Written examination in mathematics at the humanities faculties of Moscow State University in 1970

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A written exam in mathematics is taken upon admission to the following humanities faculties of Moscow State University: psychological, economics and the department of structure and applied linguistics of the philological faculty *). The level of mathematics requirements for applicants entering these faculties can be judged from the version of the 1970 written exam given below, as well as from the 1969 version **).

At the Department of Structure and Applied Linguistics of the Faculty of Philology, 5 hours (300 minutes) were given for written work. To get an A, you had to solve the first three problems and any two of the last four. Practical grades were also given for slightly worse work, and in any case, those applicants who solved at least two problems, at least one of which was among the first three problems, received a positive mark.

The version of the written exam at the Department of Economic Cybertech of the Faculty of Economics is not given here; it is better to get acquainted with the level of requirements in mathematics at this department using the 1969 version.

At the Faculty of Psychology and at the Department of Political Economy of the Faculty of Economics, 4 hours (240 minutes) were allotted for written work, and to get an A, it was necessary to solve all 4 problems. Practical marks of five were given for slightly worse work, and a positive mark guaranteed the solution of any two problems. Below we will analyze the solutions to some problems in detail, and for some data only the answers. All the options below are real (not prefabricated).

Faculty of Philology

Option

1. In modern conditions, three kayaks. The first kayak will continue every 100 m 2 seconds faster than the second and 3 seconds faster than the third. How long does it take the first kayak to travel one kilometer if every 30 seconds the second kayak is 60/13m ahead of the third?

2. Find all x for which 0.5<x<2.5 and which satisfy $\log_3x-x^2$ (3a-ax) < 1 for all a from the interval 0<a<2.

*) At the Department of Economic Cybertech of the Faculty of Economics and at the Department of Structure and Applied Linguistics of the Faculty of Philology, an oral exam in mathematics is also taken.
**) They can be found in the article “Humanities students pass mathematics” («Kvant» No. 4, 1970).

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3. Find all $x$ that satisfy the condition $\frac{\pi}{2} < |3x - \frac{\pi}{2}| \leq \pi$ and are a solution to the equation $1 + \cos x + \cos 2x = \sin x + \sin 2x + \sin 3x$.

4. Prove that $11n^2 - 14n + 3 \geq 0$ for all integers $n$.

5. Do not use tables, prove that $tg 208^\circ < \sin 492^\circ$.

6. A triangle with sides 7, 24, 25 is inscribed in a circle. Calculate the area of ​​a circular segment subtended by a chord whose length is 7.

7. How many ways is there the amount of 4 rubles? 96 kopecks can it be made from coins of 2 and 15 kopecks?

Problem solutions

1. Answer: the first kayak covered 1 km in 230 seconds.

2. First of all, we prove that for $0,5 < x < 2,5$ the inequality $3x - x^2 > 1$ holds. Indeed, $3x - x^2 = \frac{9}{4} - \left( x - \frac{3}{2} \right)^2$, but under our restrictions $\left( x - \frac{3}{2} \right)^2 < 1$, therefore, $3x - x^2 > \frac{5}{4} > 1$. Since the base of the logarithm is greater than 1, the inequality can be rewritten as $3x - x^2 > 3a - ax$ or

$$x (3 - x) > a (3 - x).$$

(*)

We must find all $x$ from the interval $0,5 < x < 2,5$ that are solutions to inequality () for all $a$ from the interval $0 < a < 2$. For any $x$ from the interval $0,5 < x < 2,5$ the inequality $3 - x > 0$ holds, and therefore inequality () is equivalent to the inequality $x > a$.

For all $a$ from the interval $0 < a < 2$ the inequality $x > a$ is satisfied by all $x \geq 2$; taking into account the inequality $0,5 < x < 2,5$, we get the answer: $2 \leq x < 2,5$.

3. $1 + \cos x + \cos 2x = \sin x + \sin 2x + \sin 3x$,

$$\cos x (1 + 2 \cos x) = \sin 2x (1 + 2 \cos x), \quad \cos x (1 + 2 \cos x) (1 - 2 \sin x) = 0.$$

The last equation is equivalent to the original one. We will solve it and check the condition $\frac{\pi}{2} < |3x - \frac{\pi}{2}| \leq \pi$ for each series of answers.

