Moscow Institute of Electronic Engineering

74

5. Find all values ​​of $x$ belonging to the interval $[-\pi; \pi]$ and satisfying relations

$$|\sin 2x| = \cos^4 \frac{x}{2} - \sin^4 \frac{x}{2}.$$

Physics

Oral exam objectives

1. A bullet flying at a certain speed penetrates into a solid barer to a depth of $l=10$ cm. How deep into the same barer will a bullet with twice the speed go?

2. A bus with a mass of $m=4$ t starts from rest and on a path $l=100$ m acquires a speed of $v=20$ m/s. Find the friction coefficient if the traction force is $F=10$ kN.

3. A ball flying at a speed of $v_1=20$ m/s is thrown back by a racket strike in the opposite direction with a speed of $v_2=15$ m/s. Find what the change in momentum of the ball is if the change in its kinetic energy is equal to $\Delta E=8,75$ J.

4. A car weighing $m=2000$ kg slows down and stops after $t=4$ s, it will travel a distance of $l=40$ m. What work is done by the traction force?

5. A cylinder with a volume of $V=0,2$ m³ contains helium under a pressure of $p_1=10^5$ Pa at a temperature of $t_1=17^\circ\text{C}$. After pumping helium, the pressure increased to $p_2=3\cdot10^5$ Pa, and the temperature became equal to $t_2=47^\circ\text{C}$. How much has the mass of helium increased? The universal gas constant is $R=8,31$ J/(mol K), the molar mass of helium is $=4\cdot10^{-3}$ kg/mol.

6. It took $\tau_1=15$ min to heat a certain amount of water from $0^\circ\text{C}$ to boiling temperature using an electric heater. After this, $\tau_2=1$ hours 20 minutes were required to convert all the water into steam under the same conditions. Using these data, determine the specific heat of vaporization of water. Specific heat of water $c=4200$ J/(kg K).

7. Two identical metal balls with charges $q_1=9,5\cdot10^{-9}$ Kl and $q_2=54,5\times10^{-9}$ Kl were in the air at some distance from each other. Then they were connected for a while and placed in kerosene in their original place. The force of interaction between the balls did not change. Find the dielectric permeability of kerosene.

8. A resistor with a resistance of $R=6$ Ohm is connected to a current source with internal resistance $r=0,4$ Ohm. How many times does the power released in the external circuit change if another identical resistor is connected in series with the first one?

9. The electric motor of a crane operates under a voltage of $U=380$ V and consumes a current of $I=20$ A. What is the efficiency of the motor if the crane lifts a load weighing $m=1$ t uniformly to a height of $h=19$ m in a time of $t=50$ s?

10. A ray of light falls from a certain liquid onto its interface with a vacuum. Angle of incidence $a=30^\circ$. The reflected and refracted rays redistributed each other. Find the refractive index of this liquid.

Moscow Institute of Electronic Engineering

Mathematics

Written exam
Option 1

1. Solve the equation

$$\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{x-a}} = \frac{1}{\sqrt{x^2-ax}}.$$

2. Solve the equation

$$\sin x + \cos x = \sqrt{1 + \tg x}.$$

3. In a regular quadrangular prism, the side of the base is $a$. A plane is drawn through the diagonal of the lower base and the vertex of the upper base, intersecting two mixed side faces recognized as straight lines, the angle between which is equal to $a$. Determine the volume of recognition.

4. Calculate the sum of the roots of the equation

$$x^2 + 2(a^2 - 3a)x - (6a^3 - 14a^2 + 4) = 0$$

and find the values ​​of $a$ at which it takes on the most value.

5. Solve the equation

$$\left(x - \lg \frac{x}{2}\right)^{\lg^2 2x - \lg \frac{x}{5}} - 1 = 1.$$

Option 2

1. Solve the equation

$$\frac{x + 2}{3x - a} + \frac{3 - x}{3x^2 + 2ax - a^2} = \frac{3x + 2}{x + a}.$$

2. Solve the equation

$$\sqrt{2\cos^2 x + 4\sin(x + 1)} = \sqrt{4\sin x \cos 1 - \sin 2x}.$$

3. Give a definition of the functions $y = \sin x$, explain its properties and the construction of its graph with justification.

4. For what values ​​of $a$ is the smallest value of the function $y = x^2 + (a - 2)x - a$ on the interval [1; 3] equals -4?

5. a) The side of the trigon is 9, the radius of the inscribed circle is 3, one of the angles is a right angle. Determine the area of ​​the trigon.

6. The side of the trigon is 9, the radius of the inscribed circle is 3. What is the smallest value that the area of ​​the trigon can take?

