Source Unified State Exam early wave.

Option No. 3109295

Early Unified State Exam in Physics 2017, option 101

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The figure shows a graph of the projection of the speed of the body v x from time to time.

Determination of the projection of the acceleration of this body a x in inter-va-le time from 15 to 20 s. The answer is in m/s 2.

Answer:

Mass cube M= 1 kg, compressed from the sides with springs (see ri-su-nok), placed on a smooth horizontal table. The first spring is compressed by 4 cm, and the second is compressed by 3 cm. Stiffness of the first spring k 1 = 600 N/m. What is the stiffness of the second spring? k 2? The answer is in N/m.

Answer:

Two bodies are moving at the same speed. The kinetic energy of the first body is 4 times less than the kinetic energy of the second body. Determine the ratio of the masses of the bodies.

Answer:

At a distance of 510 m from the observer, workers drive piles using a pile driver. How much time will pass from the moment when the observer sees the impact of the pile driver until the moment when he hears the sound of the impact? The speed of sound in air is 340 m/s. Express your answer in p.

Answer:

The figure shows graphs of pressure dependence p from diving depth h for two liquids at rest: water and heavy liquid diiodomethane, at constant temperature.

Choose two true statements that agree with the graphs given.

1) If the pressure inside a hollow ball is equal to atmospheric pressure, then in water at a depth of 10 m the pressure on its surface from the outside and from the inside will be equal to each other.

2) The density of kerosene is 0.82 g/cm 3, a similar graph of pressure versus depth for kerosene will be between the graphs for water and diiodomethane.

3) In water at a depth of 25 m, pressure p 2.5 times more than atmospheric.

4) As the depth of immersion increases, the pressure in diiodomethane increases faster than in water.

5) Density olive oil 0.92 g/cm 3 , a similar graph of pressure versus depth for oil will be between the graph for water and the x-axis (horizontal axis).

Answer:

A massive load suspended from the ceiling on a weightless spring performs free vertical vibrations. The spring remains stretched all the time. How they behave potential energy springs and the potential energy of a load in a gravitational field when the load moves upward from its equilibrium position?

1) increases;

2) decreases;

3) does not change.

Answer:

A truck moving along a straight horizontal road at a speed v, braked so that the wheels stopped rotating. Truck weight m, friction coefficient of wheels on the road μ . Formulas A and B allow you to calculate the values ​​of physical quantities characterizing the movement of the truck.

Establish a correspondence between the formulas and physical quantities, the value of which can be calculated using these formulas.

A	B

Answer:

As a result of cooling rarefied argon, it absolute temperature decreased by 4 times. How many times did the average decrease? kinetic energy thermal motion of argon molecules?

Answer:

The working fluid of a heat engine receives from the heater an amount of heat equal to 100 J per cycle and does 60 J of work. What is the efficiency of the heat engine? Express your answer in %.

Answer:

The relative humidity of the air in a closed vessel with a piston is 50%. What will it be like relative humidity air in a vessel, if the volume of the vessel at a constant temperature is reduced by 2 times? Express your answer in %.

Answer:

The hot substance, initially in a liquid state, was slowly cooled. The heat sink power is constant. The table shows the results of measurements of the temperature of a substance over time.

Select two statements from the proposed list that correspond to the results of the measurements taken and indicate their numbers.

1) The crystallization process of the substance took more than 25 minutes.

2) Specific heat substances in liquid and solid states is the same.

3) The melting point of the substance under these conditions is 232 °C.

4) After 30 min. after the start of measurements, the substance was only in a solid state.

5) After 20 minutes. after the start of measurements, the substance was only in a solid state.

Answer:

Graphs A and B show diagrams p−T And p−V for processes 1−2 and 3−4 (hyperbola), carried out with 1 mole of helium. On the charts p- pressure, V– volume and T– absolute gas temperature. Establish a correspondence between the graphs and statements characterizing the processes depicted on the graphs. For each position in the first column, select the corresponding position in the second column and write down the selected numbers in the table under the corresponding letters.

A	B

Answer:

How is the Ampere force acting on conductor 1 from conductor 2 directed relative to the figure (to the right, left, up, down, towards the observer, away from the observer) (see figure), if the conductors are thin, long, straight, parallel to each other? ( I- current strength.) Write the answer in word (words).

