problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-1 ; 1 ; 1)$
$a: 3 x-y+2 z+4=0$
$k=\frac{1}{2}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-y+2 z+2=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$f(x)=\left\{\begin{array}{c}\frac{\ln \left(1+2 x^{2}+x^{3}\right)}{x}, x \neq 0 ; \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt[3]{n^{3}+2}}{\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}}$ | Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt[3]{n^{3}+2}}{\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+2}-\sqrt[3]{n^{3}+2}\right)}{\frac{1}{n}\left(\sqrt[7]{n+2}-\sqrt[5]{n^{5}+2}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sq... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{n^{3}\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}-\sqrt[3]{n^{8}-1}\right)\left(\left(\sqrt[3]{n^{2}\left(n^{6}+4\right)}\right)^{2}+\sqrt[3]{n^{2... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+3 x^{2}}-(1+x)}{\sqrt[3]{x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1-2 x+3 x^{2}}-(1+x)}{\sqrt[3]{x}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\sqrt{1-2 x+3 x^{2}}-(1+x)\right)\left(\sqrt{1-2 x+3 x^{2}}+(1+x)\right)}{\sqrt[3]{x}\left(\sqrt{1-2 x+3 x^{2}}+(1+x)\right)}= \\
& =\lim _{x \rightarrow 0} \frac{1-2 x+3 x... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sin ^{2} x-\tan^{2} x}{x^{4}}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& \sin x \sim x, \text { as } x \rightarrow 0 \\
& \operatorname{tg} x \sim x, \text { as } x \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin ^{2} x-\operatorname{tg}^{2} x}{x^{4}... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(e^{x}+x\right)^{\cos x^{4}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(e^{x}+x\right)^{\cos x^{4}}=\left(e^{0}+0\right)^{\cos 0^{4}}= \\
& =(1+0)^{\cos 0}=1^{1}=1
\end{aligned}
$$
## Problem Kuznetsov Limits 18-25 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2}(\cos \pi x)^{\tan(x-2)}$ | ## Solution
$\lim _{x \rightarrow 2}(\cos \pi x)^{\operatorname{tg}(x-2)}=(\cos (\pi \cdot 2))^{\operatorname{tg}(2-2)}=(\cos 2 \pi)^{\operatorname{tg} 0}=1^{0}=1$
## Problem Kuznetsov Limits 20-25 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathrm{c}^{2}$... | ## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction ... | 13574468085 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction ... | 17191616766 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction ... | 19907875186 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction ... | 18820058997 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathrm{c}^{2}$... | ## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
$$
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction ... | 21017569546 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction ... | 18546511628 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction ... | 20253968254 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction ... | 16452722063 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Determine the work (in joules) performed when lifting a satellite from the Earth's surface to a height of $H$ km. The mass of the satellite is $m$ tons, the radius of the Earth $R_{3}=6380$ km. The acceleration due to gravity $g$ at the Earth's surface is taken to be $10 \mathrm{~m} / \mathbf{c}^{2}$... | ## Solution
By definition, the elementary work $\Delta A=F(x) \Delta x$, where $F(x)=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}} ; G=6.67 \cdot 10^{-11} \mathrm{H}^{*} \mathrm{m}^{*} \mathrm{m} /($ kg*kg)
\[
\begin{aligned}
& F_{x}=G \cdot \frac{m \cdot M}{\left(R_{3}+x\right)^{2}}-\text { force of attraction ... | 17697012802 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state for the gas
$\rho V = \text{con... | ## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
... | 2700 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state of the gas
$\rho V=$ const, where ... | ## Solution
Let the piston be at a distance $x, \quad 0 \leq x \leq h$
The force with which the gas presses on the walls is: $F(x)=p(x) \cdot S$
where: $S=\pi R^{2}-$ area of the piston, $p=p(x)-$ pressure of the gas.
Since the process is isothermal, then $p V=$ Const $\Rightarrow p_{0} \cdot \pi \cdot R^{2} \cdot ... | 1800 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) done during the isothermal compression of the gas by a piston moving inward by $h$ meters (see figure).
