Let $f : [2, \infty) \to R$ be defined by $f (x) = (x - 2)^{2}$ . Determine whether $f$ is one‑to‑one on its domain and justify your answer.

easy 1 mark Class 12 · Mathematics · Relations and Functions

Answer

Yes,

f

is one‑to‑one on

[2, \infty)

The correct answer is Yes, $f$ is one‑to‑one on $[2, \infty)$ . — Since the implication holds, $f$ is one‑to‑one on $[2, \infty)$ .

Step-by-step solution

Write the definition of one‑to‑one: $f$ is one‑to‑one if $f (a) = f (b)$ implies $a = b$ for all $a, b$ in the domain.
Assume $f (a) = f (b)$ , i.e. $(a - 2)^{2} = (b - 2)^{2}$ .
Take square roots: $∣ a - 2∣ = ∣ b - 2∣$ .
Since $a, b \geq 2$ , both $a - 2$ and $b - 2$ are non‑negative, so $∣ a - 2∣ = a - 2$ and $∣ b - 2∣ = b - 2$ .
Thus $a - 2 = b - 2$ , which gives $a = b$ .
Since the implication holds, $f$ is one‑to‑one on $[2, \infty)$ .

Common mistake: Assuming a square function is never one‑to‑one without checking the domain restriction.

Concept tested: One‑to‑one function on restricted domain