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Condition of the problem
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{n-1}-\sqrt{n^{2}+1}}{\sqrt[3]{3 n^{3}+3}+\sqrt[3]{n^{5}+1}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n-1}-\sqrt{n^{2}+1}}{\sqrt[3]{3 n^{3}+3}+\sqrt{n^{5}+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n-1}-\sqrt{n^{2}+1}\right)}{\frac{1}{n}\left(\sqrt[3]{3 n^{3}+3}+\sqrt[1]{n^{5}+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \fr... | -\infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,675 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(2 n+1)!+(2 n+2)!}{(2 n+3)!}=\lim _{n \rightarrow \infty}\left(\frac{(2 n+1)!}{(2 n+3)!}+\frac{(2 n+2)!}{(2 n+3)!}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{(2 n+2)(2 n+3)}+\frac{1}{2 n+3}\right)=0+0=0
\end{aligned}
$$
## Problem Kuzne... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,676 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$\lim _{x \rightarrow 1} \frac{5 x^{2}-4 x-1}{x-1}=6$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). The number
=\frac{\varepsilon}{5} | Calculus | proof | Yes | Yes | olympiads | false | 45,678 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0 \text { (find }} \delta(\varepsilon)_{\text {): }}$ :
$f(x)=4 x^{2}-2, x_{0}=5$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0$ : $\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x-x_{0}\r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 45,679 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x+x^{2}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x+x^{2}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-2\right)(x+1)}{x(x+1)}= \\
& =\lim _{x \rightarrow-1} \frac{x^{2}-x-2}{x}=\frac{(-1)^{2}-(-1)-2}{-1}=\frac{1+1-2}{-1}=0
\end{aligned}
$$
## Problem Kuznetsov Limi... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,680 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{1-\cos 10 x}{e^{-x^{2}}-1}$ | ## Solution
Let's use the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& 1-\cos 10 x \sim \frac{(10 x)^{2}}{2}, \text{ as } x \rightarrow 0(10 x \rightarrow 0) \\
& \varepsilon^{x^{2}}-1 \sim x^{2}, \text{ as } x^{x} \rightarrow 0\left(x^{2} \rightarrow 0\right)
\end{aligned}
$$
We get:
$$
\lim _{x... | 50 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,682 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{\sqrt{x^{2}-x+1}-1}{\ln x}$ | Solution
Substitution:
$$
\begin{aligned}
& x=y+1 \Rightarrow y=x-1 \\
& x \rightarrow 1 \Rightarrow y \rightarrow 0
\end{aligned}
$$
We get:
$\lim _{x \rightarrow 1} \frac{\sqrt{x^{2}-x+1}-1}{\ln x}=\lim _{y \rightarrow 0} \frac{\sqrt{(y+1)^{2}-(y+1)+1}-1}{\ln (y+1)}=$
$=\lim _{y \rightarrow 0} \frac{\sqrt{y^{2}+2... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,683 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x-1)^{2}}{e^{\sin \pi x}-e^{-\sin 3 \pi x}}$ | ## Solution
Substitution:
$x=y+\frac{1}{2} \Rightarrow y=x-\frac{1}{2}$
$x \rightarrow \frac{1}{2} \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{1}{2}} \frac{(2 x-1)^{2}}{e^{\sin \pi x}-e^{-\sin 3 \pi x}}=\lim _{y \rightarrow 0} \frac{\left(2\left(y+\frac{1}{2}\right)-1\right... | \frac{1}{e\cdot\pi^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,684 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{-2 x}}{2 \arcsin x-\sin x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{3 x}-e^{-2 x}}{2 \arcsin x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right)-\left(e^{-2 x}-1\right)}{2 \arcsin x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{3 x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}{x}(2 \arcsin x-\sin x)}=... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,685 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{1+x \sin x-\cos 2 x}{\sin ^{2} x}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1+x \sin x-\cos 2 x}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{1+x \sin x-\left(1-2 \sin ^{2} x\right)}{\sin ^{2} x}= \\
& =\lim _{x \rightarrow 0} \frac{x \sin x+2 \sin ^{2} x}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{x \sin x}{\sin ^{2} x}+\lim _{x \r... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,686 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}(\cos \sqrt{x})^{\frac{1}{x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}(\cos \sqrt{x})^{\frac{1}{x}}=\lim _{x \rightarrow 0}\left(e^{\ln (\cos \sqrt{x})}\right)^{\frac{1}{x}}= \\
& =\lim _{x \rightarrow 0} e^{\frac{\ln (\cos \sqrt{x})}{x}}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \frac{\ln \left(1-2 \sin ^{2} \frac{\sqrt{x}}{2}\ri... | \frac{1}{\sqrt{e}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,687 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{2+x}{3-x}\right)^{x}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{2+x}{3-x}\right)^{x}=\left(\frac{2+0}{3-0}\right)^{0}=\left(\frac{2}{3}\right)^{0}=1$
## Problem Kuznetsov Limits 18-2 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,688 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow a}\left(\frac{\sin x}{\sin a}\right)^{\frac{1}{x-a}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow a}\left(\frac{\sin x}{\sin a}\right)^{\frac{1}{x-a}}=\lim _{x \rightarrow a}\left(e^{\ln \left(\frac{\sin x}{\sin a}\right)}\right)^{\frac{1}{x-a}}= \\
& =\lim _{x \rightarrow a} e^{\frac{1}{x-a} \cdot \ln \left(\frac{\sin x}{\sin a}\right)}=\exp \left\{\lim _{x \r... | e^{\cot} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,689 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{4}}(\operatorname{tg} x)^{\operatorname{ctg} x}$ | Solution
