Domain(good one)

\hspace{-16}\mathbf{(1)\;\;:: f(x)=\sqrt{x^{12}-x^9+x^4-x+1}}$\\\\ Now function $\mathbf{f(x)}$ is defined only when $\mathbf{x^{12}-x^9+x^4-x+1>0}$\\\\ Now Let $\mathbf{g(x)=x^{12}-x^9+x^4-x+1}$\\\\ \textbf{case{(a)::\;}}If $\mathbf{x\leq 0\;,}$ Then $\mathbf{g(x)>0}$\\\\ \textbf{case{(b)::\;}}If $\mathbf{0<x<1 \;,}$ Then $\mathbf{g(x)=x^{12}+x^4-x^9+1-x=x^{12}+x^4(1-x^5)+(1-x)>0}$\\\\ \textbf{case{(c)::\;}}If $\mathbf{x\geq 1 \;,}$ Then $\mathbf{g(x)=x^{12}-x^9+x^4-x+1=x^{9}(x^3-1)+x(x^3-1)+1>0}$\\\\ So $\mathbf{g(x)>0\forall x\in\mathbb{R}}$\\\\ So Domain is $\mathbf{x\in\mathbb{R}}$

\hspace{-16}\mathbf{(2)::f(x)=\sqrt{\binom{x^2+4x}{2x^2+3}}}$\\\\ Here $\mathbf{f(x)}$ is defined only when $\mathbf{\binom{x^2+4x}{2x^2+3}>0}$\\\\ *Use the face that $\mathbf{\binom{n}{r}=\frac{n!}{r!.(n-r)!}}$\\\\ Which is defined when $\mathbf{n,r\geq 0}$ and $\mathbf{n\geq r}$ and $\mathbf{n,r\in\mathbb{Z^{+}}}$\\\\ So $\mathbf{x^2+4x\geq 0}$ and $\mathbf{2x^2+3\geq 0}$ and $\mathbf{x^2+4x\geq 2x^2+3}$\\\\ $\mathbf{x(x+4)\geq 0}$ and $\mathbf{2x^2+3>0\in\mathbb{R}}$ and $\mathbf{x^2-4x+3\leq 0}$\\\\ $\mathbf{x\geq 0}$ and $\mathbf{x\in \left[1\;,3\right]}$\\\\ $\mathbf{x\in\left[1\;,3\right]}$\\\\ But $\mathbf{x^2+4x\;, 2x^2+3\in \mathbb{Z^{+}}}$\\\\ So $\mathbf{x=\left\{1\;,2\;,3\right\}}$

the only requirement is (x-1) and (x- 2{x}) must have the same sign , and x≠2{x} so x≠0

when x<0, of course 2{x}>0 so x-2{x}<0

we need x-1<0 i.e x<-1

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When x>0

x≠2{x} , when 0<x<1

when 1≤x<2, x= 1+ {x} = 2{x}, so not possible because {x}<1

when x≥2, then not possible becasue 2{x}<2

x-2{x}<0 when 0<x<1 also x-1<0

x-2{x}>0 when x≥1 also x-1≥0

so domain

x<-1,x>0

yeah, these sums are from arihant Differential calculus series by Amit M Agarwal.