C

1003034402

Level: 
C
In a class of \( 31 \) pupils, \( 18 \) pupils learn Italian, \( 16 \) pupils learn Spanish and \( 12 \) pupils learn Portuguese. Now, \( 7 \) pupils learn Italian and Spanish, \( 9 \) pupils learn Italian and Portuguese and \( 5 \) pupils learn Spanish and Portuguese. Finally, \( 3 \) students do not learn any of these three languages. How many pupils learn all these languages?
\( 3 \)
\( 5 \)
\( 2 \)
\( 1 \)

1003034401

Level: 
C
There are \( 47 \) students in a class. Among these students \( 22 \) are members of a sports club, \( 33 \) are members of a film club and \( 20 \) students joined a literary club. Further, \( 17 \) students are members of the sports club and the film club, \( 13 \) joined the film club and the literary club, \( 6 \) attend the sports club and the literary club. Only one student is a member of all the clubs. How many students from the class attend neither the sports club nor the literary club?
\( 11 \)
\( 7 \)
\( 36 \)
\( 4 \)

1003047608

Level: 
C
Choose the step to take first to efficiently evaluate the limit of the sequence \( \left( \frac{3n+2}{\sqrt{n^2-1}} \right)_{n=1}^{\infty} \).
We divide the numerator and the denominator by \( n \).
We take \( \sqrt n \) outside brackets in the numerator and the denominator.
We square the denominator.
We divide the numerator by \( n \).
We divide the denominator by \( n \).

1003047606

Level: 
C
The sequence \( \left( \sqrt n \left( \sqrt n-\sqrt{n-1} \right) \right)_{n=1}^{\infty} \) is:
convergent and \( \lim\limits_{n\to\infty} \sqrt n \left( \sqrt n-\sqrt{n-1} \right) =\frac12 \)
convergent and \( \lim\limits_{n\to\infty} \sqrt n \left( \sqrt n-\sqrt{n-1} \right) =0 \)
convergent and \( \lim\limits_{n\to\infty} \sqrt n \left( \sqrt n-\sqrt{n-1} \right) =2 \)
divergent and \( \lim\limits_{n\to\infty} \sqrt n \left( \sqrt n-\sqrt{n-1} \right) =\infty \)
divergent and it does not have an infinite limit

1003047604

Level: 
C
Choose the correct computation of the limit. \[ L=\lim\limits_{n\to\infty} \left( \sqrt{n^2+3n}-2n \right) \]
\( L=\lim\limits_{n\to\infty}n\left( \sqrt{1+\frac3n}-2 \right) = -\infty \)
\( L= \infty-\infty=0 \)
\( L=\lim\limits_{n\to\infty}⁡(n-2n)=-\infty \)
\( L=\lim\limits_{n\to\infty} \left( n^2+3n-4n^2 \right) =-3 \)
\( L=\lim\limits_{n\to\infty}⁡\frac{n^2+3n-4n^2}{\sqrt{n^2+3n}+2n}=\infty \)
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