$\lim _{x \rightarrow \frac{\pi}{4}}(\tan x)^{\cot x}=(\tan \frac{\pi}{4})^{\cot \frac{\pi}{4}}=1^{1}=1$
## Problem Kuznetsov Limits 20-2 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,690 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}} \sqrt{3 \sin x+(2 x-\pi) \sin \frac{x}{2 x-\pi}}$ | ## Solution
Since $\sin \frac{x}{2 x-\pi}$ is bounded, and $(2 x-\pi) \rightarrow 0$, as $x \rightarrow \frac{\pi}{2}$, then
$$
(2 x-\pi) \sin \frac{x}{2 x-\pi} \rightarrow{ }_{, \text{as }}^{2 x} x \rightarrow \frac{\pi}{2}
$$
Then:
$$
\lim _{x \rightarrow \frac{\pi}{2}} \sqrt{3 \sin x+(2 x-\pi) \sin \frac{x}{2 x-... | \sqrt{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,691 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{2 ; 7 ; 5\}$
$p=\{1 ; 0 ; 1\}$
$q=\{1 ;-2 ; 0\}$
$r=\{0 ; 3 ; 1\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 4p-2q+r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,692 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{2 ; 0 ;-5\}$
$b=\{1 ;-3 ; 4\}$
$c_{1}=2 a-5 b$
$c_{2}=5 a-2 b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
\begin{aligned}
& c_{1}=2 a-5 b=\{2 \cdot 2-5 \cdot 1 ; 2 \cdot 0-5 \cdot(-3) ; 2 \cdot(-5)-5 \cdot 4\}=\{-1 ; 15 ;-30\} \\
& c_{... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,693 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-4 ; 0 ; 4), B(-1 ; 6 ; 7), C(1 ; 10 ; 9)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(-1-(-4) ; 6-0 ; 7-4)=(3 ; 6 ; 3)$
$\overrightarrow{A C}=(1-(-4) ; 10-0 ; 9-4)=(5 ; 10 ; 5)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrigh... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,694 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=2 p-3 q$
$b=5 p+q$
$|p|=2$
$|q|=3$
$(\widehat{p, q})=\frac{\pi}{2}$ | ## 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=(2 p-3 q) \times(5 p+q)=2 \cdot 5 \cdot p \times p+2 \cdot p \times q-3 \cdot 5 \... | 102 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,695 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{-7 ; 10 ;-5\}$
$b=\{0 ;-2 ;-1\}$
$c=\{-2 ; 4 ;-1\}$ | ## 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.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
-7 & 10 & -5 \\
0 & -2 & -1 \\
-2 & 4 & -1
\end{array}\right|= \\
& =-7 \cdot\l... | -2\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,696 |
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(2 ; -4 ; -3) \)
\( A_{2}(5 ; -6 ; 0) \)
\( A_{3}(-1 ; 3 ; -3) \)
\( A_{4}(-10 ; -8 ; 7) \) | ## Solution
From vertex $A_{1}$, we draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{5-2 ;-6-(-4) ; 0-(-3)\}=\{3 ;-2 ; 3\} \\
& A_{1} A_{3}=\{-1-2 ; 3-(-4) ;-3-(-3)\}=\{-3 ; 7 ; 0\} \\
& A_{1} \overrightarrow{A_{4}}=\{-10-2 ;-8-(-4) ; 7-(-3)\}=\{-12 ;-4 ; 10\}
\end{aligned}
$$
According to the geome... | 73 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,697 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ; 1 ;-1)$
$M_{2}(2 ; 3 ; 1)$
$M_{3}(3 ; 2 ; 1)$
$M_{0}(-3 ;-7 ; 6)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{lll}
x-1 & y-1 & z-(-1) \\
2-1 & 3-1 & 1-(-1) \\
3-1 & 2-1 & 1-(-1)
\end{array}\right|=0
$$
Perform the transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-1 & z+1 \\
1 & 2 & 2 \\... | \frac{45}{\sqrt{17}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,698 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-3 ; 6 ; 4)$
$B(8 ;-3 ; 5)$
$C(10 ;-3 ; 7)$ | ## Solution
Let's find the vector $\overrightarrow{B C}:$
$\overrightarrow{B C}=\{10-8 ;-3-(-3) ; 7-5\}=\{2 ; 0 ; 2\}$
Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$$
\begin{aligned}
& 2 \cdot(x-(-3... | x+z-1=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,699 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-3 ; 6 ; 4)$
$B(8 ;-3 ; 5)$
$C(0 ;-3 ; 7)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{0-8 ;-3-(-3) ; 7-5\}=\{-8 ; 0 ; 2\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$-8 \cdot(x-(-3))+0 \cdot(y-6)+2 \cdo... | -4x+z-16=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,700 |
## Problem Statement
Find the angle between the planes:
$x+y+z \sqrt{2}-3=0$
$x-y+z \sqrt{2}-1=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{1 ; 1 ; \sqrt{2}\}$
$\overrightarrow{n_{2}}=\{1 ;-1 ; \sqrt{2}\}$
The angle $\phi$ between the planes is determined by the formula:
$$
\begin{align... | \frac{\pi}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,701 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(x ; 0 ; 0)$
$B(1 ; 2 ; 3)$
$C(2 ; 6 ; 10)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(1-x)^{2}+(2-0)^{2}+(3-0)^{2}}=\sqrt{1-2 x+x^{2}+4+9}=\sqrt{x^{2}-2 x+14} \\
& A C=\sqrt{(2-x)^{2}+(6-0)^{2}+(10-0)^{2}}=\sqrt{4-4 x+x^{2}+36+100}=\sqrt{x^{2}-4 x+140}
\end{aligned}
$$
Since by the condition of the problem $A B=A C$... | A(63;0;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,702 |
## Task Condition
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
A $(7 ; 0 ;-1)$
$a: x-y-z-1=0$
$k=4$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a:$
$a^{\prime}: x-y-z-4=0$
Substitute the coordinates of point $A$ into the equation ... | 4\neq0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,703 |
## problem statement
Write the canonical equations of the line.