Hint: The equation of state for the gas is
$\rho V = \text{const}$,... | ## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
... | 900 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state for the gas
$\rho V=$ const, wh... | ## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
... | 14400 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state of the gas
$\rho V=$ const, where ... | ## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
... | 97200 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state of the gas
$\rho V=$ const, where ... | ## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
... | 64800 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state of the gas
$\rho V=$ const, where ... | ## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
... | 32400 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state of the gas $\rho V=$ const, where $... | ## Solution
Piston area: $S=\pi R^{2}$
Volume of gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x \Rightarrow$
... | 144000 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
A cylinder is filled with gas at atmospheric pressure (103.3 kPa). Assuming the gas is ideal, determine the work (in joules) during the isothermal compression of the gas by a piston moving inside the cylinder by $h$ meters (see figure).
Hint: The equation of state for the gas
$\rho V=$ const, wh... | ## Solution
Area of the piston: $S=\pi R^{2}$
Volume of the gas during compression: $V(x)=S \cdot(H-x) ; 0 \leq x \leq h$
Pressure of the gas during compression: $p(x)=\frac{p_{0} \cdot S \cdot H}{V(x)}$
Force of pressure on the piston: $F(x)=p(x) \cdot S$
By definition, the elementary work $\Delta A=F(x) \Delta x... | 72000 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\ln \left(1-\sin \left(x^{3} \sin \frac{1}{x}\right)\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\arcsin (\sin t) \\
y=\arccos (\cos t)
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\arcsin (\sin t))^{\prime}=t^{\prime}=1$
$y_{t}^{\prime}=(\arccos (\cos t))^{\prime}=t^{\prime}=1$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{1}{1}=1$
## Kuznetsov Differentiation 16-4 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in the Cartesian coordinate system.
$$
y=2+\arcsin \sqrt{x}+\sqrt{x-x^{2}}, \frac{1}{4} \leq x \leq 1
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(2+\arcsin \sqrt{x}+\sqrt{x-x^{2}}\rig... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=-\arccos \sqrt{x}+\sqrt{x-x^{2}}, 0 \leq x \leq \frac{1}{4}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(-\arccos \sqrt{x}+\sqrt{x-x^{2}}\righ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
x+\arcsin \left(x^{2} \sin \frac{6}{x}\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Condition of the problem
To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=\sqrt{x}-3 \sqrt[3]{x}, x_{0}=64$ | ## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=(\sqrt{x}-3 \sqrt[3]{x})^{\prime}=\left(\sqrt{x}-3 \cdot x^{\frac{1}{3}}\right)^{\prime}=\frac{1}{2 \sqrt{x}}-3 \cdot \frac{1}{3} \cdot x^{-\frac{2}{3}}=\frac{1}{2 \sqrt{x}}-\frac{1}{\sqrt[3]{x^{2}}}
$$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{1}{2 ... | 64 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\frac{\cos x-\cos 3 x}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$\left\{\begin{array}{l}x=\cos t+\sin t \\ y=\sin 2 t\end{array}\right.$ | ## Solution
$x_{t}^{\prime}=(\cos t+\sin t)^{\prime}=-\sin t+\cos t$
$y_{t}^{\prime}=(\sin 2 t)^{\prime}=2 \cos 2 t$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2 \cos 2 t}{-\sin t+\cos t}=2 \cdot \frac{\cos ^{2} t-\sin ^{2} t}{\cos t-\sin t}=2(\sin t+\cos t) \\
