $( a ) ( \sim q \wedge ( p \rightarrow q ) ) \rightarrow \sim p$
(b) $( ( p \vee q ) \wedge \sim p ) \rightarrow q$ Then which of the following statements is correct?
(1) (b) is a tautology but not (a).
(2) (a) and (b) both are tautologies.
(3) (a) and (b) both are not tautologies.
(4) (a) is a tautology but not (b).

jee-main 2021 Q66 Proof of Equivalence or Logical Relationship Between Conditions View

The Boolean expression $( p \wedge \sim q ) \Rightarrow ( q \vee \sim p )$ is equivalent to:
(1) $q \Rightarrow p$
(2) $p \Rightarrow q$
(3) $\sim q \Rightarrow p$
(4) $p \Rightarrow \sim q$

jee-main 2021 Q67 True/False Justification View

The statement among the following that is a tautology is:
(1) $A \vee A \wedge B$
(2) $A \wedge A \vee B$
(3) $B \rightarrow A \wedge A \rightarrow B$
(4) $A \wedge A \rightarrow B \rightarrow B$

jee-main 2021 Q67 True/False Justification View

Let $F _ { 1 } ( A , B , C ) = ( A \wedge \sim B ) \vee [ \sim C \wedge ( A \vee B ) ] \vee \sim A$ and $F _ { 2 } ( A , B ) = ( A \vee B ) \vee ( B \rightarrow \sim A )$ be two logical expressions. Then :
(1) $F _ { 1 }$ is a tautology but $F _ { 2 }$ is not a tautology
(2) $F _ { 1 }$ is not a tautology but $F _ { 2 }$ is a tautology
(3) Both $F _ { 1 }$ and $F _ { 2 }$ are not tautologies
(4) $F _ { 1 }$ and $F _ { 2 }$ both are tautologies

jee-main 2021 Q68 True/False Justification View

If $P$ and $Q$ are two statements, then which of the following compound statement is a tautology?
(1) $( ( P \Rightarrow Q ) \wedge \sim Q ) \Rightarrow Q$
(2) $( ( P \Rightarrow Q ) \wedge \sim Q ) \Rightarrow \sim P$
(3) $( ( P \Rightarrow Q ) \wedge \sim Q ) \Rightarrow P$
(4) $( ( P \Rightarrow Q ) \wedge \sim Q ) \Rightarrow ( P \wedge Q )$

jee-main 2021 Q69 True/False Justification View

If the Boolean expression $( p \Rightarrow q ) \Leftrightarrow \left( q ^ { * } ( \sim p ) \right)$ is a tautology, then the Boolean expression $p ^ { * } ( \sim q )$ is equivalent to:
(1) $q \Rightarrow p$
(2) $\sim q \Rightarrow p$
(3) $p \Rightarrow \sim q$
(4) $p \Rightarrow q$

jee-main 2021 Q70 True/False Justification View

Consider the statement "The match will be played only if the weather is good and ground is not wet". Select the correct negation from the following:
(1) The match will not be played and weather is not good and ground is wet.
(2) If the match will not be played, then either weather is not good or ground is wet.
(3) The match will be played and weather is not good or ground is wet.
(4) The match will not be played or weather is good and ground is not wet.

jee-main 2022 Q67 Proof of Equivalence or Logical Relationship Between Conditions View

The statement $( p \wedge q ) \Rightarrow ( p \wedge r )$ is equivalent to
(1) $q \Rightarrow ( p \wedge r )$
(2) $p \Rightarrow ( p \wedge r )$
(3) $( p \wedge r ) \Rightarrow ( p \wedge q )$
(4) $( p \wedge q ) \Rightarrow r$

jee-main 2022 Q68 True/False Justification View

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

jee-main 2022 Q68 True/False Justification View

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) $(\wedge, \vee)$

jee-main 2022 Q68 Proof of Equivalence or Logical Relationship Between Conditions View

The statement $(p \Rightarrow (q \vee p)) \Rightarrow r$ is NOT equivalent to:
(1) $p \wedge \sim r \Rightarrow q$
(2) $\sim q \Rightarrow (\sim r \vee p)$
(3) $p \Rightarrow (q \vee r)$
(4) $p \wedge \sim q \Rightarrow r$

jee-main 2022 Q69 True/False Justification View

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

jee-main 2022 Q69 True/False Justification View

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) $\sim p \wedge \sim q$

jee-main 2022 Q69 Proof of Equivalence or Logical Relationship Between Conditions View

For $\alpha \in \mathbb{N}$, consider a relation $R$ on $\mathbb{N}$ given by $R = \{(x, y) : 3x + \alpha y \text{ is a multiple of } 7\}$. The relation $R$ is an equivalence relation if and only if
(1) $\alpha = 14$
(2) $\alpha$ is a multiple of 4
(3) 4 is the remainder when $\alpha$ is divided by 10
(4) 4 is the remainder when $\alpha$ is divided by 7

jee-main 2022 Q70 Proof of Equivalence or Logical Relationship Between Conditions View

The negation of the Boolean expression $(\sim q \wedge p) \Rightarrow (\sim p \vee q)$ is logically equivalent to
(1) $p \Rightarrow q$
(2) $q \Rightarrow p$
(3) $\sim p \Rightarrow q$
(4) $\sim q \Rightarrow p$

jee-main 2022 Q70 True/False Justification View

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) $((P \wedge (\sim R)) \wedge Q) \vee ((\sim Q) \wedge ((\sim P) \wedge R))$

jee-main 2023 Q66 Proof of Equivalence or Logical Relationship Between Conditions View

The compound statement $( \sim ( P \wedge Q ) ) \vee ( ( \sim P ) \wedge Q ) \Rightarrow ( ( \sim P ) \wedge ( \sim Q ) )$ is equivalent to
(1) $( ( \sim P ) \vee Q ) \wedge ( ( \sim Q ) \vee P )$
(2) $( \sim Q ) \vee P$
(3) $( ( \sim P ) \vee Q ) \wedge ( \sim Q )$
(4) $( \sim P ) \vee Q$

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LFM Pure

Straight Lines & Coordinate Geometry 247

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Simultaneous equations 79

Proof 711

Proof by induction 25

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Standard trigonometric equations 106

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Implicit equations and differentiation 43

Differential equations 352

3x3 Matrices 144