I. $\cos x = 0$, $x = \frac{\pi}{2} + \pi k$. Hence $\left| 3x - \frac{\pi}{2} \right| = \left| 3 \frac{\pi}{2} + 3\pi k - \frac{\pi}{2} \right| = |(3k + 1)\pi|$. For $k = 0$ $\left| 3x - \frac{\pi}{2} \right| = \pi$, that is, our condition is $\left(\frac{\pi}{2} < \pi \leq \pi\right)$, and, therefore, $x = \frac{\pi}{2} + \pi \cdot 0 = \frac{\pi}{2}$ is the answer to the problem. If $k \neq 0$, then, as is easy to see, $|(3k + 1)\pi| \geq 2\pi$ and from the series of roots of the form $\frac{\pi}{2} + k\pi$ only $x = \frac{\pi}{2}$ satisfies our condition.

II. $1 + 2\cos x = 0$, $\cos x = -\frac{1}{2}$, $x = \pm \frac{2\pi}{3} + 2k\pi$. Hence $\left| 3x - \frac{\pi}{2} \right| = \left| \pm 2\pi + 6k\pi - \frac{\pi}{2} \right| = |(3k \pm 1)2\pi - \frac{\pi}{2}|$.

For any $k$ the last expression is too large:

a) $k = 0$, $\left| (3k \pm 1)2\pi - \frac{\pi}{2} \right| = \left| \pm 2\pi - \frac{\pi}{2} \right| \geq \frac{3\pi}{2}$;

b) $k \neq 0$, $\left| (3k \pm 1)2\pi - \frac{\pi}{2} \right| \geq |(3k \pm 1)2\pi| - \left| \frac{\pi}{2} \right| \geq 4\pi - \frac{\pi}{2} = \frac{7}{2}\pi$.

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Thus, no value $x$ from the series $x = \pm \frac{2\pi}{3} + 2k\pi$ satisfies the conditions.

III. 1 - 2 sin $x = 0$; a) $x = \frac{\pi}{6} + 2k\pi$; b) $x = \frac{5\pi}{6} + 2k\pi$.

a) $|3x - \frac{\pi}{2}| = |\frac{\pi}{2} + 6k\pi - \frac{\pi}{2}| = |6k\pi|$. If $k = 0$, then $|3x - \frac{\pi}{2}| = 0 < \frac{\pi}{2}$; if $k \neq 0$, then $|3x - \frac{\pi}{2}| \geq |6\pi| > \pi$.

b) $|3x - \frac{\pi}{2}| = |\frac{5}{2}\pi + 6k\pi - \frac{\pi}{2}| = |2\pi + 6k\pi| \geq 2\pi$.

Thus, none of the solutions to the equation $\sin x = \frac{1}{2}$ satisfies the conditions. Answer: $x = \frac{\pi}{2}$.

4. Let us find for which $x$ the trinomial $11x^2 - 14x + 3$ is less than zero. This trinomial has roots $x_1 = \frac{3}{11}, x_2 = 1$. Thus, $11x^2 - 14x + 3 < 0$ at $\frac{3}{11} < x < 1$. But in the last interval there is not a single integer, which means that for all integers $n$ $11n^2 - 14n + 3 \geq 0$.

5. Obviously $tg 208^\circ = tg 28^\circ < tg 30^\circ = \frac{1}{\sqrt{3}}$; $\sin 492^\circ = \sin 48^\circ > \sin 45^\circ = \frac{1}{\sqrt{2}}$. But $\frac{1}{\sqrt{3}} < \frac{1}{\sqrt{2}}$ means $tg 208^\circ < \frac{1}{\sqrt{3}} < \frac{1}{\sqrt{2}} < \sin 492^\circ$, which is what needed to be proven.

6. U Kazan and e: by the theorem converse to the Pythagorean theorem, our triangle is right-angled $(7^2 + 24^2 = 25^2)$. Answer: $\frac{625}{4} \arcsin \frac{7}{25} - 42$.