$6 x-5 y+3 z+8=0$
$6 x+5 y-4 z+4=0$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vect... | \frac{x+1}{5}=\frac{y-\frac{2}{5}}{42}=\frac{z}{60} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,704 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x+3}{0}=\frac{y-2}{-3}=\frac{z+5}{11}$
$5 x+7 y+9 z-32=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x+3}{0}=\frac{y-2}{-3}=\frac{z+5}{11}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-3 \\
y=2-3 t \\
z=-5+11 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$5 \cdot(-3)+7(2-3 t)+9(-5+11 t)-32=0$... | (-3,-1,6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,705 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(3 ; 3 ; 3)$
$8 x+6 y+8 z-22=0$ | ## Solution
Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$. Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:
$\vec{s}=\vec{n}=\{8 ; 6 ; 8\}$
Then the equation of the desired line is:
$\f... | M^{\}(-1;0;-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,706 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{4+2 n}{1-3 n}, a=-\frac{2}{3}$ | ## Solution
By the definition of the limit:
$\forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{12+6 n+2-6 n}{3(1-3 n)}\right| \\
& \left.\frac{14}{3(1-3 n)} \right\rvert\, \\
& \left|\frac{14}{3(3 n-1)}\right|
\end{aligned}\right.
$
S... | N(\varepsilon)=[\frac{14+12\varepsilon}{9\varepsilon}] | Calculus | proof | Yes | Yes | olympiads | false | 45,707 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(2 n-3)^{3}-(n+5)^{3}}{(3 n-1)^{3}+(2 n+3)^{3}}
$$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{(2 n-3)^{3}-(n+5)^{3}}{(3 n-1)^{3}+(2 n+3)^{3}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((2 n-3)^{3}-(n+5)^{3}\right)}{\frac{1}{n^{3}}\left((3 n-1)^{3}+(2 n+3)^{3}\right)}=$
$=\lim _{n \rightarrow \infty} \frac{\left(2-\frac{3}{n}\right)^{3}-\left(1+\frac... | \frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,708 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}-7}+\sqrt[3]{n^{2}+4}}{\sqrt[4]{n^{5}+5}+\sqrt{n}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}-7}+\sqrt[3]{n^{2}+4}}{\sqrt[4]{n^{5}+5}+\sqrt{n}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt[3]{n^{3}-7}+\sqrt[3]{n^{2}+4}\right)}{\frac{1}{n}\left(\sqrt[4]{n^{5}+5}+\sqrt{n}\right)}= \\
& =\lim _{n \rightarrow \infty} \f... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,709 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n\left(n^{5}+9\right)}-\sqrt{\left(n^{4}-1\right)\left(n^{2}+5\right)}}{n}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n\left(n^{5}+9\right)}-\sqrt{\left(n^{4}-1\right)\left(n^{2}+5\right)}}{n}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\sqrt{n\left(n^{5}+9\right)}-\sqrt{\left(n^{4}-1\right)\left(n^{2}+5\right)}\right)\left(\sqrt{n\left(n^{5}+9\right)}+\sqr... | -\frac{5}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,710 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\frac{n+2}{1+2+3+\ldots+n}-\frac{2}{3}\right)
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n+2}{1+2+3+\ldots+n}-\frac{2}{3}\right)=\lim _{n \rightarrow \infty}\left(\frac{n+2}{\frac{(1+n) n}{2}}-\frac{2}{3}\right)= \\
& =\lim _{n \rightarrow \infty}\left(\frac{2 n+4}{n+n^{2}}-\frac{2}{3}\right)=\lim _{n \rightarrow \infty} \frac{2 n+4}... | -\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,711 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{n+3}{n+5}\right)^{n+4}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n+3}{n+5}\right)^{n+4}=\lim _{n \rightarrow \infty}\left(\frac{n+5}{n+3}\right)^{-n-4}= \\
& =\lim _{n \rightarrow \infty}\left(\frac{n+3+2}{n+3}\right)^{-n-4}=\lim _{n \rightarrow \infty}\left(1+\frac{2}{n+3}\right)^{-n-4}= \\
& =\lim _{n \right... | e^{-2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,712 |
## Task Condition
Prove that (find $\delta(\varepsilon)$ ):
$\lim _{x \rightarrow \frac{1}{3}} \frac{6 x^{2}+x-1}{x-\frac{1}{3}}=5$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). The number
_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$f(x)=5 x^{2}+1, x_{0}=7$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$. $\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x-x_{0}\r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 45,714 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2} \frac{x^{3}-6 x^{2}+12 x-8}{x^{3}-3 x^{2}+4}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{x^{3}-6 x^{2}+12 x-8}{x^{3}-3 x^{2}+4}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^{2}-4 x+4\right)}{(x-2)\left(x^{2}-x-2\right)}= \\
& =\lim _{x \rightarrow 2} \frac{x^{2}-4 x+4}{x^{2}-x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \ri... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,715 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 4} \frac{\sqrt[3]{16 x}-4}{\sqrt{4+x}-\sqrt{2 x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 4} \frac{\sqrt[3]{16 x}-4}{\sqrt{4+x}-\sqrt{2 x}}=\lim _{x \rightarrow 4} \frac{(\sqrt[3]{16 x}-4)\left(\sqrt[3]{(16 x)^{2}}+\sqrt[3]{16 x} \cdot 4+16\right)}{(\sqrt{4+x}-\sqrt{2 x})\left(\sqrt[3]{(16 x)^{2}}+\sqrt[3]{16 x} \cdot 4+16\right)}= \\
& =\lim _{x \right... | -\frac{4\sqrt{2}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,716 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{2 \sin (\pi(x+1))}{\ln (1+2 x)}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$\ln (1+2 x) \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$
$\sin \pi x \sim \pi x$, as $x \rightarrow 0(\pi x \rightarrow 0)$
We obtain:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{2 \sin (\pi(x+1))}{\ln (1+2 x)}=\left\{\frac{0}{0}\... | -\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,717 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln 2 x-\ln \pi}{\sin \left(\frac{5 x}{2}\right) \cos x}$ | ## Solution
