The domain of f(x) = sin^{-1}left[log_{2}left | vedclass

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

Question

(C) The given function is $f(x) = \sin^{-1}\left[\log_{2}\left(\frac{x}{2}\right)\right]$.For the function $\sin^{-1}(u)$ to be defined,the argument $

Vedclass · Accepted Answer

$1 \leq x \leq 4$

Vedclass · Answer

(C) The given function is $f(x) = \sin^{-1}\left[\log_{2}\left(\frac{x}{2}\right)\right]$.For the function $\sin^{-1}(u)$ to be defined,the argument $u$ must satisfy $-1 \leq u \leq 1$.Therefore,we must have $-1 \leq \log_{2}\left(\frac{x}{2}\right) \leq 1$.Applying the definition of the logarithm,this inequality is equivalent to $2^{-1} \leq \frac{x}{2} \leq 2^{1}$.This simplifies to $\frac{1}{2} \leq \frac{x}{2} \leq 2$.Multiplying the entire inequality by $2$,we get $1 \leq x \leq 4$.Thus,the domain of the function is $x \in [1, 4]$.

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

Similar Questions

The range of the real valued function $f(x) = \operatorname{Cos}^{-1}\left(\frac{3}{\sqrt{9x^2 - 12x + 22}}\right)$ is

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

Range of the function $f(x) = \cot^{-1} \left( \log_{4/5} (5x^2 - 8x + 4) \right)$ is:

The domain of the derivative of the function $f(x) = \operatorname{Cos}^{-1}(2x - 5) - \operatorname{Sin}^{-1}(x - 2)$ is

The range of $\tan^{-1} x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module