problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
## Task Condition
Find the derivative.
$y=\lg \ln (\operatorname{ctg} x)$ | ## Solution
$y^{\prime}=(\lg \ln (\operatorname{ctg} x))^{\prime}=\frac{1}{\ln (\operatorname{ctg} x) \cdot \ln 10} \cdot(\ln (\operatorname{ctg} x))^{\prime}=$
$=\frac{1}{\ln (\operatorname{ctg} x) \cdot \ln 10} \cdot \frac{1}{\operatorname{ctg} x} \cdot \frac{-1}{\sin ^{2} x}=-\frac{1}{\ln (\operatorname{ctg} x) \c... | -\frac{2}{\ln(\operatorname{ctg}x)\cdot\ln10\cdot\sin2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,381 |
## Task Condition
Find the derivative.
$$
y=\frac{\sqrt[5]{\operatorname{ctg} 2} \cdot \cos ^{2} 18 x}{36 \sin 36 x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\sqrt[5]{\operatorname{ctg} 2} \cdot \cos ^{2} 18 x}{36 \sin 36 x}\right)^{\prime}=\frac{\sqrt[5]{\operatorname{ctg} 2}}{36} \cdot\left(\frac{\cos ^{2} 18 x}{\sin 36 x}\right)^{\prime}= \\
& =\frac{\sqrt[5]{\operatorname{ctg} 2}}{36} \cdot\left(\frac{\cos ^{2} 18... | -\frac{\sqrt[5]{\operatorname{ctg}2}}{4\sin^{2}18x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,382 |
## Task Condition
Find the derivative.
$$
y=\frac{(1+x) \operatorname{arctg} \sqrt{x}-\sqrt{x}}{x}
$$ | ## Solution
$y^{\prime}=\left(\frac{(1+x) \operatorname{arctg} \sqrt{x}-\sqrt{x}}{x}\right)^{\prime}=\left(\left(\frac{1}{x}+1\right) \operatorname{arctg} \sqrt{x}-\frac{1}{\sqrt{x}}\right)^{\prime}=$
$=-\frac{1}{x^{2}} \cdot \operatorname{arctg} \sqrt{x}+\left(\frac{1}{x}+1\right) \cdot \frac{1}{1+(\sqrt{x})^{2}} \cd... | -\frac{1}{x^{2}}\cdot\operatorname{arctg}\sqrt{x}+\frac{1}{x\sqrt{x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,383 |
## Task Condition
Find the derivative.
$y=\frac{1}{\sqrt{8}} \arcsin \frac{3+\operatorname{ch} x}{1+3 \operatorname{ch} x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{\sqrt{8}} \arcsin \frac{3+\operatorname{ch} x}{1+3 \operatorname{ch} x}\right)^{\prime}= \\
& =\frac{1}{\sqrt{8}} \cdot \frac{1}{\sqrt{1-\left(\frac{3+\operatorname{ch} x}{1+3 \operatorname{ch} x}\right)^{2}}} \cdot\left(\frac{3+\operatorname{ch} x}{1+3 \opera... | -\frac{9\operatorname{sh}x}{8(1+3\operatorname{ch}x)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,384 |
## Task Condition
Find the derivative.
$y=\left(x^{2}+1\right)^{\cos x}$ | ## Solution
$y=\left(x^{2}+1\right)^{\cos x}$
$\ln y=\ln \left(x^{2}+1\right)^{\cos x}=\cos x \cdot \ln \left(x^{2}+1\right)$
$\frac{y^{\prime}}{y}=-\sin x \cdot \ln \left(x^{2}+1\right)+\cos x \cdot \frac{1}{x^{2}+1} \cdot 2 x=$
$=\frac{2 x \cdot \cos x}{x^{2}+1}-\sin x \cdot \ln \left(x^{2}+1\right)$
$$
\begin{a... | (x^{2}+1)^{\cosx}\cdot(\frac{2x\cdot\cosx}{x^{2}+1}-\sinx\cdot\ln(x^{2}+1)) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,385 |
## Problem Statement
Find the derivative.
