JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023(S1) (p⟹q)∨(p∧(∼q)) is a tautology
(S2) ((∼p)⟹(∼q))∧((∼p)∨q) is a contradiction.
Then
Choose the correct answer:
- A.
both (S1) and (S2) are correct
- B.
only (S1) is correct
(Correct Answer) - C.
only (S2) is correct
- D.
both (S1) and (S2) are wrong
only (S1) is correct
Explanation
The Solution
-
(S1): Note that p⟹q is equivalent to ∼p∨q.
The expression becomes (∼p∨q)∨(p∧∼q).
Since (p∧∼q) is the negation of (∼p∨q), this is of the form X∨∼X, which is always True. (S1 is correct).
-
(S2): (∼p⟹∼q) is equivalent to p∨∼q.
The expression is (p∨∼q)∧(∼p∨q).
If both p and q are True, the expression is (T∨F)∧(F∨T)=T∧T=True. Since it can be true, it is not a contradiction. (S2 is wrong).
Correct Answer: Only (S1) is correct (Option 2)
Explanation
The Solution
-
(S1): Note that p⟹q is equivalent to ∼p∨q.
The expression becomes (∼p∨q)∨(p∧∼q).
Since (p∧∼q) is the negation of (∼p∨q), this is of the form X∨∼X, which is always True. (S1 is correct).
-
(S2): (∼p⟹∼q) is equivalent to p∨∼q.
The expression is (p∨∼q)∧(∼p∨q).
If both p and q are True, the expression is (T∨F)∧(F∨T)=T∧T=True. Since it can be true, it is not a contradiction. (S2 is wrong).
Correct Answer: Only (S1) is correct (Option 2)