Entrance exams in mathematics at the Faculty of Mathematics of Moscow State Pedagogical Institute named after. V.I. Lenin in 1970

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ENTRANCE EXAMINATIONS
IN MATHEMATICS IN MATHEMATICS
FACULTY OF MSPI
them. V. I. LENIN IN 1970
E. A. Shegolkov

The written examination in mathematics at our department in 1969 (the report on it can be read in «Kvant» No. 3 for 1970) was somewhat unusual. The unusualness of the state is in some imitation of competitive exams at Moscow State University. As a result, out of 475 applicants, 224, that is, about half, wrote the paper with a bad mark and dropped out of the further “struggle.” This method of exchanging applicants does not seem suitable for a pedagogical institute; too many candidates ended up without personal acquaintance with the examiners.

In 1970, the selection committee of the Faculty of Mathematics decided on almost different paths. The written work included four tasks to get to know the student well.

1. Stereometric problem with application of trigonometry.
2. The task of composing a quadratic equation with physical or other content.
3. Algebraic equations or inequality.
4. Trigonometric equation.

All tasks did not require any special knowledge or training. But at the same time, they assumed a thorough knowledge of school mathematics and known abilities of applicants. 4 hours were allotted to complete the written work.

When assessing written work, “five” was given only if all tasks were completed; “four” - if all tasks were completed, but with minor shortcomings. “Two” was given, for example, if the first tasks were not solved due to a group of errors. At the same time, the rating of “three” acquired a greater share and did not exclude the applicant from participation in the competition.

In 1970, 508 people entered the Faculty of Mathematics, and there were 175 places; the competition was about three people per place.

Of the 508 applicants for the written work, 120 wrote it as “four” and “five”, 233 as “three” and 155 as “two”.
As an example, below we will analyze the written work option and point out the typical mistakes made by applicants when solving it.

Option

1. The length of the generatrix of the cone is equal to l, the angle of the generatrix with the plane of the base is equal to α. Find the volume of a primid circumscribed around a cone whose base is a rhombus with an acute angle β.
2. In n hours, the tractor plows p hectors more than the horse. How many hectors can a horse plow in n hours and how many hectors can a tractor plow in n hours, if a tractor plows one hector in t hours faster than a horse.
3. Solve inequality
   $$\sqrt{1+2x}-\sqrt{x+4}<1.$$
4. Solve the equation
   $$\frac{3(\cos 2x+\mathrm{ctg} 2x)}{\mathrm{ctg} 2x-\cos 2x}-2(\sin 2x+1)=0.$$

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As you already noticed, when solving this problem I had to solve right-angled triangles three times. The applicants made mistakes here too. Solid ability to solve right-angled triangles is a prerequisite for successfully solving most stereometric problems.

2. Let the tractor turn $x$ ($2a$) in $n$ hours, therefore $1$ ($2a$) in $\frac{n}{x}$ hours.

A horse turns in $n$ hours $(x-p)$ ($2a$), and $1$ ($2a$) in $\frac{n}{x-p}$ hours.

According to the conditions of the problem, $\frac{n}{x}$ is less than $\frac{n}{x-p}$ by $t$ hours, that is, $\frac{n}{x} + t = \frac{n}{x-p}$ ($x \neq 0, x \neq p$). After the transformation we obtain the equation $tx^2 - ptx - np = 0$. Solving it, we find $x_{1,2} = \frac{pt \pm \sqrt{(pt)^2 + 4ptn}}{2t}$.

Which of these two roots satisfies the conditions of the problem? By condition, $n > 0, t > 0, p > 0$, so $\sqrt{(pt)^2 + 4ptn}$ has a real value. $(pt)^2 + 4ptn > (pt)^2$ and therefore $\sqrt{(pt)^2 + 4ptn} > pt$.

So the root

$$x_1 = \frac{pt + \sqrt{(pt)^2 + 4ptn}}{2t}$$

gives the answer to the question of the problem, exactly how many hectares does a tractor turn in $n$ hours, and a horse turns in $n$ hours

$$x - p = \frac{-pt + \sqrt{(pt)^2 + 4ptn}}{2t}$$.

The second root of the equation is negative and disappears according to the meaning of the problem.

When compiling the initial equation from the value $\frac{n}{x}, \frac{n}{x-p}, t$, some of the applicants for some reason subtracted more from less and put the equation $\frac{n}{x-p} + t = \frac{n}{x}$.

This is the result of not paying attention.

When solving a problem with specific physical or other content, you should be especially careful and always check your calculations. Formal statements must correspond to the actual state of affairs. In addition, it should be remembered that the whole solution to this type of problem is not the solution of the compiled equation, but obtaining an answer to the question posed in the task.

