problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
236. Write the number $z=4\left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)$ in algebraic and exponential forms. | Solution. Since the argument $\varphi$ of the given number is $4 \pi / 3$, the number $z$ corresponds to a point on the complex plane located in the III quadrant. Using the reduction formulas, we find
$$
\begin{aligned}
\cos \frac{4 \pi}{3}=\cos \left(\pi+\frac{\pi}{3}\right)= & -\cos \frac{\pi}{3}=-\frac{1}{2}, \sin ... | -2-2i\sqrt{3},\4e^{\frac{4\pi}{3}i} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,853 |
254. Given complex numbers $z_{1}=3\left(\cos 330^{\circ}+i \sin 330^{\circ}\right)$ and $z_{2}=2\left(\cos 60^{\circ}+i \sin 60^{\circ}\right) . \quad$ Find: a) $z_{1} z_{2} ;$ b) $z_{1} / z_{2} ;$ c) $z_{2}^{4}$; d) $\sqrt[3]{z_{1}}$ | Solution. a) According to formula (5), we get
$$
\begin{gathered}
z_{1} z_{2}=\left[3\left(\cos 330^{\circ}+i \sin 330^{\circ}\right)\right] \cdot\left[2\left(\cos 60^{\circ}+i \sin 60^{\circ}\right)\right]=3 \cdot 2\left[\operatorname { c o s } \left(330^{\circ}+\right.\right. \\
\left.\left.+60^{\circ}\right)+i \sin... | \begin{aligned}&z_{1}z_{2}=3\sqrt{3}+3i,\\&\frac{z_{1}}{z_{2}}=-1.5i,\\&z_{2}^{4}=-8-8i\sqrt{3},\\&\sqrt[3]{z_{1}}=\sqrt[3]{3}(\cos110^{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,854 |
255. Given: $z_{1}=3\left(\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4}\right), \quad z_{2}=5\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$. Find: a) $z_{1} z_{2}$; b) $z_{1} / z_{2}$; c) $z_{1}^{5}$; d) $\sqrt{z_{1}}$. | Solution. a) We have $\left|z_{1} z_{2}\right|=3.5=15 ; \quad \arg \left(z_{1} z_{2}\right)=\frac{5 \pi}{4}+\frac{\pi}{2}=$ $=\frac{7 \pi}{4} .3$ so
$$
z_{1} z_{2}=15\left(\cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4}\right)
$$
Using the reduction formulas:
$$
\begin{aligned}
\cos \frac{7 \pi}{4}=\cos \left(2 \pi-\fra... | 7.5\sqrt{2}-7.5\sqrt{2}i,-0.3\sqrt{2}+0.3\sqrt{2}i,121.5\sqrt{2}+121.5\sqrt{2}i,\sqrt{5}(\cos\frac{5\pi}{8}+i\sin\frac{5\pi}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,855 |
264. Find $\sqrt[3]{z}$, if $z=1-i$. | Solution. Let's write the complex number $z$ in trigonometric form. Find $r=\sqrt{a^{2}+b^{2}}=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$. Since $a=1$ and $b=-1$, the point corresponding to this number is located in the IV quadrant.
Form the ratios
$$
\cos \varphi=a / r=1 / \sqrt{2}=\sqrt{2} / 2 ; \quad \sin \varphi=b / r=-1 / ... | z^{(1)}=\sqrt[6]{2}(\cos\frac{7\pi}{12}+i\sin\frac{7\pi}{12}),z^{(2)}=\sqrt[6]{2}(\cos\frac{15\pi}{4}+i\sin\frac{15\pi}{4}),z^{(3)}=\sqrt[6]{2}(\cos | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,856 |
265. Find $z^{6}$, if $z=-\sqrt{3}+i$. | Solution. Let's write the number $z$ in trigonometric form, considering that $a=-\sqrt{3}, b=1$. We find $r=\sqrt{a^{2}+b^{2}} ; r=\sqrt{3+1}=2$. The point $z$ is located in the second quadrant. We form the ratios
$$
\cos \varphi=a / r=\sqrt{3} / 2, \sin \varphi=b / r=1 / 2
$$
Considering that the point $z$ is locate... | -64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,857 |
266. Calculate $z=\sqrt[4]{-16}$. | Solution. Let's write the number -16 in trigonometric form. We find \( r = \sqrt{a^2 + b^2} = \sqrt{(-16)^2 + 0} = 16 \). Here, \( a = -16 \), \( b = 0 \), and thus the point is located on the negative part of the \( O x \) axis; therefore, \( \varphi = \pi \).
To find the roots, we use the formula
\[
\sqrt[4]{z} = \... | z^{(1)}=\sqrt{2}+i\sqrt{2},\quadz^{(2)}=-\sqrt{2}+i\sqrt{2},\quadz^{(3)}=-\sqrt{2}-i\sqrt{2},\quadz^{(4)}=\sqrt{2}-i\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,858 |
18. Given points $A(3 ; 2), B(-1 ; 5), C(0 ; 3)$. Find the coordinates of the vectors $\overrightarrow{A B}, \overrightarrow{B C}, \overrightarrow{A C}$. | Solution. To find the coordinates of a vector, we subtract the coordinates of the beginning from the coordinates of the end. Then we get: $\overrightarrow{A B}=(-1-3$; $5-2)=(-4 ; 3) ; \overrightarrow{B C}=(0-(-1) ; 3-5)=(1 ;-2) ; \overrightarrow{A C}=(0-3 ; 3-2)=$ $=(-3 ; 1)$. | \overrightarrow{AB}=(-4;3),\overrightarrow{BC}=(1;-2),\overrightarrow{AC}=(-3;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,866 |
21. Given vectors $\vec{a}=(3 ; 5), \vec{b}=(2 ;-7)$. Find: a) $\vec{a}+\vec{b}$; b) $\vec{a}-\vec{b}$; c) $4 \vec{a}$; d) $-0.5 \vec{b}$. | Solution. According to the given rules, we get:
a) $\vec{a}+\vec{b}=(3+2 ; 5-7)=(5 ;-2)$;
b) $\vec{a}-\vec{b}=(3-2 ; 5-(-7))=(1 ; 12)$;
c) $4 \vec{a}=(4 \cdot 3 ; 4 \cdot 5)=(12 ; 20)$;
d) $-0.5 \vec{b}=(-0.5 \cdot 2 ;-0.5 \cdot(-7))=(-1 ; 3.5)$. | )(5;-2);b)(1;12);)(12;20);)(-1;3.5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,867 |
25. Find the length of the vector: a) $\vec{a}=(5 ; 12) ;$ b) $\vec{b}=(7 ;-1)$. | Solution. According to formula (1), we have:
a) $|\vec{a}|=\sqrt{x^{2}+y^{2}}=\sqrt{25+144}=\sqrt{169}=13$;
b) $|\vec{b}|=\sqrt{x^{2}+y^{2}}=\sqrt{49+1}=\sqrt{50}=5 \sqrt{2}$. | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,868 |
30. On the x-axis, find a point that is 5 units away from the point $M(1 ; 3)$. | Solution. Let the required point be denoted as \( A(x ; 0) \) (since by the condition it lies on the x-axis). Then the length of the segment \( AM \) can be expressed by the formula \( |AM| = \sqrt{(x_M - x_A)^2 + (y_M - y_A)^2} \), from which, substituting the coordinates of the points and the known distance, we have
