The domain of the function f(x) = sin^{-1}lef | vedclass

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

Question

(C) $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is defined if $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$. Since $|x|+5 > 0$ and $x^2+1 > 0$,the left

Vedclass · Accepted Answer

$\frac{1 + \sqrt{17}}{2}$

Vedclass · Answer

(C) $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is defined if $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$. Since $|x|+5 > 0$ and $x^2+1 > 0$,the left inequality $\frac{|x|+5}{x^2+1} \geq -1$ is always true. We only need to solve $\frac{|x|+5}{x^2+1} \leq 1$. $|x|+5 \leq x^2+1$ $x^2 - |x| - 4 \geq 0$. Let $t = |x|$,where $t \geq 0$. Then $t^2 - t - 4 \geq 0$. The roots of $t^2 - t - 4 = 0$ are $t = \frac{1 \pm \sqrt{1 - 4(1)(-4)}}{2} = \frac{1 \pm \sqrt{17}}{2}$. Since $t \geq 0$,we have $t \geq \frac{1 + \sqrt{17}}{2}$. Thus,$|x| \geq \frac{1 + \sqrt{17}}{2}$,which implies $x \in \left(-\infty, -\frac{1 + \sqrt{17}}{2}\right] \cup \left[\frac{1 + \sqrt{17}}{2}, \infty\right)$. Comparing this with $(-\infty, -a] \cup [a, \infty)$,we get $a = \frac{1 + \sqrt{17}}{2}$.

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

Similar Questions

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

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

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

Domain of the function $f(x) = \frac{1}{\sqrt{\ln(\cot^{-1}x)}}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module