1708\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 $\mathbf{x\in\mathbb{R}}$
1708
man111 singh
·2011-11-08 06:41:54
\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 $\mathbf{x=\left\{1\;,2\;,3\right\}}$
71These are from arihant I think. Yeah These are good.
1Thanks MAN111. These solutions are good. Even my teacher could not provide me solution to these questions.
Yeah Vivek,these are absolutely from Arihant.
1708
man111 singh
·2011-11-08 22:52:32
your most welcome Abhinav
$Abhinav (3)\; is $\mathbf{\sqrt{\frac{x-1}{x-2\left\{x\right\}}}}$
21the 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
- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -
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
7yeah, these sums are from arihant Differential calculus series by Amit M Agarwal.
