Exams › JEE Main › Maths › Mathematical Reasoning
118 questions with worked solutions.
Answer: If T(k) is true, then T(k + 1) is also true
The true sum is 1+3+...+(2k-1) = k^2, so T(k): k^2 = k^2 + 10 is never true (T(1): 1 = 11 is false). However, assuming T(k), adding (2k+1) gives sum = k^2+10+(2k+1) = (k+1)^2+10 = T(k+1), so the implication 'T(k) => T(k+1)' is correct even though no T(n) actually holds.
Answer: S(k) implies S(k + 1).
If S(k): 1+3+...+(2k-1)=k^2+3 is assumed, adding (2k+1) gives k^2+3+2k+1=(k+1)^2+3=S(k+1), so S(k) implies S(k+1). However S(1) gives 1 = 4 which is false, so the inductive step holds but the statement is never actually true.
Q3. For every natural number n, which of the following inequalities is true?
Answer: n⁴ < 10ⁿ
2^n<n fails at n=1; n^2>2n fails at n=1,2; 8^n>7n+1 fails at n=1 (8=8). Only n^4<10^n holds for every natural n (1<10, 16<100, 81<1000, and 10^n grows far faster thereafter).
Q4. Let p, q, and r be arbitrary logical propositions. Which of the following equivalences is correct?
Answer: ~[p ∨ (~q)] ≡ (~p) ∧ q
The correct option is valid because it applies De Morgan's laws, which state that the negation of a disjunction is equivalent to the conjunction of the negations. Thus, negating the expression 'p or not q' results in 'not p and q', confirming the equivalence.
Q5. The proposition ~(p → q) → [(~p) ∨ (~q)] is best classified as
Answer: a tautology
The proposition is a tautology because it is true for all possible truth values of p and q, meaning that the implication holds regardless of the truth values assigned to the variables.
Answer: Q ∧ R → P
If m is prime and m divides n^2, then a prime dividing a square must divide the base, so m divides n. Thus Q ^ R -> P is the valid implication; the others fail (e.g. m=4,n=2 breaks Q->P).
Answer: False
(p->r) => (q v r) is false only if (p->r) is true and (q v r) is false (q,r both false). With r false, p->r is true only when p is false. Hence p is False.
Answer: p implies q is true
If x,y are both odd then xy is odd, so p => q is true. But ~q (xy even) does not imply p (both odd) - e.g. x=2,y=3 gives xy even yet p false. Hence only statement (a) holds.
Q9. What is the dual of the statement (p ∧ q) ∨ ∼ q = p ∨ ∼ q?
Answer: (p ∨ q) ∧ ∼ q = p ∧ ∼ q
The dual is obtained by interchanging AND and OR (and T,F). (p AND q) OR ~q = p OR ~q becomes (p OR q) AND ~q = p AND ~q.
Q10. Which statement is the converse of: if x < y, then x² < y²?
Answer: If x² < y², then x < y
The converse of 'if P then Q' is 'if Q then P'. Here P: x<y, Q: x^2<y^2, so the converse is 'if x^2 < y^2, then x < y'.
Answer: false
With p=T,q=T,r=F,s=F: p v ~r = T v T = T, so (p v ~r) v (~q v s) = T. Negation gives F. The expression is false.
Q12. What is the contrapositive of the statement p → ¬(q → ¬r)?
Answer: None of these
The contrapositive of a statement has the form ~q → ~p, where q is the conclusion and p is the hypothesis. In this case, the original statement's structure does not yield any of the provided options as a valid contrapositive, confirming that 'None of these' is the correct choice.
Q13. The statement ~(p ⇒ q) ⇔ (~p ∨ ~q) is:
Answer: Neither a tautology nor a contradiction
~(p=>q) is equivalent to p AND ~q. Compare with (~p OR ~q): for p=T,q=F both sides are true; for p=F,q=F left is false while right is true. Since the biconditional is true for some rows and false for others, the statement is neither a tautology nor a contradiction.
Q14. Identify the statement that is incorrect.
Answer: The proposition p ∧ ~(p ∨ q) is always T.
Take p=T: ~(p∨q)=F, so p∧F=F; with p=F the whole thing is F too. Hence p∧~(p∨q) is always FALSE, so the claim that it is always T is the incorrect statement. (A is De Morgan-valid, B evaluates to T, C is De Morgan-valid.)
Answer: TFFT
(~p->~q)/\(~q->~p) = (~p<->~q) = (p<->q). For (p,q) = TT, TF, FT, FF the biconditional gives T, F, F, T, so the final column is TFFT.
Answer: p OR (NOT p) = c
p OR (NOT p) is always true, so it equals t (a tautology), not c. Hence 'p OR (NOT p) = c' is the statement that is NOT true. The other three identities (p OR t = t, p AND t = p, p AND c = c) are all valid.
Q17. Which of the following statements is logically equivalent to p ⇔ q?
