Practice MCQs for JEE Mathematics Mathematical Reasoning
Review structured MCQ sets for JEE Mathematics Mathematical Reasoning. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.
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Multiple Choice Questions
(a) \( \sim S(p,q,r) \)
(b) \( S(p,q,r) \)
(c) \( p\vee(q\wedge r) \)
(d) \( p\vee(q\vee r) \)
Show Answer & Explanation
Answer: (d) \( p\vee(q\vee r) \)
Explanation:
1. Replace p, q, r by their negations: S = p ∨ ~(~q ∧ ~r).
2. By De Morgan, ~(~q ∧ ~r) = q ∨ r.
3. So S(~p,~q,~r) = p ∨ (q ∨ r).
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) If a number is not a prime then it is odd
(b) If a number is not a prime then it is not odd
(c) If a number is not odd then it is not a prime
(d) If a number is not odd then it is a prime
Show Answer & Explanation
Answer: (b) If a number is not a prime then it is not odd
Explanation:
1. For p implies q, the inverse is ~p implies ~q.
2. Here p is 'prime' and q is 'odd', so the inverse is: if a number is not prime then it is not odd.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) Open the door.
(b) Do your home work.
(c) Switch on the fan.
(d) Two plus two is four.
Show Answer & Explanation
Answer: (d) Two plus two is four.
Explanation:
1. A statement (proposition) is a sentence that is either true or false, but not both.
2. The first three are commands, which have no truth value.
3. 'Two plus two is four' is true, so it is a statement.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) T
(b) F
(c) both T and F
(d) 54
Show Answer & Explanation
Answer: (a) T
Explanation:
1. 3 + 3 = 33 is false and 1 + 2 = 12 is false, so the 'or' statement is false.
2. The negation of a false statement is true, so the given statement is true.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( (\sim p\vee\sim q)\vee(p\vee\sim q) \)
(b) \( (p\Rightarrow q)\vee(p\wedge\sim q) \)
(c) \( (\sim p\wedge q)\wedge(\sim q) \)
(d) \( (\sim p\wedge q)\vee(\sim q) \)
Show Answer & Explanation
Answer: (c) \( (\sim p\wedge q)\wedge(\sim q) \)
Explanation:
1. A contradiction is false for every truth assignment.
2. In (c), q and ~q cannot both be true, so the statement is always false.
3. The other options are true for some assignments (for instance (b) is a tautology).
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) Please do me a favour
(b) 2 is an even integer
(c) 2 + 1 = 3
(d) The number 17 is prime
Show Answer & Explanation
Answer: (a) Please do me a favour
Explanation:
1. A request has no truth value, so it is not a statement.
2. The other three are true statements.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) Paris is in England and London is in France
(b) Paris is not in France or London is not in England
(c) Paris is in England or London is in France
(d) none of these
Show Answer & Explanation
Answer: (b) Paris is not in France or London is not in England
Explanation:
1. By De Morgan, ~(p ∧ q) = ~p ∨ ~q.
2. So the negation is: Paris is not in France or London is not in England.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( (\sim q\wedge r)\to\sim p \)
(b) \( (q\to r)\to\sim p \)
(c) \( (q\vee\sim r)\to\sim p \)
(d) none of these
Show Answer & Explanation
Answer: (a) \( (\sim q\wedge r)\to\sim p \)
Explanation:
1. The contrapositive of A implies B is ~B implies ~A.
2. Here B = (~q implies ~r), and ~B = ~q ∧ r.
3. So the contrapositive is (~q ∧ r) implies ~p.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( p\wedge(\sim q) \)
(b) \( p\wedge q \)
(c) \( (\sim p)\wedge q \)
(d) \( (\sim p)\wedge(\sim q) \)
Show Answer & Explanation
Answer: (a) \( p\wedge(\sim q) \)
Explanation:
1. 'It is snowing' is p.
2. 'It is not that I am cold' is ~q.
3. Joined by 'and': p ∧ ~q.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) New Delhi is not a city
(b) It is false that New Delhi is a city
(c) It is not the case that New Delhi is a city
(d) none of these
Show Answer & Explanation
Answer: (d) none of these
Explanation:
1. All three given sentences say the same thing: that New Delhi is not a city.
2. So each is a negation of p, which means 'none of these' is the correct choice for 'not a negation'.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( \sqrt3 \) is a prime
(b) \( \sqrt2 \) is irrational
(c) Mathematics is interesting
(d) 5 is an even integer
Show Answer & Explanation
Answer: (c) Mathematics is interesting
Explanation:
1. A proposition must be definitely true or false.
2. 'Mathematics is interesting' is a matter of opinion, so it has no definite truth value.
