Entrance exams in mathematics at the Moscow Textile Institute in 1970

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Entrance exams to MIT are conducted in full accordance with the mathematics program for applicants to universities. As in all technical schools, mathematics is a professional subject for admission to MIT.

At the faculties of energy-mechanical and technical engineering, applicants will take two exams in mathematics: written and oral. At the mechanical-technological, chemical-technological and engineering-economic faculties there is one exam in mathematics - an oral one. In all faculties, medalists will take an exam in mathematics.

Let’s look at one of the options offered in the written exam.

Option

1. Solve the equation
   $$\text{ctg} x - 1 = \frac{\cos 2x}{\text{tg} x + 1}.$$

2. Solve the equation
   $$\log_5^4 (x - 1)^2 + \log_5^2 (x - 1)^3 = 25.$$

3. Simplify with $0 < a \leq 1$:
   $$\frac{1}{\sqrt{5}} \left[ 4 (a + 1) + \left( \sqrt[3]{a \sqrt{a}} - 1 \right)^2 - \left( \frac{\sqrt[6]{ab^2} + \sqrt{a}}{\sqrt[3]{a} + \sqrt[3]{b}} \right)^3 \right]^{\frac{1}{2}}.$$

4. The base of the example is a right triangle with a hypotenuse $c$ and an acute angle $\alpha$, the side edges of the example are inclined to the plane of the base at an angle $\phi$. Find volume examples.

Analysis of varnat problems

1. We find the ODZ of the equation $x \neq \frac{\pi}{2} l, x \neq \frac{\pi}{4} (4m - 1) (l, m = 0, \pm 1, \dots)$. Now let’s transform the left side: $(\text{ctg} x - 1) (\text{tg} x + 1) = \cos 2x,$
   $$\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \cos 2x,$$
   $$\cos 2x = \frac{1}{2} \cos 2x \sin 2x,$$
   $$\cos 2x \left( 1 - \frac{1}{2} \sin 2x \right) = 0.$$ (*)

Since the second factor in (*) is nonzero ($\sin 2x \neq 2$), then $\cos 2x = 0$ and
$$x = \frac{\pi}{4} (2k + 1) \quad (k = 0, \pm 1, \dots).$$

Taking into account the ODZ, we exclude the value $x$, which is obtained for odd $k$.

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

Solving this example did not cause any particular difficulties, but a lot of extra roots came up, since the ODZ was not taken into account.

2. The ODZ equation is: $x > 1$. Let’s transform the left side using the property of logarithms:
   $$16 \log_5^4 (x - 1) + 9 \log_5^2 (x - 1) = 25.$$

Solving the resulting biquadratic equation

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nechne, which is equivalent to the given one, we find

$$\log_5^2 (x-1) = 1 \text{ или } \log_5^2 (x-1) = -\frac{25}{16}.$$

The last equality is impossible; what remains is $\log_5 (x-1) = \pm 1$, hence $x_1 = 6, x_2 = \frac{6}{5}$.

A typical mistake: many applicants forgot to raise indicators 2 and 3 to the corresponding power when placing them outside the logarithm sign.

3. First we find the valid values ​​of $a$ and $b$: $a \neq -b$. Let’s transform the original expression:

$$\frac{1}{\sqrt{5}} \left[ 4 (a+1) + (\sqrt{a}-1)^2 - \left( \frac{\sqrt[6]{a} (\sqrt[3]{b} + \sqrt[3]{a})}{\sqrt[3]{a} + \sqrt[3]{b}} + \sqrt[6]{a} \right)^3 \right]^{\frac{1}{2}} = \frac{1}{\sqrt{5}} \left[ 4 (a+1) + a - 2 \sqrt{a} + 1 - (2 \sqrt[6]{a})^3 \right]^{\frac{1}{2}} = \left| \sqrt{a}-1 \right| = 1-\sqrt{a}.$$

When solving this example, applicants did not always pay attention to the condition $0 < a \leq 1$ and therefore made a mistake when extracting the square root.