7. Note that each method is completely determined by the number of 15-kopeck coins. Let’s see what this number could be. Since the total amount (496 kopecks) is even and any amount made up of 2-kopeck coins is also even, then the number of 15-kopeck coins must be even. In addition, this number is non-negative and not too large, since in any method the amount made up of 15-kopeck coins is non-negative and does not exceed 496, that is, $0 \leq 2n \cdot 15 \leq 496$. It is easy to see that there are no other numbers from 15-kopeck coins: if we make up any amount from an even number of 15-kopeck coins that does not exceed 496 kopecks, then an insufficient amount (it is even) can always be made from 2-kopeck coins. It remains to solve the inequality $0 \leq 2n \cdot 15 \leq 496$, or $0 \leq n \leq 16 \frac{16}{30}$ in integers. It is clear that this inequality has 17 integer solutions: $n = 0, 1, 2, \dots, 16$.

Answer: 17 ways.

Problems 2 and 3 caused minor difficulties in this version, most likely due to the unusual formulation. Judging by the work, some applicants did not fully summarize the conditions of these tasks. So, for example, in problem 3, some applicants simply solved (and not always correctly) a trigonometric equation, without taking into account the additional condition on $x$. Other applicants, having received an infinite number of solutions to a trigonometric equation and the concept that it is necessary to check the fulfillment of an additional condition, were unable to complete the check. Having “opened the model” in additional conditions, many could not be fully verified.

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Faculty of Psychology (full-time department)

Option 1

1. Two identical steamships depart from two piers: the first steamer from pier $A$ downstream, the second from pier $B$ upstream. Each ship, one to the final destination, stays there for 45 minutes and returns back. If steamships depart from their initial points at the same time, then on the way back they meet at point $K$, which is 2 times closer to $A$ than $B$. If a steamer departs from $A$ 1 hour later than the second steamer departs from $B$, then on the way back they meet 20 km from $A$. If a steamship departs from $A$ 30 minutes earlier than the second one departs from $B$, then on the way back the steamships meet 5 km above $K$. Find the speed of the river and the time it takes for the second steamer to reach from $A$ to $K$.

2. Find all real values ​​of $a$ for which the equation

$$x^8 + ax^4 + 1 = 0$$

has rfno four real roots forming an arithmetic progression

3. Find all $x$ satisfying the inequality

$$\log_{\cos x^2} \left( \frac{3}{x} - 2x \right) < \log_{\cos x^2} (2x - 1).$$

4. 3 balls are placed in a regular triangular paramid such that the 1st ball touches all the side faces of the paramid and the 2nd ball, and the 2nd ball touches the side faces and the 3rd ball, and the 3rd ball touches the side faces, the base of the paramid and the 2nd ball. What fraction of the volume of the paramid is occupied by three balls if its lateral faces are inclined to the base at an angle of $\alpha^2$?

Problem solutions

1. According to the conditions of the problem, the steamships are the same, which means that the speeds of the steamships in still water are the same. Let us denote this speed by $v$ ($\kappa m/\mu$). Let us denote by $v_0$ ($\kappa m/\mu$) the speed of the river, and by $S$ ($\kappa m$) the distance between $A$ and $B$. Below we will measure time in hours and will omit the dimensions of all quantities. Comparing the times that the ships spent on the journey from the moment of departure to the moment of meeting on the way back (including parking), we compose a system of three equations with unknowns:

$$\begin{align*} \frac{S}{v + v_0} + \frac{\frac{2S}{3}}{v - v_0} &= \frac{S}{v - v_0} + \frac{\frac{S}{3}}{v + v_0}, \\ \frac{S}{v + v_0} + \frac{S - 20}{v - v_0} + 1 &= \frac{S}{v - v_0} + \frac{20}{v + v_0}, \\ \frac{S}{v + v_0} + \frac{\frac{2S}{3} + 5}{v - v_0} &= \frac{S}{v - v_0} + \frac{\frac{S}{3} - 5}{v + v_0} + \frac{1}{2}. \end{align*}$$

From the first equation we find that $v = 3v_0$, and from the first and third we find: $v_0 = 7,5$; $v = 22,5$. Substituting these values ​​$v$ and $v_0$ into the second equation, we find $S = 30$, hence $AK = 10$. This means that the second steamer reaches from $A$ to $K$ in $\frac{10}{30} = \frac{1}{3}$ hours. Answer: The speed of the river is 7.5 $\kappa m/\mu$, and the second boat reaches from $A$ to $K$ in 20 minutes.

