Q.The domain and range of real function $f$ defined by $f(x) = \sqrt{x - 1}$ is given by
(A) Domain $= (1,\ \infty)$, Range $= (0,\ \infty)$
(B) Domain $= [1,\ \infty)$, Range $= (0,\ \infty)$
(C) Domain $= [1,\ \infty)$, Range $= [0,\ \infty)$
(D) Domain $= [1,\ \infty)$, Range $= [0,\ \infty)$

Concept: Domain of a square‑root function requires the radicand to be non‑negative; the range follows from the principal (non‑negative) square root.

1. Domain: $x - 1 \ge 0 \implies x \ge 1$. So domain is $[1, \infty)$. …

The key to this problem is understanding what a square root function actually does — not just mechanically, but conceptually. A real-valued square root is only defined when the quantity inside is non-negative. That’s not a rule to memorise; it’s because the square root of a negative number isn’t a real number, and we’re working with real functions.

Similarly, the range of a square root function is never negative. The principal square root always returns a value $\ge 0$. So both domain and range involve closed intervals at zero — that’s the subtle point many students miss.

Let’s walk through it step by step.

1. Domain: what $x$ values are allowed?

   The function is $f(x) = \sqrt{x - 1}$. For this to be a real number, the radicand $x - 1$ must satisfy $x - 1 \ge 0$.

   Solve: $x \ge 1$.

   So the domain is all real numbers from $1$ onward, including $1$ itself. In interval notation: $[1, \infty)$.

   Watch out

   A common mistake is to write $(1, \infty)$ instead of $[1, \infty)$. But $x = 1$ gives $\sqrt{0} = 0$, which is perfectly valid — so the bracket must be square, not round.

2. Range: what $y$ values come out?

   When $x = 1$, $f(1) = \sqrt{0} = 0$.

   As $x$ increases beyond $1$, $x-1$ becomes positive and grows without bound, so $\sqrt{x-1}$ also grows without bound.

   The square root function never outputs a negative number — it’s defined as the non-negative root. So the smallest output is $0$, and there is no largest output. …