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln 2 x-\ln \pi}{\sin \left(\frac{5 x}{2}\right) \cos x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln \frac{2 x}{\pi}}{\sin \left(\frac{5 x}{2}\right) \cos x}=$
Substitution:
$x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2}$
$x \rightarrow \frac{\pi}{2} \Right... | \frac{2\sqrt{2}}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,718 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 3} \frac{\ln (2 x-5)}{e^{\sin \pi x}-1}
$$ | ## Solution
Substitution:
$x=y+3 \Rightarrow y=x-3$
$x \rightarrow 3 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 3} \frac{\ln (2 x-5)}{e^{\sin \pi x}-1}=\lim _{y \rightarrow 0} \frac{\ln (2(y+3)-5)}{e^{\sin \pi(y+3)}-1}= \\
& =\lim _{y \rightarrow 0} \frac{\ln (2 y+6-5)}{e^{\sin ... | -\frac{2}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,719 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{7^{3 x}-3^{2 x}}{\tan x+x^{3}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{7^{3 x}-3^{2 x}}{\tan x+x^{3}}=\lim _{x \rightarrow 0} \frac{\left(343^{x}-1\right)-\left(9^{x}-1\right)}{\tan x+x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 343}\right)^{x}-1\right)-\left(\left(e^{\ln 9}\right)^{x}-1\right)}{\tan x+x^{3}... | \ln\frac{7^{3}}{3^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,720 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{\sin 2 x}-e^{\sin x}}{\tan x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{\sin 2 x}-e^{\sin x}}{\tan x}=\lim _{x \rightarrow 0} \frac{\left(e^{\sin 2 x}-1\right)-\left(e^{\sin x}-1\right)}{\tan x}=$
$=\lim _{x \rightarrow 0} \frac{e^{\sin 2 x}-1}{\tan x}-\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{\tan x}=$
Using the substitution of equivalent... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,721 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(2-e^{x^{2}}\right)^{1 / \ln \left(1+\operatorname{tg}^{2}\left(\frac{\pi x}{3}\right)\right)}$ | ## Solution
$\lim _{x \rightarrow 0}\left(2-e^{x^{2}}\right)^{1 / \ln \left(1+\operatorname{tg}^{2}\left(\frac{\pi x}{3}\right)\right)}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln \left(2-e^{x^{2}}\right)}\right)^{1 / \ln \left(1+\operatorname{tg}^{2}\left(\frac{\pi x}{3}\right)\right)}=$
$$
\begin{aligned}
& =\lim _{x ... | e^{-\frac{9}{\pi^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,722 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{2^{2 x}-1}{x}\right)^{x+1}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{2^{2 x}-1}{x}\right)^{x+1}=\left(\lim _{x \rightarrow 0} \frac{2^{2 x}-1}{x}\right)^{\lim _{x \rightarrow 0} x+1}= \\
& =\left(\lim _{x \rightarrow 0} \frac{2^{2 x}-1}{x}\right)^{0+1}=\left(\lim _{x \rightarrow 0} \frac{2^{2 x}-1}{x}\right)^{1}= \\
& ... | 2\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,723 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}\left(2 e^{x-1}-1\right)^{\frac{x}{x-1}}
$$ | ## Solution
$\lim _{x \rightarrow 1}\left(2 e^{x-1}-1\right)^{\frac{x}{x-1}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(2 e^{x-1}-1\right)}\right)^{\frac{x}{x-1}}=$
$=\lim _{x \rightarrow 1} e^{\frac{x}{x-1} \cdot \ln \left(2 e^{x-1}-1\right)}=\exp \left\{\lim _{x \rightarrow 1} \frac{x}{x-1} \cdot \ln \left(2 e^{x-1}-... | e^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,724 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{4}}(\sin x+\cos x)^{\frac{1}{\tan x}}$ | ## Solution
$\lim _{x \rightarrow \frac{\pi}{4}}(\sin x+\cos x)^{\frac{1}{\operatorname{tg} x}}=\left(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right)^{\frac{1}{\operatorname{tg} \frac{\pi}{4}}}=\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right)^{\frac{1}{1}}=\sqrt{2}$
## Problem Kuznetsov Limits 20-17 | \sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,725 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \sqrt{\operatorname{arctg} x \cdot \sin ^{2}\left(\frac{1}{x}\right)+5 \cos x}$ | ## Solution
Since $\sin ^{2}\left(\frac{1}{x}\right)_{\text { is bounded, and }} \operatorname{arctg} x \rightarrow 0$, as $x \rightarrow 0$, then
$$
\operatorname{arctg} x \cdot \sin ^{2}\left(\frac{1}{x}\right) \rightarrow 0 \underset{\text { as } x \rightarrow 0}{ }
$$
Then:
$\lim _{x \rightarrow 0} \sqrt{\opera... | \sqrt{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,726 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{3 ; 1 ; 3\}$
$p=\{2 ; 1 ; 0\}$
$q=\{1 ; 0 ; 1\}$
$r=\{4 ; 2 ; 1\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | -3p+q+2r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,727 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{-9 ; 5 ; 3\}$
$b=\{7 ; 1 ;-2\}$
$c_{1}=2 a-b$
$c_{2}=3 a+5 b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
\begin{aligned}
& c_{1}=2 a-b=\{2 \cdot(-9)-7 ; 2 \cdot 5-1 ; 2 \cdot 3-(-2)\}=\{-25 ; 9 ; 8\} \\
& c_{2}=3 a+5 b=\{3 \cdot(-9)+5... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,728 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-1 ; 2 ;-3), B(0 ; 1 ;-2), C(-3 ; 4 ;-5)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-(-1) ; 1-2 ;-2-(-3))=(1 ;-1 ; 1)$
$\overrightarrow{A C}=(-3-(-1) ; 4-2 ;-5-(-3))=(-2 ; 2 ;-2)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\be... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,729 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=p+3 q$
$b=3 p-q$
$|p|=3$
$|q|=5$
$(\widehat{p, q})=\frac{2 \pi}{3}$ | ## 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=(p+3 q) \times(3 p-q)=3 \cdot p \times p-p \times q+3 \cdot 3 \cdot q \times ... | 75\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,730 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{3 ; 1 ; 0\}$
$b=\{-5 ;-4 ;-5\}$
$c=\{4 ; 2 ; 4\}$ | ## 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.
