R= {(a, b): a ≤ b²} is neither reflexive nor symmetric nor transitive

Solution: We have, $R=\{(a,b): a\leq b^2\}$ where a, b ∈ R

Reflexivity:- Obiviously, $\frac{1}{2}$ is a real number and $\frac{1}{2}\leq (\frac{1}{2})^2$ is not true.

$\Rightarrow \left(\frac{1}{2},\frac{1}{2}\right)\notin R$

Therefore, R is not reflexive.

Symmetric:- Let the real numbers 1 and 2

such that $(1,2)\in R \Rightarrow 1\leq 2^2$

$\Rightarrow 2 \leq 1^2$ is not true

Thus $(2,1)\notin R$

Hence, R is not symmetric.

Transitive:- Let $a = 2, b = -2, c = 1$

$(2,-2)\in R$ and $(-2, 1)\in R \quad \Rightarrow 2 \leq (-2)^2$ and $-2\leq (1)^2$

$\quad \Rightarrow 2 \leq (1)^2$ is not true

$\Rightarrow (2,1)\notin R$

Hence, R is not transitive.

Class 12 ncert solution math exercise 1.1