2. It is not difficult to prove that the original equation is quadratic. An equation of this type is discussed in a school textbook. Let’s put $x^4 = t$. Then the equation will be rewritten as $t^2 + at + 1 = 0$. Let $t_1$ and $t_2$ be the real roots of this equation. It is easy to see that $t_1$ and $t_2$ must be positive and distinct. For the sake of definition, assume that $0 < t_1 < t_2$. Then the original equation has the following roots (ordered in ascending order): $x_1 = -\sqrt[4]{t_2}$, $x_2 = -\sqrt[4]{t_1}$, $x_3 = \sqrt[4]{t_1}$, $x_4 = \sqrt[4]{t_2}$.

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For these four roots to form an arthmetic progression, it is necessary and sufficient that $x_2-x_1=x_3-x_2=x_4-x_3$ or $\sqrt[4]{t_2}-\sqrt[4]{t_1}=2\sqrt[4]{t_1}=\sqrt[4]{t_2}-\sqrt[4]{t_1}$, that is, $2\sqrt[4]{t_1}=\sqrt[4]{t_2}-\sqrt[4]{t_1}, 3\sqrt[4]{t_1}=\sqrt[4]{t_2}, 81t_1=t_2$.

Now let’s use Vieta’s theorem. If $81t_1=t_2$, then $81t_1^2=t_1\cdot t_2$.

But by Vieta’s theorem $t_1t_2=1$, therefore $81t_1^2=1, t_1=\pm\frac{1}{9}, t_2=\pm9$. But we require that $t_1$ and $t_2$ be positive, so $t_1=\frac{1}{9}, t_2=9$. Thus, the found values ​​of $t_1$ and $t_2$ are the only possible ones. Let’s see if such roots can be obtained for some values ​​of $a$. Again, by Vieta’s theorem, $a=-(t_1+t_2)=-9\frac{1}{9}$ is found. But there is a theorem converse to Vieta’s theorem that if $x_1x_2=q$ and $x_1+x_2=-p$, then the numbers $x_1$ and $x_2$ are the roots of the equation $x^2+px+q=0$. This means that the numbers $\frac{1}{9}$ and $9$ are indeed the roots of the equation $t^2-\left(\frac{1}{9}+9\right)t+\frac{1}{9}\cdot 9=0$.

Answer: $a=-9\frac{1}{9}$.

Most of the applicants did not use Vieta’s theorem, but wrote down the roots in the form

$$t_1=\frac{-a-\sqrt{a^2-4}}{2}, \quad t_2=\frac{-a+\sqrt{a^2-4}}{2},$$

which is where $a$ is located. The solution turned out to be more complicated and longer.

3. It is easy to see that the ODZ is given by three inequalities:

1. $\frac{3}{x}-2x>0$, 2) $2x-1>0$, 3) $0<\cos x^2<1$.

For such $x$ the original inequality is equivalent to the inequality

$$\frac{3}{x}-2x>2x-1.$$

It remains to solve the system of four inequalities:

1. $\frac{3}{x}-2x>0$, 2) $2x-1>0$, 3) $0<\cos x^2<1$, 4) $\frac{3}{x}-2x>2x-1$.

From the second inequality $2x-1>0$, $x>\frac{1}{2}$. Now we can multiply the first inequality by $x$: $3-2x^2>0,x^2<\frac{3}{2}$. Since $\pi>3$ and $x^2<\frac{3}{2}<\frac{\pi}{2}$, then $0<x^2<\frac{\pi}{2}$ and the inequality $0<\cos x^2<1$ holds automatically. The last inequality can also be multiplied by $x$: $3-4x^2+x>0$, hence $-\frac{3}{4}<x<1$. Matching three inequalities: $x>\frac{1}{2}$, $x^2<\frac{3}{2}$ and $-\frac{3}{4}<x<1$, we find the answer: $\frac{1}{2}<x<1$.

