C

9000063303

Část: 
C
Derivace funkce \(f\colon y = \sqrt{\sin x}\) je rovna:
\(f'(x) = \frac{\cos x} {2\sqrt{\sin x}},\ x\in \mathop{\mathop{\bigcup }}\nolimits _{k\in \mathbb{Z}}\left (2k\pi ;\pi + 2k\pi \right )\)
\(f'(x) = \frac{\sin x} {2\sqrt{\cos x}},\ x\in \mathop{\mathop{\bigcup }}\nolimits _{k\in \mathbb{Z}}\left (2k\pi ; \frac{\pi } {2} + 2k\pi \right )\)
\(f'(x) = \frac{1} {2\sqrt{\sin x}},\ x\in \mathop{\mathop{\bigcup }}\nolimits _{k\in \mathbb{Z}}\left (2k\pi ;\pi + 2k\pi \right )\)
\(f'(x) = \frac{\cos x} {2\sqrt{\sin x}},\ x\in \mathop{\mathop{\bigcup }}\nolimits _{k\in \mathbb{Z}}\left \langle 2k\pi ; \frac{\pi } {2} + 2k\pi \right \rangle \)

9000063305

Část: 
C
Derivace funkce \(f\colon y = \sqrt{\frac{x-1} {x+1}}\) je rovna:
\(f'(x) = \frac{1} {(x+1)^{2}} \sqrt{\frac{x+1} {x-1}},\ x\in (-\infty ;-1)\cup (1;\infty )\)
\(f'(x) = \frac{\sqrt{x-1}} {(x-1)^{2}\sqrt{x+1}},\ x\in (-\infty ;-1)\cup \langle 1;\infty )\)
\(f'(x) = \frac{x-1} {2\sqrt{(x+1)^{3}}} ,\ x\neq - 1\)
\(f'(x) = \frac{x-1} {\sqrt{(x+1)^{3}}} ,\ x\in (-\infty ;-1)\cup \langle 1;\infty )\)

9000039110

Část: 
C
Za předpokladu $z\in\mathbb{C}$, vyřešte danou rovnici. \[ \left (1 + \mathrm{i}\sqrt{3}\right )z = 1 -\mathrm{i}\sqrt{3}\]
\(z = -\frac{1} {2} -\frac{\sqrt{3}} {2} \mathrm{i}\)
\(z = \frac{\sqrt{3}} {2} + \frac{1} {2}\mathrm{i}\)
\(z = -\frac{1} {2} + \frac{\sqrt{3}} {2} \mathrm{i}\)
\(z = -\frac{\sqrt{3}} {2} + \frac{1} {2}\mathrm{i}\)

9000039108

Část: 
C
Rovnice \(2z -\mathrm{i}\, \overline{z} = 1 -\mathrm{i}\) má v \(\mathbb{C}\) řešení:
\(z = \frac{1} {3} -\frac{1} {3}\mathrm{i}\)
\(z = 1 + \mathrm{i}\)
\(z = -\frac{3} {5} + \frac{6} {5}\mathrm{i}\)
\(z = -\frac{1} {5} -\frac{3} {5}\mathrm{i}\)