A = [2, 3, 5, 7] (given in the question).

The given statement is: ∃ x ∈ A such that x + 3 > 9.

So, we need to see whether there exists ‘x’ which belongs to ‘A’, such that x + 3 > 9.

When x = 7 ∈ A,

x + 3 = 7 + 3 = 10 > 9

So, ∃ x ∈ A and x + 3 > 9.

So, the given statement is TRUE.

(ii) A = [2, 3, 5, 7] (given in the question).

The given statement is ∃ x ∈ A such that x is even.

So, we need to see whether there exists ‘x’ which belongs to ‘A’, such that x is even.

In the set A = [2, 3, 5, 7]

x = 2, is an even number and 2 ∈ A.

∴ ∃ x ∈ A such that x is even.

So, the given statement is TRUE.

(iii) A = [2, 3, 5, 7] (given in the question).

The given statement is: ∃ x ∈ A such that x + 2 = 6.

So, we need to see whether there exists ‘x’ which belongs to ‘A’, such that x + 2 = 6.

x = 2 → x + 2 = 4 ≠ 6

x = 3 → x + 2 = 5 ≠ 6

x = 5 → x + 2 = 7 ≠ 6

x = 7 → x + 2 = 9 ≠ 6

So, the given statement is FALSE.

(iv) A = [2, 3, 5, 7] (given in the question).

The given statement is: ∀ x ∈ A, x is prime.

So, we need to see whether for all ‘x’ which belongs to ‘A’, such that x is a prime number.

All ‘x’ which belongs to A = [2, 3, 5, 7] is a prime number.

∴∀ x ∈ A, x is prime.

So, the given statement is TRUE.

(v) A = [2, 3, 5, 7] (given in the question).

The given statement is: ∀ x ∈ A, x + 2 < 10.

So, we need to see whether for all ‘x’ which belongs to ‘A’, such that x + 2 < 10.

A = [2, 3, 5, 7]

x = 2 → x + 2 = 4 < 10

x = 3 → x + 2 = 5 < 10

x = 5 → x + 2 = 7 < 10

x = 7 → x + 2 = 9 < 10

∀ x ∈ A, x + 2 < 10, is a TRUE statement.

(vi) A = [2, 3, 5, 7] (given in the question).

The given statement is: ∀ x ∈ A, x + 4 ≥ 11.

So, we need to see whether for all ‘x’ which belongs to ‘A’, such that x + 4 ≥ 11.

A = [2, 3, 5, 7]

x = 2 → x + 4 = 6 ≥ 11

x = 3 → x + 4 = 7 ≥ 11

x = 5 → x + 4 = 9 ≥ 11

x = 7 → x + 4 = 11 ≥ 11

Only for x = 7, x + 4 = 11 ≥ 11.

∀ x ∈ A, x + 4 ≥ 11, is a FALSE statement.