\documentclass[11pt,a4paper]{article}
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{geometry}
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\usepackage[croatian, english]{babel}
\usepackage{esint, hvfloat, multicol}
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\def\du{{\mbox{\rm d$u$}}}
\def\dx{{\mbox{\rm d$x$}}}
\def\df{{\mbox{\rm d$f$}}}
\def\dr{{\mbox{\rm d$r$}}}
\def\dt{{\mbox{\rm d$t$}}}
\def\dy{{\mbox{\rm d$y$}}}
\def\dz{{\mbox{\rm d$z$}}}
\def\dfi{{\mbox{\rm d$\varphi$}}}
\def\dro{{\mbox{\rm d$\rho$}}}
\def\ds{{\mbox{\rm d$s$}}}
\def\dS{{\mbox{\rm d$S$}}}
\def\tg{{\mbox{\rm tg\,}}}

\def\B{{\mbox{$\mathcal B$}}}



\def\R{{\mbox{$\mathbb R$}}}


\def\arctg{{\mbox{\rm arc\,tg\,}}}
\def\arcsin{{\mbox{\rm arc\,sin\,}}}
\def\arccos{{\mbox{\rm arc\,cos\,}}}
\def\ctg{{\mbox{\rm ctg\,}}}
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  {\thmname{#1} \thmnumber{\Large #2.}\thmnote{\,(#3)}}       %Thm head spec  
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\theoremstyle{zaFiguru}
\newtheorem{zad}{}
\setlength{\textheight}{270mm}
\setlength{\textwidth}{183mm}



\begin{document}
\raggedright





\renewcommand{\thesection}{}
\section{Grupa A - Pismeni ispit iz Matematike, 13.02.2014.}
\setcounter{zad}{0}



\textbf{Pravila: Ispit pisati isključivo hemiskom olovkom, obratiti pažnju na matematičku pismenost}


\begin{multicols}{2}


\begin{zad}
Date su dvije baze $\B$ i $\B'$ vektorskog prostora $\R^3$. Vektor $v\in\R^3$ u odnosu na bazu $\B$ ima koordinate 
$\left[\begin{matrix} 
4\\ -1\\ 7
\end{matrix}\right]$
(gdje su $\mathcal{B}=\left\{
\left[\begin{matrix} 
1\\ 2\\ 6
\end{matrix}\right],
\left[\begin{matrix} 
0\\ 2\\ 6
\end{matrix}\right],
\left[\begin{matrix} 
0\\ 0\\ -3
\end{matrix}\right]
\right\}$ i $\mathcal{B}'=\left\{
\left[\begin{matrix} 
0\\ 0\\ 3
\end{matrix}\right],
\left[\begin{matrix} 
0\\ 2\\ 3
\end{matrix}\right],
\left[\begin{matrix} 
1\\ 2\\ 3
\end{matrix}\right]
\right\}$). 
Odrediti koordinate vektora $v$ u odnosu na bazu $\mathcal{B}'$.
\end{zad}



\begin{zad}
Ispitati funkciju i nacrtati njen grafik \hspace*{20mm}$y=\displaystyle{ \ln(2x-x^3) }$.
\end{zad}



\begin{zad}
Primjenom određenog integrala izračunati površinu figure koju ograničavaju linije $x+2y-5=0$, $2x+y-7=0$ i $y=x+1$.
\end{zad}






\begin{zad}
Rješiti diferencijalnu jednačinu $\displaystyle{\frac{\dy}{\dx} + \frac{y}{x} = y^2 }$.
\end{zad}







\end{multicols}














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\vspace*{1mm}


\renewcommand{\thesection}{}
\section{Grupa B - Pismeni ispit iz Matematike, 13.02.2014.}
\setcounter{zad}{0}


\textbf{Pravila: Ispit pisati isključivo hemiskom olovkom, obratiti pažnju na matematičku pismenost}




\begin{multicols}{2}

\begin{zad}
Date su dvije baze $\B$ i $\B'$ vektorskog prostora $\R^3$. Vektor $v\in\R^3$ u odnosu na bazu $\B$ ima koordinate 
$\left[\begin{matrix} 
5\\ -1\\ 3
\end{matrix}\right]$
(gdje su 
$\mathcal{B}=\left\{
\left[\begin{matrix} 
2\\ 1\\ 2
\end{matrix}\right],
\left[\begin{matrix} 
0\\ -1\\ 1
\end{matrix}\right],
\left[\begin{matrix} 
0\\ 0\\ 2
\end{matrix}\right]
\right\}$
i $\mathcal{B}'=\left\{
\left[\begin{matrix} 
1\\ 1\\ 1
\end{matrix}\right],
\left[\begin{matrix} 
1\\ 0\\ 0
\end{matrix}\right],
\left[\begin{matrix} 
0\\ 0\\ 1
\end{matrix}\right]
\right\}$).
Odrediti koordinate vektora $v$ u odnosu na bazu $\mathcal{B}'$.
\end{zad}




