Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.5 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.5

Question 1.

Use qualifiers to convert each of the following open sentences defined on N, into a true statement:

(i) x^{2} + 3x – 10 = 0

Solution:

∃ x ∈ N, such that x^{2} + 3x – 10 = 0 is a true statement

(x = 2 ∈ N satisfy x^{2} + 3x – 10 = 0)

(ii) 3x – 4 < 9

Solution:

∃ x ∈ N, such that 3x – 4 < 9 is a true statement.

(x = 1, 2, 3, 4 ∈ N satisfy 3x – 4 < 9)

(iii) n^{2} ≥ 1

Solution:

∀ n ∈ N, n^{2} ≥ 1 is a true statement.

(All n ∈ N satisfy n^{2} ≥ 1)

(iv) 2n – 1 = 5

Solution:

∃ x ∈ N, such that 2n – 1 = 5 is a true statement.

(n = 3 ∈ N satisfy 2n – 1 = 5)

(v) y + 4 > 6

Solution:

∃ y ∈ N, such that y + 4 > 6 is a true statement.

(y = 3, 4, 5, … ∈ N satisfy y + 4 > 6

(vi) 3y – 2 ≤ 9

Solution:

∃ y ∈ N, such that 2y ≤ 9 is a true statement.

(y = 1, 2, 3 ∈ N satisfy 3y – 2 ≤ 9).

Question 2.

If B = {2, 3, 5, 6, 7}, determine the truth value of each of the following:

(i) ∀ x ∈ B, x is a prime number.

Solution:

(i) x = 6 ∈ B does not satisfy x is a prime number.

So, the given statement is false, hence its truth value is F.

(ii) ∃ n ∈ B, such that n + 6 > 12.

Solution:

Clearly n = 7 ∈ B satisfies n + 6 > 12.

So, the given statement is true, hence its truth value is T.

(iii) ∃ n ∈ B, such that 2n + 2 < 4.

Solution:

No element n ∈ B satisfy 2n + 2 < 4.

So, the given statement is false, hence its truth value is F.

(iv) ∀ y ∈ B, y^{2} is negative.

Solution:

No element y ∈ B satisfy y^{2} is negative.

So, the given statement is false, hence its truth value is F.

(v) ∀ y ∈ B, (y – 5) ∈ N.

Solution:

y = 2 ∈ B, y = 3 ∈ B and y = 5 ∈ B do not satisfy (y – 5) ∈ N.

So, the given statement is false, hence its truth value is F.