... | A_1(-3;0),A_2(5;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,869 |
33. Point $K$ divides the segment $M N$ in the ratio $|M K|:|K N|=$ $=2: 3$. Find the coordinates of point $K$, if $M(7 ; 4) ; N(-3 ; 9)$. | Solution. Substituting $\lambda=2 / 3$ and the coordinates of points $M$ and $N$ into formula (3), we find
$$
\begin{gathered}
x_{K}=\frac{x_{M}+\lambda x_{N}}{1+\lambda}=\frac{7+\frac{2}{3}(-3)}{1+\frac{2}{3}}=\frac{7-2}{\frac{5}{3}}=3 \\
y_{K}=\frac{y_{M}+\lambda y_{N}}{1+\lambda}=\frac{4+\frac{2}{3} \cdot 9}{1+\fra... | K(3;6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,870 |
34. Divide the segment $A B$, given by the points $A(5 ; 1)$ and $B(-4 ;-14)$, into three equal parts. | Solution. Let $M$ and $N$ be the points of division (Fig. 40). We will form the ratios for these points. We have $|A M|:|M B|=1: 2$, i.e., $\lambda_{M}=1 / 2 ; \quad|A N|:|N B|=2: 1$, i.e., $\lambda_{N}=2$. Now, substituting these ratios and the coordinates of points $A$ and $B$ into formula (3), we find
$$
x_{M}=\fra... | M(2,-4),N(-1,-9) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,871 |
40. In an equilateral triangle $ABC$ with a side length of 6 (Fig. 42), find the scalar product of the vectors: a) $\overrightarrow{A B}$ and $\overrightarrow{A C} ;$ b) $\overrightarrow{A B}$ and $\overrightarrow{B C}$. | Solution. a) Since the angle $\varphi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$ (and their directions) is $60^{\circ}$, for the scalar product of these vectors we get
$$
\overrightarrow{A B} \cdot \overrightarrow{A C}=|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \cdot \cos B \widehat... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,872 |
42. Two vectors are given such that $|\vec{a}|=5,|\vec{b}|=3$, and the angle between them is $45^{\circ}$. Find $(\vec{a}+\vec{b})^{2}$. | Solution. We have $(\vec{a}+\vec{b})^{2}=\vec{a}^{2}+2 \vec{a} \vec{b}+\vec{b}^{2}$. Since $|\vec{a}|=5$ and $|\vec{b}|=3$, then $\vec{a}^{2}=|\vec{a}|^{2}=25, \vec{b}^{2}=|\vec{b}|^{2}=9$, from which $2 \vec{a} \vec{b}=2|\vec{a}| \cdot|\vec{b}| \cos 45^{\circ}=$ $=2 \cdot 5 \cdot 3 \cdot 0.5 \sqrt{2}=15 \sqrt{2}$. The... | 34+15\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,873 |
45. Find the scalar product of vectors $\vec{a}=(3 ; 5)$ and $\vec{b}=(-2 ; 7)$. | Solution. Here $x_{a}=3, x_{b}=-2, y_{a}=5, y_{b}=7$. Using formula (3), we get
$$
\vec{a} \vec{b}=3 \cdot(-2)+5 \cdot 7=-6+35=29
$$
46-51. Find the scalar product of the vectors: | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,874 |
52. Find the angle between the vectors: a) $\vec{a}=(4 ; 0)$ and $\vec{b}=(2 ;-2)$; b) $\vec{a}=(5 ;-3)$ and $\vec{b}=(3 ; 5)$. | Solution. a) Using formula (5), we find
$\cos \varphi=\frac{4 \cdot 2-0 \cdot(-2)}{\sqrt{16+0} \cdot \sqrt{4+4}}=\frac{2}{2 \sqrt{2}}=\frac{\sqrt{2}}{2} ; \varphi=\arccos \frac{\sqrt{2}}{2}=\frac{\pi}{4}$.
b) We have
$$
\cos \varphi=\frac{5 \cdot 3+(-3) \cdot 5}{\sqrt{25+9} \cdot \sqrt{9+25}}=\frac{0}{34}=0 ; \varphi... | \frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,875 |
71. Determine whether the points $A(2 ; 5)$ and $B(1 ; 2.2)$ lie on the line given by the equation $3 x-5 y+8=0$. | Solution. Substituting the coordinates of point $A$ into the equation, we get $3 \cdot 2 - 5 \cdot 5 + 8 \neq 0, 6 - 25 + 8 \neq 0$. Therefore, point $A$ does not belong to the given line.
Substituting the coordinates of point $B: 3 \cdot 1 - 5 \cdot 2.2 + 8 = 0; 11 - 11 = 0$. Therefore, point $B$ lies on the given li... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,876 |
74. Points $A(x ; 3)$ and $B(-5 ; y)$ belong to the line given by the equation $7 x+2 y=41$. Find the unknown coordinates of the points. | Solution. Since points $A$ and $B$ belong to the given line, their coordinates satisfy the equation of this line. Substituting the known coordinates into this equation, we get an equation with one unknown:
$$
\begin{gathered}
A(x ; 3) ; 7 x+2 \cdot 3=41 ; 7 x=35 ; x=5, \text { i.e. } A(5 ; 3) \\
B(-5 ; y) ; 7 \cdot(-5... | A(5;3),B(-5;38) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,877 |
75. Parametric equations of a line are given: $x=R \cos t$, $y=R \sin t$. Convert them to an equation with two variables. | Solution. From the second equation
Fig. 44, we express $\sin t=\frac{y}{R}$. Knowing that $\cos t=$ $=\sqrt{1-\sin ^{2} t}$, we find $\cos t=\sqrt{1-\frac{y^{2}}{R^{2}}}$.
Now substitute ... | x^{2}+y^{2}=R^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,878 |
76. Parametric equations of the line $x=2 \cos t, y=3 \sin t$, where $0 \leqslant t \leqslant 2 \pi$, convert to an equation with two variables. | Solution. Expressing from the second equation $\sin t=\frac{y}{3}$ and considering that $\cos t=\sqrt{1-\sin ^{2} t}$, we find $\cos t=\sqrt{1-\frac{y^{2}}{9}}$. Next, substituting this expression into the first equation, we have
$$
x=2 \sqrt{1-\frac{y^{2}}{9}} ; x^{2}=4\left(1-\frac{y^{2}}{9}\right) ; \frac{x^{2}}{4}... | \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,879 |
80. A point $M_{1}(7 ; -8)$ and a normal vector of the line $\vec{n}=(-2 ; 3)$ are known. Form the equation of the line. | Solution. $1^{0}$. Choose a point $M(x ; y)$.
