C

9000064009

Level: 
C
Find the limit of the following sequence. \[ {\left({\Bigl (\frac{\root{n}\of{2}} {n} + \root{n}\of{2}\Bigr )}^{n}\right)}_{ n=1}^{\infty } \] Hint: The limit of the sequence \({\bigl ({\bigl (1 + \frac{1} {n}\bigr )}^{n}\bigr )}_{n=1}^{\infty }\) is the Euler number \(\mathrm{e}\).
\(2\mathrm{e}\)
\(\mathrm{e}^{2}\)
\(\mathrm{e} + 2\)
\(\infty \)

9000064010

Level: 
C
Find the limit of the following sequence. \[ {\left({\Bigl (\frac{2n + 1} {n} \Bigr )}^{n}\right)}_{ n=1}^{\infty } \] Hint: The limit of the sequence \({\bigl ({\bigl (1 + \frac{1} {n}\bigr )}^{n}\bigr )}_{n=1}^{\infty }\) is the Euler number \(\mathrm{e}\).
\(\infty \)
\(2\mathrm{e}\)
\(\mathrm{e}^{2}\)
\(\mathrm{e} + 2\)

9000063303

Level: 
C
Differentiate the following function. \[ f(x) = \sqrt{\sin x} \]
\(f'(x) = \frac{\cos x} {2\sqrt{\sin x}},\ x\in \mathop{\mathop{\bigcup }}\nolimits _{k\in \mathbb{Z}}\left (2k\pi ;\pi + 2k\pi \right )\)
\(f'(x) = \frac{\sin x} {2\sqrt{\cos x}},\ x\in \mathop{\mathop{\bigcup }}\nolimits _{k\in \mathbb{Z}}\left (2k\pi ; \frac{\pi } {2} + 2k\pi \right )\)
\(f'(x) = \frac{1} {2\sqrt{\sin x}},\ x\in \mathop{\mathop{\bigcup }}\nolimits _{k\in \mathbb{Z}}\left (2k\pi ;\pi + 2k\pi \right )\)
\(f'(x) = \frac{\cos x} {2\sqrt{\sin x}},\ x\in \mathop{\mathop{\bigcup }}\nolimits _{k\in \mathbb{Z}}\left [ 2k\pi ; \frac{\pi } {2} + 2k\pi \right ] \)

9000039110

Level: 
C
Assuming \(z\in \mathbb{C}\), solve the following equation. \[ \left (1 + \mathrm{i}\sqrt{3}\right )z = 1 -\mathrm{i}\sqrt{3} \]
\(z = -\frac{1} {2} -\frac{\sqrt{3}} {2} \mathrm{i}\)
\(z = \frac{\sqrt{3}} {2} + \frac{1} {2}\mathrm{i}\)
\(z = -\frac{1} {2} + \frac{\sqrt{3}} {2} \mathrm{i}\)
\(z = -\frac{\sqrt{3}} {2} + \frac{1} {2}\mathrm{i}\)

9000039108

Level: 
C
Assuming \(z\in \mathbb{C}\), solve the following equation. By \(\overline{z }\) the complex conjugate of \(z \) is denoted. \[ 2z -\mathrm{i}\, \overline{z} = 1 -\mathrm{i} \]
\(z = \frac{1} {3} -\frac{1} {3}\mathrm{i}\)
\(z = 1 + \mathrm{i}\)
\(z = -\frac{3} {5} + \frac{6} {5}\mathrm{i}\)
\(z = -\frac{1} {5} -\frac{3} {5}\mathrm{i}\)