Question

The domain of the function defined by f(x) = \(\cos^{-1}\sqrt{x-1}\) is

• A. [1, 2]
• B. [0, 2]
• C. [-1, 1]
• D. [0, 1]

Answer 1

The Correct Option is A

We are given the function \(f(x) = \cos^{-1}(\sqrt{x-1})\) and need to find its domain.

There are two conditions that must be satisfied for the function to be defined:

1. The expression inside the square root must be non-negative: \(x - 1 \ge 0 \implies x \ge 1\)
2. The argument of the inverse cosine function must be between -1 and 1 (inclusive): \(-1 \le \sqrt{x-1} \le 1\)

Since \(\sqrt{x-1}\) is always non-negative, we can rewrite the second condition as:

\(0 \le \sqrt{x-1} \le 1\)

Squaring all parts, we get:

\(0 \le x - 1 \le 1\)

Adding 1 to all parts, we get:

\(1 \le x \le 2\)

Combining both conditions \(x \ge 1\) and \(1 \le x \le 2\), we find that the domain is \(1 \le x \le 2\).

Thus, the correct option is (A) \([1, 2]\).

Answer 2

The function is \( f(x) = \cos^{-1}(\sqrt{x - 1}) \).  For \( \sqrt{x - 1} \) to be real, we require: \[ x - 1 \geq 0 \quad \Rightarrow \quad x \geq 1 \] The range of \( \cos^{-1}(y) \) is \( [0, \pi] \), so the argument \( \sqrt{x - 1} \) must lie within the range \( [0, 1] \). Thus, we need: \[ 0 \leq \sqrt{x - 1} \leq 1 \] Squaring both sides: \[ 0 \leq x - 1 \leq 1 \] This simplifies to: \[ 1 \leq x \leq 2 \] 

The domain of the function is \( [1, 2] \).