$$
y=\ln \left(e^{5 x}+\sqrt{e^{10 x}-1}\right)+\arcsin \left(e^{-5 x}\right)
$$ | ## Solution
$y^{\prime}=\left(\ln \left(e^{5 x}+\sqrt{e^{10 x}-1}\right)+\arcsin \left(e^{-5 x}\right)\right)^{\prime}=$
$=\frac{1}{e^{5 x}+\sqrt{e^{10 x}-1}} \cdot\left(5 e^{5 x}+\frac{1}{2 \sqrt{e^{10 x}-1}} \cdot 10 e^{10 x}\right)+\frac{1}{\sqrt{1-e^{-10 x}}} \cdot\left(-5 e^{-5 x}\right)=$
$=\frac{1}{e^{5 x}+\s... | 5e^{5x}\cdot\sqrt{1-e^{-10x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,386 |
## Problem Statement
Find the derivative.
$$
y=\sqrt{x^{2}+1}-\frac{1}{2} \ln \frac{\sqrt{x^{2}+1}-x}{\sqrt{x^{2}+1}+1}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sqrt{x^{2}+1}-\frac{1}{2} \ln \frac{\sqrt{x^{2}+1}-x}{\sqrt{x^{2}+1}+1}\right)^{\prime}= \\
& =\frac{1}{2 \sqrt{x^{2}+1}}-\frac{1}{2} \cdot \frac{\sqrt{x^{2}+1}+1}{\sqrt{x^{2}+1}-x} \cdot\left(\frac{\sqrt{x^{2}+1}-x}{\sqrt{x^{2}+1}+1}\right)^{\prime}=\frac{1}{2 \sqrt{... | \frac{2\sqrt{x^{2}+1}+x+2}{2(\sqrt{x^{2}+1}+1)\cdot\sqrt{x^{2}+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,387 |
## Task Condition
Find the derivative.
$$
y=\frac{5^{x}(2 \sin 2 x+\cos 2 x \cdot \ln 5)}{4+\ln ^{2} 5}
$$ | ## Solution
$y^{\prime}=\left(\frac{5^{x}(2 \sin 2 x+\cos 2 x \cdot \ln 5)}{4+\ln ^{2} 5}\right)^{\prime}=$
$$
\begin{aligned}
& =\frac{1}{4+\ln ^{2} 5} \cdot\left(5^{x} \cdot \ln 5 \cdot(2 \sin 2 x+\cos 2 x \cdot \ln 5)+5^{x} \cdot(4 \cos 2 x-2 \sin 2 x \cdot \ln 5)\right)= \\
& =\frac{5^{x}}{4+\ln ^{2} 5} \cdot\lef... | 5^{x}\cdot\cos2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,388 |
## Condition of the problem
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\frac{1}{\ln t} \\
y=\ln \frac{1+\sqrt{1-t^{2}}}{t}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\frac{1}{\ln t}\right)^{\prime}=-\frac{1}{\ln ^{2} t} \cdot \frac{1}{t}=-\frac{1}{t \cdot \ln ^{2} t}$
$$
\begin{aligned}
& y_{t}^{\prime}=\left(\ln \frac{1+\sqrt{1-t^{2}}}{t}\right)^{\prime}=\frac{t}{1+\sqrt{1-t^{2}}} \cdot \frac{\frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t) \cdot t-\left(... | \frac{\ln^{2}}{\sqrt{1-^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,389 |
## problem statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
$\left\{\begin{array}{l}x=\frac{t+1}{t} \\ y=\frac{t-1}{t}\end{array}\right.$
$t_{0}=-1$ | ## Solution
Since $t_{0}=-1$, then
$x_{0}=\frac{-1+1}{-1}=0$
$y_{0}=\frac{-1-1}{-1}=2$
Find the derivatives:
$x_{t}^{\prime}=\left(\frac{t+1}{t}\right)^{\prime}=\frac{1 \cdot t-(t+1) \cdot 1}{t^{2}}=\frac{t-t-1}{t^{2}}=-\frac{1}{t^{2}}$
$y_{t}^{\prime}=\left(\frac{t-1}{t}\right)^{\prime}=\frac{1 \cdot t-(t-1) \cd... | -x+2x+2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,390 |
## Task Condition
Find the $n$-th order derivative.