The publication was prepared by V. A. Tonin

75

Moscow Institute of Physics and Technology
Matematika
Option 1

1. $0 < x < 1$.
2. $\frac{\pi}{4} + \pi k, \frac{\pi}{6} + 2\pi k (k \in \mathbb{Z})$.
3. $\frac{7}{\sqrt{3}}$. Note. Let $AK = a, \angle KAO = \varphi$, point $K'$ be the orthogonal projection of point $K$ onto line $AC$ (Fig. 1). From the conditions of the problem it follows that $OK' = 1$. Then
   $$\begin{cases} a \cos \varphi = 3, \\ a \cos \left(\frac{\pi}{3} - \varphi\right) = 2, \end{cases}$$
   whence $tg \varphi = \frac{1}{3\sqrt{3}}$ and $a = 2\sqrt{\frac{7}{3}}$.
4. $(0; \frac{5}{2})$. Note. The set $M$ is defined by the system of inequalities
   $$|a| < 1, \quad (3a - 4b + 15)(7a - 24b + 35) < 0.$$
   This set is a graption with vertices at points $A\left(1; \frac{9}{2}\right), B\left(1; \frac{7}{4}\right), C\left(-1; \frac{7}{6}\right), D(-1; 3)$, and $AB + CD = AD + BC$. Let the ordinate of the center of the circle be $b$. Since the radius of the circle is equal to 1 and the straight line $AD$ intersects the ordinate axis at the point $\left(0; \frac{15}{4}\right)$, then $b$ is the equation $\left(\frac{15}{4} - b\right)\cos \left(\arctg \frac{3}{4}\right) = 1$.
5. $\frac{63}{4}\sqrt{\frac{3}{2}}$. Solution. In a pentagon $KK_1K_2K_3K_4$ (Fig. 2) $K_1K_2 \cos \angle KK_1M = \frac{7}{2} \cos (\arctg \sqrt{55}) = \frac{7}{4\sqrt{14}}$. Note that $KK_1 - K_2K_3 = \frac{7}{4\sqrt{14}}$ is therefore $(KK_3) \perp (KK_1)$. Let $AB = a, \angle CKK_1 = \varphi$, points $K'_2$ and $K'_3$ be orthogonal projections onto the plane $ABC$ of points $K_2$ and $K_3$ (Fig. 3). Applying the theorem of sines to the triangle $KCK_1, AK'_2K'_2, KBK'_3$ (taking into account that $\angle K_1KK'_3$ is a straight line), we obtain:
   $$\frac{\sqrt{14}}{2} \sin \left(\frac{\pi}{3} + \varphi\right) = \frac{a}{2} \sin \frac{\pi}{3},$$
   $$\frac{3}{8} \sqrt{14} \sin \left(\frac{\pi}{3} + \varphi\right) = AK'_2 \sin \frac{\pi}{3},$$
   $$\frac{a}{2} \cos \varphi = (a - AK'_2) \sin \left(\frac{\pi}{6} + \varphi\right).$$
   From (1) and (2) we obtain that $AK'_2 = \frac{3}{8} a$, taking this into account from (3) it follows that $\varphi = \arctg \frac{\sqrt{3}}{5}$. Next we find $a = 3\sqrt{2}, K_1K'_2 = \frac{7}{2\sqrt{2}}$. The height of the lines is $\frac{7}{2\sqrt{2}}$.
   Option 2
6. $(-1)^n \frac{\pi}{6} + \pi n, n \in \mathbb{Z}$.
7. $2\sqrt{30}$. Note. Let $AB = a$. Using the bisector property, we obtain $KC = \frac{7}{13} a$. Then $KD = \frac{2\sqrt{30}}{13} a, AK = \frac{17}{13} a$.
8. 4 hours.
9. $p \leq 2$. Note. The tangent drawn to the graph of the function at point $(x_0; y_0), x_0 \geq 0$ intersects the ordinate axes at point $(0; a)$, where $a = -3x_0^2 + \sqrt{3}x_0^2 + \frac{21p^2}{2^p + 3}$. The function $a(x_0)$ takes its greatest value at point $\sqrt{\frac{\sqrt{3}}{6}}$, it is equal to $\frac{1}{4} + \frac{21p^2}{2^p + 3}$.
10. $AB = 10\sqrt{13}, MN = 30$, the distance from point $N$ to plane $ABCD$ is $10\sqrt{\frac{14}{13}}$. Solution. Let $AB = a, AM = BK = b, \angle ABN = \varphi$, planes $ABCD$ and KLMN intersect at angle $\varphi$ (Fig. 4). We find $BN = \frac{a}{2} (\sin \varphi + \cos \varphi)$. Then from the condition that the point $N$ is equidistant from the point $B$ and $C$ we obtain

