What is the domain of $\sec^{-1}(2x)$ ?

\left(-\infty, -\frac{1}{2}\right] \cup \left[\frac{1}{2}, \infty\right)

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\left[-\frac{1}{2}, \frac{1}{2}\right]

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(-\infty, -2] \cup [2, \infty)

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\mathbb{R} - \{0\}

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Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

The input of the inverse function must lie in its domain: $\lvert 2x \rvert \ge 1$ .
$\lvert x \rvert \ge \frac{1}{2}$ .

Worth noting

$\left[-\frac{1}{2}, \frac{1}{2}\right]$ is the domain of $\cos^{-1}(2x)$ ; $\sec^{-1}$ needs the outside.

Ten questions from Inverse Trigonometric Functions, at your own pace.

Ten questions from Inverse Trigonometric Functions, at your own pace.

What is the domain of sec⁡−1(2x)\sec^{-1}(2x)sec−1(2x)?