It is enough to add (2, 2), (3, 3) and (5, 5) to make R reflexive.
To make R symmetric, (5, 2) needs to be included.
∴ Minimum number of ordered pairs required are (5, 2), (2,2), (3,3) and (5,5).

Let \(y={ -x }^{ \left( \frac { 1 }{ 3 } \right) }\)

Then \(y={ -x }^{ \frac { 1 }{ 3 } }\) is the reflection of the graph of \(y = { -x }^{ \left( \frac { 1 }{ 3 } \right) }\) about the x-axis.

Given \(f(x)=\sqrt{x+2}\)and \(g(x)=\sqrt{4-x^2}\) Clearly,f(x) is defined for all x, satisfying x + 2 > 0 ⇒x > -2 ⇒ x ∈[-2, ∞)
∴ Domain of f = [-2, ∞)
Also, g(x) is defined for all x satisfying 4 - x² ≥ 0 ⇒ x² - 4 < 0 ⇒ (x - 2) (x + 2) < 0
⇒ x ∈ [-2, 2]
∴ Domain of g = [-2, 2]
(f + g) : [-2, 2] ⟶ R is given by \((f+ g)(x) =f(x) + g(x) =\sqrt{x+2}+\sqrt{4-x^2}\)

The relation R on the set of all polygons is defined as R = {(P₁, P₂): P₁ and P₂ have same number of sides}
Reflexivity : Let P be any polygon in A. Then P and P have same number of sides.
⇒ (P, P) ∈ R
⇒ R is reflexive on A.
Symmetry : Let P₁ and P₂ be two polygons in A such that (P_1, P₂) ∈ R
(P₁, P₂) ∈ R ⇒ P₁ and P₂ have same number of sides.
⇒ P₂ and P₁ have same number of sides.
⇒ (P₂, P₁)∈ R
∴ R is symmetric on A
Transitivity : Let P₁, P₂, P₃ be three polygons in A such that (P₁, P₂) E Rand (P₂, P₃) ∈ R.
⇒ P₁ and P₂ have same number of sides and P2 and P3 have same number of sides.
⇒ P₁ and P₃ have same number of sides.
⇒ (P₁, P₃) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.

Venn diagram

From (ii) and (iv), it is clear that A ∩ (B ∩ C) = (A ∩ B) ∩ C.