$2^{0}$. Find the vector $\overrightarrow{M_{1} M}=(x-7 ; y+8)$.
$3^{0}$. The normal vector $\vec{n}=(2 ; 3)$, i.e., $A=2, B=3$.
$4^{0}$. Write the equation of the desired line:
$$
2(x-7)+3(y+8)=0
$$
from which
$$
2 x-14+3 y+24=0 ; 2 x+3 y+10=0
$$
- the desired equat... | 2x+3y+10=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,881 |
81. Form the equation of the height $A D$ of the triangle defined by the points $A(-5 ; 3), B(3 ; 7), C(4 ;-1)$. | Solution. $1^{0}$. The height $A D$ passes through points $A(-5 ; 3)$ and point $D(x ; y)$ with unknown coordinates.
$2^{0}$. We find the vector $\overrightarrow{A D}=(x+5 ; y-3)$.
$3^{\circ}$. We find the vector $\overrightarrow{B C}$, defined by points $B(3 ; 7)$ and $C(4 ;-1)$; we have $\overrightarrow{B C}=(1 ;-8... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,882 |
88. Form the equation of the line passing through the point $A(3, -2)$ and having the direction vector $\vec{n}=(-5, 3)$. | Solution. $1^{0}$. Choose a point $M(x ; y) \in l$.
$2^{0}$. Find the vector $\overrightarrow{A M}=(x-3 ; y+2)$.
$3^{0}$. Direction vector $\vec{n}=(-5 ; 3)$.
$4^{\circ}$. Write the equation of the line:
$$
\frac{x-3}{-5}=\frac{y+2}{3}
$$
from which $3 x-9=-5 y-10 ; 3 x+5 y+1=0$ - the desired equation in general f... | 3x+5y+1=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,883 |
89. The triangle is defined by points $A(5 ; 2), B(-1 ;-4), C(-5$; -3). Form the equation of the line passing through point $B$ and parallel to $A C$. | Solution. $1^{0}$. Choose a point $M(x ; y)$.
$2^{0}$. Find the vector $\overrightarrow{B M}=(x+1 ; y+4)$.
$3^{0}$. Find the vector defined by points $A(5 ; 2)$ and $C(-5 ;-3)$; we have $\overrightarrow{A C}=(-10 ;-5)$.
$4^{0}$. Since the desired line and line $A C$ are parallel, their direction vectors are collinear... | x-2y-7=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,884 |
112. Find the angle between the lines $x+5 y-3=0$ and $2 x-3 y+$ $+4=0$. | Solution. Let's find the coordinates of the normal vectors of the given lines: $\vec{n}_{1}=(1 ; 5), \vec{n}_{2}=(2 ;-3)$. According to formula (5), we get
$$
\begin{gathered}
\cos \varphi=\left|\frac{\vec{n}_{1} \vec{n}_{2}}{\left|\vec{n}_{1}\right| \vec{n}_{2} \mid}\right|=\left|\frac{1 \cdot 2+5 \cdot(-3)}{\sqrt{1+... | \frac{\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,887 |
118. Compose the equation of a circle with center $O(3; -2)$ and radius $r=5$. | Solution. Substituting $a=3, b=-2$ and $r=5$ into equation (1), we get $(x-3)^{2}+(y+2)^{2}=25$. | (x-3)^{2}+(y+2)^{2}=25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,888 |
119. Form the equation of a circle with the center at the origin and radius $r$. | Solution. In this case $a=0 ; b=0$. Then the equation will take the form $x^{2}+y^{2}=r^{2}$. | x^{2}+y^{2}=r^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,889 |
121. Construct the circle $x^{2}+y^{2}+6 x-4 y-3=0$. | Solution. To construct a circle, it is necessary to find its center and radius. For this, we need to complete the squares in the equation, i.e., transform the equation into the form $(x-a)^{2}+(y-b)^{2}=r^{2}$. We have
$$
x^{2}+y^{2}+6 x-4 y-3=0 ;\left(x^{2}+6 x\right)+\left(y^{2}-4 y\right)=3
$$
By adding the sum $9... | (x+3)^{2}+(y-2)^{2}=16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,890 |
123. Find the coordinates of the foci, the lengths of the axes, and the eccentricity of the ellipse given by the equation $2 x^{2}+y^{2}=32$. | Solution. We will bring the equation of the ellipse to the canonical form (2). For this, we divide all its terms by 32:
$$
\frac{2 x^{2}}{32}+\frac{y^{2}}{32}=\frac{32}{32} ; \frac{x^{2}}{16}+\frac{y^{2}}{32}=1
$$
We see that $a^{2}=16, b^{2}=32$, from which $a=4, b=4 \sqrt{2}$.
Since $b>a$, the foci of the ellipse ... | F_{1}(0;4),F_{2}(0;-4),2b=8\sqrt{2},2a=8,\varepsilon=\frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,891 |
124. Form the canonical equation of an ellipse for which the minor axis $2 b=6$, and the distance between the foci $\left|F_{1} F_{2}\right|=8$. | Solution. Since the minor axis $2b=6$, the foci are located on the $Ox$ axis. We have $b=3, c=4$. From the relation $a^{2}-c^{2}=b^{2}$, we find $a^{2}=b^{2}+c^{2}=9+16=25$, i.e., $a=5$. Therefore, the canonical equation of the ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$. | \frac{x^{2}}{25}+\frac{y^{2}}{9}=1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,892 |
131. Find the coordinates of the foci, the lengths of the axes, the eccentricity, and the equations of the asymptotes of the hyperbola given by the equation $16 x^{2}-$ $-25 y^{2}=400$. | Solution. We will bring the equation to the canonical form (3). For this, we divide all its terms by 400:
$$
\frac{16 x^{2}}{400}-\frac{25 y^{2}}{400}=\frac{400}{400} ; \quad \frac{x^{2}}{25}-\frac{y^{2}}{16}=1
$$
From this equation, it is clear that the foci of the hyperbola are located on the x-axis, and we can wri... | F_{1}(-\sqrt{41};0);F_{2}(\sqrt{41};0),2a=10,2b=8,\varepsilon=\sqrt{41}/5,\(4/5)x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,893 |
132. Form the canonical equation of a hyperbola with foci on the x-axis, given that the eccentricity $\varepsilon=1.5$, and the focal distance is 6. | Solution. Since $\left|F_{1}, F_{2}\right|=6$, then $2c=6$, i.e., $c=3$. Next, substituting the known values of $\varepsilon$ and $c$ into the formula $\varepsilon=c / a$, we get $1.5=3 / a$, from which $a=2$. Knowing $a$ and $c$, from the relation $c^{2}=a^{2}+b^{2}$ we find $9=4+b^{2}$, from which $b^{2}=5$.
