The domain of the function defined by f(x) = | vedclass

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

Question

(A) We have the function $f(x) = \cos^{-1} \sqrt{x-1}$.For the function $\cos^{-1}(u)$ to be defined,the argument $u$ must satisfy $u \in [-1, 1]$.Sin

Vedclass · Accepted Answer

$[1, 2]$

Vedclass · Answer

(A) We have the function $f(x) = \cos^{-1} \sqrt{x-1}$.For the function $\cos^{-1}(u)$ to be defined,the argument $u$ must satisfy $u \in [-1, 1]$.Since $\sqrt{x-1}$ is always non-negative,the condition becomes $0 \leq \sqrt{x-1} \leq 1$.Squaring the inequality,we get $0 \leq x-1 \leq 1$.Adding $1$ to all parts,we obtain $1 \leq x \leq 2$.Therefore,the domain of $f(x)$ is $[1, 2]$.

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

Similar Questions

The value of $\cos \left(\sin ^{-1} \frac{\pi}{3}+\cos ^{-1} \frac{\pi}{3}\right)$ is

The domain of $f(x) = \sin^{-1}\left[\log_{2}\left(\frac{x}{2}\right)\right]$ is

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

The domain of the function $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is $(-\infty, -a] \cup [a, \infty)$. Then $a$ is equal to

If $\cos ^{-1} x = y$,then . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module