Given the following two statements: $(\mathrm{S}_{1}):(\mathrm{q} \vee \mathrm{p}) \rightarrow(\mathrm{p} \leftrightarrow \sim \mathrm{q})$ is a tautology. $(\mathrm{S}_{2}): \sim \mathrm{q} \wedge(\sim \mathrm{p} \leftrightarrow \mathrm{q})$ is a fallacy. Then:

(C) both $(\mathrm{S}_{1})$ and $(\mathrm{S}_{2})$ are incorrect

(A) both $(\mathrm{S}_{1})$ and $(\mathrm{S}_{2})$ are correct

(B) only $(\mathrm{S}_{2})$ is correct

(D) only $(\mathrm{S}_{1})$ is correct

Mathematical Reasoning * — JEE Main · 2020 · 04 Sep 2020 · Shift 1 Question with Solution

Given the following two statements:

(S_{1}) : (q \lor p) \to (p \leftrightarrow\sim q)

is a tautology.

(S_{2}) :\sim q \land (\sim p \leftrightarrow q)

is a fallacy. Then:

Answer: (C)

Test

S_{1}

: taking

p = T, q = T

gives

(q \lor p) = T

while

(p \leftrightarrow\sim q) = (T \leftrightarrow F) = F

, so

S_{1} = T \to F = F

; a statement that is sometimes false is not a tautology, so the claim about

S_{1}

is wrong. Test

S_{2}

: taking

p = T, q = F

gives

\sim q = T

and

(\sim p \leftrightarrow q) = (F \leftrightarrow F) = T

, so

S_{2} = T \land T = T

; a statement that is sometimes true is not a fallacy, so the claim about

S_{2}

is also wrong. Both statements are incorrect.

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option