3. Let’s find the range of acceptable values ​​for $x$ (OD3):

$$\begin{cases} 1 + 2x \geq 0 \\ x + 4 \geq 0 \end{cases}$$ or $$\begin{cases} x \geq -\frac{1}{2} \\ x \geq -4. \end{cases}$$

Consequently, OD3 is determined by the condition

$$x \geq -\frac{1}{2}.$$ (1)

Let us transform the original inequality as follows:

$$\sqrt{1 + 2x} < 1 + \sqrt{x + 4}.$$

In marked OD3, both sides of the last inequality are non-negative; squaring them we get

$$\sqrt{x + 4} > \frac{x}{2} - 2.$$ (2)

The left side of inequality (2) is positive, the right side is negative, and the left side is positive.

1. $\frac{x}{2} - 2 < 0$ therefore

$$-\frac{1}{2} \leq x < 4.$$

For such values ​​of $x$ inequality (2) depends negatively, and the left one is positive.

2. $\frac{x}{2} - 2 \geq 0$ and by condition

$$\sqrt{x + 4} > \frac{x}{2} - 2.$$

Solving the first of these inequalities, we obtain $x \geq 4$. To solve the second inequality, we square both sides (this can be done,

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since both sides are non-negative).
We get
$$x^2 - 12x < 0.$$ (3)

Solving (3), we find $0 < x < 12$. Finally, in the second case we obtain $4 \leq x < 12$. Combining both cases, we write down the solutions to the original inequality:
$$-\frac{1}{2} \leq x < 12.$$

A typical mistake when solving such inequalities is the non-contradictory squaring of the inequality. It is known that the resulting inequality may turn out to be unequal to the original one. Therefore, when squaring, one should take into account the signs of inequality and, if necessary, consider different cases.

4. Let’s write the following equation:

$$\frac{3 \left( \cos 2x + \frac{\cos 2x}{\sin 2x} \right)}{\cos 2x - \cos 2x} - 2 (\sin 2x + 1) = 0.$$ (1)

Now it’s easy for us to determine DL. In fact, at $\sin 2x = 0$ $\frac{\cos 2x}{\sin 2x}$ loses its meaning, and at $\cos 2x = 0$ the denominator of the fraction is zero and the entire fraction loses its meaning. Consequently, the ODZ is determined by conditions $\sin 2x \neq 0$, $\cos 2x \neq 0$. After reduction by $\cos 2x$ and simplification, equation (1) takes the form

$$3(\sin 2x + 1) - 2(\sin 2x + 1) (1 - \sin 2x) = 0$$ (2)

(the denominator was the set $1 - \sin 2x$, which does not vanish in the ODZ, since $\cos 2x \neq 0$). After taking the common factor out of the bracket, equation (2) splits into two:

1. $1 + 2 \sin 2x = 0$;
2. $1 + \sin 2x = 0$.

Solving the first equation, we find

$$\sin 2x = -\frac{1}{2}; x = (-1)^{n+1} \frac{\pi}{12} + \frac{\pi k}{2}$$

($k = 0, \pm 1, \pm 2, \dots$).

All these values ​​are contained in the ODZ.

Solving the second equation, we obtain $\sin 2x = -1$, whence $\cos 2x = 0$, which eliminates the ODZ.

So, the solutions to the original equation are only the values

$$x = (-1)^{n+1} \frac{\pi}{12} + \frac{\pi k}{2}$$

($k = 0, \pm 1, \pm 2, \dots$).

When solving this equation, many people did not take into account the ODZ and included the values ​​$\cos 2x = 0$, that is, $x = -\frac{\pi}{4} + \pi k$, in the solution.

Some were unable to write down the general form of the solution.

Of the 68 works of this option, 3 works are rated “five”, 8 works are rated “four”, 39 works are rated “three” and 18 works are rated “two”.

Let us note the mistakes that applicants made when solving problems in other options, as well as general shortcomings when performing written work.

When solving a stereometric problem, in many cases the drawings were unsuccessful (unclear, dirty) or simply incorrect. This happened because the person solving the problem incorrectly imagined or incorrectly depicted the position of a given figure, body, section and other elements, especially when it came to described or inscribed bodies.

The drawing and subsequent construction in many cases were not accompanied by the necessary explanations, and no justification was given for the conclusions obtained as a result of the construction.

The absence of errors in the constructions on the drawing and the substantiation of the conclusions entailed a reduction in the grade for solving the stereometric problem.

In a problem with content on composing an equation, some do not sum

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Can you create an equation based on the conditions of the problem?

Others, having correctly composed the equation and solved it, did not examine the resulting roots from the point of view of their suitability as a solution to the problem.

Still others, although they examined the obtained roots of the equation, were unable to choose the desired solution.

When solving algebraic and trigonometric equations, they were often reduced to expressions containing unknowns. As a result, the roots of the equation were rubbed. In other cases, they did not pay attention to the ODZ and because of this, the roots were built.

A few words about the oral exam. The exam papers contained two questions. The first is in geometry, the second is in algebra or trigonometry. The wording of the exam questions almost verbatim repeated the wording of the 1970 university entrance exam program.