4. Answer: $\frac{4\pi\sqrt{3}}{9\mathrm{tg}\alpha}\left(\mathrm{tg}^3\frac{\alpha}{2}+\mathrm{tg}^9\frac{\alpha}{2}+\mathrm{tg}^{15}\frac{\alpha}{2}\right)$.

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Faculty of Economics (full-time)

Option

1. The volume of a block in the shape of a rectangular paralleliped is 150 cubic meters. cm, the total surface area is 280 cubic meters. cm, the perimeter of the base is 40 cm. Find the dimensions of the block.

2. In the square $ABCD$ of area $J$ the side $AD$ is taken as the point $D$ and on the continuation the point $O$ is taken at a distance of 3 from the point $D$. Two rays are drawn from point $O$. The first ray intersects the segment $CD$ at the point $M$ and the segment $AB$ at the point $N$, and the length of the segment $ON$ is equal to $a$. The second ray intersects the segment $CD$ at point $L$ and the segment $BC$ at point $K$, and at $\Rightarrow BKL = \alpha$. Find the area of ​​the polygon $BKLMN$.

3. Solve equation $\log_2\log_2 x = \log_4\log_4 2x$.

4. Solve equation $|z| \cdot i - z = 1 + 2i$.

Problem solutions

1. Let us denote the sides of the base of the bar by $x$ ($cm$) and $y$ ($cm$), and the height of the bar by $z$ ($cm$). Let for definition $x \geq y$. Then

$$\begin{cases} xyz = 150, \\ 2(xy + yz + xz) = 280, \\ 2(x + y) = 40, \\ x \geq y. \end{cases}$$

(*)

From the first and third equations $xy = \frac{150}{z}$ and $x + y = 20$. Substituting these expressions into the second equation, we obtain the equation for $z$:

$$\frac{15}{z} + 2z = 14,$$

whence $z_{1,2} = \frac{7 \pm \sqrt{19}}{2}$. Now we know the sum $x + y$ and the product $xy = \frac{150}{z}$ of the numbers $x$ and $y$, which means that $x$ and $y$ are the roots of the quadratic equation $t^2 - 20t + \frac{300}{7 \pm \sqrt{19}} = 0$. Solving it, we find

$$t_{1,2} = 10 \pm \sqrt{100 - \frac{300}{7 \pm \sqrt{19}}}.$$

For $z = \frac{7 - \sqrt{19}}{2}$ the discriminant of this equation is negative: 100 – $\frac{300}{7 - \sqrt{19}} < 100 - \frac{300}{7 - 4} = 0$, and for $z = \frac{7 + \sqrt{19}}{2}$ the discriminant is positive and less than 100. In this case, the values ​​$z$ $t_1 > 0$, $t_2 > 0$, that is, $x > 0$, $y > 0$. It is easy to see that the numbers $z = \frac{7 + \sqrt{19}}{2}$, $x = 10 + \sqrt{100 - \frac{300}{7 + \sqrt{19}}}$, $y = 10 - \sqrt{100 - \frac{300}{7 + \sqrt{19}}}$ are indeed a solution to system (*). Answer: the sides of the base are $10 \pm \sqrt{100 - \frac{300}{7 + \sqrt{19}}}$, the height is $\frac{7 + \sqrt{19}}{2}$.

Almost none of the applicants had any difficulty in composing the correct system of equations for this problem. However, the solution to the resulting system turned out to be beyond our power. Some applicants, having found two values ​​of $z$, proceeded to find the corresponding value $x$ and $y$, without caring that in one of the answers they provided, $x$ and $y$ are not real numbers.

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2. Answer: $S_{BKLMN} = 1 + \frac{(1 + 3\log a)^2}{2\log a} - \frac{7}{8} \sqrt{a^2 - 16}$.

3. $\log_2 \log_2 x = \log_4 \log_4 2x, \log_4 (\log_2 x)^2 = \log_4 \log_4 2x, (\log_2 x)^2 = \log_4 2 + \frac{1}{2} \log_2 x,$

$$(\log_2 x - 1) \left( \log_2 x + \frac{1}{2} \right) = 0.$$

(*)

Equation (), as is easy to see, is a consequence of the original one, that is, every root of the original equation satisfies (). Find the roots (*) and check them by substituting them into the original equation, we get the answer: $x = 2$.