Answer:

A direct current flows through a section of the circuit (see figure) I= 4 A. What current will be shown by an ideal ammeter connected to this circuit if the resistance of each resistor r= 1 Ohm? Express your answer in amperes.

Answer:

In an observation experiment electromagnetic induction a square frame made of one turn of thin wire is in a uniform magnetic field perpendicular to the plane of the frame. Induction magnetic field increases uniformly from 0 to maximum value IN max per time T. In this case, an induced emf equal to 6 mV is excited in the frame. What induced emf will occur in the frame if T reduce by 3 times, and IN Reduce max by 2 times? Express your answer in mV.

Answer:

A uniform electrostatic field is created by a uniformly charged extended horizontal plate. The field strength lines are directed vertically upward (see figure).

From the list below, select two correct statements and indicate their numbers.

1) If to the point A place a test point negative charge, then a force directed vertically downward will act on it from the side of the plate.

2) The plate has a negative charge.

3) Potential electrostatic field at the point IN lower than at point WITH.

5) The work of the electrostatic field to move a test point negative charge from a point A and to the point IN equal to zero.

Answer:

An electron moves in a circle in a uniform magnetic field. How will the Lorentz force acting on the electron and its period of revolution change if its kinetic energy is increased?

For each quantity, determine the corresponding nature of the change:

1) will increase;

2) will decrease;

3) will not change.

Write down the selected numbers for each in the table. physical quantity. The numbers in the answer may be repeated.

Answer:

The figure shows a DC circuit. Establish a correspondence between physical quantities and formulas by which they can be calculated ( ε – EMF of the current source, r – internal resistance current source, R– resistor resistance).

For each position in the first column, select the corresponding position in the second column and write down the selected numbers in the table under the corresponding letters.

PHYSICAL QUANTITIES		FORMULAS
A) current strength through the source with switch K open B) current strength through the source with the key K closed

Answer:

Two monochromatic waves propagate in a vacuum electromagnetic waves. The energy of a photon of the first wave is 2 times greater than the energy of a photon of the second wave. Determine the ratio of the lengths of these electromagnetic waves.

Answer:

How will they change when β − − decay mass number of the nucleus and its charge?

For each quantity, determine the corresponding nature of the change:

1) will increase

2) will decrease

3) will not change

Write down the selected numbers for each physical quantity in the table. The numbers in the answer may be repeated.

Answer:

Determine the voltmeter readings (see figure), if the error direct measurement voltage is equal to the division value of the voltmeter. Give your answer in volts. In your answer, write down the value and error together without a space.

Answer:

To conduct laboratory work to detect the dependence of the resistance of a conductor on its length, the student was given five conductors, the characteristics of which are indicated in the table. Which two of the following guides should a student take to conduct this study?

Task 1

A pack of chips costs \(170\) rubles. Which greatest number packs of chips can be bought for \(1100\) rubles during the sale, when the discount is \(20\%\)?

During the sale, a pack of chips costs \(170\cdot (1 - 0.2) = 136\) rubles. According to the conditions of the problem, we need to find the largest integer, when multiplied by \(136\), the result will remain no more than \(1100\) . This number is obtained after rounding down the result of dividing \(1100\) by \(136\) and is equal to \(8\) .

Answer: 8

Task 2

The graph shows the process of warming up the engine of an old motorcycle. The x-axis shows the time in minutes that has passed since the engine started, and the y-axis shows the engine temperature in degrees Fahrenheit. Determine from the graph how many minutes the engine warmed up from temperature \(60^\circ F\) to temperature \(100^\circ F\) .

The engine warmed up to a temperature of \(60^\circ F\) \(3\) minutes after starting, and to \(100^\circ F\) \(8\) minutes after the start. From \(60^\circ F\) to \(100^\circ F\) the engine warmed up for \(8 - 3 = 5\,\) minutes.

