A

1103061201

Level: 
A
From the following list, choose parametric equations, that do not represent the straight line passing through the points \( A \) and \( B \) (see the picture).
$\begin{aligned} p\colon x&=2+4t, \\ y&=6+2t;\ t\in\mathbb{R} \end{aligned}$
$\begin{aligned} p\colon x&=2+2t, \\ y&=1+t;\ t\in\mathbb{R} \end{aligned}$
$\begin{aligned} p\colon x&=6+4t, \\ y&=3+2t;\ t\in\mathbb{R} \end{aligned}$
$\begin{aligned} p\colon x&=2-2t, \\ y&=1-t;\ t\in\mathbb{R} \end{aligned}$
$\begin{aligned} p\colon x&=4+4t, \\ y&=2+2t;\ t\in\mathbb{R} \end{aligned}$

1003047309

Level: 
A
The sequence \[ \left(\frac{3n^5+2n^3+1}{n^3+3}\right)_{n=1}^{\infty} \]
is divergent and \( \lim\limits_{n\to\infty}\frac{3n^5+2n^3+1}{n^3+3}=\infty \).
is convergent and \( \lim\limits_{n\to\infty}\frac{3n^5+2n^3+1}{n^3+3}=0 \).
is convergent and \( \lim\limits_{n\to\infty}\frac{3n^5+2n^3+1}{n^3+3}=3 \).
is divergent and \( \lim\limits_{n\to\infty}\frac{3n^5+2n^3+1}{n^3+3}=-\infty \).
has no limit.

1003047308

Level: 
A
Choose the best first step to take in order to evaluate the limit of the following sequence. \[ \left(\frac{3n^2-2n+4}{8n^2+13n+2}\right)_{n=1}^{\infty} \]
We divide the numerator and the denominator by \( n^2 \).
We divide the numerator and the denominator by \( n \).
We substitute \( n=\infty \).
We factor out \( n \) in the numerator and in the denominator separately.
We factor out \( 8 \) in the numerator and in the denominator separately.

1003047306

Level: 
A
Which of the following expressions describes the correct calculation of the limit? \[ L=\lim\limits_{n\to\infty}\frac{7n^4+6n^3-5n^2}{8n^5-7n^4+6} \]
\( L=\lim\limits_{n\to\infty}\frac{\frac7n+\frac6{n^2}-\frac5{n^3}}{8-\frac7n+\frac6{n^5}}=0 \)
\( L=\lim\limits_{n\to\infty}\frac{7+\frac6n-\frac5{n^2}}{8n-7+\frac6{n^4} }=-1 \)
\( L=\lim\limits_{n\to\infty}\frac{7+6-5n^2}{8n-7n+6}=-\infty \)
\( L=\lim\limits_{n\to\infty}\frac{7+\frac6n-\frac5{n^2}}{8-\frac7n+\frac6{n^5}}=\frac78 \)
\( L=\lim\limits_{n\to\infty}\frac{\frac7n+\frac6{n^2}-\frac5{n^3}}{8n-7+\frac6{n^4}}=0 \)

1003047305

Level: 
A
The sequence \[ \left(\frac{12n^3+5n+1}{2n^3-6}\right)_{n=1}^{\infty} \]
is convergent and \( \lim\limits_{n\to\infty}\frac{12n^3+5n+1}{2n^3-6}=6 \).
is convergent and \( \lim\limits_{n\to\infty}\frac{12n^3+5n+1}{2n^3-6}=0 \).
is convergent and \( \lim\limits_{n\to\infty}\frac{12n^3+5n+1}{2n^3-6}=12 \).
is divergent and \( \lim\limits_{n\to\infty}\frac{12n^3+5n+1}{2n^3-6}=\infty \).
has no limit.