Thus, t... | x^{2}/4-y^{2}/5=1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,894 |
138. Form the equation of an equilateral hyperbola with foci on the $O x$ axis, passing through the point $M(4; -2)$. | Solution. Since point $M$ belongs to the hyperbola, its coordinates satisfy the canonical equation of the hyperbola:
$$
x^{2}-y^{2}=a^{2} ; 4^{2}-(-2)^{2}=a^{2} ; a^{2}=12
$$
Therefore, the desired equation is $x^{2}-y^{2}=12$. | x^{2}-y^{2}=12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,895 |
142. Find the coordinates of the focus and the equation of the directrix of the parabola given by the equation $y^{2}=8 x$. | Solution. From the given canonical equation of the parabola, it follows that $2 p=8$, i.e., $p=4$, from which $p / 2=2$. Therefore, the point $F(2 ; 0)$ is the focus of the parabola, and $x=2$ is the equation of its directrix. | F(2;0),x=-2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,896 |
143. Find the canonical equation of the parabola and the equation of its directrix, given that the vertex of the parabola lies at the origin, and the focus has coordinates $(0; -3)$. | Solution. According to the condition, the focus of the parabola is located on the negative half-axis $O y$, i.e., its equation has the form $x^{2}=-2 p y$. Since $-p / 2=-3$, then $p=6$, from which $2 p=12$. Therefore, the equation of the parabola is $x^{2}=-12 p y$, and the equation of the directrix is $y=3$ or $y-3=0... | x^{2}=-12y,\3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,897 |
144. Form the equation of a parabola with its vertex at the origin, symmetric about the $O x$ axis, and passing through the point $A(-3; -6)$. | Solution. From the condition, we conclude that the equation of the parabola should be sought in the form $y^{2}=-2 p x$. Since point $A$ belongs to the parabola, its coordinates satisfy this equation:
$$
36=-2 p(-3) ; 36=3 \cdot 2 p ; 2 p=12
$$
Thus, the equation of the parabola has the form $y^{2}=-12 x$. | y^{2}=-12x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,898 |
1. According to Boyle-Mariotte's law, during an isothermal process $P V=C$, where $P$ is the pressure of the gas, and $V$ is the volume it occupies. Indicate in this formula the variables and constant quantities. | Solution. Here the quantity $C$ is a constant for a given gas and a given temperature; the quantities $P$ and $V$ are variables. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,899 |
2. The period of small oscillations $T$ of a mathematical pendulum is calculated by the formula $T=2 \pi \sqrt{\frac{l}{g}}$, where $l$ is the length of the pendulum, $g$ is the acceleration due to gravity. Which of the quantities in this formula are constants, and which are variables? | Solution. Here $g$ is a constant, which does not change at this point on the Earth's surface; 2 and $\pi$ are absolute constants; $l$ and $T$ are variables. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,900 |
4. What do the notations mean: a) $(a, \infty) ;[a, \infty) ;$ b) $(-\infty, b)$; $(-\infty, b] ?$ | Solution. a) The notation $(a, \infty)$ should be understood as the set of real numbers greater than the number $a$, and the notation $[a, \infty)$ as the set of real numbers greater than or equal to (i.e., not less than) $a$.
b) The notation $(-\infty, b)$ means the set of real numbers less than the number $b$, and t... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,901 |
5. Which of the following statements are incorrect: $3.4 \in [2 ; 3.4) ; 3.4 \in [2 ; 3.4] ; 3.4 \in [2,3] ; 3.4 \in (2,5) ; 3.4 \in [3.4 ; 5)$ ? | Solution. The records $3.4 \in [2 ; 3.4)$ and $3.4 \in [2,3]$ are incorrect, since the number 3.4 does not satisfy the inequalities $2 \leqslant 3.4 < 3.4$ and $2 \leqslant 3.4 \leqslant 3$. | 3.4\in[2;3.4)3.4\in[2,3] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,902 |
7. The path traveled by a freely falling body is expressed by the formula $s=\frac{g t^{2}}{2}$, where $g-$ is the acceleration of a freely falling body, a constant for a given latitude. Indicate the independent and dependent variables. | Solution. By assigning different values to time $t$, we can determine the path $s$ for any given time interval $t$. Thus, here $t$ is the independent variable, and $s$ is the dependent variable on $t$. | istheindependentvariable,isthedependentvariableon | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,903 |
8. The volume of a sphere is determined by the formula $V=\frac{4}{3} \pi R^{3}$. Indicate the independent and dependent variables. | Solution. Here $\frac{4}{3} \pi$ is a constant value. By assigning different values to the radius $R$, we can find the volume of the sphere for each of the given radius values. Thus, the radius $R$ is the independent variable, and the volume of the sphere $V$ is the dependent variable.
The independent variable, i.e., ... | R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 31,904 |
9. Determine the value of the function $f(x)=2 x^{2}-1$ at $x=3$. | Solution. Find $f(3)=y_{x=3}=2 \cdot 3^{2}-1=17$. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,905 |
11. Find $\varphi(\pi / 4)$, if $\varphi(t)=\frac{2 t}{1+\sin ^{2} t}$. | Solution. $\varphi(\pi / 4)=\frac{2 \cdot \pi / 4}{1+\sin ^{2}(\pi / 4)}=\frac{\pi / 2}{1+1 / 2}=\frac{\pi}{3}$. | \frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,906 |
14. Find the domain of the function $y=x^{2}$. | Solution. Obviously, for any real value of $x$, the function $y$ is also expressed as a real number. Therefore, the given function is defined for any value of $x \in(-\infty, \infty)$. This result can be written as $x \in \mathbb{R}$.
Let's note the peculiarities of finding the domain of some functions. | x\in\mathbb{R} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,907 |
15. $y=\frac{1}{x}$. | Solution. The denominator becomes zero at $x=0$. The given function takes real values for all $x$, except $x=0$. Therefore, the domain of the given function consists of the intervals $(-\infty, 0)$ and $(0, \infty)$.
$\begin{array}{ll}\text { 16. } y=\frac{2}{1-x} & \text { 17. } y=\frac{3}{x-4}\end{array}$ | (-\infty,0)\cup(0,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,908 |
18. \( y=\frac{1}{2 x-5} \). | Solution. Here $2 x-5 \neq 0$, hence $x \neq 2.5$. Thus, we get the answer: $(-\infty ; 2.5)$ and $(2.5 ; \infty)$.