Answer: 5

Task 3

On checkered paper with a cell size \(1\times 1\) the angle \(AOB\) is depicted. Find the tangent of this angle.

\[\mathrm(tg)\,(\beta - \alpha) = \dfrac(\mathrm(tg)\,\beta - \mathrm(tg)\,\alpha)(1 + \mathrm(tg)\, \alpha\cdot \mathrm(tg)\,\beta)\] Angle \(AOB\) can be represented as

\[\angle AOB = \beta - \alpha,\] Then \[\mathrm(tg)\, AOB = \mathrm(tg)\,(\beta - \alpha) = \dfrac(\mathrm(tg)\,\beta - \mathrm(tg)\,\alpha)( 1 + \mathrm(tg)\,\alpha\cdot \mathrm(tg)\,\beta) = \dfrac(2 - \frac(1)(3))(1 + \frac(1)(3)\ cdot 2) = 1\,.\]

Answer: 1

Task 4

The factory sews hats. On average, \(7\) hats out of \(40\) have hidden defects. Find the probability that the purchased hat will be free of defects.

On average, \(40 - 7 = 33\) hats out of forty have no defects, therefore, the probability of buying a hat without defects is equal to \[\dfrac(33)(40) = \dfrac(330)(400) = \dfrac(82.5)(100) = 0.825\,.\]

Answer: 0.825

Task 5

Find the root of the equation \

ODZ: \

On ODZ: \ therefore, on the ODZ the equation has the form: \[\sqrt(13x - 13) = 13\quad\Rightarrow\quad 13x - 13 = 13^2\quad\Rightarrow\quad 13x = 182\quad\Rightarrow\quad x = 14\]– fits according to ODZ.

Answer: 14

Task 6

IN right triangle\(ABC\) angle \(C\) is equal to \(90^\circ\) , \(AB = 6\) , \(\mathrm(tg)\, A = \dfrac(1)(2\sqrt(2))\). Find \(BC\) .

Let us denote \(BC = x\) , then \(AC = 2\sqrt(2)x\)

According to the Pythagorean theorem: \ whence \(x = 2\) (since we are only interested in \(x > 0\)).

Answer: 2

Task 7

The straight line \(y = 2x - 1\) is tangent to the graph of the function \(y = x^3 + 6x^2 + 11x - 1\) . Find the abscissa of the tangent point.

At the point of tangency between the straight line \(y = 2x - 1\) and the graph of the function \(y = x^3 + 6x^2 + 11x - 1\), the derivative of this function coincides with slope\(k\) is a straight line, which in this case is equal to \(2\) .

Then \ The roots of the last equation are: \

Let's check for which of the obtained \(x\) the straight line and the graph have a common point:

at \(x = -3\) :
the ordinate of a point on a straight line is equal to \(2\cdot(-3) - 1 = -7\) , and the ordinate of a point on a graph is equal to \[(-3)^3 + 6\cdot(-3)^2 + 11\cdot(-3) - 1 = -7,\] that is, the straight line and the graph pass through the point \((-3; -7)\) and the derivative of the function at the point \(x = -3\) coincides with the slope of the straight line, therefore, they touch at this point.

for \(x = -1\) :
the ordinate of a point on a line is equal to \(2\cdot(-1) - 1 = -3\) , and the ordinate of a point on a graph is equal to \[(-1)^3 + 6\cdot(-1)^2 + 11\cdot(-1) - 1 = -7,\] that is, the ordinates of these points are different, therefore, when \(x = -1\) the straight line and the graph have no common point.

Total: \(-3\) is the required abscissa.

Answer: -3

Task 8

Find the surface area of ​​the polyhedron shown in the figure (all dihedral angles straight).

The surface area of ​​a given polyhedron is equal to the surface area rectangular parallelepiped with dimensions \(10\times 12\times 13\) and is thus equal \(2\cdot(10\cdot 12 + 12\cdot 13 + 10\cdot 13) = 812\).

