The domain of the function cos^{-1}left(frac{ | vedclass

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

Question

(D) For the function $\cos^{-1}(u)$ to be defined,we must have $-1 \leq u \leq 1$. Thus,$-1 \leq \frac{2 \sin^{-1}(\frac{1}{4x^2-1})}{\pi} \leq 1$. Mu

Vedclass · Accepted Answer

$(-\infty, -\frac{1}{\sqrt{2}}] \cup [\frac{1}{\sqrt{2}}, \infty) \cup \{0\}$

Vedclass · Answer

(D) For the function $\cos^{-1}(u)$ to be defined,we must have $-1 \leq u \leq 1$. Thus,$-1 \leq \frac{2 \sin^{-1}(\frac{1}{4x^2-1})}{\pi} \leq 1$. Multiplying by $\frac{\pi}{2}$,we get $-\frac{\pi}{2} \leq \sin^{-1}(\frac{1}{4x^2-1}) \leq \frac{\pi}{2}$. Since the range of $\sin^{-1}(y)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$,this inequality is always satisfied for all $y$ in the domain of $\sin^{-1}$,which is $[-1, 1]$. Therefore,we need $-1 \leq \frac{1}{4x^2-1} \leq 1$. Case $1$: $\frac{1}{4x^2-1} \leq 1 \implies \frac{1 - (4x^2-1)}{4x^2-1} \leq 0 \implies \frac{2-4x^2}{4x^2-1} \leq 0 \implies \frac{2(1-2x^2)}{(2x-1)(2x+1)} \geq 0$. This holds for $x \in (-\infty, -\frac{1}{\sqrt{2}}] \cup (-\frac{1}{2}, \frac{1}{2}) \cup [\frac{1}{\sqrt{2}}, \infty)$. Case $2$: $\frac{1}{4x^2-1} \geq -1 \implies \frac{1 + 4x^2 - 1}{4x^2-1} \geq 0 \implies \frac{4x^2}{4x^2-1} \geq 0$. This holds for $x \in (-\infty, -\frac{1}{2}) \cup (\frac{1}{2}, \infty) \cup \{0\}$. Taking the intersection of both cases,we get $x \in (-\infty, -\frac{1}{\sqrt{2}}] \cup [\frac{1}{\sqrt{2}}, \infty) \cup \{0\}$.

The domain of the function $\cos^{-1}\left(\frac{2 \sin^{-1}\left(\frac{1}{4x^2-1}\right)}{\pi}\right)$ is

Similar Questions

The domain of $\sin^{-1} \left[ \log_3 \left( \frac{x}{3} \right) \right]$ is

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

Domain of $\sin^{-1}$ is . . . . . . .

If the domain of the function $f(x) = \sin^{-1}\left(\frac{2}{x^2-2x-2}\right)$ is $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$,then $\alpha + \beta + \gamma + \delta$ is equal to

Let $f$ be a real valued function defined by $f(x) = \sin^{-1} \left( \frac{1 - |x|}{3} \right) + \cos^{-1} \left( \frac{|x| - 3}{5} \right)$. Then the domain of $f(x)$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module