$(a, b, c)=\left|\begin{array}{ccc}3 & 1 & 0 \\ -5 & -4 & -5 \\ 4 & 2 & 4\end{array}\right|=$
$=3 \cdot\left|\begin{array}{cc}-4 & -5 ... | -18\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,731 |
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(3 ; 10 ;-1) \)
\( A_{2}(-2 ; 3 ;-5) \)
\( A_{3}(-6 ; 0 ;-3) \)
\( A_{4}(1 ;-1 ; 2) \) | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-2-3 ; 3-10 ;-5-(-1)\}=\{-5 ;-7 ;-4\} \\
& \vec{A}_{1} A_{3}=\{-6-3 ; 0-10 ;-3-(-1)\}=\{-9 ;-10 ;-2\} \\
& \overrightarrow{A_{1} A_{4}}=\{1-3 ;-1-10 ; 2-(-1)\}=\{-2 ;-11 ; 3\}
\end{aligned}
$$
According to... | 45.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,732 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(-2 ; 0 ;-4)$
$M_{2}(-1 ; 7 ; 1)$
$M_{3}(4 ;-8 ;-4)$
$M_{0}(-6 ; 5 ; 5)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-(-2) & y-0 & z-(-4) \\
-1-(-2) & 7-0 & 1-(-4) \\
4-(-2) & -8-0 & -4-(-4)
\end{array}\right|=0
$$
Perform the transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+2 & y & z+4 \\
1 ... | \frac{23\sqrt{2}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,733 |
## Task Condition
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(1; -5; -2)$
$B(6; -2; 1)$
$C(2; -2; -2)$ | ## Solution
Let's find the vector $\overrightarrow{BC}$:
$\overrightarrow{BC}=\{2-6 ;-2-(-2) ;-2-1\}=\{-4 ; 0 ;-3\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$-4 \cdot(x-1)+0 \cdot(y-(-5))-3 \cdo... | 4x+3z+2=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,734 |
## Task Condition
Find the angle between the planes:
$4 x+3 z-2=0$
$x+2 y+2 z+5=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{4 ; 0 ; 3\}$
$\overrightarrow{n_{2}}=\{1 ; 2 ; 2\}$
$
$B(8 ; 1 ;-7)$
$C(10 ;-2 ; 1)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$A B=\sqrt{(8-x)^{2}+(1-0)^{2}+(-7-0)^{2}}=\sqrt{64-16 x+x^{2}+1+49}=\sqrt{x^{2}-16 x+114}$
$A C=\sqrt{(10-x)^{2}+(-2-0)^{2}+(1-0)^{2}}=\sqrt{100-20 x+x^{2}+4+1}=\sqrt{x^{2}-20 x+105}$
Since by the condition of the problem $A B=A C$, then
$$
\begin{aligned}
& \... | A(-2.25;0;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,736 |
## 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 $a$?
$A(-1 ; 1 ;-2)$
$a: 4 x-y+3 z-6=0$
$k=-\frac{5}{3}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0$ and the coefficient $k$, the plane transitions to
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a:$
$a^{\prime}: 4 x-y+3 z+10=0$
Substitute the coordinates of point $A$ into the equatio... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,737 |
## Task Condition
Write the canonical equations of the line.
$6 x-7 y-z-2=0$
$x+7 y-4 z-5=0$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- are coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction ... | \frac{x-1}{35}=\frac{y-\frac{4}{7}}{23}=\frac{z}{49} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,738 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z-8}{0}$
$5 x+9 y+4 z-25=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-3}{1}=\frac{y+2}{-1}=\frac{z-8}{0}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=3+t \\
y=-2-t \\
z=8
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$5(3+t)+9(-2-t)+4 \cdot 8-25=0$
$15+5 t-18-9... | (4,-3,8) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,739 |
## Task Condition
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(0; -3; -2)$
$2x + 10y + 10z - 1 = 0$ | ## Solution
Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$. Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:
$\vec{s}=\vec{n}=\{2 ; 10 ; 10\}$
Then the equation of the desired line is:
$... | M^{\}(1;2;3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,740 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\cos 2 \phi
$$ | ## Solution
Construct the graph of the given curve. Since the polar radius is non-negative, i.e.,
$\mathrm{r} \geq 0$, then $\cos 2 \phi \geq 0, -\frac{\pi}{2} + 2 \pi \mathrm{k} \leq 2 \phi \leq \frac{\pi}{2} + 2 \pi \mathrm{k}; \mathrm{k} \in \mathrm{Z}_{0}$.
$-\frac{\pi}{4} + \pi \mathrm{k} \leq \phi \leq \frac{\... | \frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,742 |
## Problem Statement
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\sqrt{3} \cos \phi, r=\sin \phi,\left(0 \leq \phi \leq \frac{\pi}{2}\right)
$$ | ## Solution
$\sqrt{3} \cos \phi=\sin \phi ; \operatorname{tg} \phi=\sqrt{3} ; \phi=\frac{\pi}{3}$.
