Calculus - Domain(good one)

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}}$$

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1708

man111 singh ·Nov 8 '11 at 6:41

$\hspace{-16}\mathbf{(2)::f(x)=\sqrt{\binom{x^2+4x}{2x^2+3}}}$\\\\ 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\}}$$

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Vivek @ Born this Way ·Nov 8 '11 at 7:02

These are from arihant I think. Yeah These are good.

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Abhinav Gupta ·Nov 8 '11 at 9:32

Thanks MAN111. These solutions are good. Even my teacher could not provide me solution to these questions.

Yeah Vivek,these are absolutely from Arihant.

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man111 singh ·Nov 8 '11 at 22:52

your most welcome Abhinav

$Abhinav (3)\; is $\mathbf{\sqrt{\frac{x-1}{x-2\left\{x\right\}}}}$

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Shubhodip ·Nov 9 '11 at 0:01

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

xâ‰ 2{x} , when 0<x<1

when 1â‰¤x<2, x= 1+ {x} = 2{x}, so not possible because {x}<1

when xâ‰¥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 xâ‰¥1 also x-1â‰¥0

so domain

x<-1,x>0

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Sigma ·Dec 15 '11 at 10:36

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

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fahadnasir nasir ·Dec 16 '11 at 2:46

IR-[-1,0]

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Domain(good one)

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