$y=\lg (1+x)$ | ## Solution
$$
\begin{aligned}
& y=\lg (1+x) \\
& y^{\prime}=(\lg (1+x))^{\prime}=\frac{1}{(1+x) \ln 10}=\frac{1}{\ln 10} \cdot(1+x)^{-1} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{\ln 10} \cdot(1+x)^{-1}\right)^{\prime}=-\frac{1}{\ln 10} \cdot(1+x)^{-2} \\
& y^{\prime \prime \prime}=\left(... | y^{(n)}=\frac{(-1)^{n-1}\cdot(n-1)!}{\ln10\cdot(1+x)^{n}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,391 |
## Task Condition
Find the derivative of the specified order.
$$
y=\frac{1}{x} \cdot \sin 2 x, y^{\prime \prime \prime}=?
$$ | ## Solution
$y^{\prime}=\left(\frac{1}{x} \cdot \sin 2 x\right)^{\prime}=-\frac{1}{x^{2}} \cdot \sin 2 x+\frac{2}{x} \cdot \cos 2 x$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-\frac{1}{x^{2}} \cdot \sin 2 x+\frac{2}{x} \cdot \cos 2 x\right)^{\prime}=$
$=\frac{2}{x^{3}} \cdot \sin 2 x-\frac{2}{x^{2}} ... | \frac{12x^{2}-6}{x^{4}}\cdot\sin2x+\frac{12-8x^{2}}{x^{3}}\cdot\cos2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,392 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\sin t \\
y=\ln (\cos t)
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\sin t)^{\prime}=\cos t$
$y_{t}^{\prime}=(\ln (\cos t))^{\prime}=\frac{1}{\cos t} \cdot(-\sin t)=-\frac{\sin t}{\cos t}$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{\sin t}{\cos t}\right) / \cos t=-\frac{\sin t}{\cos ^{2} t} \\
& \le... | -\frac{1+\sin^{2}}{\cos^{4}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,393 |
## Task Condition
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{-19 ;-1 ; 7\}$
$p=\{0 ; 1 ; 1\}$
$q=\{-2 ; 0 ; 1\}$
$r=\{3 ; 1 ; 0\}$ | ## 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:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 2p+5q-3r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,395 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{1 ;-2 ; 5\}$
$b=\{3 ;-1 ; 0\}$
$c_{1}=4 a-2 b$
$c_{2}=b-2 a$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
It is not hard to notice that $c_{1}=-2(b-2 a)=-2 c_{2}$ for any $a$ and $b$.
That is, $c_{1}=-2 \cdot c_{2}$, which means the vectors $c_{1}... | c_{1}=-2\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,396 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$$
A(-3 ; -7 ; -5), B(0 ; -1 ; -2), C(2 ; 3 ; 0)
$$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(0-(-3) ;-1-(-7) ;-2-(-5))=(3 ; 6 ; 3) \\
& \overrightarrow{A C}=(2-(-3) ; 3-(-7) ; 0-(-5))=(5 ; 10 ; 5)
\end{aligned}
$$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,397 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=2 p-q$
$b=p+3 q$
$|p|=3$
$|q|=2$
$(\widehat{p, q})=\frac{\pi}{2}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(2 p-q) \times(p+3 q)=2 \cdot p \times p+2 \cdot 3 \cdot p \times q-q \times p-3 ... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,398 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{4 ; 3 ; 1\}$
$b=\{1 ;-2 ; 1\}$
$c=\{2 ; 2 ; 2\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
4 & 3 & 1 \\
1 & -2 & 1 \\
2 & 2 & 2
\end{array}\right|= \\
& =4 \cdot\left|\be... | -18\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,399 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(5 ; 2 ; 0) \)
\( A_{2}(2 ; 5 ; 0) \)
\( A_{3}(1 ; 2 ; 4) \)
\( A_{4}(-1 ; 1 ; 1) \) | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{2-5 ; 5-2 ; 0-0\}=\{-3 ; 3 ; 0\} \\
& A_{1} A_{3}=\{1-5 ; 2-2 ; 4-0\}=\{-4 ; 0 ; 4\} \\
& A_{1} \overrightarrow{A_{4}}=\{-1-5 ; 1-2 ; 1-0\}=\{-6 ;-1 ; 1\}
\end{aligned}
$$
According to the geometric meanin... | 2\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,400 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ; 2 ;-3)$
$M_{2}(1 ; 0 ; 1)$
$M_{3}(-2 ;-1 ; 6)$
$M_{0}(3 ;-2 ;-9)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-1 & y-2 & z-(-3) \\
1-1 & 0-2 & 1-(-3) \\
-2-1 & -1-2 & 6-(-3)
