2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

Students can Download Basic Maths Question Bank Chapter 6 Mathematical Logic Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

2nd PUC Basic Maths Mathematical Logic One Mark Questions and Answers

Question 1.
Write the verbal form of the following compound proposition given:
Answer:
p: x is a real number
q: Mathematics is easy
r: Today is a holiday

  • (p ∨ q) = x is a real number or mathematics is easy
  • (~p ∧ r) = x is not a real number and maths is easy
  • (q ∧ ~ p) = Maths is easy and today is not a holiday
  • (p ↔ q) ∨ = r x is a real number if and only if mathematics is easy or today
    is a holiday
  • ((p ∨ q) → q) =  If x is a real number or today is a holiday then maths is easy
  • (p ∨ r) ∧ ~ q = If x is a real number or today is a holiday then mathematics is
    not easy.

Question 2.
Symbolise the following propositions.
(i) 3x = 9 and x < 7
Answer:
Let p : 3x = 9, q : x < 1
The given proposition in symbols is p ∧ q

(ii) 33 ÷ − 11 ≠ 3 or 8 – 6 = 2
Answer:
= ~ p or q ⇒~p v q

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

(iii) If oxygen is a gas then gold is a compound
Answer:
p → q

(iv) y + 4 ≠ 4 or e is not a vowel
Answer:
~ pv ~ q

(v) If p :  \(\sqrt[3]{3}\) is an integer
Answer:
q : 2 is an odd number
r : 5 is a prime number.

Question 3.
What is the truth values of the following?

(i) (p ↔ q) ∧ r
Answer:
Here p = F, q – F, r = T
(F↔ F) ∧ T = T ∧ T = T
∴ Given pro position is true.

(ii) (p → q)∧ ~ r
Answer:
(F → F) ∧ ~ T = T ∧ F = F

(iii) (p ∧ q) → r
Answer:
(F ∧ F) ⇒ T = F → T = T

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

(iv) (~p ∨ q) ∧  r
Answer:
(~F ∨ F) ∧ T = (T ∨ F) ∧ T = T ∧ T = T

Question 4.
Negate the following.
(i) p ∨ ~q
Answer:
~(p ∨ ~q) ≡ ~p ∧ ~ (~q) = ~ p ∧ q

(ii) (~ p → q)
Answer:
– (~ p → q) =~ P∧ ~ q                   ,

(iii) ~ p ∧ ~q
Answer:
~ (~ p∧ ~ q) = p ∨ q

(iv) p∧ ~ q
Answer:
~(p∧ ~ q)=~ p ∨ q

(v) 4 is an even integer or 7 is a prime number.
Answer:
Given = p ∨ q
~(p ∨ q) ≡ ~ p ∧ ~ q
4 is not an even integer and 7 is not a prime number.

(vi) He likes to run and he does not like to sit.
Answer:
He does not likes to run or he likes to sit.

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

(vii) He likes mathematics and he does not like logic.
Answer:
He does not like mathematics or he likes logic.

(viii) If two triangles are similar then their areas are equal.
Answer:
Given (p → q)
~(p → q) ≡ p∧ ~ q
Two triangles are similar and their areas are not equal.

2nd PUC Basic Maths Mathematical Logic Two Marks Questions and Answers

Question 1.
If p, q and r are propositions with truth values F, T, and F respectively, then find the truth values of the following compound propositions.

(i) (~ p → q) ∨ r
Answer:
(~ F → T) ∨ F
≡ (T→ T) ∨ F ≡ T ∨ F = T

(ii) (p ∧ ~ q) →  r
Answer:
(F∧~T) → F
(F∧F) → F ≡ F → F ≡ T

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

(iii) P→ (q → r)
Answer:
F → (T→ F)
F→ (F) = T

(iv) ~{p~>q)v ~(p ↔ q)
Answer:
~(F → T) ∨ – (F↔T)
~(T) ∨ ~(F)
14 is not a divisor of 48 or 28 is divisible by 82.

2nd PUC Basic Maths Mathematical Logic Three Marks Questions and Answers

Question 1.
If the compound proposition p → (q ∨ r) is False.
Answer:
Then find the truth values of p, q and r.
Given p → (q ∨ r) is False
⇒ p is true and (q v r) is. False
p is true and ⇒ q is False and r is False
∴ p = T,q = F,r = F.