3. The others are definitely true or false.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( (\sim p\vee\sim q)\equiv(p\wedge q) \)
(b) \( (p\to q)\equiv(\sim q\to\sim p) \)
(c) \( \sim(p\to\sim q)\equiv(p\wedge\sim q) \)
(d) \( \sim(p\vee q)\equiv(p\to q)\wedge(q\to p) \)
Show Answer & Explanation
Answer: (b) \( (p\to q)\equiv(\sim q\to\sim p) \)
Explanation:
1. (a) is false: by De Morgan ~p ∨ ~q = ~(p ∧ q).
2. (b) is the contrapositive law, which always holds.
3. (c) is false since ~(p implies ~q) = p ∧ q. (d) is false since the right side is p iff q.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( p\to(p\to q) \)
(b) \( p\to(p\vee q) \)
(c) \( p\to(p\wedge q) \)
(d) \( p\to(p\leftrightarrow q) \)
Show Answer & Explanation
Answer: (b) \( p\to(p\vee q) \)
Explanation:
1. p → (q → p) = ~p ∨ (~q ∨ p), which is always true, a tautology.
2. p → (p ∨ q) = ~p ∨ p ∨ q is also always true.
3. The other options are false when p is true and q is false.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( \sim(p\vee\sim q)\to\sim r \)
(b) \( (\sim p\wedge q)\to\sim r \)
(c) \( (\sim p\vee q)\to\sim r \)
(d) none of these
Show Answer & Explanation
Answer: (c) \( (\sim p\vee q)\to\sim r \)
Explanation:
1. The inverse of A implies B is ~A implies ~B.
2. ~(p ∧ ~q) = ~p ∨ q by De Morgan.
3. So the inverse is (~p ∨ q) implies ~r.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) F, F
(b) T, T
(c) T, F
(d) F, T
Show Answer & Explanation
Answer: (c) T, F
Explanation:
1. An implication is false only when the antecedent is true and the consequent false.
2. So p is T and (~p ∨ q) is F, which needs ~p = F (true since p = T) and q = F.
3. Thus p = T and q = F.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) Raju is not intelligent.
(b) Raju is not tall or he is intelligent
(c) Raju is not tall and he is intelligent
(d) Raju is not tall implies he is intelligent
Show Answer & Explanation
Answer: (b) Raju is not tall or he is intelligent
Explanation:
1. ~p is 'Raju is not tall'.
2. The symbol ∨ means 'or'.
3. So ~p ∨ q reads: Raju is not tall or he is intelligent.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) It rains and I shall go to school
(b) It rains and I shall not go to school
(c) It does not rain and I shall go to school
(d) none of these
Show Answer & Explanation
Answer: (b) It rains and I shall not go to school
Explanation:
1. The negation of p implies q is p ∧ ~q.
2. So: it rains and I shall not go to school.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) T, F, F
(b) F, F, F
(c) F, T, T
(d) T, T, F
Show Answer & Explanation
Answer: (a) T, F, F
Explanation:
1. An implication is false only when the antecedent is true and the consequent is false.
2. So p = T and q ∨ r = F, which means q = F and r = F.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( \sim P\vee Q \)
(b) \( \sim P\wedge\sim Q \)
(c) \( \sim P\wedge Q \)
(d) \( \sim(P\vee Q) \)
Show Answer & Explanation
Answer: (d) \( \sim(P\vee Q) \)
Explanation:
1. 'He is either intelligent or strong' is P ∨ Q.
2. 'It is not true that' negates the whole statement: ~(P ∨ Q).
3. Option (b) is equivalent by De Morgan but the direct translation of the sentence is ~(P ∨ Q).
Teacher's Note:
1. Translate the sentence literally first, then simplify if needed.
2. Several options can be logically equivalent; choose the direct symbolic form.
(a) If Ashok works hard then he gets good grade
(b) If Ashok does not work hard then he gets good grade
(c) If Ashok does not work hard then he does not get good grade
(d) Ashok works hard if and only if he gets good grade
Show Answer & Explanation
Answer: (b) If Ashok does not work hard then he gets good grade
Explanation:
1. ~p means 'Ashok does not work hard' and q means 'Ashok gets good grade'.
2. The arrow means 'if ... then', so the statement reads: If Ashok does not work hard then he gets good grade.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) F, T
(b) F, F
(c) T, T
(d) T, F
Show Answer & Explanation
Answer: (d) T, F
Explanation:
1. The implication is false only when p = T and (~p ∨ q) = F.
2. With p = T, ~p is F, so ~p ∨ q = F requires q = F.
3. So p = T and q = F.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( (p\wedge q)\vee(p\vee\sim r) \)
(b) \( (p\wedge q)\vee(p\wedge\sim r) \)
(c) \( (p\wedge q)\wedge(p\wedge\sim r) \)
(d) \( p\vee q \)
Show Answer & Explanation
Answer: (c) \( (p\wedge q)\wedge(p\wedge\sim r) \)
Explanation:
1. The negation of A implies B is A ∧ ~B.
2. ~B = ~(~p ∨ r) = p ∧ ~r.
