Option No. 3109295
Early Unified State Examination in Physics 2017, option 101
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On the ri-sun-ke, a graph is given for-vi-si-mo-sti of the projection of the speed of the body v x from time.
Define-de-li-te projection of the acceleration of this body a x in in-ter-va-le time-me-no from 15 to 20 s. The answer is you-ra-zi-te in m / s 2.
Answer:
Mas-soy cube M\u003d 1 kg, compressed from the sides by spring-on-mi (see ri-su-nok), in-ko-it-sya on a smooth go-ri-zone-tal table. The first spring is compressed by 4 cm, and the second is compressed by 3 cm. The stiffness of the first spring k 1 = 600 N/m. What is the stiffness of the second spring k 2? Answer you-ra-zi-te 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 long will it take from the moment when the observer sees the impact of a copra 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
Answer:
The figure shows graphs of the dependence of pressure p from the depth of immersion h for two liquids at rest: water and the heavy liquid diiodomethane, at a constant temperature.
Choose two true statements that are consistent with the given graphs.
1) If inside a hollow ball the pressure is equal to atmospheric pressure, then in water at a depth of 10 m the pressures 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) With increasing depth of immersion, the pressure in diiodomethane increases faster than in water.
5) The density of olive oil is 0.92 g/cm 3 , a similar graph of pressure versus depth for oil will be between the graph for water and the abscissa (horizontal axis).
Answer:
A massive load suspended from the ceiling on a weightless spring performs vertical free oscillations. The spring remains stretched all the time. How do the potential energy of the spring and the potential energy of the load behave in the gravitational field when the load moves upward from the equilibrium position?
1) increases;
2) decreases;
3) does not change.
Answer:
A truck moving along a straight horizontal road at a speed of v braked so that the wheels stopped turning. Truck weight m, coefficient of friction of the 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 formulas and physical quantities, the value of which can be calculated using these formulas.
| BUT | B |
Answer:
As a result of cooling rarefied argon, its absolute temperature decreased by a factor of 4. How many times did the average kinetic energy of the thermal motion of argon molecules decrease in this case?
Answer:
The working body of a heat engine receives an amount of heat from the heater equal to 100 J per cycle, and performs work of 60 J. What is the efficiency of a heat engine? Express your answer in %.
Answer:
The relative humidity of the air in a closed vessel with a piston is 50%. What will be the relative humidity of the air in the vessel if the volume of the vessel at a constant temperature is doubled? Express your answer in %.
Answer:
The hot substance, which was originally 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.
Choose from the proposed list two statements that correspond to the results of the measurements, and indicate their numbers.
1) The process of crystallization of the substance took more than 25 minutes.
2) The specific heat capacity of a substance 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 beginning of the measurements, the substance was only in the solid state.
5) After 20 min. after the beginning of the measurements, the substance was only in the 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 is the absolute temperature of the gas. Establish a correspondence between the graphs and statements that characterize the processes depicted on the graphs. For each position of the first column, select the corresponding position of the second column and write down the selected numbers in the table under the corresponding letters.
| BUT | B |
Answer:
How is the Ampère force directed relative to the figure (to the right, left, up, down, towards the observer, away from the observer), acting on conductor 1 from the side of conductor 2 (see figure), if the conductors are thin, long, straight, parallel to each other? ( I- current strength.) Write down the answer in a word (s).
Answer:
A direct current flows through a section of the circuit (see figure) I\u003d 4 A. What current strength will the ideal ammeter included in this circuit show if the resistance of each resistor r= 1 ohm? Express your answer in amperes.
Answer:
In an experiment on the observation of electromagnetic induction, a square frame of one turn of a thin wire is placed in a uniform magnetic field perpendicular to the plane of the frame. The magnetic field induction increases uniformly from 0 to the maximum value AT max per time T. In this case, an induction EMF equal to 6 mV is excited in the frame. What EMF of induction will appear in the frame if T decrease by 3 times AT max decrease 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 upwards (see figure).