Answer: 812

Task 9

Find the meaning of the expression \[\sqrt(48)\sin^2 \dfrac(\pi)(12) - 2\sqrt(3)\]

Let's use the double angle cosine formula: \(\cos 2x = 1 - 2\sin^2x\), then with \(x = \dfrac(y)(2)\) we have: \[\cos y = 1 - 2\sin^2\dfrac(y)(2)\qquad\Rightarrow\qquad \sin^2\dfrac(y)(2) = \dfrac(1 - \cos y)( 2)\,.\]

Substituting \(y = \dfrac(\pi)(6)\) we get: \[\sin^2\dfrac(\pi)(12) = \dfrac(1 - \cos \frac(\pi)(6))(2) = \dfrac(1 - \frac(\sqrt(3) )(2))(2)\,.\]

Since \(\sqrt(48) = 4\sqrt(3)\) , the original expression can be rewritten as \

Answer: -3

Task 10

A truck drags a car with a force of \(120\,\) kN directed under acute angle\(\alpha\) to the horizon. The truck's work (in kilojoules) over a section of length \(l = 150\,\) m is calculated using the formula \(A = Fl\cos\alpha\) . At what maximum angle \(\alpha\) (in degrees) will the work done be at least \(9000\,\) kJ?

According to the conditions of the problem we have: \

Considering that \(\alpha\in\), we find that \(\alpha\leqslant 60^\circ\) (this is easy to see by looking at the trigonometric circle).

Thus, the answer is: at \(\alpha = 60^\circ\) .

Answer: 60

Task 11

The first and second pumps fill the pool in \(9\) minutes, the second and third in \(15\) minutes, and the first and third in \(10\) minutes. How many minutes will it take these three pumps to fill the pool working together?

The first and second pumps fill \(\dfrac(1)(9)\) part of the pool in a minute,

the second and third pumps fill \(\dfrac(1)(15)\) part of the pool in a minute,

the first and third pumps fill \(\dfrac(1)(10)\) part of the pool in a minute, then \[\dfrac(1)(9) + \dfrac(1)(15) + \dfrac(1)(10) = \dfrac(25)(90)\]- part of the pool filled per minute by all three pumps, if the contribution of each pump is taken into account twice. Then \[\dfrac(1)(2)\cdot\dfrac(25)(90) = \dfrac(25)(180)\]- part of the pool that is filled in a minute by all three pumps.

Therefore, all three pumps fill the pool in \(\dfrac(180)(25) = 7.2\) minutes.

Answer: 7.2

Task 12

Find smallest value functions \ on a segment

ODZ: \ Let's decide on ODZ:

1) \

Let's find critical points (that is, internal points of the function's domain of definition at which its derivative is equal to \(0\) or does not exist): \[\dfrac(121x - 1)(x) = 0\qquad\Leftrightarrow\qquad x = \dfrac(1)(121)\]

The derivative of the function \(y\) does not exist for \(x = 0\) , but \(x = 0\) is not included in the ODZ. In order to find the largest/smallest value of a function, you need to understand how its graph looks schematically.

2) Let’s find intervals of constant sign \(y"\) :

3) Find intervals of constant sign \(y"\) on the segment under consideration \(\left[\dfrac(1)(242); \dfrac(5)(242)\right]\):

4) Sketch of a graph on a segment \(\left[\dfrac(1)(242); \dfrac(5)(242)\right]\):

Thus, the smallest value on the segment \(\left[\dfrac(1)(242); \dfrac(5)(242)\right]\) the function \(y\) reaches in \(x = \dfrac(1)(121)\) :

Total: \(4\) – the smallest value of the function \(y\) on the segment \(\left[\dfrac(1)(242); \dfrac(5)(242)\right]\).

Answer: 4

Task 13

a) Solve the equation \[\cos x(2\cos x + \mathrm(tg)\, x) = 1\,.\]

b) Find all the roots of this equation, belonging to the segment \(\left[-\pi; \dfrac(\pi)(2)\right]\).

a) ODZ: \[\cos x\neq 0\qquad\Leftrightarrow\qquad x \neq \dfrac(\pi)(2) + \pi k,\ k\in\mathbb(Z)\]

On ODZ: \[\cos x(2\cos x + \mathrm(tg)\, x) = 1\quad\Leftrightarrow\quad 2\cos^2 x + \sin x = 1\quad\Leftrightarrow\quad 2 - 2\ sin^2 x + \sin x = 1\]

Let's make a replacement \(t = \sin x\) : \

The roots of the last equation are: \ whence \(\sin x = 1\) or \(\sin x = -\dfrac(1)(2)\)

1) \(\sin x = 1\) , therefore, \(x = \dfrac(\pi)(2) + 2\pi n\)– do not qualify for DL.

