what is the domain of the following functions

1. 1/sqrt(4x-|x^2 -10x +9|)

2. sin-1 (1/|x^2 -1|) + 1/sqrt(sin^2 x +sinx +1)

3. squt(sin-1 (3x-4x^3)) +squt(cos-1 (x))

cos-1 is cos inverse of sin-1 is sin inverse of ^represents to the power

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{y}=\frac{1}{\sqrt{4\mathrm{x}-\left|{\mathrm{x}}^2-10\mathrm{x}+9\right|}} \\ =\frac{1}{\sqrt{4\mathrm{x}-\left|{\mathrm{x}}^2-\mathrm{x}-9\mathrm{x}+9\right|}} \\ =\frac{1}{\sqrt{4\mathrm{x}-\left|\mathrm{x}\left(\mathrm{x}-1\right)-9\left(\mathrm{x}-1\right)\right|}} \\ =\frac{1}{\sqrt{4\mathrm{x}-\left|\left(\mathrm{x}-1\right)\left(\mathrm{x}-9\right)\right|}} \\ \mathrm{By} \mathrm{method} \mathrm{of} \mathrm{interval}: \\ +++++ 1 ---- 9 ++++ \\ \left(\mathrm{x}-1\right)\left(\mathrm{x}-9\right)>0 \mathrm{when} \mathrm{x}<1 \mathrm{or} \mathrm{x}>9 \\ \left(\mathrm{x}-1\right)\left(\mathrm{x}-9\right)<0 \mathrm{when} \mathrm{x}\in \left[1,9\right] \\ \Rightarrow \left|\left(\mathrm{x}-1\right)\left(\mathrm{x}-9\right)\right|=\left\{\begin{array}{ll}\left(\mathrm{x}-1\right)\left(\mathrm{x}-9\right)={\mathrm{x}}^2-10\mathrm{x}+9& \mathrm{x}<1 \mathrm{or} \mathrm{x}>9\\ -\left(\mathrm{x}-1\right)\left(\mathrm{x}-9\right)=-\left({\mathrm{x}}^2-10\mathrm{x}+9\right)& 1\le \mathrm{x}\le 9\end{array}\right. \\ \mathbf{Case}\mathbf{ }\left(\mathbf{i}\right)\mathbf{ }\mathbf{x}\mathbf{<}\mathbf{1}\mathbf{ }\mathbf{or}\mathbf{ }\mathbf{x}\mathbf{>}\mathbf{9} \\ \mathrm{y}=\frac{1}{\sqrt{4\mathrm{x}-\left({\mathrm{x}}^2-10\mathrm{x}+9\right)}} \\ =\frac{1}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2+10\mathrm{x}-9}} \\ =\frac{1}{\sqrt{-{\mathrm{x}}^2+14\mathrm{x}-9}} \\ \mathrm{For} \mathrm{root} \mathrm{to} \mathrm{be} \mathrm{defined} \\ -{\mathrm{x}}^2+14\mathrm{x}-9>0 \\ -\left({\mathrm{x}}^2-14\mathrm{x}+9\right)>0 \\ \Rightarrow {\mathrm{x}}^2-14\mathrm{x}+9<0 \\ \Rightarrow {\mathrm{x}}^2-14\mathrm{x}+49-49+9<0 \\ \Rightarrow {\left(\mathrm{x}-7\right)}^2-40<0 \\ \Rightarrow {\left(\mathrm{x}-7\right)}^2-{\left(2\sqrt{10}\right)}^2<0 \\ \Rightarrow \left(\mathrm{x}-7-2\sqrt{10}\right)\left(\mathrm{x}-7+2\sqrt{10}\right)<0 \\ \Rightarrow \left\{\mathrm{x}-\left(7+2\sqrt{10}\right)\right\}\left\{\mathrm{x}-\left(7-2\sqrt{10}\right)\right\}<0 \\ \Rightarrow \mathrm{x}\in \left(7-2\sqrt{10},7+2\sqrt{10}\right) \\ \mathrm{Taking} \mathrm{intersection} \left(\mathrm{common}\right) \mathrm{with} \mathrm{x}<1 \mathrm{or} \mathrm{x}>9, \mathrm{we} \mathrm{get}: \\ \mathrm{x}\in \left(7-2\sqrt{10},1\right)\cup \left(9,7+2\sqrt{10}\right)...\left(\mathrm{i}\right) \\ \mathbf{Case}\left(\mathbf{ii}\right)\mathbf{ }\mathbf{1}\mathbf{\le }\mathbf{x}\mathbf{\le }\mathbf{9} \\ \mathrm{y}=\frac{1}{\sqrt{4\mathrm{x}-\left\{-\left({\mathrm{x}}^2-10\mathrm{x}+9\right)\right\}}} \\ =\frac{1}{\sqrt{4\mathrm{x}+{\mathrm{x}}^2-10\mathrm{x}+9}} \\ =\frac{1}{\sqrt{{\mathrm{x}}^2-6\mathrm{x}+9}} \\ =\frac{1}{\sqrt{{\left(\mathrm{x}-3\right)}^2}} \\ \mathrm{Now} {\left(\mathrm{x}-3\right)}^2>0 \mathrm{for} \mathrm{all} \mathrm{values} \mathrm{of} \mathrm{x}, \mathrm{except} \mathrm{x}=3 \\ \Rightarrow \mathrm{x}\in \mathrm{R}-\left\{3\right\} \\ \mathrm{Taking} \mathrm{intersection} \mathrm{with} 1\le \mathrm{x}\le 9, \mathrm{we} \mathrm{get}: \\ \mathrm{x}\in \left[1,9\right]-\left\{3\right\}...\left(\mathrm{ii}\right) \\ \mathrm{Taking} \mathrm{union} \mathrm{of} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}: \\ \mathrm{x}\in \left(7-2\sqrt{10},7+2\sqrt{10}\right)-\left\{3\right\} \\ \mathrm{Please} \mathrm{ask} \mathrm{only} \mathrm{one} \mathrm{question} \mathrm{in} \mathrm{one} \mathrm{thread}. \end{gathered}$$

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