& \lef... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}-3 n}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}-3 n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((n+1)^{3}+(n-1)^{3}\right)}{\frac{1}{n^{3}}\left(n^{3}-3 n\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\rig... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{n!+(n+2)!}{(n-1)!+(n+2)!}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n!+(n+2)!}{(n-1)!+(n+2)!}=\lim _{n \rightarrow \infty} \frac{(n-1)!(n+n(n+1)(n+2))}{(n-1)!(1+n(n+1)(n+2))}= \\
& =\lim _{n \rightarrow \infty} \frac{n+n(n+1)(n+2)}{1+n(n+1)(n+2)}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}(n+n(n+1)(n+2))}{\frac... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x^{2}+x^{5}}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x^{2}+x^{5}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{1^{3}+3 \cdot 1^{2} \cdot x+3 \cdot 1 \cdot x^{2}+x^{3}-1-3 x}{x^{2}\left(1+x^{3}\right)}=$
$=\lim _{x \rightarrow 0} \frac{1+3 x+3 x^{2}+x^{3}-1-3 x}{x^{2}\left(1+x^{3}\right)}=\lim _{x ... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{\sqrt[3]{x^{3}+8}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{\sqrt[3]{x^{3}+8}}=\lim _{x \rightarrow-2} \frac{(\sqrt[3]{x-6}+2)\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}{\sqrt[3]{x^{3}+8}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \frac{(\sqrt[3]{x-6}+... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{1-\sqrt{x^{2}+1}}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$\ln \left(1+x^{2}\right) \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$
We get:
$\lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{1-\sqrt{x^{2}+1}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{x^{2}}{1-\sqr... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{\sin 2 x-\sin x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{\sin 2 x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{x}-1\right)}{\sin 2 x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{x}-1\right)\right)}{\frac{1}{x}(\sin 2 x-\sin x)}=$
$=\frac{\lim _{... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Condition of the problem
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{tg}^{2} x}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{tg}^{2} x}=\left(6-\frac{5}{\cos 0}\right)^{\operatorname{tg}^{2} 0}= \\
& =\left(6-\frac{5}{1}\right)^{0^{2}}=1^{0}=1
\end{aligned}
$$
## Problem Kuznetsov Limits $18-28$ | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2} \\
& x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0
\end{aligned}
$$
We obtain:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}=\lim _{y \rightarrow 0}\le... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}(\sqrt[3]{x}+x-1)^{\sin \left(\frac{\pi x}{4}\right)}
$$ | ## Solution
$\lim _{x \rightarrow 1}(\sqrt[3]{x}+x-1)^{\sin \left(\frac{\pi x}{4}\right)}=(\sqrt[3]{1}+1-1)^{\sin \left(\frac{\pi \cdot 1}{4}\right)}=(1)^{\frac{\sqrt{2}}{2}}=1$
## Problem Kuznetsov Limits 20-28 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1 / \sqrt{2}} \frac{d x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{1 / \sqrt{2}} \frac{d x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& x=\sin t ; d x=\cos t d t \\
& x=0 \Rightarrow t=\arcsin 0=0 \\
& x=\frac{1}{\sqrt{2}} \Rightarrow t=\arcsin \frac{1}{\sqrt{2}}=\frac{\pi}{4}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a_{\text {and }} b$.
$$
\begin{aligned}
& a=5 p-q \\
& b=p+q \\
& |p|=5 \\
& |q|=3 \\
& (\widehat{p, q})=\frac{5 \pi}{6}
\end{aligned}
$$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(5 p-q) \times(p+q)=5 \cdot p \times p+5 \cdot p \times q-q \times p-q \times q=$... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{1 ;-1 ; 4\}$
$b=\{1 ; 0 ; 3\}$
$c=\{1 ;-3 ; 8\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$(a, b, c)=\left|\begin{array}{ccc}1 & -1 & 4 \\ 1 & 0 & 3 \\ 1 & -3 & 8\end{array}\right|=$
$=1 \cdot\left|\begin{array}{cc}0 & 3 \\ -... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+3}-\sqrt[3]{8 n^{3}+3}\right)}{\frac{1}{n}\left(\sqrt[4]{n+4}-\sqrt[5]{n^{5}+5}\right)}= \\