2) \(\sin x = -\dfrac(1)(2)\)

where \(x_1 = -\dfrac(\pi)(6) + 2\pi k\), \(x_2 = \dfrac(7\pi)(6) + 2\pi k\), \(k\in\mathbb(Z)\) – suitable for DL.

b) \(-\pi \leqslant -\dfrac(\pi)(6) + 2\pi k \leqslant \dfrac(\pi)(2)\) equivalent \(-\dfrac(5\pi)(6) \leqslant 2\pi k \leqslant \dfrac(4\pi)(6)\), which is equivalent \(-\dfrac(5)(12) \leqslant k \leqslant \dfrac(1)(3)\), but \(k\in\mathbb(Z)\) , therefore, among these solutions only the solution for \(k = 0\) is suitable: \(x = -\dfrac(\pi)(6)\)

\(-\pi \leqslant \dfrac(7\pi)(6) + 2\pi k \leqslant \dfrac(\pi)(2)\) equivalent \(-\dfrac(13\pi)(6) \leqslant 2\pi k \leqslant -\dfrac(4\pi)(6)\), which is equivalent \(-\dfrac(13)(12) \leqslant k \leqslant -\dfrac(1)(3)\), but \(k\in\mathbb(Z)\) , therefore, among these solutions only the solution for \(k = -1\) is suitable: \(x = -\dfrac(5\pi)(6)\) .

Answer:

A) \(-\dfrac(\pi)(6) + 2\pi k, \dfrac(7\pi)(6) + 2\pi k, k\in\mathbb(Z)\)

b) \(-\dfrac(\pi)(6), -\dfrac(5\pi)(6)\)

Task 14

In a regular quadrangular prism \(ABCDA_1B_1C_1D_1\) the point \(M\) divides the lateral edge \(AA_1\) in the ratio \(AM: MA_1 = 1: 3\) . Through the points \(B\) and \(M\) a plane \(\alpha\) is drawn parallel to the line \(AC\) and intersecting the edge \(DD_1\) at the point \(N\) .

a) Prove that the plane \(\alpha\) divides the edge \(DD_1\) in the ratio \(D_1N: DD_1 = 1: 2\) .

b) Find the cross-sectional area if it is known that \(AB = 5\) , \(AA_1 = 8\) .

a) Because If the prism is regular, then it is straight and its base is a square \(ABCD\) .

Let us denote \(AM=x\) , then \(MA_1=3x\) . Because \(\alpha\parallel AC\), then \(\alpha\) will intersect the plane \(ACC_1\) in which the straight line \(AC\) lies along the straight line \(MK\) parallel to \(AC\) . So, \(CK=x, KC_1=3x\) .

It is necessary to prove that the point \(N\) is the midpoint of \(DD_1\) .

Let \(MK\cap BN=O\) , \(AC\cap BD=Q\) . The planes \(BDD_1\) and \(ACC_1\) intersect along the straight line \(QQ_1\) passing through the intersection points of the diagonals of the faces \(ABCD\) and \(A_1B_1C_1D_1\) and parallel to \(AA_1\). Because \(BN\in BDD_1\) , \(MK\in ACC_1\) , then the point \(O\) lies on \(QQ_1\) , therefore, \(OQ\parallel AA_1 \Rightarrow OQ\perp (ABC)\). Thus, \(OQ=AM=x\) .

\(\triangle OQB\sim \triangle NDB\) at two corners ( \(\angle D=\angle Q=90^\circ, \angle B\)- general), therefore,

\[\dfrac(ND)(OQ)=\dfrac(DB)(QB) \Leftrightarrow \dfrac(ND)x= \dfrac(2QB)(QB) \Rightarrow ND=2x\]

But the entire edge is \(DD_1=AA_1=4x\) , therefore, \(N\) is the middle of \(DD_1\) .

b) By the theorem of three perpendiculars ( \(OQ\perp (ABC), \text(projection ) BQ\perp AC\)) oblique \(BO\perp AC \Rightarrow BO\perp MK\)(since \(AC\parallel MK\) ). So, \(BN\perp MK\) .

