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**NCERT mathematics class 12 chapter Relations and Functions Exercise 1.1 Problem 5 Solution**. Solutions for other problems are available at Exercise 1.1 SolutionsExercise 1.1 Problem 5 Solution

5. Check whether the relation R in R defined by R = {(a, b) : a \le b^3} is reflexive, symmetric or transitive.

To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a) ∈ R, for every a ∈ A.

When {a = -2}, we’ve {a^3 = -8}. But -2 \nleq (-2)^3

∴ R is not Reflexive

To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2) ∈ R implies that (a_2, a_1) ∈ R, for all a_1, a_2 ∈ A

Consider {a = 1} and {b = 2}

we’ve 1 ≤ 2³ but 2 \nleq 1^3

⇒ When (a, b) ∈ R, we have (b, a) ∉ R

∴ R is not symmetric.

To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2) ∈ R and (a_2, a_3) ∈ R implies that (a_1, a_3) ∈ R, for all a_1, a_2, a_3 ∈ A

Consider the case, {a = 2}, {b = 3} and {c = 4}

We’ve {b^3 = 3^3 = 27} and 4³ = 64

So, 2 ≤ 3³, 3 ≤ 4³ and 2 ≤ 4³.

However, when {a = 4}, {b = \dfrac{7}{4}} and {c = \dfrac{5}{4}}

We’ve {b^3 = \left(\dfrac{7}{4}\right)^{3} = 5.359} and {c^3 = \left(\dfrac{5}{4}\right)^{3} = 1.953}

In this case, 4 ≤ \left(\dfrac{7}{4}\right)^3 and \dfrac{7}{4} ≤ \left(\dfrac{5}{4}\right)^3, but {4 \nleq \left(\dfrac{5}{4}\right)^3}

∴ R is not Transitive

Hence, in this case, R is neither reflexive nor symmetric nor transitive.