$\begin{array}{ll}\text { 19. } y=\frac{x-1}{x+1} & \text { 20. } y=\frac{2 x+1}{3 x-1} \text {. }\end{array}$ | (-\infty;2.5)\cup(2.5; | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,909 |
21. $y=\frac{3}{x^{2}-4}$. | Solution. By setting the denominator to zero, we solve the obtained equation: $x^{2}-4=0 ;(x+2)(x-2)=0 ; x_{1}=-2, x_{2}=2$. Therefore, the denominator equals zero at the values $x=-2$ and $x=2$, which cannot belong to the domain of the given function. Excluding them, we obtain three intervals ( $-\infty,-2$ ), $(-2,2)... | (-\infty,-2)\cup(-2,2)\cup(2,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,910 |
24. $y=\frac{x+2}{x^{2}-5 x+6}$.
24. $y=\frac{x+2}{x^{2}-5 x+6}$.
The above text has been translated into English, maintaining the original formatting and line breaks. | Solution. The function is defined for all real values of $x$, except for those for which $x^{2}-5 x+6=0$, i.e., the roots of the quadratic trinomial $x^{2}-5 x+6$; they are the numbers $x_{1}=2 ; x_{2}=3$. Therefore, the function is defined on the intervals $(-\infty, 2),(2,3)$ and $(3, \infty)$. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,911 |
27. $y=\frac{x^{2}-5 x+4}{x^{2}+x+1}$.
27. $y=\frac{x^{2}-5 x+4}{x^{2}+x+1}$. | Solution. By setting the denominator of the fraction to zero and solving the resulting quadratic equation $x^{2}+x+1=0$, we can verify that its roots are complex numbers: $x=-\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$. The denominator does not become zero for any real value of $x$. Therefore, the function is defined for all... | (-\infty,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,912 |
30. $y=\sqrt{x-4}$. | Solution. Note that this function only makes sense if the expression under the root is greater than or equal to zero. If the expression under the root is negative, then $y$ is an imaginary number. Therefore, $x-4 \geqslant 0$ or $x \geqslant 4$. Thus, this function is defined only if $x \geqslant 4$, i.e., $x \in [4, \... | x\in[4,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,913 |
33. $y=\sqrt{x}+\sqrt{x-1}$.
The above text is translated into English as follows, retaining the original text's line breaks and format:
33. $y=\sqrt{x}+\sqrt{x-1}$. | Solution. Let's find the domain of definition of each term separately. The common part of these domains will be the domain of definition of the function. For $\sqrt{x}$ we have $x \geqslant 0$, and for $\sqrt{x-1}$ we have $x \geqslant 1$. Then for the sum $\sqrt{x}+\sqrt{x-1}$ the domain of definition is $x \geqslant ... | [1,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,914 |
36. $y=\sqrt{2 x^{2}-6 x}$. | Solution. The domain of the function is found from the condition $2 x^{2}-6 x \geqslant 0$ or $2 x(x-3) \geqslant 0$. The solutions to this inequality are $x \leqslant 0, x \geqslant 3$. Therefore, the domain of the function consists of the half-intervals $(-\infty, 0]$ and $[3, \infty)$. This can be illustrated on a n... | x\leqslant0,x\geqslant3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,915 |
41. $y=\lg (x-2)$.
The above text is translated into English as follows, retaining the original text's line breaks and format:
41. $y=\lg (x-2)$. | Solution. Since the expression under the logarithm sign must be positive, then $x-2>0$, from which $x>2$, i.e., the given function exists only for $x \in(2, \infty)$. | x\in(2,\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,916 |
46. $z=\log _{3}\left(x^{2}-9\right)$.
46. $z=\log _{3}\left(x^{2}-9\right)$.
(Note: The equation is already in English and does not require translation. The provided translation is identical to the original text.) | Solution. The logarithmic function $z$ is defined only for positive values of its argument, so $x^{2}-9>0$. Solving this inequality, we get $|x|>3$, from which it follows that the domain of the function $z$ consists of two infinite intervals ( $-\infty,-3$ ) and $(3, \infty)$. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,917 |
53. $y=\arcsin \frac{x-2}{3}$. | Solution. The function is defined if
$$
\left\{\begin{array}{l}
\frac{x-2}{3} \geqslant-1 \\
\frac{x-2}{3} \leqslant 1
\end{array}\right.
$$
The solution to the system of inequalities are values $x \geqslant-1$ and $x \leqslant 5$. Therefore, the domain of the function is the interval $[-1,5]$. | [-1,5] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,918 |
58. Indicate the intervals of increase of the function
| $x$ | -2 | -1 | 0 | 1 | 2 |
| :--- | ---: | ---: | :--- | :--- | :--- |
| $y$ | 9 | 2 | 0 | 2 | 9 | | Solution. In this case, although the function is given by a table, for clarity, we construct a graph (Fig. 68). It is evident that the function is increasing on the interval $(0, \infty)$.
The most convenient method for defining a function is the third one - the analytical method.
In the analytical method, the depend... | (0,\infty) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,919 |
59. Plot the graph of the function $y=\frac{2}{x}$. | Solution. Let's create a table of function values:
| -4 | -2 | -1 | $-\frac{1}{2}$ | $-\frac{1}{3}$ | $\frac{1}{3}$ | $\frac{1}{2}$ | 1 | 2 | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $-\frac{1}{2}$ | -1 | -2 | -4 | -6 | 6 | 4 | 2 | 1 | $\frac{1}{2}$ |
According to the ta... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,920 |
61. Prove that the function $\varphi(x)=\frac{\cos 2 x}{\sqrt[3]{x}+3 x}$ is odd. | Solution. We find
$$
\varphi(-x)=\frac{\cos (-2 x)}{\sqrt[3]{-x}-3 x}=\frac{\cos 2 x}{-\sqrt[3]{x}-3 x}=-\frac{\cos 2 x}{\sqrt[3]{x}+3 x}=-\varphi(x),
$$
i.e., the given function is odd. | proof | Algebra | proof | Yes | Yes | olympiads | false | 31,922 |
62. Determine whether the function $g(x)=2^{x}-3 x+1$ is even or odd. | Solution. We have $g(-x)=2^{-x}-3(-x)+1=2^{-x}+3 x+1$. As can be seen, in this case, the conditions for evenness and oddness are not met. Therefore, the function $g(x)$ is neither even nor odd.
63-74. Determine which of the given functions is even and which is odd: | neitherevennorodd | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,923 |
75. Prove that the function $f(x)=\sin 3 x$ is periodic with period $l=2 \pi / 3$. | Solution. Since $\sin 3\left(x+\frac{2 \pi}{3}\right)=\sin (3 x+2 \pi)=\sin 3 x$, the period of the function $f(x)$ is $2 \pi / 3$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 31,924 |
80. Given the function $y=2 x+3, x \in[-1.5 ; 1]$. Find the function that is the inverse of the given one. | Solution. Solving the given equation for $x$, we have $2 x=$ $=y-3$, from which $x=0.5 y-1.5$. Transitioning to conventional notation, i.e., replacing $x$ with $y$ and $y$ with $x$ in the last equation, we obtain the inverse function: $y=0.5 x-1.5, x \in[0,5]$.