& =\lim _{n \rightarrow \infty} \f... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!-(2 n+2)!}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!-(2 n+2)!}=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3) \cdot(2 n+2)!-(2 n+2)!}=$
$$
=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+2)!((2 n+3)-1)}=\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+2)!\cd... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2} \frac{x^{3}-3 x-2}{x-2}$ | ## Solution
$\lim _{x \rightarrow 2} \frac{x^{3}-3 x-2}{x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^{2}+2 x+1\right)}{x-2}=$
$=\lim _{x \rightarrow 2}\left(x^{2}+2 x+1\right)=2^{2}+2 \cdot 2+1=4+4+1=9$
## Problem Kuznetsov Limits 10-20 | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-5 x}}{2 \sin x-\tan x}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)}{2 \sin x-\tan x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{-5 x}-1\right)\right)}{\frac{1}{x}(2 \sin x-\tan x)}=$
$=\frac{... | 7 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{2+\ln \left(e+x \sin \left(\frac{1}{x}\right)\right)}{\cos x+\sin x}$ | ## Solution
Since $\sin \left(\frac{1}{x}\right)_{\text {- is bounded as }} x \rightarrow 0$, then
$x \sin \left(\frac{1}{x}\right) \rightarrow 0 \quad$ as $x \rightarrow 0$
Then:
$\lim _{x \rightarrow 0} \frac{2+\ln \left(e+x \sin \left(\frac{1}{x}\right)\right)}{\cos x+\sin x}=\frac{2+\ln (e+0)}{\cos 0+\sin 0}=\f... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Find the angle between the planes
$x-y \sqrt{2}+z-1=0$
$x+y \sqrt{2}-z+3=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{1 ;-\sqrt{2} ; 1\}$
$\overrightarrow{n_{2}}=\{1 ; \sqrt{2} ;-1\}$
The angle $\phi$ between the planes is determined by the formula:
$$
\begin{align... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A\left(\frac{1}{2} ; \frac{1}{3} ; 1\right)$
$a: 2 x-3 y+3 z-2=0$
$k=1.5$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 2 x-3 y+3 z-3=0$
Substitute the coordinates of point $A$ into the equ... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+6}-\sqrt{n^{2}-5}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+6}-\sqrt{n^{2}-5}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+6}-\sqrt{n^{2}-5}\right)}{\frac{1}{n}\left(\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \fra... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{2^{n}+7^{n}}{2^{n}-7^{n-1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{2^{n}+7^{n}}{2^{n}-7^{n-1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{7^{n}}\left(2^{n}+7^{n}\right)}{\frac{1}{7^{n}}\left(2^{n}-7^{n-1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\frac{2}{7}\right)^{n}+1}{\left(\frac{2}{7}\right)^{n}-\... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt{x}-4)^{2}}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt{x}-4)^{2}}}=\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt[4]{x}-2)^{2}(\sqrt[4]{x}+2)^{2}}}= \\
& =\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{(\sqrt[4]{x}-2)^{\frac{2}{3}} \sqrt[3]{(\sqrt[4]{x}+2)^{2}}}=\l... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\tan\left(\frac{\pi}{4}-x\right)\right)^{\left(e^{x}-1\right) / x}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\tan\left(\frac{\pi}{4}-x\right)\right)^{\left(e^{x}-1\right) / x}=\left(\lim _{x \rightarrow 0} \tan\left(\frac{\pi}{4}-x\right)\right)^{\lim _{x \rightarrow 0}\left(e^{x}-1\right) / x}= \\
& =\left(\tan\left(\frac{\pi}{4}-0\right)\right)^{\lim _{x \righta... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}(\cos x+1)^{\sin x}$ | ## Solution
$\lim _{x \rightarrow \frac{\pi}{2}}(\cos x+1)^{\sin x}=\left(\cos \frac{\pi}{2}+1\right)^{\sin \frac{\pi}{2}}=(0+1)^{1}=1^{1}=1$
## Problem Kuznetsov Limits 20-27 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
2 x^{2}+x^{2} \cos \frac{1}{9 x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0):$
$f(x)=\left\{\begin{array}{c}\tan\left(x^{3}+x^{2} \sin \left(\frac{2}{x}\right)\right), x \neq 0 \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Condition of the problem
