Q117 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
$\text { Let } f(x)=\cos ^{-1}(3 x-1) \text {, then domain of } f(x) \text { is equal to }$
- A$\left(0, \dfrac{2}{3}\right)$
- B$\left[-\dfrac{2}{3}, \dfrac{2}{3}\right]$
- C$\left(-\dfrac{2}{3}, \dfrac{2}{3}\right)$
- Correct answerD$\left[0, \dfrac{2}{3}\right]$
Solution
The function $f(x) = \cos^{-1}(3x - 1)$ is defined when the argument of the inverse cosine function lies in the interval $[-1, 1]$.
Therefore, we set the inequality:
$-1 \le 3x - 1 \le 1$
Adding $1$ to all parts of the inequality:
$-1 + 1 \le 3x - 1 + 1 \le 1 + 1$
$0 \le 3x \le 2$
Dividing by $3$ throughout:
$0 \le x \le \dfrac{2}{3}$
Thus, the domain of $f(x)$ is $[0, \dfrac{2}{3}]$.
Answer: $\left[0, \dfrac{2}{3}\right]$
Therefore, we set the inequality:
$-1 \le 3x - 1 \le 1$
Adding $1$ to all parts of the inequality:
$-1 + 1 \le 3x - 1 + 1 \le 1 + 1$
$0 \le 3x \le 2$
Dividing by $3$ throughout:
$0 \le x \le \dfrac{2}{3}$
Thus, the domain of $f(x)$ is $[0, \dfrac{2}{3}]$.
Answer: $\left[0, \dfrac{2}{3}\right]$
Hence, the correct answer is option: 'D'.