JEE Advanced Maths: Mathematical Reasoning questions with solutions

Q1. Given that p is a proposition, t represents a tautology, and c denotes a contradiction, which of the following statements is accurate?

1. p OR (NOT p) equals c
2. p OR t results in t
3. p AND t is equivalent to p
4. p AND c is equal to c

Answer: p OR t results in t

The statement p OR t results in t because when combined with a tautology, the result is always true, regardless of the value of p.

Q2. Which of the following pairs of statements correctly represent a logical dual?

1. (p OR q) AND (r OR s), (p AND q) OR (r AND s)
2. [p OR (NOT q) AND (NOT p)] AND (NOT q) OR (NOT p)
3. (p AND q) OR r, (p OR q) AND r
4. (p OR q) OR s, AND (p AND q) OR s

Answer: (p AND q) OR r, (p OR q) AND r

The pair of statements (p AND q) OR r and (p OR q) AND r are logical duals because they have the same structure but with the operators AND and OR interchanged.

Q3. Which of the following statements is accurate?

1. p → q is equivalent to ∼p ∨ q in logic.
2. If p, q, and r have truth values T, F, and T respectively, then (p ∨ q) ∧ (q ∨ r) evaluates to T.
3. The expression ∼(p ∨ q ∨ r) is equal to ∼p ∧ ∼q ∧ ∼r.
4. The logical value of p ∧ ∼(p ∨ q) is always T.

Answer: p → q is equivalent to ∼p ∨ q in logic.

The statement p → q is equivalent to ∼p ∨ q in logic because the implication p → q is true if p is false or q is true, which is the same as ∼p ∨ q.

Q4. Which of the following correctly represents the logical equivalence of [(p ∨ q) ∧ (¬q)] ∨ (¬p)?

1. The dual of [(p ∨ q) ∧ (¬q)] ∨ (¬p) is [(p ∧ q) ∨ (¬q)] ∧ (¬p).
2. The logical equivalence of [(p ∨ q) ∧ (¬q)] ∨ (¬p) is [(p ∧ ¬q) ∨ (q ∧ ¬q)] ∨ (¬p) or [(p ∧ ¬q) ∨ (¬p)].
3. The negation of [(p ∨ q) ∧ (¬q)] ∨ (¬p) is ¬[(p ∨ q) ∧ (¬q)] ∧ (¬p) or [(¬p ∧ ¬q) ∨ q] ∧ (¬p).
4. The contrapositive of [(p ∨ q) ∧ (¬q)] ∨ (¬p) is ¬[(p ∨ q) ∧ (¬q)] or ¬[(¬p ∧ ¬q) ∨ q].

Answer: The logical equivalence of [(p ∨ q) ∧ (¬q)] ∨ (¬p) is [(p ∧ ¬q) ∨ (q ∧ ¬q)] ∨ (¬p) or [(p ∧ ¬q) ∨ (¬p)].

The logical equivalence simplifies to [(p ∧ ¬q) ∨ (¬p)] because the distributive property and negation rules are applied to the original expression. This matches the given equivalence.

Q5. A function F satisfies F(0) = 2, F(1) = 3, and the recurrence relation F(x + 2) = 2F(x) - F(x + 1) for all integers x >= 0. Find F(5).

1. -7
2. -3
3. 17
4. 13

Answer: 13

Using the recurrence iteratively: F(2) = 2F(0) - F(1) = 1. F(3) = 2F(1) - F(2) = 5. F(4) = 2F(2) - F(3) = -3. F(5) = 2F(3) - F(4) = 10 - (-3) = 13.

Q6. Find the coefficient of x⁴ in the power series expansion of (e^x - 1)².

1. 3/8
2. 5/12
3. 7/12
4. 5/24

Answer: 7/12

Expanding (e^x - 1)² = e^(2x) - 2e^x + 1. Coefficients of x⁴: from e^(2x) -> (2x)⁴/4! = 16/24 = 2/3; from -2e^x -> -2/4! = -2/24 = -1/12; from 1 -> 0. Total = 2/3 - 1/12 = 8/12 - 1/12 = 7/12.

Q7. Which one of the following statements about the propositions p and q is correct? (1) p AND q is true when at least one of p, q is true. (2) p -> q is true when p is true and q is false. (3) p <-> q is true only when both p and q are true. (4) NOT(p OR q) is true only when both p and q are false.