From the list below, select two correct statements and indicate their numbers.
1) If to the point BUT place a test point negative charge, then a force directed vertically downwards will act on it from the side of the plate.
2) The plate has a negative charge.
3) The potential of the electrostatic field at a point AT lower than point FROM.
5) The work of the electrostatic field on the movement of a test point negative charge from a point BUT and to the point AT equals zero.
Answer:
An electron moves in a circle in a uniform magnetic field. How will the Lorentz force acting on the electron and the period of its revolution change if its kinetic energy is increased?
For each value, determine the appropriate nature of the change:
1) increase;
2) decrease;
3) will not change.
Write in the table the selected numbers for each physical quantity. 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 is the internal resistance of the current source, R is the resistance of the resistor).
For each position of the first column, select the corresponding position of the second column and write down the selected numbers in the table under the corresponding letters.
| PHYSICAL QUANTITIES | FORMULA | |
| A) current through the source with the key open K B) current through the source with the key closed K |
Answer:
Two monochromatic electromagnetic waves propagate in a vacuum. The photon energy of the first wave is twice as much as the photon energy 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 value, determine the appropriate nature of the change:
1) increase
2) decrease
3) will not change
Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.
Answer:
Determine the readings of the voltmeter (see figure), if the error of direct voltage measurement is equal to the division value of the voltmeter. Give your answer in volts. In your answer, write down the value and the error together without a space.
Answer:
To conduct laboratory work to detect the dependence of the resistance of the 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 the student take in order to conduct this study?
Exercise 1
A pack of chips costs \(170\) rubles. What is the largest number of packs of chips that 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 condition of the problem, it is necessary 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 equals \(8\) .
Answer: 8
Task 2
The graph shows the process of warming up the engine of an old motorcycle. The abscissa shows the time in minutes that has elapsed since the engine was started, and the ordinate shows the engine temperature in degrees Fahrenheit. Determine from the graph how many minutes the engine warmed up from the temperature \(60^\circ F\) to the temperature \(100^\circ F\) .
The engine warmed up to \(60^\circ F\) after \(3\) minutes after starting, and to \(100^\circ F\) after \(8\) minutes after starting. From \(60^\circ F\) to \(100^\circ F\) the engine warmed up \(8 - 3 = 5\,\) minutes.
Answer: 5
Task 3
On checkered paper with cell size \(1\times 1\) the angle \(AOB\) is shown. Find the tangent of this angle.
\[\mathrm(tg)\,(\beta - \alpha) = \dfrac(\mathrm(tg)\,\beta - \mathrm(tg)\,\alpha)(1 + \mathrm(tg)\, \alpha\cdot \mathrm(tg)\,\beta)\] The 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, \(7\) caps from \(40\) 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: \[\sqrt(13x - 13) = 13\quad\Rightarrow\quad 13x - 13 = 13^2\quad\Rightarrow\quad 13x = 182\quad\Rightarrow\quad x = 14\]- suitable for ODZ.
Answer: 14
Task 6
In a right triangle \(ABC\) the angle \(C\) is equal to \(90^\circ\) , \(AB = 6\) , \(\mathrm(tg)\, A = \dfrac(1)(2\sqrt(2))\). Look for \(BC\) .
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 line \(y = 2x - 1\) is tangent to the graph of the function \(y = x^3 + 6x^2 + 11x - 1\) . Find the abscissa of the point of contact.
At the point of contact of the line \(y = 2x - 1\) and the graph of the function \(y = x^3 + 6x^2 + 11x - 1\), the derivative of this function coincides with the slope \(k\) of the line, which in the given case is equal to \(2\) .
Then \ The roots of the last equation: \
Let's check for which of the obtained \(x\) the line and the graph have a common point:
for \(x = -3\) :
the ordinate of a point on the line is \(2\cdot(-3) - 1 = -7\) , and the ordinate of a point on the graph is \[(-3)^3 + 6\cdot(-3)^2 + 11\cdot(-3) - 1 = -7,\] that is, the 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 line, therefore, they touch at this point.
for \(x = -1\) :
the ordinate of a point on the line is \(2\cdot(-1) - 1 = -3\) , and the ordinate of a point on the graph is \[(-1)^3 + 6\cdot(-1)^2 + 11\cdot(-1) - 1 = -7,\] that is, the ordinates of these points are different, therefore, for \(x = -1\) the line and the graph do not have a common point.