3. So the negation is (p ∧ q) ∧ (p ∧ ~r).
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( \sim[p\wedge(\sim q)]\equiv(\sim p)\wedge q \)
(b) \( \sim[(p\vee q)\wedge(\sim r)]\equiv(\sim p)\vee(\sim q)\vee(\sim r) \)
(c) \( \sim[p\vee(\sim q)]\equiv(\sim p)\wedge q \)
(d) \( \sim[p\vee(\sim q)]\equiv(\sim p)\wedge\sim q \)
Show Answer & Explanation
Answer: (c) \( \sim[p\vee(\sim q)]\equiv(\sim p)\wedge q \)
Explanation:
1. By De Morgan, ~[p ∨ ~q] = ~p ∧ ~(~q) = ~p ∧ q, so (c) is true and (d) is false.
2. (a) is false since ~[p ∧ ~q] = ~p ∨ q.
3. (b) is false since the left side is (~p ∧ ~q) ∨ r.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) a tautology
(b) a contradiction
(c) both a tautology and a contradiction
(d) neither a tautology nor a contradiction
Show Answer & Explanation
Answer: (b) a contradiction
Explanation:
1. p ∧ ~p is always false, so the whole conjunction is always false.
2. A statement that is always false is a contradiction.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( p\wedge q \) is true
(b) \( p\vee\sim q \) is true
(c) \( q\wedge p \) is true
(d) \( p\Rightarrow q \) is true
Show Answer & Explanation
Answer: (d) \( p\Rightarrow q \) is true
Explanation:
1. p ∧ q = F ∧ T = F, and q ∧ p = F.
2. p ∨ ~q = F ∨ F = F.
3. p ⇒ q with a false antecedent is true.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) tautology and contradiction
(b) neither tautology nor contradiction
(c) contradiction
(d) tautology
Show Answer & Explanation
Answer: (c) contradiction
Explanation:
1. p ⇒ ~p is equivalent to ~p, and ~p ⇒ p is equivalent to p.
2. The conjunction is p ∧ ~p, which is always false.
3. So it is a contradiction.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) an ellipse is a circle
(b) an ellipse is not a circle
(c) a circle is not an ellipse
(d) a circle is an ellipse
Show Answer & Explanation
Answer: (c) a circle is not an ellipse
Explanation:
1. The negation of a statement says the opposite.
2. So: a circle is not an ellipse.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( p\vee\sim p \) is a tautology
(b) \( \sim(\sim p)\leftrightarrow p \) is a tautology
(c) \( p\wedge\sim p \) is a contradiction
(d) \( ((p\wedge q)\to q)\to p \) is a tautology
Show Answer & Explanation
Answer: (d) \( ((p\wedge q)\to q)\to p \) is a tautology
Explanation:
1. (p ∧ q) → q is always true, so the statement in (d) reduces to T → p, which equals p.
2. p is not true for every assignment, so (d) is not a tautology.
3. (a), (b), (c) are standard true statements.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) TTFF
(b) FFFF
(c) TTTT
(d) FTFT
Show Answer & Explanation
Answer: (a) TTFF
Explanation:
1. TT: (T) iff (F ∨ T = T) gives T.
2. TF: (F) iff (F ∨ F = F) gives T.
3. FT: (F) iff (T ∨ T = T) gives F. FF: (F) iff (T) gives F.
4. The column is T, T, F, F.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( (p\wedge q)\wedge\sim(p\vee q) \)
(b) \( p\vee(\sim p\wedge q) \)
(c) \( (p\Rightarrow q)\Rightarrow p \)
(d) none of these
Show Answer & Explanation
Answer: (a) \( (p\wedge q)\wedge\sim(p\vee q) \)
Explanation:
1. If p ∧ q is true then p ∨ q is true, so ~(p ∨ q) is false.
2. If p ∧ q is false the conjunction is already false.
3. So (a) is false always, a contradiction.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) a tumbler is half empty and half full
(b) a tumbler is half empty if and only if it is half full
(c) both (a) and (b)
(d) none of the above
Show Answer & Explanation
Answer: (b) a tumbler is half empty if and only if it is half full
Explanation:
1. 'p if and only if q' is the biconditional p ⟺ q.
2. Reading it in words: a tumbler is half empty if and only if it is half full.
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
(a) \( r\Rightarrow(p\vee q) \)
(b) \( \sim r\Rightarrow(p\vee q) \)
(c) \( \sim r\Rightarrow\sim p\wedge\sim q \)
(d) \( p\Rightarrow(q\vee r) \)
Show Answer & Explanation
Answer: (c) \( \sim r\Rightarrow\sim p\wedge\sim q \)
Explanation:
1. The contrapositive of A implies B is ~B implies ~A.
2. ~(p ∨ q) = ~p ∧ ~q by De Morgan.
3. So: ~r implies (~p ∧ ~q).
Teacher's Note:
1. Build the truth table or use the standard laws (De Morgan, contrapositive) to compare statements.
2. Remember: p implies q is false only when p is true and q is false.
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