In Fig. 79, the graphs of the given function and its inve... | 0.5x-1.5,x\in[0,5] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,925 |
85. The complex function $y=\sin u$, where $u=\lg v, v=\sqrt{x}$, write as a single equation. | Solution. Substituting the value $v=\sqrt{x}$ into the equation $u=\lg v$, we get $u=\lg \sqrt{x}$. Next, substituting the obtained value for $u$ into the equation $y=\sin u$; then the given composite function will take the form $y=$ $=\sin (\lg \sqrt{x})$. | \sin(\lg\sqrt{x}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,926 |
88. Write the complex function $y=(2 x-5)^{10}$ as a chain of equalities. | Solution. Let's denote $2 x-5$ by $u$; then we get $y=u^{10}$, where $u=2 x-5$. | u^{10},whereu=2x-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,927 |
4. Find the limit of the variable quantity $x=\frac{a z+1}{z}$ as $z \rightarrow \infty$. | Solution. Transform the variable by dividing all terms of the numerator by the denominator. We get: $x=a+\frac{1}{z} ; x-a=\frac{1}{z}$.
Notice that the larger $z$ is, the closer the values of the variable $x$ are to the constant $a$, since the condition $|x-a|<\varepsilon$ is satisfied, where $\mathbf{\varepsilon}-$ ... | a | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,928 |
92. Show that as $t \rightarrow \infty$ the limit of the variable
$x=\frac{6 t^{3}-9 t+1}{2 t^{3}-3 t}$ is 3. | Solution. We find the difference between the variable $x$ and the number 3:
$$
\begin{gathered}
x-3=\frac{6 t^{3}+9 t+1}{2 t^{3}+3 t}-3=\frac{6 t^{3}+9 t+1-3\left(2 t^{3}+3 t\right)}{2 t^{3}+3 t}=\frac{6 t^{3}+9 t+1-6 t^{3}-9 t}{2 t^{3}+3 t}= \\
=\frac{1}{2 t^{3}+3 t}
\end{gathered}
$$
If $t \rightarrow \infty$, then... | 3 | Calculus | proof | Yes | Yes | olympiads | false | 31,929 |
94. Find $\lim _{x \rightarrow 2}\left(3 x^{2}-2 x\right)$. | Solution. Using properties 1, 3, and 5 of limits sequentially, we get
$$
\begin{gathered}
\lim _{x \rightarrow 2}\left(3 x^{2}-2 x\right)=\lim _{x \rightarrow 2}\left(3 x^{2}\right)-\lim _{x \rightarrow 2}(2 x)=3 \lim _{x \rightarrow 2} x^{2}-2 \lim _{x \rightarrow 2} x= \\
=3\left(\lim _{x \rightarrow 2} x\right)^{2}... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,930 |
95. Find $\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}$. | Solution. To apply the property of the limit of a quotient, we need to check whether the limit of the denominator is not zero when $x=4$. Since $\lim _{x \rightarrow 4}(x-3)=$[^3]$=\lim _{x \rightarrow 4} x-\lim 3=4-3=1 \neq 0$, in this case, we can use property 4:
$$
\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}=\fra... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,931 |
99. Find the increments of the argument and the function $y=2 x^{2}+1$, if the argument $x$ changes from 1 to 1.02. | Solution. $1^{0}$. Find the increment of the argument: $\Delta x=1.02-1=$ $=0.02$.
$2^{0}$. Find the value of the function at the old value of the argument, i.e., at $x=1: y=2 \cdot 1^{2}+1=3$.
$3^{\circ}$. Find the value of the function at the new value of the argument, i.e., at $x=1+0.02=1.02$:
$$
y_{\mathrm{N}}=y... | 0.0808 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,932 |
104. Prove the continuity of the function $y=a x^{2}+b x+c$ at the point $x$. | Solution. $1^{0}$. Assign the argument $x$ an increment $\Delta x$.
$2^{0}$. Find the increment of the function:
$\Delta y=a(x+\Delta x)^{2}+b(x+\Delta x)+c-\left(a x^{2}+b x+c\right)=2 a x \Delta x+a(\Delta x)^{2}+b \Delta x$.
$3^{0}$. Find the limit of the function as $\Delta x \rightarrow 0$:
$$
\lim _{\Delta x \... | proof | Calculus | proof | Yes | Yes | olympiads | false | 31,934 |
105. Investigate the continuity of the function $y=x^{2}$. | Solution. Let the increment of the argument $x$ be $\Delta x$; then the function $y$ will receive some increment $\Delta y$. We have
$$
y+\Delta y=(x+\Delta x)^{2}=x^{2}+2 x \Delta x+(\Delta x)^{2}
$$
from which
$$
\Delta y=2 x \Delta x+(\Delta x)^{2}=\Delta x(2 x+\Delta x)
$$
It is obvious that for any fixed value... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,935 |
114. Calculate $\lim _{x \rightarrow 1} \frac{2 x^{3}+4 x-3}{x+4}$. | Solution. When $x=1$, the fraction $\frac{2 x^{3}+4 x-3}{x+4}$ is defined, since its denominator is not zero. Therefore, to compute the limit, it is sufficient to substitute the argument with its limiting value. Then we get
$$
\lim _{x \rightarrow 1} \frac{2 x^{3}+4 x-3}{x+4}=\frac{2 \cdot 1^{3}+4 \cdot 1-3}{1+4}=\fra... | \frac{3}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,936 |
115. Find $\lim _{x \rightarrow 3} \frac{x^{2}-9}{3-x}$. | Solution. A direct transition to the limit is impossible here, since the limit of the denominator is zero: $\lim _{x \rightarrow 3}(3-x)=3-3=0$. The limit of the dividend is also zero: $\lim _{x \rightarrow 3}\left(x^{2}-9\right)=9-9=0$. Thus, we have an indeterminate form of $0 / 0$. However, this does not mean that t... | -6 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,937 |
116. Find $\lim _{x \rightarrow \infty} \frac{3}{x+5}$. | Solution. When $x \rightarrow \infty$, the denominator $x+5$ also tends to infinity, and its reciprocal $\frac{1}{x+5} \rightarrow 0$. Therefore, the product $\frac{1}{x+5} \cdot 3=\frac{3}{x+5}$ tends to zero if $x \rightarrow \infty$. Thus, $\lim _{x \rightarrow \infty} \frac{3}{x+5}=0$.