4. The height of the examples (see figure) will pass through the center $O$ of the circle described near the base, since the side-

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This inequality is equivalent to the system of inequalities
$$\begin{cases} x^2 - 6 > 0, \\ 0 < \log_3(x^2 - 6) < 1 \end{cases}$$

or
$$\begin{cases} x^2 - 6 > 0, \\ 1 < x^2 - 6 < 3. \end{cases}$$

The second inequality includes the first, from which we find $7 < x^2 < 9$, $\sqrt{7} < |x| < 3$.

Answer:
$$\sqrt{7} < x < 3$$ or $-3 < x < -\sqrt{7}$.

In this problem, many applicants did not take into account that the base of the external logarithm is less than 1, and some forgot about the modulus sign and lost the second part of the answer.

In conclusion, we present one version of the oral exams for independent decision.

Option

Solve equations:

1. $$\cos \frac{x}{2} \cos \frac{3}{2}x - \sin x \sin 3x - \sin 2x \sin 3x = 0.$$
2. $$2\lg 2 + \left(1 + \frac{1}{2x}\right)\lg 3 - \lg \left(\frac{x}{\sqrt{3}} + 27\right) = 0.$$
3. Simplify:
   $$\frac{a + x}{\sqrt{a^2} - \sqrt{x^2}} + \frac{\sqrt[3]{ax^2} - \sqrt[3]{a^2}x}{\sqrt[3]{a^2} - 2\sqrt[3]{ax} + \sqrt[3]{x^2}} - \sqrt[6]{a} - \sqrt[6]{x}.$$
4. The base side of a regular triangular rectangle is equal to $a$, the side faces are inclined to the plane of the base at an angle $\phi$. Find the volume of the straight line.

Individual tasks

Solve the equations:
5. $$\sin^4 x + \sin^4 \left(\frac{\pi}{2} - x\right) = \sin 2x.$$

6. $$\frac{\sqrt{3}}{2} \sin 3x - \frac{1}{2} \cos 3x = \cos 7x.$$
7. $$\tg x + \frac{\cos x}{1 + \sin x} = 2.$$
8. Convert to product:
   $$\sec^2 \alpha - \tg^2 \alpha + \mathrm{tg} \left(\frac{\pi}{2} - \frac{\alpha}{2}\right) + \mathrm{ctg} \left(\frac{\pi}{4} - \frac{\alpha}{2}\right).$$
9. Given: $\tg \alpha = \frac{4}{3}$, $\pi < \alpha < \frac{3}{2}$ π.

Find $\tg \frac{\alpha}{2}$.

Solve equations:
10. $$\sqrt[3]{64} - \sqrt[3]{2^{3x+3}} + 12 = 0.$$
11. $$10^{\lg^3 x} + x^{\lg x} = 20.$$
12. $$\lg^2 10x + \lg x = 19.$$
13. $$4^{\log_3 x} + x^2 = 8.$$
14. $$\log_3 (3^x - 1) \log_3 (3^{x+1} - 3) = 6.$$
Solve inequalities:
15. $$\frac{0,2^{x-0,5}}{\sqrt{5}} > 5 \cdot 0,04^{x-1}.$$
16. $$0,8^{\frac{3-x}{1-x}} < \frac{16}{25}.$$
17. $$\lg x^2 < \lg^2 x.$$
18. $(x^2 - 4) \log_{0,5} x > 0.$
19. $$\log_{0,5} (2x + 6) > \log_{0,5} (x + 8).$$
20. $$\frac{|2x - 3| - x}{x - 2} > 0.$$
21. For what values ​​of $a$ does the equation $(5a - 1)x^2 - (5a + 2)x + (3a - 2) = 0$ have equal roots?
22. Simplify:
$$4x^4 \left(\frac{x}{a} + \sqrt[5]{a}x^{-0,6}\right) - \frac{1}{2} - 5^{0,4} \sqrt[5]{a}x^{-0,6} + \frac{3x}{a} \sqrt[5]{a^{-0,2}} \sqrt{ax^{-1}}$$
23. Prove that in a right angled triangle, the bisector of the right angle bisects the angle between the median and the altitude, lowered to the hypotenuse.
24. The area of ​​an equilateral triangle constructed on the hypotenuse of a right angle triangle is twice the area of ​​the latter. Determine the right angles of the triangle.