To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=\frac{4 x-x^{2}}{4}, x_{0}=2$ | ## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=\left(\frac{4 x-x^{2}}{4}\right)^{\prime}=\frac{4-2 x}{4}=\frac{2-x}{2}
$$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{2-x_{0}}{2}=\frac{2-2}{2}=0$
Since $y^{\prime}\left(x_{0}\right)=0$, the equation of the normal line is:
$x=x_{0}$
$x=2$
Thus, th... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
3^{x^{2} \sin \frac{2}{x}}-1+2 x, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left... | -2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}+(n-1)^{2}}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}+(n-1)^{2}}=\lim _{n \rightarrow \infty} \frac{n^{3}+3 n^{2}+3 n+1-n^{3}+3 n^{2}-3 n+1}{n^{2}+2 n+1+n^{2}-2 n+1}=$
$=\lim _{n \rightarrow \infty} \frac{6 n^{2}+2}{2 n^{2}+2}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(3 n^{2}+1\... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}\right)}{\frac{1}{n^{2}}(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}= \\
& =\lim _{n \rightarrow \infty}... | -4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1-2+3-4+\ldots+(2 n-1)-2 n}{\sqrt[3]{n^{3}+2 n+2}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1-2+3-4+\ldots+(2 n-1)-2 n}{\sqrt[3]{n^{3}+2 n+2}}= \\
& =\{1-2=3-4=\ldots=(2 n-1)-2 n=-1\}= \\
& =\lim _{n \rightarrow \infty} \frac{-1 \cdot n}{\sqrt[3]{n^{3}+2 n+2}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n} \cdot(-1) \cdot n}{\frac{1}{n} \sqr... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{x}-e^{-2 x}}{x+\sin x^{2}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{x}-e^{-2 x}}{x+\sin x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)}{x+\sin x^{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}{x}\left(x+\sin x^{2}... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\lim _{x \rightarrow 0} \frac{1}{x+6}}=$
$=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{0+6}}=\left(\lim _{x \rightarrow 0} \frac... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-2; -1; 1)$
$a: x-2y+6z-10=0$
$k=\frac{3}{5}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: x-2 y+6 z-6=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}}{(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}}{(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}\right)}{\frac{1}{n^{2}}(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}= \\
& =\lim _{n \rightarrow \... | -3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{1+2+3+\ldots+n} \cdot \frac{(1+(2 n-1)) n}{2}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{1+2+3+\ldots+n} \cdot n^{2}=\lim _{n \rightarrow \infty} \frac{1}{\frac{(1+n)... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)^{2}}{x^{4}+2 x+1}$ | Solution
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)^{2}}{x^{4}+2 x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-1\right)^{2}(x+1)^{2}}{\left(x^{3}-x^{2}+x+1\right)(x+1)}=$
$=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-1\right)^{2}(x+1)}{x^{3}-x^{2}+x+1}=\frac{\left((-1)^{2}-(-... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)}{\operatorname{arctg} x-x^{2}}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\frac{1}{x}\left... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}=\lim _{x \rightarrow 0} \frac{\left(e^{\alpha x}-1\right)-\left(e^{\beta x}-1\right)}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}=$
$=\lim _{x \rightarrow 0} \frac{e^{\alpha x}-1}{2 \sin \frac{x(\alpha... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{x^{2}+4}{x+2}\right)^{x^{2}+3}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{x^{2}+4}{x+2}\right)^{x^{2}+3}=\left(\frac{0^{2}+4}{0+2}\right)^{0^{2}+3}=$
$=\left(\frac{4}{2}\right)^{3}=2^{3}=8$
