C

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. \[ 2z -\mathrm{i}\, \overline{z} = 1 -\mathrm{i} \]
\(z = \frac{1} {3} -\frac{1} {3}\mathrm{i}\)
\(z = 1 + \mathrm{i}\)
\(z = -\frac{3} {5} + \frac{6} {5}\mathrm{i}\)
\(z = -\frac{1} {5} -\frac{3} {5}\mathrm{i}\)

9000046505

Level: 
C
Identify the optimal first step convenient to solve the following trigonometric equation. Do not consider the step which is possible but does not help to solve the equation. \[ \sin x = 1 +\cos x \]
\(\sin ^{2}x = 1 + 2\cos x +\cos ^{2}x\)
\(\sin ^{2}x = 1 +\cos ^{2}x\)
substitution \( 1 +\cos x = z\)
\(\sin x -\cos x = z\)

9000046507

Level: 
C
Identify the optimal first step convenient to solve the following trigonometric equation. Do not consider the step which is possible but does not help to solve the equation. \[ \sqrt{3}\cos x = 1 -\sin x \]
\(3\cos ^{2}x = (1 -\sin x)^{2}\)
\(3\cos ^{2}x = 1 -\sin ^{2}x\)
substitution \( 1 -\sin x = z\)
substitution \( \cos x = z\)

9000046508

Level: 
C
Identify the optimal first step convenient to solve the following trigonometric equation. Do not consider the step which is possible but does not help to solve the equation. \[ \sqrt{3}\sin x = 2 -\cos x \]
\(3\sin ^{2}x = 4 - 4\cos x +\cos ^{2}x\)
substitution \( 2 -\cos x = z\)
\(3\sin ^{2}x = 4 -\cos ^{2}x\)
\(3\sin ^{2}x = 1 - 2\cos x +\cos ^{2}x\)

9000038701

Level: 
C
The box is on the slope as in the picture. The angle of the slope is \(\alpha \). The forces acting on the box are the force of gravity \(\vec{F_{G}}\) and the friction \(\vec{F_{t}}\). The force of gravity can be replaced by two components \(\vec{F_{1}}\) and \(\vec{F_{n}}\). (The force \(\vec{F_{1}}\) is parallel to the slope and \(\vec{F_{n}}\) is perpendicular to the slope.) The friction \(F_{t}\) is given by the formula \(F_{t} = fF_{n}\), where \(f\) is the coefficient of the friction.What is the influence of the increasing angle \(\alpha \) on the forces acting on the box?
\(F_{1}\) becomes bigger and \(F_{t}\) becomes smaller
both \(F_{1}\) and \(F_{t}\) become smaller
\(F_{1}\) becomes bigger, \(F_{t}\) does not change
\(F_{1}\) becomes smaller, \(F_{t}\) does not change
both \(F_{1}\) and \(F_{t}\) become bigger
\(F_{1}\) becomes smaller and \(F_{t}\) becomes bigger

9000038702

Level: 
C
The box is on the slope as in the picture. The angle of the slope is \(\alpha \). The forces acting on the box are the force of gravity \(\vec{F_{G}}\) and the friction \(\vec{F_{t}}\). The force of gravity can be replaced by two components \(\vec{F_{1}}\) and \(\vec{F_{n}}\). (The force \(\vec{F_{1}}\) is parallel to the slope and \(\vec{F_{n}}\) is perpendicular to the slope.) Find \(F_{1}\).
\(F_{1} = F_{G}\sin \alpha \)
\(F_{1} = \frac{F_{G}} {\sin \alpha } \)
\(F_{1} = F_{G}\mathop{\mathrm{tg}}\nolimits \alpha \)
\(F_{1} = \frac{F_{G}} {\mathop{\mathrm{tg}}\nolimits \alpha } \)
\(F_{1} = F_{G}\cos \alpha \)
\(F_{1} = \frac{F_{G}} {\cos \alpha } \)

9000038703

Level: 
C
The box is on the slope as in the picture. The angle of the slope is \(\alpha \). The forces acting on the box are the force of gravity \(\vec{F_{G}}\) and the friction \(\vec{F_{t}}\). The force of gravity can be replaced by two components \(\vec{F_{1}}\) and \(\vec{F_{n}}\). (The force \(\vec{F_{1}}\) is parallel to the slope and \(\vec{F_{n}}\) is perpendicular to the slope.) Find \(F_{p}\).
\(F_{p} = F_{G}\cos \alpha \)
\(F_{p} = \frac{F_{G}} {\cos \alpha } \)
\(F_{p} = F_{G}\mathop{\mathrm{tg}}\nolimits \alpha \)
\(F_{p} = \frac{F_{G}} {\mathop{\mathrm{tg}}\nolimits \alpha } \)
\(F_{p} = F_{G}\sin \alpha \)
\(F_{p} = \frac{F_{G}} {\sin \alpha } \)