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

(1) ${1+\sqrt{17}\over2}$

(2) ${\sqrt{17}-1\over2}$

(3) ${\sqrt{17}\over2}$

(4) ${\sqrt{17}\over2}+1$
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Sets,relations and functions,
Introduction of functions

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Ans. (1) $1+\sqrt{17}\over2$

Sol. $\left|{|x|+5\over x^2+1}\right|\leq1$

$|x|+5\leq x^2+1$

$x^2-|x|-4\geq0$

Let $|x|=t\implies t^2-t-4\geq0$

$\left(|x|+{\sqrt{17}-1\over2}\right)\left(|x|-{\sqrt{17}+1\over2}\right)\geq0$

$x\in\left(-\infty,-{\sqrt{17}+1\over2}\right]\cup\left[{\sqrt{17}+1\over2},\infty\right)$

$a={1+\sqrt{17}\over2}$

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