4. Let us write the complex number $z$ in the form $z = x + iy$, where $x, y$ are real numbers. By definition, $|z| = \sqrt{x^2 + y^2}$, in particular, $|z|$, is a real number. Let’s rewrite the equation in the form $|z| \cdot i = (x + iy) = 1 + 2i$. Complex numbers are equal if and only if they have equal real and minimum parts. Therefore, the equation being solved is equivalent to the system

$$\begin{cases} -x = 1, \\ |z| - y = 2. \end{cases}$$

From the first equation $x = -1$, therefore $|z| = \sqrt{x^2 + y^2} = 1 \sqrt{1 + y^2}$, and to determine $y$ we obtain the equation $\frac{1}{1 + y^2} = y + 2$. By squaring both sides and checking, we find $y = -\frac{3}{4}$, that is, $z = -1 - \frac{3}{4} i$.

Although solving Problem 4 does not require knowledge beyond the scope of the “Program of entrance examinations in mathematics for applicants to higher educational institutions of the USSR,” the applicants coped with this task very poorly. The majority of applicants found independent definitions of the complex number model and the equality of complex numbers.

Faculty of Psychology (evening department)

Option

1. Find all real values ​​of $a$ for which the roots $x_1$ and $x_2$ of equation $x^2 + ax + a = 0$ are real and satisfy the corresponding $x_1 < x_2, x_1^2 x_2 = a^2$.

2. Solve inequality

$$\log \frac{x-1}{x} \log \frac{x+1}{x} \frac{x}{3} \leq \log \frac{x-1}{x} \log \frac{x+1}{x} \left( \frac{x}{2} - 1 \right).$$

3. The gardener, when the work is done, the first day of work is like this: half a day he dug a hole, the next half day he planted a bush; earned 3 rubles. The next day he spent half a day planting a bush and half a day mowing the grass; I earned 2.5 rubles that day. On the third day he spent half a day digging a hole and half a day mowing the grass; earned 3.5 rubles. How much money will he earn if he mows grass all day?

4. A ball is inscribed in a body formed by the rotation of a right triangle around the hypotenuse. Find the ratio of its surface area to the surface area of ​​the body, knowing that one of the acute angles of the triangle is equal to $a$.

Problem solutions

1. By Vieta’s theorem $x_1 x_2 = a$. Therefore, if $x_1 (x_1 x_2) = a^2$, then $x_1 a = a^2$, or $a (x_1 - a) = 0$. It is clear that $a \neq 0$, since at $a = 0$ $x_1 = x_2 = 0$ and condition $x_1 < x_2$ is not satisfied. So, $x_1 - a = 0$, $x_1 = a$. But

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$x_1 x_2 = a \neq 0$, therefore $x_2 = 1$. Again according to Vieta’s theorem

$$x_1 + x_2 = -a,$$

poseable $1 + a = -a$, from where $a = -\frac{1}{2}$. But the solution to the problem is not over.

Examination. For $a = -\frac{1}{2}$ the roots of our equation are 1 and $-\frac{1}{2}$. Let’s take $x_1 = -\frac{1}{2}$ and $x_2 = 1$. Then $x_1 < x_2$ and $(-\frac{1}{2})^2 \cdot 1 = (-\frac{1}{2})^2$, that is, $x_1^2 x_2 = a^2$ and the value $a = -\frac{1}{2}$ satisfy all the conditions of the problem. Answer: $a = -\frac{1}{2}$.

2. For both sides of the inequality to make sense, it is necessary and sufficient to satisfy the following inequalities:

1. $\frac{x}{3} > 0$; 2) $\frac{x}{2} - 1 > 0$;
2. $\frac{x+1}{x} > 0$, $\frac{x+1}{x} \neq 1$;
3. $\frac{x-1}{x} > 0$, $\frac{x-1}{x} \neq 1$;
4. $\log_{\frac{x+1}{x}} \frac{x}{3} > 0$;
5. $\log_{\frac{x+1}{x}} \left( \frac{x}{2} - 1 \right) > 0$.

