Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 1 Mathematical Logic Ex 1.4 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.4
(ii) p ∧ ~q
Solution:
The negation of p ∧ ~q is
~(p ∧ ~q) ≡ ~p ∨ ~(~q) … (Negation of conjunction)
≡ ~ p ∨ q … (Negation of negation)
(iii) p ∨ ~q
Solution:
The negation of p ∨ ~ p is
~ (p ∨ ~(q) ≡ ~p ∧ ~(~(q) … (Negation of disjunction)
≡ ~ p ∧ q … (Negation of negation)
(iv) (p ∨ ~q) ∧ r
Solution:
The negation of (p ∨ ~ q) ∧ r is
~[(p ∨ ~q) ∧ r] ≡ ~(p ∨ ~q) ∨ ~r … (Negation of conjunction)
≡ [ ~p ∧ ~(~q)] ∨ ~ r… (Negation of disjunction)
≡ (~ p ∧ q) ∧ ~ r … (Negation of negation)
(v) p → (p ∨ ~q)
Solution:
The negation of p → (p ∨ ~q) is
~ [p → (p ∨ ~q)] ≡ p ∧ ~ (p ∧ ~p) … (Negation of implication)
≡ p ∧ [ ~ p ∧ ~ (~(q)] … (Negation of disjunction)
≡ p ∧ (~ p ∧ q) (Negation of negation)
(vi) ~(p ∧ q) ∨ (p ∨ ~q)
Solution:
The negation of ~(p ∧ q) ∨ (p ∨ ~q) is
~[~(p ∧ q) ∨ (p ∨ ~q)] ≡ ~[~(p ∧ q)] ∧ ~(p ∨ ~q) … (Negation of disjunction)
≡ ~[~(p ∧ q)] ∧ [ p ∧ ~(~q)] … (Negation of disjunction)
≡ (p ∧ q) ∧ (~ p ∧ q) … (Negation of negation)
(vii) (p ∨ ~q) → (p ∧ ~q)
Solution:
The negation of (p ∨ ~q) → (p ∧ ~q) is
~[(p ∨ ~q) → (p ∧ ~q)]
≡ (p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of implication)
≡ (p ∨ ~q) ∧ [ ~p ∨ ~(~q)] … (Negation of conjunction)
≡ (p ∨ ~q) ∧ (~p ∨ q) … (Negation of negation)
(viii) (~ p ∨ ~q) ∨ (p ∧ ~q)
Solution:
The negation of (~ p ∨ ~q) ∨ (p ∧ ~ q) is
~ [(~p ∨ ~q) ∨ (p ∧ ~ q)]
≡ ~(~p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of disjunction)
≡ [~(~p) ∧ ~(~q)] ∧ [~p ∨ ~(~q)] … (Negation of disjunction and conjunction)
≡ (p ∧ q) ∧ (~p ∨ q) … (Negation of negation)
(ii) It 2 is a rational number then is irrational number.
Solution:
2 is not a rational number or is irrational number.
(iii) It f(2) = 0 then f(x) is divisible by (x – 2).
Solution:
f(2) ≠ 0 or f(x) is divisible by (x – 2).
(ii) (p ∨ q) ∧ (p ∨ ~q) ≡ p
Solution:
LHS = (p ∨ q) ∧ (p ∨ ~q)
≡ p ∨ (q ∧ ~q) … (Distributive Law)
≡ p ∨ F … (Complement Law)
≡ p … (Identity Law)
≡ RHS.
(iii) (p ∧ q) ∨ (~ p ∧ q) ∨ (p ∧ ~q) ≡ p ∨ q
Solution:
LHS = (p ∧ q) v (~p ∧ q) ∨ (p ∧ ~q)
≡ [(p ∨ ~p) ∧ q] ∨ (p ∧ ~q) … (Distributive Law)
≡ (T ∧ q) ∨ (p ∧ ~q) … (Complement Law)
≡ q ∨ (p ∧ ~q) … (Identity Law)
≡ (q ∨ p) ∧ (q ∨ ~q) … (Distributive Law)
≡ (q ∨ p) ∧ T.. (Complement Law)
≡ q ∨ p … (Identity Law)
≡ p ∨ q … (Commutative Law)
≡ RHS.
(iv) ~[(p ∨ ~q) → (p ∧ ~q)] ≡ (p ∨ ~q) ∧ (~p ∨ q)
Solution:
LHS = ~[(p ∨ ~q) → (p ∧ ~q)]
≡ (p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of implication)
≡ (p ∨ ~q) ∧ [~p ∨ ~(~q)] … (Negation of conjunction)
≡ (p ∨ ~ q) ∧ (~p ∨ q)… (Negation of negation)
≡ RHS.