$\mathrm{S}=\frac{1}{2} \int_{0}^{\frac{\pi}{3}} \sin ^{2} \phi \mathrm{d} \phi+\frac{3}{2... | \frac{5\pi}{24}-\frac{\sqrt{3}}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,743 |
## Problem Statement
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=2 \cos \phi, r=2 \sqrt{3} \sin \phi,\left(0 \leq \phi \leq \frac{\pi}{2}\right)
$$ | ## Solution
## Construct
graphsIntegrals $16-5$
of the
functions:
Find the points
of intersection:
$$
2 \cos \varphi = 2 \sqrt{3} \sin \varphi
$$
$$
\begin{aligned}
& \operatornam... | \frac{5\pi}{6}-\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,745 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\sin 3 \phi
$$ | ## Solution
We will use the formula for calculating the area of a region in polar coordinates:
$$
S=\frac{1}{2} \int_{\alpha}^{\beta} r^{2}(\phi) d \phi
$$
Clearly, to find the area of the... | \frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,746 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\cos 3 \phi
$$ | ## Solution
## Integral Calculus. Option 8
## Problem 16.
Calculate the area of the figure bounded by the lines given by the equations in polar coordinates
$\mathrm{r}=\cos 3 \varphi$
Let's plot the graph of the given function in the polar coordinate system (see fig.)
,\left(-\frac{\pi}{4} \leq \phi \leq \frac{\pi}{2}\right)
$$ | ## Solution
We will construct the graphs of the given functions:
Find the points of intersection:
$$
\begin{aligned}
& \cos \varphi=\sqrt{2} \cos \left(\varphi-\frac{\pi}{4}\right) \\
& ... | \frac{\pi}{4}-\frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,749 |
## Problem Statement
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\sin \phi, r=\sqrt{2} \cos \left(\phi-\frac{\pi}{4}\right),\left(0 \leq \phi \leq \frac{3 \pi}{4}\right)
$$ | ## Solution
Let's construct the graphs
of the given functions:
Let's find the points of intersection:
$$
\begin{aligned}
& \sin \varphi=\sqrt{2} \cos \left(\varphi-\frac{\pi}{4}\right) ... | \frac{\pi}{4}-\frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,750 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\frac{1}{2}+\sin \phi
$$ | ## Solution
Since the figure is symmetrical, we calculate the area in the I and IV quadrants (i.e., for $\left[-\frac{\pi}{2} ; \frac{\pi}{2}\right]$) and multiply by 2:
$$
\begin{aligned... | \frac{3\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,752 |
## Problem Statement
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\cos \phi, r=\sin \phi,\left(0 \leq \phi \leq \frac{\pi}{2}\right)
$$ | ## Solution
Let's construct the graphs of the given functions:
Since the figure is symmetric, we will find the area of half, i.e., the upper limit will be $\frac{\pi}{4}$ instead of $\frac{\pi}{2}$.
Then we get:
, r=\sqrt{2} \sin \left(\phi-\frac{\pi}{4}\right),\left(\frac{\pi}{4} \leq \phi \leq \frac{3 \pi}{4}\right)
$$ | ## Solution
Let's find the points of intersection:
$$
\begin{aligned}
& \sqrt{2} \cos \left(\phi-\frac{\pi}{4}\right)=\sqrt{2} \sin \left(\phi-\frac{\pi}{4}\right) \\
& \operatorname{tg}\left(\phi-\frac{\pi}{4}\right)=1 \\
& \phi-\frac{\pi}{4}=\frac{\pi}{4}+\pi n, n \in \mathbb{Z} \\
& \phi=\frac{\pi}{2}+\pi n, n \in... | \frac{\pi+2}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,754 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\cos \phi, r=2 \cos \phi
$$ | ## Solution
## Integrals $16-15$
Let's construct the graphs of the given functions in polar coordinates.
The area under the curve in polar coordinates can be found as an integral:
$$
\... | \frac{3\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,755 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\sin \phi, r=2 \sin \phi
$$ | ## Solution
$$
S=\frac{1}{2} \int_{\alpha}^{\beta}\left(r_{1}^{2}(\varphi)-r_{2}^{2}(\varphi)\right) d \varphi
$$
Where $\alpha=-\frac{\pi}{2} ; \beta=\frac{\pi}{2}$, and then we get:
$$
\begin{aligned}
& S=\frac{1}{2} \int_{-\pi / 2}^{\pi / 2}\left(4 \sin ^{2} \varphi-\sin ^{2} \varphi\right) d \varphi= \\
& =\frac... | \frac{3\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,756 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=1+\sqrt{2} \cos \phi
$$ | ## Solution
Since the figure is symmetrical, we calculate the area in the I and II quadrants (i.e., for $[0 ; \pi]$) and multiply by 2:
\[
\begin{aligned}
& S=2 \cdot \frac{1}{2} \cdot \i... | 2\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,757 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\frac{1}{2}+\cos \phi
$$ | ## Solution
Since the figure is symmetrical, we calculate the area in the I and II quadrants (i.e.,
for $[0 ; \pi]$) and multiply by 2:
\[
\begin{aligned}
& S=2 \cdot \frac{1}{2} \cdot \... | \frac{3\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,758 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=1+\sqrt{2} \sin \phi
$$ | ## Solution
Since the figure is symmetrical,
we calculate the area in the I and IV quadrants (i.e., for $\left[-\frac{\pi}{2} ; \frac{\pi}{2}\right]$) and multiply by 2:
$$
\begin{aligned}... | 2\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,759 |
## Problem Statement
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\frac{5}{2} \sin \phi, r=\frac{3}{2} \sin \phi
$$ | ## Solution
$$
S=\frac{1}{2} \int_{\alpha}^{\beta}\left(r_{1}^{2}(\varphi)-r_{2}^{2}(\varphi)\right) d \varphi
$$
Where $\alpha=-\frac{\pi}{2} ; \beta=\frac{\pi}{2}$, and then we get:
$... | \pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,760 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=4 \cos 4 \phi
$$ | ## Solution
Let's draw the graph of the function $r=4 \cos 4 \varphi:$
We will use
the formula
for calculating the area
in polar
coordinates:
$$
S=\frac{1}{2} \int_{\alpha}^{\beta} ... | 8\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,762 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\sin 6 \phi
$$ | ## Solution
We will consider one petal.