## Problem Kuznetsov Limits 18-6 | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}-(2 n+3)^{3}}{(2 n+1)^{2}+(2 n+3)^{2}}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}-(2 n+3)^{3}}{(2 n+1)^{2}+(2 n+3)^{2}}=\lim _{n \rightarrow \infty} \frac{8 n^{3}+3 \cdot 4 n^{2}+3 \cdot 2 n+1-8 n^{3}-3 \cdot 3 \cdot 4 n^{2}-3 \cdot 3^{2} \cdot 2 n-3^{3}}{(2 n+1)^{2}+(2 n+3)^{2}}=$
$=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}}{(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}}{(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}\right)}{\frac{1}{n^{2}}(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}= \\
& =\lim _{n \rightarrow \inft... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}-\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}\right)
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}-\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}-\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}\right... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n}{3 n^{2}-5 n+7}\right)^{n+1}$ | ## Solution
$\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n}{3 n^{2}-5 n+7}\right)^{n+1}=\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n+7}{3 n^{2}-5 n}\right)^{-n-1}=$
$=\lim _{n \rightarrow \infty}\left(1+\frac{7}{3 n^{2}-5 n}\right)^{-n-1}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{3 n^{2}-5... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Condition of the problem
Calculate the limit of the function:
$$
\lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x^{2}+2 x+1}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x^{2}+2 x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+1)\left(x^{2}-x-2\right)}{(x+1)^{2}}= \\
& =\lim _{x \rightarrow-1} \frac{x^{2}-x-2}{x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+1)(x-2)}{x+1}= \\
& ... | -3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{h \rightarrow 0} \frac{a^{x+h}+a^{x-h}-2 a^{x}}{h}$ | Solution
$\lim _{h \rightarrow 0} \frac{a^{x+h}+a^{x-h}-2 a^{x}}{h}=\lim _{h \rightarrow 0} \frac{a^{x+h}-a^{x}+a^{x-h}-a^{x}}{h}=$
$=\lim _{h \rightarrow 0} \frac{a^{x+h}-a^{x}}{h}+\lim _{h \rightarrow 0} \frac{a^{x-h}-a^{x}}{h}=$
$=\lim _{h \rightarrow 0} \frac{a^{x}\left(a^{h}-1\right)}{h}+\lim _{h \rightarrow 0}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{x+8}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{x+8}}=\left(\lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\lim _{x \rightarrow 0} \frac{3}{x+8}}=$
$=\left(\lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{0+8}}=\le... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}\left(\ln ^{2} e x\right)^{\frac{1}{x^{2}+1}}
$$ | ## Solution
$$
\lim _{x \rightarrow 1}\left(\ln ^{2} e x\right)^{\frac{1}{x^{2}+1}}=\left(\ln ^{2}(e \cdot 1)\right)^{\frac{1}{1^{2}+1}}=\left(1^{2}\right)^{\frac{1}{2}}=1
$$
## Problem Kuznetsov Limits 20-21 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=10 p+q$
$b=3 p-2 q$
$|p|=4$
$|q|=1$
$(\widehat{p, q})=\frac{\pi}{6}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(10 p+q) \times(3 p-2 q)=10 \cdot 3 \cdot p \times p+10 \cdot(-2) \cdot p \times ... | 46 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Based on the definition of the derivative, find $f^{\prime}(0)$ :
$$
f(x)=\left\{\begin{array}{c}
\sqrt[3]{1-2 x^{3} \sin \frac{5}{x}}-1+x, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$$
f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}
$$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=7 p-2 q$
$b=p+3 q$
$|p|=\frac{1}{2}$
$|q|=2$
$(\widehat{p, q})=\frac{\pi}{2}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(7 p-2 q) \times(p+3 q)=7 \cdot p \times p+7 \cdot 3 \cdot p \times q-2 \cdot q \... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(7 ; 2 ; 4)$
$M_{2}(7 ;-1 ;-2)$
$M_{3}(-5 ;-2 ;-1)$
$M_{0}(10 ; 1 ; 8)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-7 & y-2 & z-4 \\
7-7 & -1-2 & -2-4 \\
-5-7 & -2-2 & -1-4
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-7 & y-2 & z-4 \\
0 & -3 & -6 \\
-12 &... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A\left(\frac{1}{3} ; 1 ; 1\right)$