\begin{zad}
Ispitati funkciju i nacrtati njen grafik \hspace*{20mm}$y=\displaystyle{\frac{3x-1}{(x^2+1)^2}}$.
\end{zad}





\begin{zad}
Primjenom određenog integrala izračunati površinu figure koju ograničavaju linije $-2x-y+8=0$, $-x-2y+7=0$ i $y=x+2$.
\end{zad}



\begin{zad}
Rješiti diferencijalnu jednačinu $\displaystyle{\frac{\dy}{\dx} + \frac{1}{3}y = e^xy^4 }$.
\end{zad}





\end{multicols}




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%%\vspace*{1mm}


\renewcommand{\thesection}{}
\section{Grupa C - Pismeni ispit iz Matematike, 13.02.2014.}
\setcounter{zad}{0}



\textbf{Pravila: Ispit pisati isključivo hemiskom olovkom, obratiti pažnju na matematičku pismenost}




\begin{multicols}{2}


\begin{zad}
Date su dvije baze $\B$ i $\B'$ vektorskog prostora $\R^3$. Vektor $v\in\R^3$ u odnosu na bazu $\B$ ima koordinate 
$\left[\begin{matrix} 
7\\ -3\\ 5
\end{matrix}\right]$
(gdje su
$\mathcal{B}=\left\{
\left[\begin{matrix} 
1\\ 1\\ 0
\end{matrix}\right],
\left[\begin{matrix} 
0\\ 1\\ 1
\end{matrix}\right],
\left[\begin{matrix} 
1\\ 0\\ 1
\end{matrix}\right]
\right\}$
i $\mathcal{B}'=\left\{
\left[\begin{matrix} 
1\\ 0\\ 0
\end{matrix}\right],
\left[\begin{matrix} 
1\\ 1\\ 0
\end{matrix}\right],
\left[\begin{matrix} 
1\\ 1\\ 1
\end{matrix}\right]
\right\}$).
Odrediti koordinate vektora $v$ u odnosu na bazu $\mathcal{B}'$.
\end{zad}







\begin{zad}
Ispitati funkciju i nacrtati njen grafik \hspace*{20mm}$y=\displaystyle{\frac{e^x}{e^x+e^{-x}}}$.
\end{zad}




\begin{zad}
Primjenom određenog integrala izračunati površinu figure koju ograničavaju linije $y+2x+7=0$, $x+2y+5=0$ i $y=x-1$.
\end{zad}


\begin{zad}
Rješiti diferencijalnu jednačinu $\displaystyle{x\frac{\dy}{\dx} + y = xy^3 }$.
\end{zad}


\end{multicols}


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%%\vspace*{1mm}


\renewcommand{\thesection}{}
\section{Grupa D - Pismeni ispit iz Matematike, 13.02.2014.}
\setcounter{zad}{0}

\textbf{Pravila: Ispit pisati isključivo hemiskom olovkom, obratiti pažnju na matematičku pismenost}




\begin{multicols}{2}

\begin{zad}
Date su dvije baze $\B$ i $\B'$ vektorskog prostora $\R^3$. Vektor $v\in\R^3$ u odnosu na bazu $\B$ ima koordinate 
$\left[\begin{matrix} 
5\\ -1\\ 3
\end{matrix}\right]$
(gdje su 
$\mathcal{B}=\left\{
\left[\begin{matrix} 
2\\ 1\\ 2
\end{matrix}\right],
\left[\begin{matrix} 
0\\ -1\\ 1
\end{matrix}\right],
\left[\begin{matrix} 
0\\ 0\\ 2
\end{matrix}\right]
\right\}$
i $\mathcal{B}'=\left\{
\left[\begin{matrix} 
1\\ 1\\ 1
\end{matrix}\right],
\left[\begin{matrix} 
1\\ 0\\ 0
\end{matrix}\right],
\left[\begin{matrix} 
0\\ 0\\ 1
\end{matrix}\right]
\right\}$).
Odrediti koordinate vektora $v$ u odnosu na bazu $\mathcal{B}'$.
\end{zad}




\begin{zad}
Ispitati funkciju i nacrtati njen grafik \hspace*{20mm}$y=\displaystyle{\frac{3x-1}{(x^2+1)^2}}$.
\end{zad}





\begin{zad}
Primjenom određenog integrala izračunati površinu figure koju ograničavaju linije $-2x-y+8=0$, $-x-2y+7=0$ i $y=x+2$.
\end{zad}



\begin{zad}
Rješiti diferencijalnu jednačinu $\displaystyle{\frac{\dy}{\dx} + \frac{1}{3}y = e^xy^4 }$.
\end{zad}





\end{multicols}




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\end{document}