The domain of the function f(x) = sqrt{2 - se | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the function $f(x) = \sqrt{2 - \sec^{-1}x}$ to be defined,the expression inside the square root must be non-negative:$2 - \sec^{-1}x \ge 0$$\s

Vedclass · Accepted Answer

$\left( -\infty, \sec 2 \right] \cup \left[ 1, \infty \right)$

Vedclass · Answer

(C) For the function $f(x) = \sqrt{2 - \sec^{-1}x}$ to be defined,the expression inside the square root must be non-negative:$2 - \sec^{-1}x \ge 0$$\sec^{-1}x \le 2$Also,the domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$.Since $\sec^{-1}x \le 2$,and the range of $\sec^{-1}x$ is $[0, \pi] \setminus \{\pi/2\}$,we consider the inequality $\sec^{-1}x \le 2$.For $x \ge 1$,$\sec^{-1}x \in [0, \pi/2)$,so $\sec^{-1}x \le 2$ is always true for all $x \in [1, \infty)$.For $x \le -1$,$\sec^{-1}x \in (\pi/2, \pi]$. We need $\sec^{-1}x \le 2$. Since $2 Taking the secant of both sides (noting that $\sec$ is increasing on $[0, \pi/2)$ and decreasing on $(\pi/2, \pi]$ is not directly applicable,but we know $\sec^{-1}x \le 2 \implies x \le \sec 2$ for $x \le -1$ because $\sec 2$ is negative).Thus,the domain is $(-\infty, \sec 2] \cup [1, \infty)$.

The domain of the function $f(x) = \sqrt{2 - \sec^{-1}x}$ is:

Similar Questions

The domain of the function $f(x) = \sqrt{\cos^{-1}\left(\frac{1-|x|}{2}\right)}$ is

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f(x) = \cos^{-1} \left( \frac{4x+2[x]}{3} \right)$ is $[\alpha, \beta]$, then $12(\alpha + \beta)$ is equal to:

The range of $\sin^{-1}\left(\frac{1+x^2}{2+x^2}\right)$ is:

Domain of function $f(x) = \sin^{-1}(5x)$ is

Let $[x]$ denote the greatest integer less than or equal to $x$. Then the domain of $f(x) = \sec^{-1}(2[x] + 1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module