1. (1)
2. (2)
3. (3)
4. (4)

Answer: (4)

Statement (4) is the correct truth-table fact: NOT(p OR q) is true only when both p and q are false. The others misstate AND, implication, and biconditional.

Q8. The compound statement ~(p OR q) OR (~p AND q) is logically equivalent to:

1. p
2. ~p
3. q
4. ~q

Answer: ~p

Using De Morgan, ~(p OR q) = ~p AND ~q; combining with ~p AND q and factoring ~p gives ~p AND (~q OR q) = ~p.

Q9. Which of the following compound statements is logically equivalent to (p AND q)?

1. p -> ~q
2. ~p OR ~q
3. ~(p -> ~q)
4. ~(~p AND ~q)

Answer: ~(p -> ~q)

Since p -> ~q is equivalent to ~p OR ~q, its negation ~(p -> ~q) becomes p AND q by De Morgan's law.

Q10. If the implication p -> (q OR r) is false, then the truth values of p, q, r respectively are:

1. T, F, F
2. F, F, F
3. F, T, T
4. T, T, F

Answer: T, F, F

p -> (q OR r) is false requires p true and (q OR r) false, so p=T, q=F, r=F.

Q11. Classify the compound statement (p AND (NOT q)) AND ((NOT p) OR q):

1. a tautology
2. a contradiction
3. neither a tautology nor a contradiction
4. None of these

Answer: a contradiction

p AND NOT q requires p true and q false, but (NOT p OR q) then requires p false or q true - a direct conflict - so the statement is always false: a contradiction.

Q12. What is the negation of the statement p → (q ∧ r)?

1. ~p → ~(q ∧ r)
2. ~p ∨ (q ∧ r)
3. (q ∧ r) → p
4. p ∧ (~q ∨ ~r)

Answer: p ∧ (~q ∨ ~r)

Negating an implication: ~(p -> (q ∧ r)) = p ∧ ~(q ∧ r) = p ∧ (~q ∨ ~r).

Q13. Consider the statement: 'If x = 5 and y = -2, then x - 2y = 9.' Which of the following is its contrapositive?

1. If x - 2y != 9, then x != 5 or y != -2
2. If x - 2y != 9, then x != 5 and y != -2
3. If x - 2y = 9, then x = 5 and y = -2
4. None of these

Answer: If x - 2y != 9, then x != 5 or y != -2

The contrapositive negates both parts and reverses them; negating the conjunction gives an 'or', so it reads: if x-2y != 9 then x != 5 or y != -2.

Q14. Which one of the following statements is FALSE?

1. p ^ ~(p v q) is a tautology (always true)
2. p -> q is logically equivalent to ~p v q
3. (p v q) ^ (q v r) is true when p, q, r are T, F, T respectively
4. ~(p ^ (q v r)) is equivalent to (~p v ~q) ^ (~p v ~r)

Answer: p ^ ~(p v q) is a tautology (always true)

p ^ ~(p v q) = p ^ ~p ^ ~q, which contains p ^ ~p (a contradiction), so it is always false, not always true; hence that statement is wrong.

Q15. Let S*(p, q, r) be the dual of the compound statement S(p, q, r), where S(p, q, r) = (~p) AND (~(q OR r)). Then S*(~p, ~q, ~r) is logically equivalent to:

1. S(p, q, r)
2. ~S(~p, ~q, ~r)
3. ~S(p, q, r)
4. S*(p, q, r)

Answer: ~S(p, q, r)

Taking the dual and substituting negated variables and simplifying with De Morgan's laws gives the negation of the original statement, ~S(p, q, r).

Q16. What is the negation of the statement: "If a quadrilateral is a square, then it is a rhombus"?

1. If a quadrilateral is not a square then it is a rhombus
2. If a quadrilateral is a square then it is not a rhombus
3. A quadrilateral is a square and it is not a rhombus
4. A quadrilateral is not a square and it is a rhombus

Answer: A quadrilateral is a square and it is not a rhombus

Since ~(p -> q) is equivalent to p AND ~q, the negation is 'a quadrilateral is a square and it is not a rhombus'.