Let A = [2, 3, 5, 7]. Examine whether the statements given below are true or false.

Solution:

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

The given statement is: $\exists x \in A$ such that $x+3>9$.

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

When $x=7 \in A$,

$x+3=7+3=10>9$

So, $\exists x \in A$ and $x+3>9$

So, the given statement is TRUE.

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

The given statement is $\exists x \in A$ 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 $2 \in A$.

$\therefore \exists \mathrm{x} \in \mathrm{A}$ such that $\mathrm{x}$ is even.

So, the given statement is TRUE.

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

The given statement is: $\exists x \in A$ such that $x+2=6$.

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

$x=2 \rightarrow x+2=4 \neq 6$

$x=3 \rightarrow x+2=5 \neq 6$

$x=5 \rightarrow x+2=7 \neq 6$

$x=7 \rightarrow x+2=9 \neq 6$

So, the given statement is FALSE.

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

The given statement is: $\forall x \in A, x$ 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.

$\therefore \forall \mathrm{x} \in \mathrm{A}, \mathrm{x}$ is prime.

So, the given statement is TRUE.

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

The given statement is: $\forall x \in A, x+2<10$.

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

$A=[2,3,5,7]$

$x=2 \rightarrow x+2=4<10$

$x=3 \rightarrow x+2=5<10$

$x=5 \rightarrow x+2=7<10$

$x=7 \rightarrow x+2=9<10$

$\forall x \in A, x+2<10$, is a TRUE statement.

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

The given statement is: $\forall x \in A, x+4 \geq 11$.

So, we need to see whether for all ' $x$ ' which belongs to ' $A$ ', such that $x+4 \geq 11$.

$A=[2,3,5,7]$

$x=2 \rightarrow x+4=6 \geq 11$

$x=3 \rightarrow x+4=7 \geq 11$

$x=5 \rightarrow x+4=9 \geq 11$

$x=7 \rightarrow x+4=11 \geq 11$

Only for $x=7, x+4=11 \geq 11$

$\forall x \in A, x+4 \geq 11$, is a FALSE statement.