From 1) and 2) it follows that $x > 2$. From $x > 2$ it follows that $\frac{x+1}{x} > 1$ and $\frac{x-1}{x} < 1$; This means that 3) and 4) are performed automatically. Since for $x > 2$ $\frac{x+1}{x} > 1$, and 6) is equivalent to the inequality $\frac{x}{2} - 1 > 1$ or $x > 4$. Thus, the ODZ is given by the inequality $x > 4$.

Since $\frac{x-1}{x} < 1$ and $\frac{x+1}{x} > 1$ for $x > 4$, the original inequality is equivalent to the inequality

$$\log_{\frac{x+1}{x}} \frac{x}{3} \geq \log_{\frac{x+1}{x}} \left( \frac{x}{2} - 1 \right),$$

and the last inequality is equivalent to the inequality

$$\frac{x}{3} \geq \frac{x}{2} - 1,$$

solving which, we find $x \leq 6$. Taking into account the ODZ, we get the final answer: $4 < x \leq 6$.

3. Answer: 3 rubles.
4. Answer: $\frac{2\sin 2\alpha}{(\sin \alpha + \cos \alpha)^3}$.

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Faculty of Economics (evening department)
Option

1. Two people come out from different ends of the carriage of a moving train towards each other. The first one moves towards the head carriage of the train and has a relative speed of 0.5 m/sec. The second has a relative ground speed of 63 km/h. 5 seconds after the start of movement, the distance between them is reduced to one-fifth of the length of the carriage and is one fortieth of the distance covered by the train in 5 seconds. Find the speed of the train.

2. In the rectangle ABCD, points M and N are taken on sides AB=6 and BC=8 so that the segment MN is parallel to the segment AC. It is known that the perimeter of the polygon AMNCD is related to the perimeter of the triangle MBN as 7 : 3. Find the length of the segment MN.

3. Solve the equation sin x+sin 4x+sin 7x=0.

4. Solve the inequality 2^{2x-1}>1-2^{x-1}.

Problem solutions

1. Answer: train speed $\frac{170}{9}$ m/sec ≈ 18.9 m/sec.

2. According to the conditions of the problem MN || AC, so △MBN∼△ABC. In similar triangles, similar sides are proportional: $\frac{MB}{BN} = \frac{AB}{BC}$. But $\frac{AB}{BC} = \frac{3}{4}$, so it is convenient to take MB = 3x, then BN = $\frac{4}{3}$ MB = 4x. Expressing the perimeters of AMNCD and MBN in terms of x, we obtain the equation $\frac{28-2x}{12x} = \frac{7}{3}$, from which MN = 5x = $\frac{14}{3}$. Answer: MN = $\frac{14}{3}$.

3. Answer: $x = \frac{\pi n}{4} (n=0, \pm 1, \dots)$, $x = \frac{\pi}{3} \pm \frac{\pi}{9} + \frac{2\pi k}{3} (k=0, \pm 1, \dots)$.

4. Answer: $x > 0$.

More common errors in this version were errors in solving the simplest trionometric and exponential equations to which problems 3 and 4 are reduced. For example, polygons incorrectly solved the equation sin 4x=0 in problem 3 and the equation 2^{x}=-2 in problem 4.

In conclusion, we offer several tasks from other options for independent solution.

Full-time department of the Faculty of Psychology

1. Find all non-zero real values ​​of a for which the equation $x (x^{12}-ax^6+a^4)=0$ has the same path of real roots that form an arithmetic progression.

2. Find all values ​​of x satisfying $\log_{\cos x^2} \left( \frac{5}{2x}-2x \right) > \log_{\cos x^2} (2x-1)$.

Evening department of the Faculty of Psychology

3. Find all real values ​​of a for which the roots $x_1$ and $x_2$ of equation $x^2+ax+1=0$ are real and satisfy the corresponding $x_1<x_2$, $x_1+2x_2=4$.

4. Solve inequality
   $$\log \sqrt{x+\frac{1}{4}} \log \frac{1}{\sqrt{x}+1} \left( x-\frac{3}{4} \right) \leq \log \sqrt{x+\frac{1}{4}} \log \frac{1}{\sqrt{x}+1} (2-x).$$