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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
(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
(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
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
(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