76

Dem

$$g_M = \frac{F}{m} \cos \alpha = \left( \frac{2\pi}{T} \right)^2 l \cos \alpha \approx 3,8 \text{ m/c}^2.$$

2. The state of 1 mole of an ideal gas on the right side of the vessel at the initial moment is described by the Mendeleev-Clapeyron equation:

$$p_1 V_1 = RT_1.$$

Here and henceforth, the index “1” will refer to the gas parameters on the right side of the vessel, and “2” - on the left. When we supply a small amount of heat $\Delta Q$ to the right side of the vessel, the parameters of the gas change, but will still satisfy the equation of state:

$$(p_1 + \Delta p_1)(V_1 + \Delta V_1) = R(T_1 + \Delta T_1).$$

Subtracting one equality from the other term by term, we get:

$$p_1 \Delta V_1 + V_1 \Delta p_1 + \Delta p_1 \Delta V_1 = R \Delta T_1.$$

As usual, assuming that in the process we are conducting all gas parameters change by small values ​​($\Delta T_1 \ll T_1$, $\Delta p_1 \ll p_1$, $\Delta V_1 \ll V_1$), we can neglect the small term $\Delta p_1 \Delta V_1$ compared to the others and write:

$$p_1 \Delta V_1 + V_1 \Delta p_1 = R \Delta T_1.$$

We can obtain a similar expression for 1 mole of gas on the left side of the vessel:

$$p_2 \Delta V_2 + V_2 \Delta p_2 = 0.$$

By definition of heat capacity

$$C = \frac{\Delta Q}{\Delta T}.$$

Using the first law of thermodynamics, for the gas on the right side of the container we can write:

$$\Delta Q = C \Delta T_1 = \Delta U_1 + p_1 \Delta V_1,$$

or

$$C \Delta T_1 = \frac{3}{2} R \Delta T_1 + p_1 \Delta V_1.$$

Since the pressures in both parts of the vessel remain equal all the time: $p_1 = p_2 = p$ and $\Delta p_1 = \Delta p_2 = \Delta p$, and the changes in volumes are related by the estimating relation $\Delta V_1 = -\Delta V_2$, from equalities (1) and (2) it is easy to obtain that

$$p_1 \Delta V_1 = \frac{R \Delta T_1}{1 + V_1/V_2},$$

and therefore

$$C = \frac{3}{2} R + \frac{R}{1 + V_1/V_2}.$$

At the moment when the piston divides the vessel in half, $V_1 = V_2$, from where

78

Interpreting taking into account the initial conditions, we obtain
$$I = I_0 - \frac{\mathcal{E}_2}{L} t = \pm 0,6 \mathcal{E}_1 \sqrt{\frac{C}{L}} - 0,8 \frac{\mathcal{E}_1}{L} t,$$
from where we find two solutions:

1. $-1,8 \mathcal{E}_1 \sqrt{\frac{C}{L}} = 0,6 \mathcal{E}_1 \sqrt{\frac{C}{L}} - 0,8 \frac{\mathcal{E}_1}{L} t_1 \Rightarrow t_1 = 3 \sqrt{LC};$
2. $-1,8 \mathcal{E}_1 \sqrt{\frac{C}{L}} = -0,6 \mathcal{E}_1 \sqrt{\frac{C}{L}} - 0,8 \frac{\mathcal{E}_1}{L} t_2 \Rightarrow t_2 = 3/2 \sqrt{LC}.$