Obviously, the total area will be 12 times larger.
$$
\begin{aligned}
& \sin 6 \phi=0 \\
& 6 \phi=\pi k, k \in \mathbb{Z} \\
& \phi=\frac{\pi k}{6}, k \in \mathbb{Z}
\end{aligned}
$$
For the first petal (see figure):
$$
\frac{\pi}{6} \geq \phi \geq 0
$$
Then the desired are... | \frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,763 |
## Problem Statement
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=2 \cos \phi, r=3 \cos \phi
$$ | ## Solution
Integrals $16-24$
$$
S=\frac{1}{2} \int_{\alpha}^{\beta}\left(r_{2}^{2}(\varphi)-r_{1}^{2}(\varphi)\right) d \varphi
$$
| $\varphi$ | $\frac{3 \pi}{2}$ | $\frac{5 \pi}{3}$ |... | \frac{5\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,764 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=2 \sin 4 \phi
$$ | ## Solution
Let's draw the graph of the function $r=2 \sin 4 \varphi:$
We will use
the formula
Integrals 16-26
for calculating
the area
of a region in polar
coordinates:
$$
S=\fra... | 2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,766 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=2 \cos 6 \phi
$$ | ## Solution
Let's draw the graph of the function $r=2 \cos 6 \phi$:
We will use the formula for calculating the area of a region in polar coordinates:
$$
S=\frac{1}{2} \int_{\alpha}^{\be... | 2\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,767 |
## Task Condition
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=\cos \phi-\sin \phi
$$ | ## Solution
Since $r \geq 0$, we get $\sin \left(\frac{\pi}{4}-\varphi\right) \geq 0$, hence
$$
-\frac{3 \pi}{4}+2 \pi k \leq \varphi \leq \frac{\pi}{4}+2 \pi k ; k \in Z
$$
Then the des... | \frac{\pi}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,768 |
## Problem Statement
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=3 \sin \phi, r=5 \sin \phi
$$ | ## Solution
Integrals $16-29$
$$
S=\frac{1}{2} \int_{\alpha}^{\beta}\left(r_{1}^{2}(\varphi)-r_{2}^{2}(\varphi)\right) d \varphi
$$
Where $\alpha=-\frac{\pi}{2} ; \beta=\frac{\pi}{2}$, ... | 4\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,769 |
## Problem Statement
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=2 \sin \phi, r=4 \sin \phi
$$ | ## Solution
Integrals $16-30$
$$
S=\frac{1}{2} \int_{\alpha}^{\beta}\left(r_{1}^{2}(\varphi)-r_{2}^{2}(\varphi)\right) d \varphi
$$
Where $\alpha=-\frac{\pi}{2} ; \beta=\frac{\pi}{2}$, ... | 3\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,770 |
## Problem Statement
Calculate the areas of figures bounded by lines given in polar coordinates.
$$
r=6 \sin \phi, r=4 \sin \phi
$$ | ## Solution
$$
S=\frac{1}{2} \int_{\alpha}^{\beta}\left(r_{1}^{2}(\varphi)-r_{2}^{2}(\varphi)\right) d \varphi
$$
Where $\alpha=-\frac{\pi}{2} ; \beta=\frac{\pi}{2}$, and then we get:
$... | 5\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,771 |
## Task Condition
Based on the definition of the derivative, find $f^{\prime}(0)$ :
$$
f(x)=\left\{\begin{array}{c}
e^{x \sin \frac{3}{x}}-1, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left... | f^{\}(0) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,772 |
## Problem Statement
Find the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=\frac{2 x}{x^{2}+1}, x_{0}=1$ | ## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2 x}{x^{2}+1}\right)^{\prime}=\frac{2 x^{\prime}\left(x^{2}+1\right)-2 x\left(x^{2}+1\right)^{\prime}}{\left(x^{2}+1\right)^{2}}= \\
& =\frac{2 x^{2}+2-2 x \cdot 2 x}{\left(-2 x^{2}+2\right)^{2}}=\frac{2-2 x^{2}}{\left(x^{2}+1\right)^{2}... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,773 |
## Task Condition
Find the differential $d y$.
$$
y=\ln |\cos \sqrt{x}|+\sqrt{x} \tan \sqrt{x}
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=(\ln |\cos \sqrt{x}|+\sqrt{x} \tan \sqrt{x})^{\prime} d x= \\
& =\left(\frac{1}{\cos \sqrt{x}} \cdot(-\sin \sqrt{x}) \cdot \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x}} \tan \sqrt{x}+\sqrt{x} \cdot \frac{1}{\cos ^{2} \sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}\right) d x=... | \frac{}{2\cos^{2}\sqrt{x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,774 |
## Task Condition
Approximately calculate using the differential.