Answer: (p ⇒ q) ∧ (q ⇒ p)
The statement p ⇔ q means that p is true if and only if q is true, which is precisely captured by the conjunction of the implications (p ⇒ q) and (q ⇒ p). This indicates that both conditions must hold for the biconditional to be true.
Q18. What is the inverse of the proposition (p ∧ ∼ q) → r?
Answer: (∼ p ∨ q) → ∼ r
The inverse of p->q is (~p)->(~q). Here antecedent is (p ^ ~q) and consequent is r, so the inverse is ~(p ^ ~q) -> ~r = (~p v q) -> ~r.
Q19. Consider the statement: if x = 5 and y = −2, then x − 2y = 9. Its contrapositive is:
Answer: None of these
Statement: if (x=5 AND y=-2) then x-2y=9. Contrapositive of 'if P then Q' is 'if ~Q then ~P', where ~P = (x!=5 OR y!=-2). So it reads 'If x-2y!=9, then x!=5 OR y!=-2'. The listed option uses AND instead of OR, so none of the options match -> None of these.
Answer: If I pursue engineering, then I obtain good marks
The statement is 'if not good marks (~G) then cannot pursue engineering (~E)', i.e. ~G => ~E. Its contrapositive is the negation of consequent implying negation of antecedent: E => G, that is 'If I pursue engineering, then I obtain good marks.'
Q21. Which of the following is logically equivalent to the proposition p → (q → p)?
Answer: p → (p ∨ q)
The proposition p → (q → p) asserts that if p is true, then regardless of the truth value of q, p remains true. This is logically equivalent to p → (p ∨ q), which states that if p is true, then at least one of p or q must be true, thus maintaining the truth of p.
Q22. The statement (p ∧ q) ⇒ p is
Answer: always true, i.e., a tautology
The statement (p nd q) implies p is always true because if both p and q are true, then p must be true as well, making the implication valid in all scenarios.
Q23. Given that p and q are true statements while r is a false statement, which of the following is true?
Answer: (p∨q)∧(p∨r) is true
The expression (p∨q)∧(p∨r) is true because both p and q are true, making p∨q true, and since p is true, p∨r is also true, resulting in the conjunction being true.
Q24. Which of the following pairs is NOT a correct dual pair?
Answer: (p∨q)∨s, (p∧q)∨s
The dual is obtained by interchanging ^ and v. The dual of (p v q) v s is (p ^ q) ^ s, but option D lists (p ^ q) v s, so this pair is NOT a correct dual pair.
Answer: A glass is half empty if and only if it is half full
With p = 'glass is half empty' and q = 'glass is half full', 'p if and only if q' translates literally to 'A glass is half empty if and only if it is half full.'
Answer: ~(Q↔(P∧~R))
Let P=intelligent, Q=wealthy, R=truthful so dishonest=~R. The statement 'intelligent and dishonest iff wealthy' is (P AND ~R) <=> Q. Its negation is ~(Q <=> (P AND ~R)), which is option 0.
Answer: Statement-1 is true, Statement-2 is true; Statement-2 correctly explains Statement-1.
Both logical expressions yield the same truth values under all possible interpretations of p and q, confirming their logical equivalence. Since the truth tables for both expressions match exactly, Statement-2 accurately supports Statement-1.
Q28. Which of the following statements is always true for every truth assignment?
Answer: [A ∧ (A → B)] → B
[A and (A->B)] -> B is the modus-ponens schema and is true under every assignment. A or (A and B) reduces to A and A and (A or B) reduces to A, neither a tautology; B -> [A and (A->B)] fails when B is true and A is false.
Q29. Let S(K)=1+3+5+...+(2K-1)=3+K². Then which of the following is true
Answer: S(K)⇒S(K+1)
The statement S(K)⇒S(K+1) indicates that if the formula holds for K, it also holds for K+1, which is a key aspect of mathematical induction. This relationship is essential for proving the validity of the formula for all natural numbers.
Q30. The statement p → (q → p) is equivalent to
Answer: p → (p ∨ q)
The statement p → (q → p) asserts that if p is true, then regardless of the truth value of q, p remains true. This is logically equivalent to saying that if p is true, then either p is true or q is true, which is represented by p → (p ∨ q).
Q31. Statement-1: ~(p ↔ ~q) is equivalent to p ↔ q. Statement-2: ~(p ↔ ~q) is a tautology.
Answer: Statement-1 is true, Statement-2 is false
Statement-1 is true because the negation of a biconditional can be shown to be equivalent to the original biconditional when both sides are negated. However, Statement-2 is false because ~(p ↔ ~q) is not always true for all truth values of p and q, thus it is not a tautology.
Answer: Every rational number x ∈ S satisfies x ≤ 0.
The correct negation of statement P asserts that all rational numbers in the subset S must be less than or equal to zero, which directly contradicts the original claim that there exists at least one positive rational number in S.