Total: \(-3\) - the desired abscissa.
Answer: -3
Task 8
Find the surface area of the polyhedron shown in the figure (all dihedral angles are right).
The surface area of a given polyhedron is equal to the surface area of a cuboid with dimensions \(10\times 12\times 13\) and thus equals \(2\cdot(10\cdot 12 + 12\cdot 13 + 10\cdot 13) = 812\).
Answer: 812
Task 9
Find the value of an expression \[\sqrt(48)\sin^2 \dfrac(\pi)(12) - 2\sqrt(3)\]
We use the double angle cosine formula: \(\cos 2x = 1 - 2\sin^2x\) , then for \(x = \dfrac(y)(2)\) 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 pulls a car with a force \(120\,\) kN directed at an acute angle \(\alpha\) to the horizon. The work of a truck (in kilojoules) on a section of length \(l = 150\,\) m is calculated by the formula \(A = Fl\cos\alpha\) . At what maximum angle \(\alpha\) (in degrees) will the work done be at least \(9000\,\) kJ?
By the condition of the problem, we have: \
Given that \(\alpha\in\), we get that \(\alpha\leqslant 60^\circ\) (this can be easily verified by looking at the trigonometric circle).
So the answer is: with \(\alpha = 60^\circ\) .
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 \[\dfrac(1)(9) + \dfrac(1)(15) + \dfrac(1)(10) = \dfrac(25)(90)\] is the part of the pool filled per minute by all three pumps, if the contribution of each pump is taken into account twice. Then \[\dfrac(1)(2)\cdot\dfrac(25)(90) = \dfrac(25)(180)\]- the part of the pool filled per 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 the smallest value of the function \ on the segment
ODZ: \ Let's decide on ODZ:
1) \
Let's find the critical points (that is, the internal points of the domain of the function, in 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) Find the intervals of constant sign \(y"\) :
3) Find the intervals of constant sign \ (y "\) on the segment under consideration \(\left[\dfrac(1)(242);\dfrac(5)(242)\right]\):
4) Sketch of the graph on the 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 that belong 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 \(t = \sinx\) : \
The roots of the last equation: \ whence \(\sin x = 1\) or \(\sin x = -\dfrac(1)(2)\)
1) \(\sin x = 1\) , therefore, \(x = \dfrac(\pi)(2) + 2\pi n\)- do not fit the ODZ.
2) \(\sin x = -\dfrac(1)(2)\)
where \(x_1 = -\dfrac(\pi)(6) + 2\pi k\), \(x_2 = \dfrac(7\pi)(6) + 2\pi k\), \(k\in\mathbb(Z)\) – fit according to ODZ.
b) \(-\pi \leqslant -\dfrac(\pi)(6) + 2\pi k \leqslant \dfrac(\pi)(2)\) is tantamount to \(-\dfrac(5\pi)(6) \leqslant 2\pi k \leqslant \dfrac(4\pi)(6)\), 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: \(x = -\dfrac(\pi)(6)\)
\(-\pi \leqslant \dfrac(7\pi)(6) + 2\pi k \leqslant \dfrac(\pi)(2)\) is tantamount to \(-\dfrac(13\pi)(6) \leqslant 2\pi k \leqslant -\dfrac(4\pi)(6)\), 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 side edge \(AA_1\) in relation to \(AM: MA_1 = 1: 3\) . 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 \(DD_1\) with respect to \(D_1N: DD_1 = 1: 2\) .
b) Find the cross-sectional area if it is known that \(AB = 5\) , \(AA_1 = 8\) .
a) Because the prism is regular, then it is a straight line and its base is a square \ (ABCD \) .
