B

1003082303

Level: 
B
Given the complex numbers \( a=6\sqrt2\left(\cos\frac{\pi}3+\mathrm{i}\cdot\sin\frac{\pi}3\right) \), \( b=3\sqrt2\left(\cos\frac56\pi+\mathrm{i}\cdot\sin\frac56\pi\right) \) and \( c=2\left(\cos240^{\circ}+\mathrm{i}\cdot\sin240^{\circ}\right) \), evaluate \( \frac a{b\cdot c} \).
\( \cos\frac{\pi}6+\mathrm{i}\cdot\sin\frac{\pi}6 \)
\( \cos\frac{11}6\pi+\mathrm{i}\cdot\sin\frac{11}6\pi \)
\( 4\left(\cos\frac{\pi}6\pi+\mathrm{i}\cdot\sin\frac{\pi}6\pi\right) \)
\( 4\left(\cos⁡\frac{11}6\pi+\mathrm{i}\cdot\sin\frac{11}6\pi\right) \)

1003082302

Level: 
B
Given the complex numbers \( a=10\left(\cos\frac43\pi+\mathrm{i}\cdot\sin\frac43\pi\right) \), \( b=7\left(\cos150^{\circ}+\mathrm{i}\cdot\sin150^{\circ}\right) \) and \( c=5\left(\cos⁡\frac74\pi+\mathrm{i}\cdot\sin\frac74\pi \right) \), evaluate \( \frac{a\cdot b}c \).
\( 14\left(\cos⁡\frac5{12}\pi+\mathrm{i}\cdot\sin\frac5{12}\pi\right) \)
\( 14\left(\cos\frac14\pi+\mathrm{i}\cdot\sin\frac14\pi\right) \)
\( 14\left(\cos\frac{23}{12}\pi+\mathrm{i}\cdot\sin\frac{23}{12}\pi\right) \)
\( 14\left(\cos\frac54\pi+\mathrm{i}\cdot\sin\frac54\pi\right) \)

1003082301

Level: 
B
Given the complex numbers \( a=\sqrt2\left(\cos⁡160^{\circ}+\mathrm{i}\cdot\sin⁡160^{\circ}\right) \), \( b=3\sqrt2\left(\cos⁡150^{\circ}+\mathrm{i}\cdot\sin150^{\circ}\right) \) and \( c=2\left(\cos240^{\circ}+\mathrm{i}\cdot\sin240^{\circ}\right) \), evaluate \( a\cdot b\cdot c \).
\(12\left(\cos190^{\circ}+\mathrm{i}\cdot\sin190^{\circ}\right) \)
\(12\left(\cos10^{\circ}+\mathrm{i}\cdot\sin10^{\circ}\right) \)
\(12\left(\cos⁡10^{\circ}-\mathrm{i}\cdot\sin10^{\circ}\right) \)
\(12\left(\cos⁡190^{\circ}-\mathrm{i}\cdot\sin190^{\circ}\right) \)

1003047409

Level: 
B
The sequence \( \left(\frac{2\cdot3^n+4^n+5}{4\cdot3^n-1}\right)_{n=1}^{\infty} \) is:
divergent and \( \lim\limits_{n\rightarrow\infty}\frac{2\cdot3^n+4^n+5}{4\cdot3^n-1}=\infty \)
convergent and \( \lim\limits_{n\rightarrow\infty}\frac{2\cdot3^n+4^n+5}{4\cdot3^n-1}=\frac12 \)
convergent and \( \lim\limits_{n\rightarrow\infty}\frac{2\cdot3^n+4^n+5}{4\cdot3^n-1}=\frac14 \)
convergent and \( \lim\limits_{n\rightarrow\infty}\frac{2\cdot3^n+4^n+5}{4\cdot3^n-1}=0 \)
divergent and it does not have an infinite limit

1003047408

Level: 
B
Choose the best first step to simplify and calculate the limit \(\lim\limits_{n\rightarrow\infty}\frac{3^n+4^{n-1}}{3^n+4^{n+1}} \).
We divide the numerator and the denominator by \( 4^n \).
We divide the numerator and denominator by \( 3^n \).
We substitute \(n=\infty \).
We take \( 3^n \) outside the brackets in the numerator and the denominator.
We take \( 4 \) outside the brackets in the numerator and the denominator.

1003047406

Level: 
B
Choose the correct formula to calculate the limit. \[ L=\lim\limits_{n\rightarrow\infty}\frac{3^{n+1}+4^n}{2^n} \]
\( L=\lim\limits_{n\rightarrow\infty}\left(3\cdot\left(\frac32\right)^n+2^n\right)=\infty \)
\( L=\lim\limits_{n\rightarrow\infty}\frac{2^n\left(3\cdot\left(\frac32\right)^n+2^n \right)}{2^n}=0 \)
\( L=\lim\limits_{n\rightarrow\infty}\frac{3^n \left(3+\left(\frac43\right)^n\right)}{2^n}=0 \)
\( L=\lim\limits_{n\rightarrow\infty}\frac{7^{n+1}}{2^n}=\infty \)
\( L=\frac{3^{\infty+1}+4^{\infty}}{2^{\infty}} =\frac72 \)

1003047405

Level: 
B
The sequence \( \left(\frac{3^n-4^{n-1}}{4^n}\right)_{n=1}^{\infty} \) is:
convergent and \(\lim\limits_{n\rightarrow\infty}⁡\frac{3^n-4^{n-1}}{4^n}=-\frac14 \)
convergent and \(\lim\limits_{n\rightarrow\infty}⁡\frac{3^n-4^{n-1}}{4^n}=\frac14 \)
convergent and \(\lim\limits_{n\rightarrow\infty}⁡\frac{3^n-4^{n-1}}{4^n}=-1 \)
convergent and \(\lim\limits_{n\rightarrow\infty}⁡\frac{3^n-4^{n-1}}{4^n}=0 \)
divergent