4. The conditions of the problem are met by a system of two intrinsic lenses, each of which produces a real image (Fig. 10). Using the thin lens formula for lens $J_1$, we find $f_1 = \frac{F_1 d_1}{d_1 - F_1}$ and notice
   $$\Gamma_{11} = \frac{f_1}{d_1} = \frac{F_1}{d_1 - F_1}.$$
   For lens $J_2$ we get
   $$d_2 = L - f_1, f_2 = \frac{F_2 d_2}{d_2 - F_2}$$
   and notice
   $$\Gamma_{12} = \frac{F_2}{d_2 - F_2} = \frac{F_2}{L - \frac{d_1 F_1}{d_1 - F_1} - F_2}.$$
   The total notification of a two-lens system is
   $$\Gamma_1 = \Gamma_{11} \cdot \Gamma_{12} = \frac{F_1}{d_1 - F_1} \times \frac{F_2}{L - \frac{d_1 F_1}{d_1 - F_1} - F_2}.$$
   In our case, it is convenient to rewrite this expression in the form:
   $$\frac{1}{\Gamma_1} = \frac{1}{\Gamma_{11}} \cdot \frac{1}{\Gamma_{12}} = \frac{d_1(L - (F_1 + F_2)) + F_1 F_2}{F_1 F_2} - \frac{L}{F_2}.$$
   After rearranging the lenses, the system notification is given by a similar expression:
   $$\frac{1}{\Gamma_2} = \frac{1}{\Gamma_{21}} \cdot \frac{1}{\Gamma_{22}} = \frac{d_1(L - (F_2 + F_1)) + F_2 F_1}{F_2 F_1} - \frac{L}{F_1}.$$
   From here
   $$\frac{1}{\Gamma_1} - \frac{1}{\Gamma_2} = L \left( \frac{1}{F_1} - \frac{1}{F_2} \right).$$
   Since the optical power of the lens is $D = 1/F$, then
   $$D_2 - D_1 = \frac{1}{L} \left( \frac{1}{\Gamma_1} - \frac{1}{\Gamma_2} \right) = 3 \text{ дитр.}$$
   Moscow State Pedagogical Institute named after. V. I. Lenina
   Matematika
   Option 1
5. 25 km/h.
6. $x_1 = \frac{\pi}{4} (4k + 1), x_2 = \pi (2l + 1), x_3 = \frac{\pi}{4} (8m + 3), k, l, m \in \mathbb{Z}.$
7. $] - \infty; 0) \cup [^1/2; + \infty].$
8. $y_{\text{max}} = 79/25.$
9. $a^2(\sqrt{6} + \sqrt{15})/3.$
   Option 2
10. $y_{\text{max}} = 16; y_{\text{min}} = -4.$
11. 5/2, 6, 13/2.
12. $[-2; 1].$
13. $(-1; 0) \cup (4; + \infty).$
14. $\pm \pi/6; \pm \pi/2.$
    Physics
15. $l' = 4l = 40 \text{ см}.$
16. $\mu = \frac{F}{mg} - \frac{v^2}{2lg} = 0,05.$
17. $\Delta P = 2\Delta E/(v_1 - v_2) = 3,5 \text{ кг} \cdot \text{м}/\text{с}.$
18. $A = 2ml^2/t^2 = 4 \cdot 10^5 \text{ Дж}.$
19. $\Delta m = \frac{MV}{R} \left( \frac{p_2}{T_2} - \frac{p_1}{T_1} \right) = 0,06 \text{ кг}.$
20. $r = c(t_k - t_0)t_2/t_1 \approx 2,2 \cdot 10^6 \text{ Дж}/\text{кг}$ (here $t_k = 100^\circ \text{C}$ is the boiling point, $t_0 = 0^\circ \text{C}$ is the initial temperature of water).
21. $\varepsilon = (q_1 + q_2)^2/(4q_1q_2) \approx 2.$
22. Power will decrease by $n = (r + 2R)^2/(2(r + R)^2) \approx 1,9 \text{ раз}.$
23. $\eta = mgh/(UIt) \approx 50 \%.$
24. $n = \text{ctg} \alpha = 1,73.$

Moscow Institute of Electronic Engineering
Matematika
Option 1

1. $x = (a + 1)^2/4 \text{ при } |a| < 1$. For other a there are no roots.
2. $2\pi k; \frac{\pi}{4} + 2\pi k; -\frac{\pi}{4} + \pi k (k \in \mathbb{Z}).$
3. $V = a^3 \sqrt{2\cos a}/(2 \sin \frac{\alpha}{2}).$
4. $a = 2$. Note. The sum of the roots is $2(3a - a^2)$ subject to $D \geq 0.$
5. $[0, 5]; 5$. Note. $x - \lg \frac{x}{2} > 1$ for all $x > 0$, since the function $f(x) = x - \lg \frac{x}{2}$ has a minimum at the point $x_0 = \lg e$, and $f(x_0) = \lg \frac{2e}{lg e} > 1.$
   Option 2
6. $[-1; +(4a + 3)/8]$ at $a \neq -3, -9/4, -1/4, 1; [-9/8]$ at $a = -3; [-1]$ at $a = -9/4$ and $a = -1/4; [7/8]$ at $a = 1.$
7. $\frac{\pi}{2} + 2\pi k (k \in \mathbb{Z})$. Note. After squaring, we obtain after simplifications two equations: $\cos x = 0$ and $\cos x + \sin x = -2 \sin 1$; the second of them has no roots, since

80

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