JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Finding Logical Connectives that Form a Tautology

Q: Why test p=T, q=F specifically first?

Because p → q is only False in that single row. Testing where a component fails immediately restricts the connecting operator Δ.

Let Δ, ∇ ∈ {wedge, vee} be such that (p → q) Δ(p ∇ q) is a tautology. Then

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What feels right?

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AΔ = vee, ∇ = veeBΔ = vee, ∇ = wedgeCΔ = wedge, ∇ = veeDΔ = wedge, ∇ = wedge

Step-by-step solutionView

Correct answer

The compound statement $(p \to q) Δ (p \nabla q)$ is a tautology when $Δ = \lor$ and $\nabla = \lor$ .

Option analysis

Why each option works or fails

A · Δ = vee, ∇ = vee

Correct choice. When $p = T$ and $q = F$ , $p \to q$ is $F$ . For the whole expression to be $T$ , $Δ$ must be $\lor$ and the second term $p \nabla q$ must be $T$ , which forces $\nabla = \lor$ .

B · Δ = vee, ∇ = wedge

Believing that $p \land q$ is true when $p = T$ and $q = F$ . When $p = T$ and $q = F$ , $p \land q$ evaluates to $F$ , making $(p \to q) \lor (p \land q) = F \lor F = F$ , which is not a tautology.

C · Δ = wedge, ∇ = vee

Assuming that a conjunction $Δ = \land$ can produce a tautology when one of its conjuncts is false. When $p = T$ and $q = F$ , $p \to q = F$ . If $Δ = \land$ , the expression evaluates to $F \land (p \nabla q) = F$ , failing the tautology condition.

D · Δ = wedge, ∇ = wedge

Assuming $Δ = \land$ and $\nabla = \land$ will satisfy the condition across all truth values. For $p = T$ and $q = F$ , $p \to q = F$ and $p \land q = F$ , so $F \land F = F$ , which is not a tautology.

Reviewed route

Solution

StepWorking

01Given

The statement $(p \to q) Δ (p \nabla q)$ is a tautology, where $Δ, \nabla \in {\land, \lor}$ .

02Approach

A conditional $p \to q$ is False only when $p = T$ and $q = F$ . In that case, for the whole expression $(p \to q) Δ (p \nabla q)$ to be True, $Δ$ must be $\lor$ and $(p \nabla q)$ must be True when $p = T, q = F$ . We can analyze truth values case-by-case or test candidate combinations.

03Execute

Consider the assignment $p = T$ and $q = F$ : 1. Then $p \to q = F$ . 2. For $(p \to q) Δ (p \nabla q)$ to evaluate to $T$ , $Δ$ cannot be $\land$ (since $F \land X = F$ ). Therefore, $Δ = \lor$ . 3. With $Δ = \lor$ , we require $F \lor (p \nabla q) = T$ , so $(p \nabla q)$ must be $T$ when $p = T, q = F$ . 4. If $\nabla = \land$ , then $p \land q = T \land F = F$ , which fails. 5. If $\nabla = \lor$ , then $p \lor q = T \lor F = T$ , which satisfies the condition.

✓Verify

Verify all four truth rows for $Δ = \lor$ and $\nabla = \lor$ : - $p = T, q = T ⟹ (T) \lor (T) = T$ - $p = T, q = F ⟹ (F) \lor (T) = T$ - $p = F, q = T ⟹ (T) \lor (T) = T$ - $p = F, q = F ⟹ (T) \lor (F) = T$ All rows are True, confirming $(p \to q) \lor (p \lor q)$ is indeed a tautology.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why test

p = T, q = F

specifically first?

Because $p \to q$ is only False in that single row. Testing where a component fails immediately restricts the connecting operator $Δ$ .

Answer

The compound statement $(p \to q) Δ (p \nabla q)$ is a tautology when $Δ = \lor$ and $\nabla = \lor$ .

