Let the function ‘f’ be defined by f (x) = 5x² + 2 ∀ x ∈ R, then ‘f’ is

The correct option is many-one into function.

Let's analyze the given function step-by-step to understand why it falls under this classification.

The function is defined as:
$$f\left(x\right)=5x^2+2$$
where the domain and codomain are both the set of real numbers ($ℝ$), meaning $f:ℝ\to \mathrm{��}$.

Step 1: Check if the function is One-One (Injective) or Many-One
A function $f$ is one-one if distinct elements in the domain map to distinct elements in the codomain. If two different inputs produce the same output, the function is many-one.
Let's choose two different values for $x$, say $x_1=1$ and $x_2=-1$ (both belong to $ℝ$).
Calculating the function values:
$$f\left(1\right)=5(1{)}^2+2=5+2=7$$
$$f(-1)=5(-1{)}^2+2=5+2=7$$
Since $x_1\ne x_2$ but $f\left(x_1\right)=f\left(x_2\right)$, the function is many-one.

Step 2: Check if the function is Onto (Surjective) or Into
A function is onto if its range is equal to its codomain. If the range is a proper subset of the codomain (meaning there are elements in the codomain that are not mapped by any $x$ in the domain), the function is into.
The given codomain is $ℝ$ (all real numbers).
Let's find the range of $f\left(x\right)=5x^2+2$:
For any real number $x$, we know that:
$$x^2\ge 0$$
Multiplying by 5:
$$5x^2\ge 0$$
Adding 2 to both sides:
$$5x^2+2\ge 2$$
Therefore, $f\left(x\right)\ge 2$ for all $x\in ℝ$.
The range of the function is $[2,\infty )$, which is only a subset of the codomain $ℝ$.
For example, there is no real number $x$ such that $f\left(x\right)=0$ or $f\left(x\right)=-3$.
Since the range is not equal to the codomain, the function is into.

Conclusion:
Combining the findings from both steps, the function $f$ is a many-one into function.