Identical transformations un... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,938 |
117. Find $\lim _{x \rightarrow \infty} \frac{2 x^{3}+x}{x^{3}-1}$. | Solution. Here, the numerator and denominator do not have a limit, as both increase indefinitely. In this case, it is said that there is an indeterminate form of $\infty / \infty$. We will divide the numerator and denominator term by term by $x^{3}$ (the highest power of $x$ in this fraction):
$$
\lim _{x \rightarrow ... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,939 |
118. Find $\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}$. | Solution. We will transform this limit into the form (1). For this, we multiply the numerator and the denominator of the fraction by 2, and take the constant factor 2 outside the limit sign. We have
$$
\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}=\lim _{\alpha \rightarrow 0} \frac{2 \sin 2 \alpha}{2 \alph... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,940 |
119. Find $\lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^{x}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note is for you, the assistant, and should not be included in the output.)
119. Find $\lim _{x... | Solution. Dividing the numerator and the denominator by $x$, we get
$$
\lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^{x}=\lim _{x \rightarrow \infty}\left(\frac{1}{1+\frac{1}{x}}\right)^{x}=\lim _{x \rightarrow \infty} \frac{1}{\left(1+\frac{1}{x}\right)^{x}}=\frac{1}{\lim _{x \rightarrow \infty}\left(1+\frac... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,941 | |
120. Find $\lim _{x \rightarrow 3}\left(x^{2}-7 x+4\right)$. | Solution. To find the limit of the given function, we will replace the argument $x$ with its limiting value:
$$
\lim _{x \rightarrow 3}\left(x^{2}-7 x+4\right)=3^{2}-7 \cdot 3+4=-8
$$ | -8 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,942 |
121. Find $\lim _{x \rightarrow 2} \frac{x^{2}+x+2}{x^{2}+2 x+8}$. | Solution. Let's check if the denominator of the fraction does not turn into zero when $x=2$: we have $2^{2}+2 \cdot 2+8=16 \neq 0$. Substituting the limit value of the argument, we find
$$
\lim _{x \rightarrow 2} \frac{x^{2}+x+2}{x^{2}+2 x+8}=\frac{2^{2}+2+2}{2^{2}+2 \cdot 2+8}=\frac{8}{16}=\frac{1}{2}
$$
Now let's c... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,943 |
122. Find $\lim _{x \rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2-x}}{5 x}$. | Solution. Here the limits of the numerator and denominator as $x \rightarrow 0$ are both zero. By multiplying the numerator and denominator by the expression conjugate to the numerator, we get
$$
\begin{gathered}
\frac{\sqrt{2+x}-\sqrt{2-x}}{5 x}=\frac{(\sqrt{2+x}-\sqrt{2-x})(\sqrt{2+x}+\sqrt{2-x})}{5 x(\sqrt{2+x}+\sq... | \frac{\sqrt{2}}{10} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,944 |
124. Find $\lim _{x \rightarrow-2} \frac{2 x^{2}+7 x+6}{(x+2)^{2}}$. | Solution. Direct substitution $x=-2$ shows that there is an indeterminate form of $0 / 0$. By factoring the numerator and simplifying the fraction, we find
$$
\lim _{x \rightarrow-2} \frac{2 x^{2}+7 x+6}{(x+2)^{2}}=\lim _{x \rightarrow-2} \frac{2(x+2)\left(x+\frac{3}{2}\right)}{(x+2)^{2}}=\lim _{x \rightarrow-2} \frac... | \infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,945 |
131. Find $\lim _{x \rightarrow \infty} \frac{2 x^{3}-3 x^{2}+5 x+7}{3 x^{3}+4 x^{2}-x+2}$. | Solution. When $x \rightarrow \infty$, we have an indeterminate form of $\infty / \infty$. To resolve this indeterminacy, we divide the numerator and the denominator by $x^{3}$. Then we get
$$
\lim _{x \rightarrow \infty} \frac{2 x^{3}-3 x^{2}+5 x+7}{3 x^{3}+4 x^{2}-x+2}=\lim _{x \rightarrow \infty} \frac{2-\frac{3}{x... | \frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,946 |
133. Find $\lim _{x \rightarrow \infty} \frac{4 x^{3}+x^{2}-2}{3 x^{2}+5 x-2}$. | Solution. Dividing the numerator and the denominator by $x^{3}$ and taking the limit, we get
$$
\lim _{x \rightarrow \infty} \frac{4 x^{3}+x^{2}-2}{3 x^{2}+5 x-2}=\lim _{x \rightarrow \infty} \frac{4+\frac{1}{x}-\frac{2}{x^{3}}}{\frac{3}{x}+\frac{5}{x^{2}}-\frac{2}{x^{3}}}=\infty
$$
since the numerator of the last fr... | \infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,947 |
135. Find $\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+4}}{x}$. | Solution. As the argument $x$ tends to infinity, we have an indeterminate form of $\infty / \infty$. To resolve it, we divide the numerator and the denominator of the fraction by $x$. Then we get
$$
\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+4}}{x}=\lim _{x \rightarrow \infty} \frac{\sqrt{\frac{x^{2}+4}{x^{2}}}}{1... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,948 |
136. Find $\lim _{x \rightarrow \infty} \frac{3 x}{\sqrt{x^{2}-2 x+3}}$. | Solution. The limit transition as $x \rightarrow \infty$ can always be replaced by the limit transition as $\alpha \rightarrow 0$, if we set $\alpha=1 / x$ (the method of variable substitution).