Why each option works or fails

A · correct: Δ = vee, ∇ = vee - Correct choice. When $p = T$ and $q = F$ , $p \to q$ is $F$ . For the whole expression to be $T$ , $Δ$ must be $\lor$ and the second term $p \nabla q$ must be $T$ , which forces $\nabla = \lor$ .
B: Δ = vee, ∇ = wedge - Believing that $p \land q$ is true when $p = T$ and $q = F$ . When $p = T$ and $q = F$ , $T$ evaluates to $F$ , making $(p \to q) \lor (p \land q) = F \lor F = F$ , which is not a tautology.
C: Δ = wedge, ∇ = vee - Assuming that a conjunction $Δ = \land$ can produce a tautology when one of its conjuncts is false. When $p = T$ and $q = F$ , $p \to q = F$ . If $Δ = \land$ , the expression evaluates to $F \land (p \nabla q) = F$ , failing the tautology condition.
D: Δ = wedge, ∇ = wedge - Assuming $Δ = \land$ and $\nabla = \land$ will satisfy the condition across all truth values. For $p = T$ and $q = F$ , $p \to q = F$ and $p \land q = F$ , so $F \land F = F$ , which is not a tautology.

Step-by-step solution

Given: The statement $(p \to q) Δ (p \nabla q)$ is a tautology, where $Δ, \nabla \in {\land, \lor}$ .
Approach: A conditional $p \to q$ is False only when $p = T$ and $q = F$ . In that case, for the whole expression $(p \to q) Δ (p \nabla q)$ to be True, $Δ$ must be $\lor$ and $(p \nabla q)$ must be True when $p = T, q = F$ . We can analyze truth values case-by-case or test candidate combinations.
Execute: Consider the assignment $p = T$ and $q = F$ : 1. Then $p \to q = F$ . 2. For $(p \to q) Δ (p \nabla q)$ to evaluate to $T$ , $Δ$ cannot be $\land$ (since $F \land X = F$ ). Therefore, $Δ = \lor$ . 3. With $Δ = \lor$ , we require $F \lor (p \nabla q) = T$ , so $(p \nabla q)$ must be $T$ when $p = T, q = F$ . 4. If $Δ$ , then $p \land q = T \land F = F$ , which fails. 5. If $\nabla = \lor$ , then $p \lor q = T \lor F = T$ , which satisfies the condition.
Verify: Verify all four truth rows for $Δ = \lor$ and $\nabla = \lor$ : - $p = T, q = T ⟹ (T) \lor (T) = T$ - $p = T, q = F ⟹ (F) \lor (T) = T$ - $p = F, q = T ⟹ (T) \lor (T) = T$ - $p = F, q = F ⟹ (T) \lor (F) = T$ All rows are True, confirming $(p \to q) \lor (p \lor q)$ is indeed a tautology.

Shortcut:

When to use it: Directly testing the options by truth evaluation of the single row $p = T, q = F$ .

Concept: For an expression to be a tautology, it must be True for every truth assignment of $p$ and $q$ . The critical row where $p \to q$ is False is $p = T, q = F$ .

Option verdict: For $Δ = \lor, \nabla = \lor$ : when $p = T, q = F$ , $(p \to q) \lor (p \lor q) = F \lor T = T$ . For all other rows $p \to q = T$ , so the disjunction is always $T$ . Hence it is a tautology.

Option verdict: For $Δ = \lor, \nabla = \land$ : when $p = T, q = F$ , we get $(p \to q) \lor (p \land q) = F \lor F = F$ , not a tautology.

Option verdict: For $Δ = \land, \nabla = \lor$ : when $p = T, q = F$ , we get $(p \to q) \land (p \lor q) = F \land T = F$ , not a tautology.

Option verdict: For $Δ = \land, \nabla = \land$ : when $p = T, q = F$ , we get $(p \to q) \land (p \land q) = F \land F = F$ , not a tautology.

Discriminator: At $p = T, q = F$ , $p \to q$ is False, forcing $Δ = \lor$ and the second term to be True (which forces $\nabla = \lor$ ).

Finding Logical Connectives that Form a Tautology

What feels right?

The compound statement $(p \to q) Δ (p \nabla q)$ is a tautology when $Δ = \lor$ and $\nabla = \lor$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

✓Verify

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