If R1 = {(x, y) | y = 2x + 7, where x ∈ R and – 5 ≤ x ≤ 5}

1.	If R1 = {(x, y) \| y = 2x + 7, where x ∈ R and – 5 ≤ x ≤ 5} is a relation. Then find the domain and Range of R1.
Answer» According to the question, R₁ = {(x, y) \| y = 2x + 7, where x ∈R and – 5 ≤ x ≤ 5} is a relation The domain of R₁ consists of all the first elements of all the ordered pairs of R₁, i.e., x, It is also given – 5 ≤ x ≤ 5. Therefore, Domain of R₁ = [–5, 5] The range of R contains all the second elements of all the ordered pairs of R₁, i.e., y It is also given y = 2x + 7 Now x ∈ [–5,5] Multiply LHS and RHS by 2, We get, 2x ∈ [–10, 10] Adding LHS and RHS with 7, We get, 2x + 7 ∈ [–3, 17] Or, y ∈ [–3, 17] So, Range of R₁ = [–3, 17]

If R1 = {(x, y) | y = 2x + 7, where x ∈ R and – 5 ≤ x ≤ 5} is a relation. Then find the domain and Range of R1.

Answer»

According to the question,

R₁ = {(x, y) | y = 2x + 7, where x ∈R and – 5 ≤ x ≤ 5} is a relation

The domain of R₁ consists of all the first elements of all the ordered pairs of R₁, i.e., x,

It is also given – 5 ≤ x ≤ 5.

Therefore,

Domain of R₁ = [–5, 5]

The range of R contains all the second elements of all the ordered pairs of R₁, i.e., y

It is also given y = 2x + 7

Now x ∈ [–5,5]

Multiply LHS and RHS by 2,

We get,

2x ∈ [–10, 10]

Adding LHS and RHS with 7,

We get,

2x + 7 ∈ [–3, 17]

Or, y ∈ [–3, 17]

So,

Range of R₁ = [–3, 17]