Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.3 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.3

Question 1.

If A = {3, 5, 7, 9, 11, 12}, determine the truth value of each of the following.

(i) Ǝ x ∈ A such that x – 8 = 1

Solution:

Clearly x = 9 ∈ A satisfies x – 8 = 1. So the given statement is true, hence its truth value is T.

(ii) Ɐ x ∈ A, x^{2} + x is an even number

Solution:

For each x ∈ A, x^{2} + x is an even number. So the given statement is true, hence its truth value is T.

(iii) Ǝ x ∈ A such that x^{2} < 0

Solution:

There is no x ∈ A which satisfies x^{2} < 0. So the given statement is false, hence its truth value is F.

(iv) Ɐ x ∈ A, x is an even number

Solution:

x = 3 ∈ A, x = 5 ∈ A, x = 7 ∈ A, x = 9 ∈ A, x = 11 ∈ A do not satisfy x is an even number. So the given statement is false, hence its truth value is F.

(v) Ǝ x ∈ A such that 3x + 8 > 40

Solution:

Clearly x = 11 ∈ A and x = 12 ∈ A satisfies 3x + 8 > 40. So the given statement is true, hence its truth value is T.

(vi) Ɐ x ∈ A, 2x + 9 > 14

Solution:

For each x ∈ A, 2x + 9 > 14. So the given statement is true, hence its truth value is T.

Question 2.

Write the duals of each of the following.

(i) p ∨ (q ∧ r)

Solution:

The duals of the given statement patterns are :

p ∧ (q ∨ r)

(ii) p ∧ (q ∧ r)

Solution:

p ∨ (q ∨ r)

(iii) (p ∨ q) ∧ (r ∨ s)

Solution:

(p ∧ q) ∨ (r ∧ s)

(iv) p ∧ ~q

Solution:

p ∨ ~q

(v) (~p ∨ q) ∧ (~r ∧ s)

Solution:

(~p ∧ q) ∨ (~r ∨ s)

(vi) ~p ∧ (~q ∧ (p ∨ q) ∧ ~r)

Solution:

~p ∨ (~q ∨ (p ∧ q) ∨ ~r)

(vii) [~(p ∨ q)] ∧ [p ∨ ~(q ∧ ~s)]

Solution:

[ ~(p ∧ q)] ∨ [p ∧ ~(q ∨ ~s)]

(viii) c ∨ {p ∧ (q ∨ r)}

Solution:

t ∧ {p ∧ (q Ar)}

(ix) ~p ∨ (q ∧ r) ∧ t

Solution:

~p ∧ (q ∨ r) ∨ c

(x) (p ∨ q) ∨ c

Solution:

(p ∧ q) ∧ t

Question 3.

Write the negations of the following.

(i) x + 8 > 11 or y – 3 = 6

Solution:

Let p : x + 8 > 11, q : y — 3 = 6.

Then the symbolic form of the given statement is p ∨ q.

Since ~(p ∨ q) ≡ ~p ∧ ~q, the negation of given statement is :

‘x + 8 > 11 and y – 3 ≠ 6’ OR

‘x + 8 ≮ 11 and y – 3 ≠ 6’

(ii) 11 < 15 and 25 > 20

Solution:

Let p: 11 < 15, q : 25 > 20.

Then the symbolic form of the given statement is p ∧ q.

Since ~(p ∧ q) ≡ ~p ∨ ~q, the negation of given statement is :

’11 ≮ 15 or 25 > 20.’ OR

’11 ≯ 15 or 25 ≮ 20.’

(iii) Qudrilateral is a square if and only if it is a rhombus.

Solution:

Let p : Quadrilateral is a square.

q : It is a rhombus.

Then the symbolic form of the given statement is p ↔ q.

Since ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p), the negation of given statement is :

‘ Quadrilateral is a square but it is not a rhombus or quadrilateral is a rhombus but it is not a square.’

(iv) It is cold and raining.

Solution:

Let p : It is cold.

q : It is raining.

Then the symbolic form of the given statement is p ∧ q.

Since ~(p ∧ q) ≡ ~p ∨ ~q, the negation of the given statement is :

‘It is not cold or not raining.’

(v) If it is raining then we will go and play football.

Solution:

Let p : It is raining.

q : We will go.

r : We play football.

Then the symbolic form of the given statement is p → (q ∧ r).

Since ~[p → (q ∧ r)] ≡ p ∧ ~(q ∧ r) ≡ p ∧ (q ∨ ~r), the negation of the given statement is :

‘It is raining and we will not go or not play football.’

(vi) \(\sqrt {2}\) is a rational number.

Solution:

Let p : \(\sqrt {2}\) is a rational number.

The negation of the given statement is

‘ ~p : \(\sqrt {2}\) is not a rational number.’

(vii) All natural numbers are whole numers.

Solution:

The negation of the given statement is :

‘Some natural numbers are not whole numbers.’

(viii) Ɐ n ∈ N, n^{2} + n + 2 is divisible by 4.

Solution:

The negation of the given statement is :

‘Ǝ n ∈ N, such that n^{2} + n + 2 is not divisible by 4.’

(ix) Ǝ x ∈ N such that x – 17 < 20

Solution:

The negation of the given statement is :

‘Ɐ x ∈ N, x – 17 ≯ 20.’

Question 4.

Write converse, inverse and contrapositive of the following statements.

(i) If x < y then x^{2} < y^{2} (x, y ∈ R)

Solution:

Let p : x < y, q : x^{2} < y^{2}.

Then the symbolic form of the given statement is p → q.

Converse : q → p is the converse of p → q.

i.e. If x^{2} < y^{2}, then x < y.

Inverse : ~p → ~q is the inverse of p → q.

i.e. If x ≯ y, then x^{2} ≯ y^{2}. OR

If x ≮ y, then x^{2} ≮ y^{2}.

Contrapositive : ~q → p is the contrapositive of

p → q i.e. If x^{2} ≯ y^{2}, then x ≯ y. OR

If x^{2} ≮ y^{2}, then x ≮ y.

(ii) A family becomes literate if the woman in it is literate.

Solution:

Let p : The woman in the family is literate.

q : A family become literate.

Then the symbolic form of the given statement is p → q

Converse : q → p is the converse of p → q.

i.e. If a family become literate, then the woman in it is literate.

Inverse : ~p → ~q is the inverse of p → q.

i.e. If the woman in the family is not literate, then the family does not become literate.

Contrapositive : ~q → ~p is the contrapositive of p → q. i e. If a family does not become literate, then the woman in it is not literate.

(iii) If surface area decreases then pressure increases.

Solution:

Let p : The surface area decreases.

q : The pressure increases.

Then the symbolic form of the given statement is p → q.

Converse : q → p is the converse of p→ q.

i.e. If the pressure increases, then the surface area decreases.

Inverse : ~p → ~q is the inverse of p → q.

i.e. If the surface area does not decrease, then the pressure does not increase.

Contrapositive : ~q → ~p is the contrapositive of p → q.

i.e. If the pressure does not increase, then the surface area does not decrease.

(iv) If voltage increases then current decreases.

Solution:

Let p : Voltage increases.

q : Current decreases.

Then the symbolic form of the given statement is p → q.

Converse : q →p is the converse of p → q.

i.e. If current decreases, then voltage increases.

Inverse : ~p → ~q is the inverse of p → q.

i.e. If voltage does not increase, then-current does not decrease.

Contrapositive : ~q → ~p, is the contrapositive of p → q.

i.e. If current does not decrease, then voltage does not increase.