When answering the questions on the ticket, it was necessary to give an accurate proof of a theorem or derivation of a formula, and to accurately formulate the necessary definitions.

In addition to the questions on the exam paper, the applicant was given additional tasks or asked questions similar to these.

1. Graph the function:

a) $y = \lg \frac{1}{2}x$;
b) $y = -x^2 - x$;
c) $y = \frac{2}{x} - 3$;
d) $y = \sin 2x$;
e) $y = \cos \frac{x}{2}$.

2. Solve the equation:

a) $2^x = -1$;
b) $|2x - 3| = 1$;
c) $\cos 2x = -\frac{1}{2}$;

d) $\arcsin x = x$.

3. Solve inequality:

a) $\lg \frac{1}{2}x > 0$; 6) $x^2 > 5$;
b) $\frac{6}{x - 2} < 0$;
g) $|x - 2| \ge 1$.

4. Which sign has the numbers $\sin 3,14$; $\cos \frac{9\pi}{10}$; $\sin 2$?

5. Find $\sin \left(2\arcsin \frac{1}{3}\right)$.

6. Find the period of the function $y = \sin 3x$.

7. Construct a circle of given radius passing through two given points.

8. Give an example of two incommensurable segments.

9. Can the sides of a triangle have a ratio of 1:2:3?

10. Prove that in a right triangle, the median drawn from the vertex of the right angle is equal to half the hypotenuse.

11. Under what conditions can a ball be inscribed in a right circular cylinder?

12. Find the angle between the diagonal of the cube and the plane of its base.

Applicants showed much weaker knowledge on the following issues: generalization of the concept of exponents, complex numbers, number sequences, infinite geometric progression, method of mathematical induction, volumes and surfaces of bodies. Not all applicants gave clear definitions of trigonometric functions, circumference, area of ​​a circle, volume and surface of a cylinder and cone.

The following concepts are not well understood: irrational and real numbers, absolute magnitudes of numbers, consistent and incommensurable segments, radian measure of angles.

In geometrics, construction problems are solved incorrectly. Many applicants have an underdeveloped imagination.

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Those who entered with uncertainty solved problems involving construction both on a plane and in space. Especially noticeable among the overwhelming majority of applicants is a shallow and incomplete familiarity with the concept of function. Many had difficulty giving examples of functional dependencies; few showed a good understanding and use of the properties of monotonicity and periodicity of functions.

And finally, about the concept of limit. This most important concept is captured in most cases on the surface. And it doesn’t even come to talk about applying the limit. Thus, the sections of the school mathematics course related to the concept of limit are not sufficiently mastered.

At the same time, it is worth noting the positive aspects of schoolchildren’s knowledge of mathematics.

The mastery of formal computing apparatus and trigonometry became better, they were more confident in constructing graphs of various functions, they were better able to understand logarithms, and they were more fluent (compared to simple years) with trigonometric skills.

It should be noted that the number of excellent written papers and excellent answers in the oral examination in mathematics has increased compared to previous years. It is also encouraging that among the honors students there are many young people from rural areas.

In conclusion, we offer readers for independent exercise one genuine option, and one combined version, compiled from problems of various options offered in the exams.

Option 1

1. At the base of the primada lies an isosceles triangle with a perimeter $2p$ and an angle $\alpha$ at the base. Determine the volume of the base if its side faces are inclined to the plane of the base at an angle of $\beta$.

2. Two cars are moving uniformly in one direction along a circle of radius $R$. One of them makes a full circle $t$ seconds faster than the other. The time between two consecutive meetings is automatically equal to $a$. Determine the speed of the vehicle automatically.

3. For what real non-negative values ​​of $a$ the equation
   $$a + x - x^2 = \sqrt{a + x} + x$$
   has real non-negative roots? Find these roots.

4. Solve the equation
   $$(1 + \sin 2x)(\cos x - \sin x) = 1 - 2\sin^2 x.$$
   Option 2 (prefabricated)

5. The ball is inscribed in a right triangle. In this triangle, the perpendicular dropped from the vertex of the right angle to the hypotenuse has length $h$ and makes an angle $\alpha$ with one of the legs. Find the volume of the primed.

6. A piece of equal weight was cut from two pieces of the alloy with different percentages of copper, weighing $m$ kg and $n$ kg. Each of the cut pieces was fused with the remainder of the other piece, after which the percentage of copper in both alloys became the same. How much did each of the cut pieces weigh?

7. Solve inequality
   $$\log_{0.5}x + \sqrt{1 - 4(\log_{0.5}x)^2} < 1.$$

8. Solve the equation
   $$\frac{\cos x}{\text{ctg}^2 \frac{x}{2} - \text{tg}^2 \frac{x}{2}} = \frac{1}{8} \left(1 - \frac{2 \text{ctg} x}{1 + \text{ctg}^2 x}\right).$$