Thus, setting $x=1 / \alpha$ in this case, we find that $\alpha \rightarrow 0$ as $x \rightarrow \infty$. Therefore,
$$
\be... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,949 |
137. Find $\lim _{x \rightarrow \infty} \frac{3 x^{2}+5 x+1}{x^{2}-2}$. | Solution. Method I. Dividing the numerator and the denominator by $x^{2}$, we find
$$
\lim _{x \rightarrow \infty} \frac{3 x^{2}+5 x+1}{x^{2}-2}=\lim _{x \rightarrow \infty} \frac{3+\frac{5}{x}+\frac{1}{x^{2}}}{1-\frac{2}{x^{2}}}=\frac{3}{1}=3
$$
Method II. Let $x=1 / a$; then $a \rightarrow 0$ as $x \rightarrow \inf... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,950 |
142. Find $\lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})$. | Solution. Here, it is required to find the limit of the difference of two quantities tending to infinity (indeterminate form of $\infty-\infty$). By multiplying and dividing the given expression by its conjugate, we get
$$
\sqrt{x+1}-\sqrt{x}=\frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}}=\frac... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,951 |
143. Find $\lim _{x \rightarrow 0} \frac{\sin k x}{x}(k-$ a constant). | Solution. Let's make the substitution $k x=y$. From this, it follows that $y \rightarrow 0$ as $x \rightarrow 0$, and $x=y / k$. Then we get
$$
\lim _{x \rightarrow 0} \frac{\sin k x}{x}=\lim _{y \rightarrow 0} \frac{\sin y}{y / k}=\lim _{y \rightarrow 0} \frac{k \sin y}{y}=k \lim _{y \rightarrow 0} \frac{\sin y}{y}=k... | k | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,952 |
144. Find $\lim _{x \rightarrow 0} \frac{\sin k x}{\sin l x}$. | Solution. We have
$$
\lim _{x \rightarrow 0} \frac{\sin k x}{\sin l x}=\lim _{x \rightarrow 0} \frac{\frac{\sin k x}{x}}{\frac{\sin l x}{x}}=\frac{\lim _{x \rightarrow 0} \frac{\sin k x}{x}}{\lim _{x \rightarrow 0} \frac{\sin l x}{x}}=\frac{k}{l}
$$
Here we divided the numerator and the denominator of the fraction by... | \frac{k}{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,953 |
145. Find $\lim _{x \rightarrow 0} \frac{1-\cos 8 x}{2 x^{2}}$. | Solution. Transform the numerator to the form $1-\cos 8 x=2 \sin ^{2} 4 x$. Next, we find
$$
\begin{gathered}
\lim _{x \rightarrow 0} \frac{1-\cos 8 x}{2 x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} 4 x}{2 x^{2}}=\lim _{x \rightarrow 0}\left(\frac{\sin 4 x}{x} \cdot \frac{\sin 4 x}{x}\right)= \\
=\lim _{x \rightar... | 16 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,954 |
146. Find $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$. | Solution. Method I. Here we have an indeterminate form of the type $0 / 0$. Applying a known trigonometric formula and performing elementary transformations, we get
$$
\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x}{2}}{x^{2}}=\lim _{x \rightarrow 0} \frac{1}{2} \cdot ... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,955 |
154. Find $\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{3 x}$. | Solution. We have
$$
\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{3 x}=\lim _{x \rightarrow \infty}\left(\left(1+\frac{2}{x}\right)^{x / 2}\right)^{6}=\left(\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{x / 2}\right)^{6}
$$
Let $x / 2=y$. Then, as $x$ increases without bound, the variable $y$ wil... | e^6 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,956 |
155. Find $\lim _{x \rightarrow 0}\left(\frac{3+x}{3}\right)^{1 / x}$. | Solution. Write the base of the power as $\frac{3+x}{3}=1+\frac{x}{3}$, and the exponent as $\frac{1}{x}=\frac{1}{x} \cdot \frac{3}{3}=\frac{3}{x} \cdot \frac{1}{3}$. Therefore,
$$
\lim _{x \rightarrow 0}\left(\frac{3+x}{3}\right)^{1 / x}=\lim _{x \rightarrow 0}\left(1+\frac{x}{3}\right)^{3 / x \cdot 1 / 3}=\left(\lim... | \sqrt[3]{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,957 |
156. Find $\lim _{x \rightarrow e} \frac{\ln x-1}{x-e}$. | Solution. We have
$$
\begin{aligned}
& \lim _{x \rightarrow e} \frac{\ln x-1}{x-e}=\lim _{x \rightarrow e} \frac{\ln x-\ln e}{x-e}=\frac{1}{e} \lim _{x \rightarrow e} \frac{\ln \frac{x}{e}}{\frac{x}{e}-1}= \\
= & \frac{1}{e} \lim _{z \rightarrow 0} \frac{\ln (z+1)}{z}=\frac{1}{e} \cdot 1=\frac{1}{e}\left(\text { here ... | \frac{1}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,958 |
164. Given the function $f(x)=x^{2}-1$. Find the equation of the tangent line to its graph at $x=1$. | Solution. $1^{0}$. First, let's find the ordinate of the point of tangency: $f(1)=$ $=1^{2}-1=0$. Therefore, $(1 ; 0)$ is the point of tangency.
$2^{0}$. We will now form the equation of the line passing through the point ( $1 ; 0$ ). For this, we will use the well-known equation from analytic geometry $y-y_{1}=k\left... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,959 | |
168. Find the derivative of the function $y=5x$. | Solution. $1^{0} . y_{\mathrm{H}}=5(x+\Delta x)=5 x+5 \Delta x$.
$2^{0} . \Delta y=y_{\mathrm{k}}-y=(5 x+5 \Delta x)-5 x=5 \Delta x$.
$3^{0} . \frac{\Delta y}{\Delta x}=\frac{5 \Delta x}{\Delta x}=5$. | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,960 |
170. Differentiate the function $y=x^{2}$. | Solution. $1^{0} . y_{\text {n }}=(x+\Delta x)^{2}=x^{2}+2 x \Delta x+(\Delta x)^{2}$.
$2^{0} . \Delta y=y_{\mathrm{H}}-y=\left(x^{2}+2 x \Delta x+(\Delta x)^{2}\right)-x^{2}=2 x \Delta x+(\Delta x)^{2}$.
$3^{0} \cdot \frac{\Delta y}{\Delta x}=\frac{2 x \Delta x+(\Delta x)^{2}}{\Delta x}=\frac{2 x \Delta x}{\Delta x}... | 2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,961 |
172. Find the derivative of the function $y=x^{2}+x$. | Solution. $1^{0} . y_{\text {n }}=(x+\Delta x)^{2}+(x+\Delta x)=x^{2}+2 x \Delta x+(\Delta x)^{2}+x+\Delta x$.
$2^{0} . \Delta y=y_{\mathrm{N}}-y=\left(x^{2}+2 x \Delta x+(\Delta x)^{2}+x+\Delta x\right)-\left(x^{2}+x\right)=x^{2}+2 x \Delta x+$ $+(\Delta x)^{2}+x+\Delta x-x^{2}-x=2 x \Delta x+(\Delta x)^{2}+\Delta x$... | 2x+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,962 |
174. Find the derivative of the function $y=x^{2}-3 x+5$. | Solution. $1^{0} . y_{\text {n }}=(x+\Delta x)^{2}-3(x+\Delta x)+5=x^{2}+2 x \Delta x+(\Delta x)^{2}-$ $-3 x-3 \Delta x+5$
$\left.2^{0} . \Delta y=y_{\mathrm{r}}-y=\left(x^{2}+2 x \Delta x\right)+(\Delta x)^{2}-3 x+3 \Delta x+5\right)-\left(x^{2}-3 x+5\right)=$ $=2 x \Delta x+(\Delta x)^{2}-3 \Delta x$.
$3^{0} . \fra... | 2x-3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,963 |
175. On the curve $y=x^{2}-3 x+5$, find the point where the ordinate $y$ increases 5 times faster than the abscissa $x$. | Solution. We find the derivative $y^{\prime}=2 x-3$ (see the solution of the previous example). Since the derivative characterizes the rate of change of the ordinate $y$ compared to the change in the abscissa $x$, from the condition $y^{\prime}=2 x-3=5$ we find the abscissa of the desired point: $x=4$. We find the ordi... | (4,9) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,964 |
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