The area of ​​a convex quadrilateral whose diagonals are mutually perpendicular is equal to the half product of the diagonals, that is \(S_(MBKN)=\dfrac 12 MK\cdot BN\). Let's find \(MK\) and \(BN\) .

\(MK=AC=AB\sqrt 2=5\sqrt2\) .

According to the Pythagorean theorem \(BN=\sqrt(BD^2+ND^2)=\sqrt((5\sqrt2)^2+4^2)=\sqrt(66)\)

Means, \(S_(MBKN)=\dfrac12\cdot 5\sqrt2\cdot \sqrt(66)=5\sqrt(33)\).

Answer:

b) \(5\sqrt(33)\)

Task 15

Solve the inequality \[\log_x(\sqrt(x^2 + 4x - 5) + 3)\cdot\lg(x^2 + 4x - 4)\geqslant\log_x 6.\]

\[\begin(aligned) \begin(cases) x > 0\\ x\neq 1\\ x^2 + 4x - 5\geqslant 0\\ \sqrt(x^2 + 4x - 5) + 3 > 0 \\ x^2 + 4x - 4 > 0 \end(cases) \qquad\Leftrightarrow\qquad x > 1 \end(aligned)\]

On ODZ:
\(\log_x 6 > 0\) , therefore, the original inequality is equivalent to the inequality

\[\begin(aligned) &\dfrac(\log_x(\sqrt(x^2 + 4x - 5) + 3))(\log_x 6)\cdot\lg(x^2 + 4x - 4)\geqslant 1 \qquad\Leftrightarrow\\ \Leftrightarrow\qquad &\log_6(\sqrt(x^2 + 4x - 5) + 3)\cdot\lg(x^2 + 4x - 4)\geqslant 1 \end(aligned)\ ]

Let's make a replacement \(t = \sqrt(x^2 + 4x - 5) > 0\).

After replacement: \[\log_6(t + 3)\cdot\lg(t^2 + 1)\geqslant 1\]

When \(t > 0\) both factors on the left side increase, therefore, their product increases, and the right side is constant, then the equality \[\log_6(t + 3)\cdot\lg(t^2 + 1) = 1\] can only be reached at one point. It is easy to verify that it holds for \(t = 3\), therefore, only for \(t\geqslant 3\) will the last inequality be satisfied.

Thus, \[\sqrt(x^2 + 4x - 5)\geqslant 3,\] which in ODZ is equivalent \ from where, taking into account ODZ \

Answer:

Q.E.D.

b) Let us denote \(MA = ka\) , \(AN = a\) (then the desired quantity is \(k\)), therefore \(NB = a\) , then \(BK = 2a\) .

By the theorem about tangent segments: \

Let's write down the cosine theorem for the triangle \(MNK\) : \ Substituting known quantities, we get:

\[\begin(aligned) &(ka + 2a)^2 = (ka + a)^2 + 9a^2 - 2\cdot (ka + a)\cdot 3a\cdot 0.5\quad\Leftrightarrow\\ \Leftrightarrow\quad &a^2(k + 2)^2 = a^2(k + 1)^2 + 9a^2 - (k + 1)\cdot 3a^2\quad\Leftrightarrow\\ \Leftrightarrow\quad &(k + 2)^2 = (k + 1)^2 + 9 - 3(k + 1)\quad\Leftrightarrow\quad 5k = 3\quad\Leftrightarrow\quad k = 0.6\,. \end(aligned)\]

Answer:

b) \(0.6\)

Task 17

Timur dreams of his own small shopping center, which costs \(600\) million rubles. Timur can buy it on credit, while the Risky Bank is ready to give him this amount immediately, and Timur will have to repay the loan for \(40\) years in equal monthly payments, and he will have to pay an amount of \(180\%\) exceeding the original one. Instead, Timur can rent for a while shopping mall(rental cost - \(1\) million rubles per month), setting aside each month for the purchase of a shopping center the amount that will remain from his possible payment to the bank (according to the first scheme) after paying the rent for a rented shopping center. In this case, how long will Timur be able to save up for a shopping center, assuming that its value does not change?