Denote \(AM=x\) , then \(MA_1=3x\) . Because \(\alpha\parallel AC\) , then \(\alpha\) will intersect the plane \(ACC_1\) , which contains the line \(AC\) , along the 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 line \(QQ_1\) passing through the points of intersection 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, \(OQ\parallel AA_1 \Rightarrow OQ\perp (ABC)\). So \(OQ=AM=x\) .
\(\triangle OQB\sim \triangle NDB\) 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 whole edge is \(DD_1=AA_1=4x\) , so \(N\) is the middle of \(DD_1\) .
b) By the three perpendiculars theorem ( \(OQ\perp (ABC), \text(projection ) BQ\perp AC\)) oblique \(BO\perp AC\Rightarrow BO\perp MK\)(because \(AC\parallel MK\) ). So \(BN\perp MK\) .
The area of a convex quadrilateral whose diagonals are mutually perpendicular is equal to half the product of the diagonals, that is \(S_(MBKN)=\dfrac 12MK\cdot BN\). 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: \[\log_6(t + 3)\cdot\lg(t^2 + 1)\geqslant 1\]
For \(t > 0\) 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 see that it holds for \(t = 3\) , therefore, only for \(t\geqslant 3\) will the last inequality hold.
In this way, \[\sqrt(x^2 + 4x - 5)\geqslant 3,\] which is equivalent to ODZ \ from where, taking into account the ODZ \
Answer:
Q.E.D.
b) Denote \(MA = ka\) , \(AN = a\) (then the desired value is \(k\) ), therefore \(NB = a\) , then \(BK = 2a\) .
According to the tangent segment theorem: \
Let's write the cosine theorem for the triangle \(MNK\) : \ Substituting the known values, 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 \ (40 \) years in equal monthly payments, while he will have to pay the amount by \ (180 \% \) exceeding the original. Instead, Timur can rent a shopping center for some time (the cost of rent is \(1\) million rubles per month), setting aside every 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 rent for a rented mall. 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, in a 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 set aside \(3.5 - 1 \u003d 2.5\) 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 \(b\) is an angle whose branches are directed downwards, and the vertex is at the point \((b;2)\) .
Then the meaning of the inequality is as follows: it is necessary to find those values \(b\) for which there is at least one point \(X\) of the graph \(f(x)\) , which is below the graph of the function \(g(x)\) .
Let's find those values \(b\) when does not 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 all values \(b\) will be returned in response, except those found.
1) Consider the values \(b\) for which the vertex of the corner is between the point \(A_I\) and the point \(A_(II)\) (including these points). In this case, all graph points \(f(x)\) are not lower than the graph points \(g(x)\) . Let's find these values \(b\) :
point \(A_I\) has coordinates \((0;2)\) , hence \(b=0\) ; point \(A_(II)\) has coordinates \((1;2)\) , hence \(b=1\) . Hence, 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 corner vertex is between the points \(A_(II)\) and \(A_(III)\) , then there is always at least one point of the graph \(f(x)\) that is 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 plot \(f(x)\) are not below \(g(x)\) . 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 not lower than the points of the graph \(g(x)\) .
3) The case is considered similarly when the vertex of the corner 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 \(b\) when all points of the graph \(f(x)\) will be not lower than the points of the graph \(g(x)\) b) Could it be that initially the percentage of students who saw or heard the first line was expressed as an integer, and after the change - as a non-integer? c) What is the largest integer value that the percentage of students in the class who never heard or saw the first line of this poem can take?
a) This is possible, for example, if in the class \(25\) students and \(12\) of them heard the first line before the break.
b) This is possible, for example, if in the class \(28\) students 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 change \[\dfrac(8)(28)\cdot 100\% = \dfrac(200)(7)\%\ \text(students.)\]
c) If in the class \(25\) a person 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 and did not see the first line of this poem is equal to \[\dfrac(24)(25)\cdot 100 = 96\,.\]
Let us prove that this value could not take on a larger integer value. Indeed, 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 and did not see the first line is maximum if and only if the percentage of students who heard / saw the first line is minimal.