$a: 3 x-y+5 z-6=0$
$k=\frac{5}{6}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-y+5 z-5=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(x)$:
$f(x)=\left\{\begin{array}{c}\sin \left(e^{x^{2} \sin \frac{5}{x}}-1\right)+x, x \neq 0 \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$$
f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}
$$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\frac{x^{2}}{9}+y^{2}=1, z=y, z=0(y \geq 0)
$$ | ## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=1 \\
& y=0 \text { when } x=3
\end{aligned}
$$
That is, $x \in[-3,3], y \in[0,1]$
Consider the surface $z=y:$
$$ | ## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=2 \\
& y=0 \text { when } x=\sqrt{3}
\end{aligned}
$$
That is, $x \in[-\sqrt{3}, \sqrt{3}], y \in[0,2]$
Consider the surface $z=y \sqrt{3}$:
$$ | ## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=4 \\
& y=0 \text { when } x=\sqrt{3}
\end{aligned}
$$
That is, $x$ belongs to the interval $[-\sqrt{3}, \sqrt{3}]$, and $y \in [0,4]$
Consider the surface $z=y \sqrt{3}$:
$$
V_{z}=\int_{0}^{4} z d y=\i... | 32 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\frac{x^{2}}{27}+\frac{y^{2}}{25}=1, z=\frac{y}{\sqrt{3}}, z=0(y \geq 0)
$$ | ## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=5 \\
& y=0 \text { when } x=\sqrt{27}=3 \sqrt{3}
\end{aligned}
$$
That is,
$x \in[-3 \sqrt{3}, 3 \sqrt{3}], y \in[0,5]$
Consider the surface $z=\frac{y}{\sqrt{3}}:$
$$ | ## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=1 \\
& y=0 \text { when } x=3 \sqrt{3}
\end{aligned}
$$
That is,
$x \in[-3 \sqrt{3}, 3 \sqrt{3}], y \in[0,1]$
Therefore, the volume will be
$:
$f(x)=\left\{\begin{array}{c}\tan\left(2^{x^{2} \cos (1 /(8 x))}-1+x\right), x \neq 0 ; \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=1-\sin \varphi, -\frac{\pi}{2} \leq \varphi \leq -\frac{\pi}{6}
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$$
L=\int_{\phi_{1}}^{\phi_{2}} \sqrt{\rho^{2}+\left(\frac{d \rho}{d \phi}\right)^{2}} d \phi
$$
Let's find $\frac{d \rho}{d \phi}$:
$$
\frac{d \rho}{d \phi}=(-\cos \phi)
$$
We get:
$$
\begin{aligne... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.
$$
\rho=8(1-\cos \varphi),-\frac{2 \pi}{3} \leq \varphi \leq 0
$$ | ## Solution
The length of the arc of a curve given by an equation in polar coordinates is determined by the formula
$L=\int_{\varphi_{0}}^{\varphi_{1}} \sqrt{(\rho(\varphi))^{2}+\left(\rho^{\prime}(\varphi)\right)^{2}} d \varphi$
For the curve given by the equation $\rho=8(1-\cos \varphi)$, we find: $\rho^{\prime}=8... | 16 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-2 ; 1 ; 1), B(2 ; 3 ;-2), C(0 ; 0 ; 3)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(2-(-2) ; 3-1 ;-2-1)=(4 ; 2 ;-3) \\
& \overrightarrow{A C}=(0-(-2) ; 0-1 ; 3-1)=(2 ;-1 ; 2)
\end{aligned}
$$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{6 ; 3 ; 4\}$
$b=\{-1 ;-2 ;-1\}$
$c=\{2 ; 1 ; 2\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$(a, b, c)=\left|\begin{array}{ccc}6 & 3 & 4 \\ -1 & -2 & -1 \\ 2 & 1 & 2\end{array}\right|=$
$=6 \cdot\left|\begin{array}{cc}-2 & -1 \... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Find the angle between the planes:
$2 x-6 y+14 z-1=0$
$5 x-15 y+35 z-3=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{2 ;-6 ; 14\}$
$\overrightarrow{n_{2}}=\{5 ;-15 ; 35\}$
The angle $\phi_{\text{between the planes is determined by the formula: }}$
$$
\begin{aligned}
&... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## problem statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)(x+1)}{x^{4}+4 x^{2}-5}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)(x+1)}{x^{4}+4 x^{2}-5}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)(x+1)}{\left(x^{3}-x^{2}+5 x-5\right)(x+1)}= \\
& =\lim _{x \rightarrow-1} \frac{x^{3}-2 x-1}{x^{3}-x^{2}+5 x-5}=\frac{(-1)^{3}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.