According to the first scheme, Timur will have to pay \((1 + 1.8)\cdot 600 = 1680\) million rubles. for 40 years. Thus, per month Timur will have to pay \[\dfrac(1680)(40\cdot 12) = 3.5\ \text(million rubles)\]

Then, according to the second scheme, Timur will be able to save \(3.5 - 1 = 2.5\) million rubles. per month, therefore, he will need \[\dfrac(600\ \text(million rubles))(2.5\ \text(million rubles/month)) = 240\ \text(months),\] which is \(20\) years.

Consider two functions: \(f(x)=|x^2-x-2|\) and \(g(x)=2-3|x-b|\) . The graph of the function \(g(x)\) for each fixed \(b\) represents an angle whose branches are directed downward, and the vertex is at the point \((b;2)\) .

Then the meaning of the inequality is this: it is necessary to find those values ​​of \(b\) for which there is at least one point \(X\) of the graph \(f(x)\) located below the graph of the function \(g(x)\) .

Let's find those values ​​of \(b\) when doesn't exist such points \(X\) : that is, when all points of the graph \(f(x)\) are not lower than the points of the graph \(g(x)\) . Then the answer will include all values ​​of \(b\) except the ones found.

1) Consider the values ​​of \(b\) for which the vertex of the angle is between the point \(A_I\) and the point \(A_(II)\) (including these points). In this case, all points of the graph \(f(x)\) are not lower than the points of the graph \(g(x)\) . Let's find these values ​​\(b\) :

the point \(A_I\) has coordinates \((0;2)\) , therefore, \(b=0\) ; the point \(A_(II)\) has coordinates \((1;2)\) , therefore, \(b=1\) . This means that for all \(b\in \) all points of the graph \(f(x)\) are not lower than the points of the graph \(g(x)\) .

Note that when the vertex of the angle is between the points \(A_(II)\) and \(A_(III)\), then there is always at least one point on the graph \(f(x)\) located below the graph \(g (x)\) .

2) This happens until the vertex is at the point \(A_(III)\) - when the left branch \(g(x)\) touches the right branch \(f(x)\) at the point \(x_0 \) ; and in this case again all points of the graph \(f(x)\) are not lower than \(g(x)\) . Let's find this value \(b\) .

The right branch \(f(x)\) is given by the equation \(y=x^2-x-2, x\geqslant 2\) ; the left branch \(g(x)\) is given by the equation \(y_1=2+3(x-b), x\leqslant b\).

\((x^2-x-2)"=2x-1, \quad 2x_0-1=3 \Rightarrow x_0=2 \Rightarrow y(2)=y_1(2) \Rightarrow b=\dfrac83\).

This means that for all \(b\geqslant \dfrac83\) all points of the graph \(f(x)\) will be no lower than the points of the graph \(g(x)\) .

3) The case is similarly considered when the vertex of the angle is at the point \(A_(IV)\) or to the left (the right branch \(g(x)\) touches the left branch \(f(x)\)). In this case \(b\leqslant -\dfrac53\) .

Thus, we have found the values ​​of \(b\) when all points of the graph \(f(x)\) are not lower than the points of the graph \(g(x)\)

b) Could it have turned out that initially the percentage of students who saw or heard the first line was expressed as an integer number, and after the change - as a non-integer number?

c) What is the largest integer value possible for the percentage of students in the class who have never heard or seen the first line of this poem?

a) This is possible, for example, if there are \(25\) students in the class and \(12\) of them heard the first line before the break.

b) This is possible, for example, if there are \(28\) students in the class and \(7\) of them heard the first line before the break - then before the break the first line was heard or seen \[\dfrac(7)(28)\cdot 100\% = 25\%\ \text(students,)\] and after the break \[\dfrac(8)(28)\cdot 100\% = \dfrac(200)(7)\%\ \text(students.)\]

c) If there are \(25\) people in the class and, as a result, only one person heard/saw the first line of this poem, the percentage of students in the class who never heard or saw the first line of this poem is equal to \[\dfrac(24)(25)\cdot 100 = 96\,.\]

Let us prove that this quantity could not take a larger integer value. In fact, if the percentage of students who did not hear or see the first line is an integer, then the percentage of students who heard/saw the first line is also an integer.

It is also clear that the percentage of students who did not hear or see the first line is maximum if and only if the percentage of students who heard/saw the first line is minimum.