Question 2.
If the truth value of the proposition (p ∧ q) → (r ∨ ~ s) is false then find the truth val­ues of p, q, r and s.
Answer:
Given: (p ∧ q) → (r ∨ ~ s) is False
T → F = F
∴ p ∧ q is true and r v~ s is False
⇒ (p is true and q is true) and (r is false and ∼ s is false)
⇒ p = T, q = T, r = F, s = T.

Question 3.
Construct the truth table for the following
(i) (~ pA ~q)
Answer:

p q ~ p ~q ~ pA ~q
T T F F F
T F F T F
F T T F F
F F T T T

(ii) ~(p → q)
Answer:

p q p → q ~(p → q)
T T T F
T F F T
F T T F
F F T F

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

(iii) (p ∨ q) ∧ ∼ p
Answer:

p q p∨q ~p  (p∨q) ∧~p
T T T F F
T F T F F
F T T T T
F F F T F

(iv) (p ∨ q) ↔ (q ∧ p)
Answer:

p q p∨q q∧p  (p∨q)↔ (q~p)
T T T T T
T F T F F
F T T F F
F F F F T

(v) ∼ p (p∧∼q)
Answer:

p q ∼p ∼q p∧∼q ∼p ↔ (p∧∼q)
T F F F F T
T F F T T F
F T T F F F
F F T T F F

Question 4.
Write the converse, Inverse and contrapositive of the following,
(i) If x ∈ A ∩ B then x ∈ A and x ∈ B
Answer:
Given in symbols = p →(q ∧ r)
Converse = (q ∧ r) → p
If x ∈ A & x ∈ B Then x ∈ A∩B
Inverse =~ p → ~ (q ∧ r)
~ p → (~q∨ ~ r)
If x∉A∩B then x ∉ q or x ∉ B
Contrapositive =~ (q ∧ r) → ~ p
If x∈ A or x∉B Then x∉ A∩B

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

(ii) If the questions are easy then students score better marks.
Answer:
Converse:- If students score better marks then the questions are easy.
Inverse:- If questions are not easy then student do not score better marks. Contrapositive:- If the students do not score better marks then the questions are not easy.

(iii) If x2 =y2 then x = y.
Answer:
Convers = If x = y then x2= y2
Inverse = If x2 ≠ y2 then x ≠ y
Contrapositive = If x ≠ y then x2 ≠ y2

2nd PUC Basic Maths Mathematical Logic Five Marks Questions and Answers

I. Question 1.
Construct the truth tables for the following.
(i) p ∨(q ∧ ∼r)
Answer:

p q r ∼ r q∧∼ r p∨(q∧~r)
T T T F F T
T T F T T T
T F T F F T
T F F T F T
F T T F F F
F T F T T T
F F T F F F
F F F T F F

(ii) (∼p ∧ q) ∧ ∼ r
Answer:

p q r ~r ~p ~p∧q (~p∧q)∧~r
T T T F F F F
T T F T F F F
T F T F F F F
T F F T F F F
F T T F T T F
F T F T T T T
F F T F T F F
F F F T T F F

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

(iii) (p∧q) →(r ∨ ∼ q)
Answer:

p q r ~q p∧q rv ~ q p∧q → rv ~ q
T T T F T T T
T T F F T F F
T F T T F T T
T F F T F T T
F T T F F T T
F T F F F F T
F F T T F T T
F F F T F T T

(iv) (p∧q) ∨ r
Answer:

P q r (p∧q) (p∧q)∧r
T T T T T
T T F T T
T F T F T
T F F F F
F T T F T
F T F F F
F F T F T
F F F F F

(v) (p → q) →  ~ r
Answer:

p q r p → q ~r (p → q)→ ~r
T T T T F F
T T F T T T
T F T F F T
T F F F T F
F T T T F F
F T F T T T
F F T T F F

(vi) (p → r) ∧( p→ q)
Answer:

p q r p → r p →q (p → r) a (p → q)
T T T T T T
T T F F T F
T F T T F F
T F F F F F
F T T T T T
F T F T T T
F F T T T T
F F F T T T

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

II. Verify the following for Tautology, Contadiction or neither.

Question 1.
(~ p∨q) ∨ (P∧~q)
Answer:

p q ~P ~q ~p∨q p∧~q (~p∨q)v(p∧~q)
T T F F T F T
T F F T F T T
F T T F T F T
F F T T T F T

From last colum we conclude that it is a tautology.

