Very Short Answer Questions

1. Write the roster form of:
(i) [tex]\( A = \{x \mid x \text{ is a multiple of 5 and } x \ \textless \ 30 \} \)[/tex]
(ii) [tex]\( B = \{x \mid x = n^2 + 1, n \leq 5, n \in \mathbb{N} \} \)[/tex]



Answer :

Certainly!

### Part (i)
We need to find the set [tex]\(A\)[/tex] which consists of all multiples of 5 that are less than 30.

To determine this, we list out the multiples of 5:
[tex]\[ 5, 10, 15, 20, 25, 30, 35, \ldots \][/tex]

We only consider those that are less than 30:
[tex]\[ 5, 10, 15, 20, 25 \][/tex]

Therefore, the set [tex]\(A\)[/tex] in roster form is:
[tex]\[ A = \{5, 10, 15, 20, 25\} \][/tex]

### Part (ii)
We need to find the set [tex]\(B\)[/tex] which consists of the values of [tex]\(x\)[/tex] given by the expression [tex]\(x = n^2 + 1\)[/tex] where [tex]\(n \leq 5\)[/tex] and [tex]\(n\)[/tex] is a natural number.

Let's compute the value of [tex]\(x\)[/tex] for each [tex]\(n\)[/tex] from 1 to 5:

1. For [tex]\(n = 1\)[/tex]:
[tex]\[ x = 1^2 + 1 = 1 + 1 = 2 \][/tex]

2. For [tex]\(n = 2\)[/tex]:
[tex]\[ x = 2^2 + 1 = 4 + 1 = 5 \][/tex]

3. For [tex]\(n = 3\)[/tex]:
[tex]\[ x = 3^2 + 1 = 9 + 1 = 10 \][/tex]

4. For [tex]\(n = 4\)[/tex]:
[tex]\[ x = 4^2 + 1 = 16 + 1 = 17 \][/tex]

5. For [tex]\(n = 5\)[/tex]:
[tex]\[ x = 5^2 + 1 = 25 + 1 = 26 \][/tex]

Therefore, the set [tex]\(B\)[/tex] in roster form is:
[tex]\[ B = \{2, 5, 10, 17, 26\} \][/tex]

In summary:
- The set [tex]\(A\)[/tex] is [tex]\(\{5, 10, 15, 20, 25\}\)[/tex].
- The set [tex]\(B\)[/tex] is [tex]\(\{2, 5, 10, 17, 26\}\)[/tex].