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The domain of the function f(x)=\sin^{-1}\left ( \frac{\left | x \right |+5}{x^{2}+1} \right )\; is \; \left ( -\infty ,-a]\cup [a,\infty \right ).
then a is equal to:
Option: 1 \frac{\sqrt{17}}{2}
 
Option: 2 \frac{\sqrt{17}-1}{2}
Option: 3 \frac{1+\sqrt{17}}{2}  
Option: 4 \frac{\sqrt{17}}{2}+1

Answers (1)

best_answer

\\f(x)=\sin \left(\frac{|x|+5}{x^{2}+1}\right) \\\text{For domain :} \\-1 \leq \frac{|x|+5}{x^{2}+1} \leq 1\\ Since\; |x|+5 \;\& \;x^{2}+1\;\text{ is always positive } \\So,\frac{|x|+5}{x^{2}+1} \geq 0 \forall x \in \mathbb{R}

So for domain:

\\\frac{|x|+5}{x^{2}+1} \leq 1 \\ \Rightarrow|x|+5 \leq x^{2}+1 \\ \Rightarrow 0 \leq x^{2}-|x|-4 \\ \Rightarrow 0 \leq\left(|x|-\frac{1+\sqrt{17}}{2}\right)\left(|x|-\frac{1-\sqrt{17}}{2}\right) \\ \Rightarrow|x| \geq \frac{1+\sqrt{17}}{2} \text { or }|x| \leq \frac{1-\sqrt{17}}{2} \text { (Rejected) } \\

\\\Rightarrow x \in\left(-\infty,-\frac{1+\sqrt{17}}{2}\right] \cup\left[\frac{1+\sqrt{17}}{2}, \infty\right) \\ \text { So, } a=\frac{1+\sqrt{17}}{2} \\

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himanshu.meshram

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