JEE MainMathematics

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Mathematical Reasoning: Mathematics | JEE Main

Q: Why test specifically first?

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

Let

\Delta, \nabla \in \{\wedge, \vee\}

be such that

(p \rightarrow q) \Delta(p \nabla q)

is a tautology. Then

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\Delta = \vee, \nabla = \vee

\Delta = \vee, \nabla = \wedge

\Delta = \wedge, \nabla = \vee

\Delta = \wedge, \nabla = \wedge

Step-by-step solutionView

Correct answer

The compound statement $(p \rightarrow q) \Delta (p \nabla q)$ is a tautology when $\Delta = \vee$ and $\nabla = \vee$ .

Option analysis

Why each option works or fails

A ·

\Delta = \vee, \nabla = \vee

Correct choice. When $p = \text{T}$ and $q = \text{F}$ , $p \rightarrow q$ is $\text{F}$ . For the whole expression to be $\text{T}$ , $\Delta$ must be $\vee$ and the second term $p \nabla q$ must be $\text{T}$ , which forces $\nabla = \vee$ .

B ·

\Delta = \vee, \nabla = \wedge

Believing that $p \wedge q$ is true when $p = \text{T}$ and $q = \text{F}$ . When $p = \text{T}$ and $q = \text{F}$ , $p \wedge q$ evaluates to $\text{F}$ , making $(p \rightarrow q) \vee (p \wedge q) = \text{F} \vee \text{F} = \text{F}$ , which is not a tautology.

C ·

\Delta = \wedge, \nabla = \vee

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

D ·

\Delta = \wedge, \nabla = \wedge

Assuming $\Delta = \wedge$ and $\nabla = \wedge$ will satisfy the condition across all truth values. For $p = \text{T}$ and $q = \text{F}$ , $p \rightarrow q = \text{F}$ and $p \wedge q = \text{F}$ , so $\text{F} \wedge \text{F} = \text{F}$ , which is not a tautology.

Reviewed route

Solution

StepWorking

01given

The statement $(p \rightarrow q) \, \Delta \, (p \, \nabla \, q)$ is a tautology, where $\Delta, \nabla \in \{\wedge, \vee\}$ .

02approach

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

03execute

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

✓verify

Verify all four truth rows for $\Delta = \vee$ and $\nabla = \vee$ : - $p=\text{T}, q=\text{T} \implies (\text{T}) \vee (\text{T}) = \text{T}$ - $p=\text{T}, q=\text{F} \implies (\text{F}) \vee (\text{T}) = \text{T}$ - $p=\text{F}, q=\text{T} \implies (\text{T}) \vee (\text{T}) = \text{T}$ - $p=\text{F}, q=\text{F} \implies (\text{T}) \vee (\text{F}) = \text{T}$ All rows are True, confirming $(p \rightarrow q) \vee (p \vee q)$ is indeed a tautology.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why test

p=\text{T}, q=\text{F}

specifically first?

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

Answer

The compound statement $(p \rightarrow q) \Delta (p \nabla q)$ is a tautology when $\Delta = \vee$ and $\nabla = \vee$ .

Why each option works or fails

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

Step-by-step solution

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

Shortcut:

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

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

option_verdict: For $\Delta = \vee, \nabla = \vee$ : when $p=\text{T}, q=\text{F}$ , $(p \rightarrow q) \vee (p \vee q) = \text{F} \vee \text{T} = \text{T}$ . For all other rows $p \rightarrow q = \text{T}$ , so the disjunction is always $\text{T}$ . Hence it is a tautology.

option_verdict: For $\Delta = \vee, \nabla = \wedge$ : when $p=\text{T}, q=\text{F}$ , we get $(p \rightarrow q) \vee (p \wedge q) = \text{F} \vee \text{F} = \text{F}$ , not a tautology.

option_verdict: For $\Delta = \wedge, \nabla = \vee$ : when $p=\text{T}, q=\text{F}$ , we get $(p \rightarrow q) \wedge (p \vee q) = \text{F} \wedge \text{T} = \text{F}$ , not a tautology.

option_verdict: For $\Delta = \wedge, \nabla = \wedge$ : when $p=\text{T}, q=\text{F}$ , we get $(p \rightarrow q) \wedge (p \wedge q) = \text{F} \wedge \text{F} = \text{F}$ , not a tautology.

discriminator: At $p=\text{T}, q=\text{F}$ , $p \rightarrow q$ is False, forcing $\Delta = \vee$ and the second term to be True (which forces $\nabla = \vee$ ).

Mathematical Reasoning: Mathematics | JEE Main

What feels right?

The compound statement $(p \rightarrow q) \Delta (p \nabla q)$ is a tautology when $\Delta = \vee$ and $\nabla = \vee$ .

Why each option works or fails

Solution

01given

02approach

03execute

✓verify

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