It is possible to make the percentage of students who heard/saw the first line even smaller only in the case when exactly one student heard/saw the first line, and in the class the number of students is greater than \(25\) . Let there be \(u > 25\) students in the class, then the required percentage is \[\dfrac(1)(u)\cdot 100\,.\]

We have proven that this number must be an integer for the condition of the problem to be fulfilled, but then \(100\) must be divisible by \(u\), where \(25< u\leqslant 35\) – целое. Легко убедиться, что подходящих \(u\) нет, следовательно, окончательный ответ: \(96\) .

Answer:

In preparation for Unified State Examination for graduates It is better to use options from official sources of information support for the final exam.

To understand how to complete the exam work, you should first of all familiarize yourself with the demo versions of the KIM Unified State Examination in Physics of the current year and with the options for the Unified State Examination of the early period.

05/10/2015, in order to provide graduates with an additional opportunity to prepare for the unified state exam in physics, one version of the KIM used for conducting the Unified State Exam ahead of schedule in 2017. This real options from the exam conducted on April 7, 2017.

Early versions of the Unified State Exam in Physics 2017

Demo version of the Unified State Exam 2017 in physics

Task option + answers	variant + answer
Specification	download
Codifier	download

Demo versions of the Unified State Exam in Physics 2016-2015

Physics	Download option
2016	version of the Unified State Exam 2016
2015	variant EGE fizika

Changes in the Unified State Exam KIM in 2017 compared to 2016

The structure of part 1 of the examination paper has been changed, part 2 has been left unchanged. Tasks with a choice of one correct answer have been excluded from the examination work and tasks with a short answer have been added.

When making changes to the structure of the examination work, the general conceptual approaches to assessing educational achievements were preserved. Including remained unchanged maximum score for completing all tasks of the examination work, the distribution is preserved maximum points for tasks of different difficulty levels and approximate distribution of the number of tasks among sections school course physics and methods of activity.

A complete list of questions that can be controlled at the unified state exam 2017 is given in the codifier of content elements and requirements for the level of graduates’ training educational organizations for the 2017 Unified State Exam in Physics.

Appointment of demo version of the Unified State Exam in physics is to enable any USE participant and the general public to get an idea of ​​the structure of future CMMs, the number and form of tasks, and their level of complexity.

The given criteria for assessing the completion of tasks with a detailed answer, included in this option, give an idea of ​​the requirements for the completeness and correctness of recording a detailed answer. This information will allow graduates to develop a strategy for preparing and passing the Unified State Exam.

Approaches to selecting content and developing the structure of the KIM Unified State Examination in Physics

Each version of the examination paper includes tasks that test the mastery of controlled content elements from all sections of the school physics course, and for each section tasks of all taxonomic levels are offered. The most important from the point of view of continuing education in higher education educational institutions content elements are controlled in the same version by tasks of different levels of complexity.

The number of tasks for a particular section is determined by its content and in proportion to the teaching time allocated for its study in accordance with the approximate physics program. Various plans according to which are constructed exam options, are built on the principle of content addition so that, in general, all series of options provide diagnostics of the development of all content elements included in the codifier.

Each option includes tasks for all sections different levels difficulties that allow testing the ability to apply physical laws and formulas both in standard educational situations and in non-traditional situations that require the manifestation of a fairly high degree of independence when combining known action algorithms or creating your own plan for completing a task.

The objectivity of checking tasks with a detailed answer is ensured by uniform assessment criteria, the participation of two independent experts evaluating one work, the possibility of appointing a third expert and the presence of an appeal procedure. Single state exam in Physics is an optional exam for graduates and is intended for differentiation when entering higher education institutions.

For these purposes, the work includes tasks of three difficulty levels. Completing tasks basic level complexity allows you to assess the level of mastery of the most significant content elements of the physics course high school and mastery of the most important activities.

Among the tasks of the basic level, tasks are distinguished whose content corresponds to the standard of the basic level. The minimum number of Unified State Examination points in physics, confirming that a graduate has mastered a secondary (full) general education program in physics, is established based on the requirements for mastering the basic level standard. Use in exam paper advanced tasks and high levels complexity allows you to assess the degree of preparedness of a student to continue education at a university.