If the domain of the function f(x)= ^-1(1x^2-2x-2), is (-, ] [ , ] [ ,), then + + + is equ…

Question

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

Accepted Answer

Domain of ${\sin }^{-1}	heta $ is $-1\le 	heta \le 1$

$f(x)={\sin }^{-1}(\frac{1}{{x}^{2}-2x-2})$ is defined if $-1\le \frac{1}{{x}^{2}-2x-2}\le 1$

s.t $\frac{1}{{x}^{2}-2x-2}\ge -1$ and $\frac{1}{{x}^{2}-2x-2}\le 1$

so these two inequality and final solution will be intersection of solution of these two inequality.

solve $\frac{1}{{x}^{2}-2x-2}\ge -1$

$\begin{gathered} \frac{1}{{x}^{2}-2x-2}+1\ge 0 \ \frac{1+{x}^{2}-2x-2}{{x}^{2}-2x-2}\ge 0 \ \frac{({x}^{2}-2x-1)}{{x}^{2}-2x-2}\ge 0 \end{gathered}$

roots of ${x}^{2}-2x-1$ is $x=\frac{2\pm \sqrt{4+4}}{2}=1\pm \sqrt{2}$

roots of ${x}^{2}-2x-2$ is $x=\frac{2\pm \sqrt{4+8}}{2}=1\pm \sqrt{3}$

$\begin{gathered} x\in (-\infty,1-\sqrt{3})\cup [1-\sqrt{2},1+\sqrt{2}]\cup (1+\sqrt{3},\infty)-{S}_{1} \ \frac{1}{{x}^{2}-2x-2}\le 1 \ \frac{1}{{x}^{2}-2x-2}-1\le 0 \ \frac{1-{x}^{2}+2x+2}{{x}^{2}-2x-2}\le 0 \ \frac{3-{x}^{2}+2x}{{x}^{2}-2x-2}\le 0 \ \frac{{x}^{2}-2x-3}{{x}^{2}-2x-2}\ge 0-1 \ {x}^{2}-2x-3x=\frac{2\pm \sqrt{4+12}}{2}=3,-1 \ {x}^{2}-2x-2x=1\pm \sqrt{3} \end{gathered}$

$x\in (-\infty,-1]\cup (1-\sqrt{3},1+\sqrt{3})\cup [3,\infty)-{S}_{2}$

taking intersection of ${S}_{1}\&{S}_{2}$ for final solution.

so domain of $f(x)$ is $x\in (-\infty,-1]\cup [1-\sqrt{2},1+\sqrt{2}]\cup [3,\infty)$

compare this with given domain $x\in (-\infty,\alpha ]\cup [\beta ,\gamma ]\cup [\delta ,\infty)$

$\begin{gathered} \alpha =-1,\beta =1-\sqrt{2},\gamma =1+\sqrt{2},\delta =3 \ \alpha +\beta +\gamma +\delta =-1+1-\sqrt{2}+1+\sqrt{2}+3=4 \end{gathered}$

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