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

(A) For the function $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ to be defined,the argument must satisfy $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$.S

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For the function $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ to be defined,the argument must satisfy $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$.Since $|x|+5 > 0$ and $x^2+1 > 0$ for all $x \in \mathbb{R}$,the condition $\frac{|x|+5}{x^2+1} \geq -1$ is always satisfied.Thus,we only need to solve $\frac{|x|+5}{x^2+1} \leq 1$.Multiplying by $x^2+1$ (which is positive),we get $|x|+5 \leq x^2+1$.Rearranging the terms,we have $x^2 - |x| - 4 \geq 0$.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 $t = |x| \geq 0$,we must have $t \geq \frac{1+\sqrt{17}}{2}$.Therefore,$|x| \geq \frac{1+\sqrt{17}}{2}$,which implies $x \in \left(-\infty, -\frac{1+\sqrt{17}}{2}\right] \cup \left[\frac{1+\sqrt{17}}{2}, \infty\right)$.Comparing this with the given domain $(-\infty, -a] \cup [a, \infty)$,we find $a = \frac{1+\sqrt{17}}{2}$.

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

The domain of the function $f(x) = \sin^{-1}\left(\frac{3x^2+x-1}{(x-1)^2}\right) + \cos^{-1}\left(\frac{x-1}{x+1}\right)$ is:

The range of the real valued function $f(x) = \sin^{-1} ( \frac{1 + x^2}{2 x} ) + \cos^{-1} ( \frac{2 x}{1 + x^2} )$ 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

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

If $f(x) = \sqrt{2x - 1} + 5 \cos^{-1}\left(\frac{2x - 1}{3}\right)$,then the domain of the function $f(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module