Question 2.
(p∨~q) ∧ {~ p∨q)
Answer:

p q ~q p∧~q ~p q∨ ~p (p∧~q)∧(~p∨q)
T T F F F T F
T F T T F F F
F T F F T T F
F F T F T T F

From last column we conclude that it is a contradiction.

Question 3.
(p → q) ∨ (~p→ q)
Answer:

p q p → q ~p ~p →q (p → q) ∨ (~ p→q)
T T T F T T
T F F F T T
F T T T T T
F F T T F T

From last column we conclude that it is a Tautology.

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

Question 4.
(p → q) ↔ (~p ∨ q)
Answer:

p q p → q ~p ~p ∨ q (p → q)↔(∼ p ∨ q)
T T T F T T
T F F F F T
F T T T T T
F F T T T T

From last column we conclude that it is a Tautology.

Question 5.
[(p∧q) ∨ (p∨q) ]↔ q
Answer:

p q p∧q p∨q {p∧q)∨(p∨q) [(p∧q)∨(p∨q)]↔q
T T T T T T
T F F T T F
F T F T T T
F F F F F T

From last colum we canclude that it is neither Tautology nor Contradiction.

Question 6.
P.T. (p → q) ↔  (~ q → p) is a Tautology.
Answer:

p q p→q ~q ~P ~q→~p (p → q) (~ q→~p)
T T T F F T T
T F F T F F T
F T T F T T T
F F T T T T T

From last colum we conclude that it is a Tautology

Question 7.
P. T. [(P → q) ∧ (q→ r)→ (p →R) is a Tautology
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic 1
From last colum we canclude that it is a Tautology.

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

Question 8.
(∼p∧q) →(p∨q)
Answer:

p q ~p ~p∧q p∨q  (p∧q) → (p∨q)
T T F F T T
T F F F T T
F T T T T T
F F T F F T

Question 9.
Verify [p ∧ (p ∨ q)] → q is a Tautology or contradiction.
Answer:

p q ~ p (p∨q) ~p∧(p∨q) [~p∨(p∨q)] →  q
T T F T F T
T F F T F T
F T T T T T
F F T F F T

From last colum we canclude that it is a Tautology.

Question 10.
Verify (p → – q) ∨ (~ p ↔ q) for a Tautology or not.
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic 2
From last colum we conclude that it is neither Tautology nor contradiction.

III. Examine whether the following are logically equivalent?

Question 1.
p ↔ q and (p → q) ∧ (q → p)
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic 3
From 5th and 6th colums we get (p ↔ q) ≡ (p → q) a (q → p)

Question 2.
p → (q → r) and (p → q) → r
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic 4

Question 3.
(p ∧~q) ∨ q and p ∨ q.
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic 5

Question 4.
p ∧ q and ~ (p → ~ q)
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic 6

Question 5.
~(p ↔ q) and (p∧~q) ∨ (q∧p)
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic 7
From 5th and 8th colum we conclude ~(p ↔ q) ≡ (p∨ ~ q) ∨(q∧~p)

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic

Question 6.
p ↔ q an (∼p∨q ) ∧ (~ q v p)
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic 8
From 3rd and 8th colums we conclude that (p ↔ q) = (~ p ∨ q) ∧ (~q ∨ p)

Question 7.
S.T. ~(p ∼ q) ≡ ∼p ∧- q
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic 9
From 4th and 6th colums we get ~ (p → q) ≡ p∧ ∼ q

Question 8.
S.T. ~(p ∧ q) ≡ ~ pv ~ q
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic 10
From 4th and 7th colums we get ~ (p ∧ q) ≡ ~ pv ~ q.

Question 10.
S.T ~(p ∨ q) ≡ ~ p ∧ ~ q
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic 11
From 4th and 7th colums we get ~ (p ∨ q) ≡ (~ pv ~ q)

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