It is possible to make the percentage of students who heard/saw the first line even smaller only if exactly one student heard/saw the first line, and the number of students in the class is more than \(25\) . Let there be \(u > 25\) students in the class, then the desired percentage is \[\dfrac(1)(u)\cdot 100\,.\]
We have proved 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:
When preparing for the exam, graduates are better off using options from official sources of information support for the final exam.
To understand how to do the examination work, you should first of all familiarize yourself with the demo versions of the KIM USE in physics of the current year and with the USE options for the early period.
On May 10, 2015, in order to provide graduates with an additional opportunity to prepare for the unified state exam in physics, the FIPI website publishes one version of the KIM used to conduct the USE of the early period of 2017. These are real options from the exam held on 04/07/2017.
Early versions of the exam in physics 2017
Demonstration version of the exam 2017 in physics
| Task option + answers | option+answer |
| Specification | download |
| Codifier | download |
Demo versions of the exam in physics 2016-2015
| Physics | Download option |
| 2016 | version of the exam 2016 |
| 2015 | variant EGE fizika |
Changes in KIM USE in 2017 compared to 2016
The structure of part 1 of the examination paper has been changed, part 2 has been left unchanged. From the examination work, tasks with the choice of one correct answer were excluded and tasks with a short answer were added.
When making changes to the structure of the examination work, the general conceptual approaches to the assessment of educational achievements were preserved. In particular, the maximum score for completing all tasks of the examination paper remained unchanged, the distribution of maximum scores for tasks of different levels of complexity and the approximate distribution of the number of tasks by sections of the school physics course and methods of activity were preserved.
A complete list of questions that can be controlled at the unified state exam in 2017 is given in the codifier of content elements and requirements for the level of preparation of graduates of educational organizations for the unified state exam in 2017 in physics.
The purpose of the demonstration version of the Unified State Examination in Physics is to enable any participant in the Unified State Examination and the general public to get an idea of the structure of future KIM, the number and form of tasks, and the level of their complexity.
The given criteria for evaluating the performance of tasks with a detailed answer, included in this option, give an idea of the requirements for the completeness and correctness of writing a detailed answer. This information will allow graduates to develop a strategy for preparing and passing the exam.
Approaches to the selection of content, the development of the structure of the KIM USE in physics
Each version of the examination paper includes tasks that test the development of controlled content elements from all sections of the school physics course, while tasks of all taxonomic levels are offered for each section. The most important content elements from the point of view of continuing education in higher educational institutions are controlled in the same variant by tasks of different levels of complexity.
The number of tasks for a particular section is determined by its content content and in proportion to the study time allotted for its study in accordance with an exemplary program in physics. Various plans, according to which the examination options are constructed, are built on the principle of a content addition so that, in general, all series of options provide diagnostics for the development of all the content elements included in the codifier.
Each option includes tasks in all sections of different levels of complexity, allowing you to test the ability to apply physical laws and formulas both in typical educational situations and in non-traditional situations that require a sufficiently high degree of independence when combining known action algorithms or creating your own task execution plan .
The objectivity of checking tasks with a detailed answer is ensured by uniform evaluation criteria, the participation of two independent experts evaluating one work, the possibility of appointing a third expert and the existence of an appeal procedure. The Unified State Examination in Physics is an exam of choice for graduates and is designed to differentiate when entering higher education institutions.
For these purposes, tasks of three levels of complexity are included in the work. Completing tasks of a basic level of complexity allows assessing the level of mastering the most significant content elements of a high school physics course and mastering the most important activities.
Among the tasks of the basic level, tasks are distinguished, the content of which corresponds to the standard of the basic level. The minimum number of USE points in physics, which confirms that the graduate has mastered the program of secondary (complete) general education in physics, is set based on the requirements for mastering the basic level standard. The use of tasks of increased and high levels of complexity in the examination work allows us to assess the degree of readiness of the student to continue education at the university.
