The relation R on the set of real numbers defined by $R= \{(a,b): a≤ b^2\}$ is

neither reflexive nor symmetric nor transitive

The correct answer is Option (1) → neither reflexive nor symmetric nor transitive

Given relation: $R=\{(a,b):a\le b^{2}\}$ on $\mathbb{R}$

Reflexive: For reflexivity, $(a,a)\in R\ \forall a$ $\Rightarrow a\le a^{2}$ must hold for all real $a$. For $a=\frac{1}{2}$, $a^{2}=\frac{1}{4}$, and $a\le a^{2}$ is false. Hence, not reflexive.

Symmetric: If $(a,b)\in R$, then $a\le b^{2}$. For symmetry, need $(b,a)\in R \Rightarrow b\le a^{2}$. Take $a=1,\ b=2$: $1\le4$ (true), but $2\le1$ (false). Hence, not symmetric.

Transitive: 

A relation is said to be transitive if whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.

Given R = {(a, b) : a ≤ b²}, for transitivity we require that from
a ≤ b² and b ≤ c², it should follow that a ≤ c².

To verify, take a = 4, b = −2, c = 1:

For (4, −2): 4 ≤ (−2)² = 4 (true)
For (−2, 1): −2 ≤ 1² = 1 (true)
For (4, 1): 4 ≤ 1² = 1 (false)

Thus, (4, −2) ∈ R and (−2, 1) ∈ R, but (4, 1) ∉ R.
Hence, the relation is not transitive.

Answer: neither reflexive nor symmetric nor transitive