Given: A = {2, 3, 5, 7}

(i) ∃ x ∈ A such that x + 3 > 9.

We have to check 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) ∃ x ∈ A such that x is even.

We have to check whether there exists ‘x’ which belongs to ‘A’, such that x is even.

x = 2, is an even number and 2 ∈ A.

So, the given statement is TRUE.

(iii) ∃ x ∈ A such that x + 2 = 6.

We have to check whether there exists ‘x’ which belongs to ‘A’, such that x + 2 = 6.

At x = 2 ⇒ x + 2 = 4 ≠ 6

At x = 3 ⇒ x + 2 = 5 ≠ 6

At x = 5 ⇒ x + 2 = 7 ≠ 6

At x = 7 ⇒ x + 2 = 9 ≠ 6

None of the values satisfy the equation.

So, the given statement is FALSE.

(iv) ∀ x ∈ A, x is prime.

We have to check 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.

All are prime numbers.

So, the given statement is TRUE.

(v) ∀ x ∈ A, x + 2 < 10.

We have to check whether for all ‘x’ which belongs to ‘A’, such that x + 2 < 10.

A = {2, 3, 5, 7}

At x = 2 ⇒ x + 2 = 4 < 10

At x = 3 ⇒ x + 2 = 5 < 10

At x = 5 ⇒ x + 2 = 7 < 10

At x = 7 ⇒ x + 2 = 9 < 10

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

(vi) ∀ x ∈ A, x + 4 ≥ 11.

We have to check whether for all ‘x’ which belongs to ‘A’, such that x + 4 ≥ 11.

A = {2, 3, 5, 7}

At x = 2 ⇒ x + 4 = 6 ≥ 11

At x = 3 ⇒ x + 4 = 7 ≥ 11

At x = 5 ⇒ x + 4 = 9 ≥ 11

At x = 7 ⇒ x + 4 = 11 ≥ 11

Only for x = 7, statement is true.

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