Answer: ~(Q ↔ (P ∧ ~R))
The correct option negates the biconditional statement by applying De Morgan's laws, which states that the negation of 'A if and only if B' is equivalent to 'not A or not B'. Thus, negating 'Suman is rich if and only if Suman is brilliant and dishonest' correctly leads to the expression in option A.
Q34. The negation of the statement "If I become a teacher, then I will open a school", is:
Answer: I will become a teacher and I will not open a school.
The statement is p -> q ('teacher' -> 'open school'). Its negation is p AND (~q): 'I will become a teacher and I will not open a school.'
Answer: Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Statement-1 is true because the expression (p ∧ ~q) ∧ (~p ∧ q) is a contradiction, which is indeed a fallacy. Statement-2 is true as it represents the contrapositive equivalence in logic, which is always valid, but it does not explain why the first statement is a fallacy.
Q36. The statement ~(p ↔ ~q) is:
Answer: equivalent to p ↔ q
The expression ~(p ↔ ~q) negates the biconditional relationship between p and the negation of q, which simplifies to p being equivalent to q, thus confirming that it is equivalent to p ↔ q.
Q37. The negation of ~s∨(~r∧s) is equivalent to:
Answer: s∧r
Negation of ~s v (~r ^ s) = s ^ ~(~r ^ s) = s ^ (r v ~s) = (s^r) v (s^~s) = s ^ r.
Q38. The following statement (p→q)→[(~p→q)→q] is:
Answer: a tautology
The statement is a tautology because it holds true for all possible truth values of p and q, meaning that regardless of whether p is true or false, the overall expression evaluates to true.
Q39. For any two statements p and q, the negation of the expression p∨(~p∧q) is:
Answer: ~p∧~q
p v (~p ^ q) = (p v ~p) ^ (p v q) = p v q. Its negation is ~(p v q) = ~p ^ ~q.
Answer: (∧, ∨)
Testing all four values, (p AND q) AND (~p OR q) reduces to p AND q for all p,q. So the ordered pair is (AND, OR), i.e. option (∧, ∨).
Q41. The logical statement (p⇒q) ∧ (q⇒~p) is equivalent to:
Answer: ~p
(p=>q) AND (q=>~p) = (~p OR q) AND (~q OR ~p). A truth table gives True only when p is False (both p=F,q=T and p=F,q=F), and False when p is True. This is exactly ~p.
Q42. If p → (~p ∨ ~q) is false, then the truth values of p and q are respectively -
Answer: T, T
The expression p → (~p ∨ ~q) is only false when p is true and the consequent (~p ∨ ~q) is false. For (~p ∨ ~q) to be false, both ~p and ~q must be false, which means p must be true and q must also be true.
Answer: If the squares of two numbers are equal, then the numbers are equal
The contrapositive of a statement reverses and negates both the hypothesis and conclusion. Since the original statement asserts that unequal numbers lead to unequal squares, its contrapositive correctly states that equal squares imply equal numbers.
Q44. Which one of the following statements is not a tautology ?
Answer: ((p ∨ q) → (p ∨ (¬q)))
The statement ((p ∨ q) → (p ∨ (¬q))) is not a tautology because there are truth assignments for p and q where the implication does not hold true, specifically when p is false and q is true.
Q45. The negation of the Boolean expression ~s∨(~r∧s) is equivalent to:
Answer: s∧r
The negation of the expression ~s∨(~r∧s) can be simplified using De Morgan's laws, which state that the negation of a disjunction is the conjunction of the negations. Thus, negating the original expression results in s∧r, indicating that both s and r must be true.
Answer: T, T, F
The implication p → (~q ∨ r) is false only when p is true and the consequent (~q ∨ r) is false. For (~q ∨ r) to be false, both ~q and r must be false, which means q is true and r is false, making p true, q true, and r false.
Q47. If p ⇒ (q ∨ r) is false, then the truth values of p, q, r are respectively:
Answer: T, F, F
The implication p ⇒ (q ∨ r) is false only when p is true and (q ∨ r) is false. For (q ∨ r) to be false, both q and r must be false, which leads to the truth values of p being true and both q and r being false.
Answer: If you are not a citizen of India, then you are not born in India
The contrapositive of a conditional statement reverses and negates both the hypothesis and the conclusion. In this case, negating 'you are a citizen of India' leads to 'you are not a citizen of India,' and negating 'you are born in India' leads to 'you are not born in India,' resulting in the correct contrapositive.
Q49. The expression ~(p → q) is logically equivalent to:
Answer: p ∧ ~q
p->q is equivalent to ~p OR q. Negating gives ~(~p OR q) = p AND ~q. So the answer is p AND ~q.
Answer: If I will not catch the train, then I do not reach the station in time.
P = reach station in time, Q = catch train. The contrapositive is 'If I will not catch the train, then I do not reach the station in time.'