2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.1

Students can Download Basic Maths Exercise 6.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.1

Part – A

2nd PUC Basic Maths Mathematical Logic Ex 6.1 One Mark Questions and Answers

Question 1.
Symbolise the following propositions:
(1) 3x = 9 and x<7
(ii) 33 + 11 ≠ 3 or 8 – 6 = 2
(iii) If two numbers and equal then their squares are not equal.
(iv) If oxygen is a gas then gold is a compound
(v) y + 4 ≠ 4 ore is not a vowel
Answer:
(i) Let p:3x = 9, q = x < 7
Given in symbols is p ∧ q

(ii) Let p:33 = 11 = 3, q= 8 – 6 = 2
Given is ~p ∨ q

KSEEB Solutions

(iii) Let p: Two numbers are equal, q: Squares are equal then given is p → ~q

(iv) Let p: Oxygen is a gas
q: Gold is a compound
Then given is p → q

(v) P: y + 4 = 4, q: e is a vowel given proposition is ~p v ~q

Part – B

2nd PUC Basic Maths Mathematical Logic Ex 6.1 Two or Three 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
(ii) (p ∧ ~q) → r
(iii) p → (q → r)
(iv) ~(p → q) ∨~(p ↔ q)
(v) (p ∧ q) ∨ ~ r
(vi) ~(p ∨ r) → ~q
Answers:
(i) (~p → q)∨ r
(~F →T) ∨ r
(T →T) ∨ F
T ∨ F
= T

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

KSEEB Solutions

(iii) p → (q → r)
F →(T →F)
F →(T →F)
= T

(iv) ~(p → q) ∨~(p↔ q)
~(F →T) ∨ ~(F↔T)
~T ∨ ~F
F ∨ T = T

(v) (p ∧ q) ∨ ~ r
(F ∧ T) ∨ ~ r
F ∨ T
= T

(vi) ~(P ∨ r) → ~ q
~(F ∨ F) ∨ ~ T
~F → ~ T
T → F
= F

KSEEB Solutions

Question 2.
Answer:
(1) If the compound proposition “(p → q) ∧ (p ∧ r)” is false, then find the truth values of p, q and r.
(ii) If the compound proposition p→ (q ∨ r) is false, then find the truth values of p, q and r.
(iii) If the compound proposition p → (~q ∨ r) is false, then find the truth values of p, q and r.
(iv) If the truth value of the propositions (p ∧ q) → (r ∨ ~s) is false, then find the truth values of p, q, rand s.

Answers:
(i) Given (p → q) ∧ (p ∧ r) is false
(a) Case 1: p → q is true & p ∧ r is false
p is T q is T & p is T &ris F
p = T, q = T, r = F
Case 2(a): p = F, q = T p ∧ r= F → p = F, r = F
p = F, q = T, r = f

(b): (p →q) is F & par is true
T → F = F
T ∧ T is T
P=T, q = F, r=T

Case 3: (p → q) is F & (p ∧ r) is false
T → F = F
F ∧ F = F
F ∧ T = F
F ∧ F = F .
∴ p = T, q = F, r= F.

(ii) Given p → (q ∨ r) is false
T → F = F
∴ p = T & q ∨ r is false =
F ∨ F= F
∴ p = T, q = F & r= F

KSEEB Solutions

(iii) Given p → (q ∨ r) is false
Then T → F= F
∴ P = T, ~ q ∨ r= F
F ∨ F = F
∴ q = T, q = T, r = F.

(iv) Given (p ^ q) → (r ∨ ~s) is false
We know that T → F = F
∴ p∧q = T and r ∨ ~s = F is false
T∧ T = T
F ∨ F= F is false
∴ p = T, q = T, r = F, S = T

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