$y=\sqrt{x^{3}}, x=0.98$ | ## Solution
If the increment $\Delta x = x - x_{0}$ of the argument $x$ is small in absolute value, then
$f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x$
Choose:
$x_{0} = 1$
Then:
$\Delta x = -0.02$
Calculate:
$y(1) = \sqrt{1^{3}} = 1$
$y^{\prime... | 0.97 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,775 |
## Task Condition
Find the derivative.
$$
y=\frac{1}{(x+2) \sqrt{x^{2}+4 x+5}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{(x+2) \sqrt{x^{2}+4 x+5}}\right)^{\prime}=-\frac{1}{(x+2)^{2} \cdot\left(x^{2}+4 x+5\right)} \cdot\left((x+2) \sqrt{x^{2}+4 x+5}\right)^{\prime}= \\
& =-\frac{1}{(x+2)^{2} \cdot\left(x^{2}+4 x+5\right)} \cdot\left(\sqrt{x^{2}+4 x+5}+(x+2) \cdot \frac{1}{2 \sqr... | -\frac{2x^{2}+8x+9}{(x+2)^{2}\cdot\sqrt{(x^{2}+4x+5)^{3}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,776 |
## Problem Statement
Find the derivative.
$$
y=\ln \frac{\sqrt{1+e^{x}+e^{2 x}}-e^{x}-1}{\sqrt{1+e^{x}+e^{2 x}}-e^{x}+1}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \frac{\sqrt{1+e^{x}+e^{2 x}}-e^{x}-1}{\sqrt{1+e^{x}+e^{2 x}}-e^{x}+1}\right)= \\
& =\frac{\sqrt{1+e^{x}+e^{2 x}}-e^{x}+1}{\sqrt{1+e^{x}+e^{2 x}}-e^{x}-1} \cdot\left(\frac{\sqrt{1+e^{x}+e^{2 x}}-e^{x}-1}{\sqrt{1+e^{x}+e^{2 x}}-e^{x}+1}\right)^{\prime}= \\
& =\frac{\... | \frac{1}{\sqrt{1+e^{x}+e^{2x}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,777 |
## Task Condition
Find the derivative.
$y=\ln \left(b x+\sqrt{a^{2}+b^{2} x^{2}}\right)$ | ## Solution
$y^{\prime}=\left(\ln \left(b x+\sqrt{a^{2}+b^{2} x^{2}}\right)\right)^{\prime}=$
$=\frac{1}{b x+\sqrt{a^{2}+b^{2} x^{2}}} \cdot\left(b+\frac{1}{2 \sqrt{a^{2}+b^{2} x^{2}}} \cdot 2 b^{2} x\right)=$
$=\frac{1}{b x+\sqrt{a^{2}+b^{2} x^{2}}} \cdot \frac{b \sqrt{a^{2}+b^{2} x^{2}}+b^{2} x}{\sqrt{a^{2}+b^{2} ... | \frac{b}{\sqrt{^{2}+b^{2}x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,778 |
## Task Condition
Find the derivative.
$y=\ln \cos \frac{1}{3}+\frac{\sin ^{2} 23 x}{23 \cos 46 x}$ | ## Solution
$y^{\prime}=\left(\ln \cos \frac{1}{3}+\frac{\sin ^{2} 23 x}{23 \cos 46 x}\right)^{\prime}=0+\left(\frac{\sin ^{2} 23 x}{23 \cos 46 x}\right)^{\prime}=$
$=\frac{2 \sin 23 x \cdot \cos 23 x \cdot 23 \cdot \cos 46 x-\sin ^{2} 23 x \cdot(-\sin 46 x) \cdot 46}{23 \cos ^{2} 46 x}=$
$=\frac{\sin 46 x \cdot \co... | \frac{\sin46x}{\cos^{2}46x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,779 |
## Problem Statement
Find the derivative.
$y=\sqrt{1-x^{2}}-x \cdot \arcsin \sqrt{1-x^{2}}$ | ## Solution
$y^{\prime}=\left(\sqrt{1-x^{2}}-x \cdot \arcsin \sqrt{1-x^{2}}\right)^{\prime}=$
$$
\begin{aligned}
& =\frac{1}{2 \sqrt{1-x^{2}}} \cdot(-2 x)-\left(\arcsin \sqrt{1-x^{2}}+x \cdot \frac{1}{\sqrt{1-\left(\sqrt{1-x^{2}}\right)^{2}}} \cdot \frac{1}{2 \sqrt{1-x^{2}}} \cdot(-2 x)\right)= \\
& =-\frac{x}{\sqrt{... | -\arcsin\sqrt{1-x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,780 |
## Problem Statement
Find the derivative.
$$
y=\frac{2}{\operatorname{sh} x}-\frac{1}{3 \operatorname{sh}^{3} x}+\frac{\operatorname{sh} x}{2 \operatorname{ch}^{2} x}+\frac{5}{2} \operatorname{arctg}(\operatorname{sh} x)
$$ | ## Solution
$y^{\prime}=\left(\frac{2}{\operatorname{sh} x}-\frac{1}{3 \operatorname{sh}^{3} x}+\frac{\operatorname{sh} x}{2 \operatorname{ch}^{2} x}+\frac{5}{2} \operatorname{arctg}(\operatorname{sh} x)\right)^{\prime}=$
$=-\frac{2}{\operatorname{sh}^{2} x} \cdot \operatorname{ch} x-(-3) \cdot \frac{1}{3 \operatorna... | -\frac{2\operatorname{ch}x}{\operatorname{sh}^{2}x}+\frac{\operatorname{ch}x}{\operatorname{sh}^{4}x}+\frac{1-\operatorname{sh}^{2}x}{2\operatorname{ch}^{3}x}+\frac{5}{2\operatorname{ch}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,781 |
## Task Condition
Find the derivative.
$y=x^{e^{\cos x}}$ | ## Solution
$y=x^{e^{\cos x}}$
$\ln y=e^{\cos x} \cdot \ln x$
$\frac{y^{\prime}}{y}=\left(e^{\cos x} \cdot \ln x\right)^{\prime}=\left(e^{\cos x}\right)^{\prime} \cdot \ln x+e^{\cos x} \cdot(\ln x)^{\prime}=$
$=e^{\cos x} \cdot(\cos x)^{\prime} \cdot \ln x+e^{\cos x} \cdot \frac{1}{x}=e^{\cos x} \cdot\left(-\sin x ... | x^{e^{\cosx}}\cdote^{\cosx}\cdot(\frac{1}{x}-\sinx\cdot\lnx) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,782 |
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