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-2 & z+3 \\
0 & -2 & 4 \\
... | 2\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,401 |
## Task Condition
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(1; -1; 8)$
$B(-4; -3; 10)$
$C(-1; -1; 7)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{-1-(-4) ;-1-(-3) ; 7-10\}=\{3 ; 2 ;-3\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$3 \cdot(x-1)+2 \cdot(y-(-1))-3 \... | 3x+2y-3z+23=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,402 |
## Task Condition
Find the angle between the planes
$3 y-z=0$
$2 y+z=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$$
\begin{aligned}
& \overrightarrow{n_{1}}=\{0 ; 3 ;-1\} \\
& \overrightarrow{n_{2}}=\{0 ; 2 ; 1\}
\end{aligned}
$$
$
$B(-18 ; 1 ; 0)$
$C(15 ;-10 ; 2)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(-18-0)^{2}+(1-0)^{2}+(0-z)^{2}}=\sqrt{324+1+z^{2}}=\sqrt{z^{2}+325} \\
& A C=\sqrt{(15-0)^{2}+(-10-0)^{2}+(2-z)^{2}}=\sqrt{225+100+4-4 z+z^{2}}=\sqrt{z^{2}-4 z+329}
\end{aligned}
$$
Since according to the problem $A B=A C$, then
$... | A(0;0;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,404 |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(2 ; 0 ;-1)$
$a: x-3 y+5 z-1=0$
$k=-1$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: x-3 y+5 z+1=0$
Substitute the coordinates of point $A$ into the equat... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,405 |
## Task Condition
Write the canonical equations of the line.
$x+5 y+2 z+11=0$
$x-y-z-1=0$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text { - coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vec... | \frac{x+1}{-3}=\frac{y+2}{3}=\frac{z}{-6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,406 |
## Problem Statement
Find the point of intersection of the line and the plane.
$$
\begin{aligned}
& \frac{x-1}{-2}=\frac{y-2}{1}=\frac{z+1}{-1} \\
& x-2 y+5 z+17=0
\end{aligned}
$$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-1}{-2}=\frac{y-2}{1}=\frac{z+1}{-1}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1-2 t \\
y=2+t \\
z=-1-t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$(1-2 t)-2(2+t)+5(-1-t)+17=0$
$1-2 t-4-... | (-1,3,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,407 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$$
\begin{aligned}
& M(-2 ;-3 ; 0) \\
& \frac{x+0.5}{1}=\frac{y+1.5}{0}=\frac{z-0.5}{1}
\end{aligned}
$$ | ## Solution
We find the equation of the plane that is perpendicular to the given line and passes through point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as its normal vector:
$\vec{n}=\vec{s}=\{1 ; 0 ; 1\}$
Then the equation of the desired plane is:
$$
\beg... | M^{\}(-1;0;-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,408 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{6 ;-1 ; 7\}$
$p=\{1 ;-2 ; 0\}$
$q=\{-1 ; 1 ; 3\}$
$r=\{1 ; 0 ; 4\}$ | ## 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:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | -p-3q+4r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,409 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{0 ; 3 ;-2\}$
$b=\{1 ;-2 ; 1\}$
$c_{1}=5 a-2 b$
$c_{2}=3 a+5 b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
\begin{aligned}
& c_{1}=5 a-2 b=\{5 \cdot 0-2 \cdot 1 ; 5 \cdot 3-2 \cdot(-2) ; 5 \cdot(-2)-2 \cdot 1\}=\{-2 ; 19 ;-12\} \\
& c_{... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,410 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-1, -2, 1), B(-4, -2, 5), C(-8, -2, 2)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(-4-(-1) ;-2-(-2) ; 5-1)=(-3 ; 0 ; 4)$
$\overrightarrow{A C}=(-8-(-1) ;-2-(-2) ; 2-1)=(-7 ; 0 ; 1)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\o... | \frac{1}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,411 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a_{\text {and }} b$.
\[
\begin{aligned}
& a=4 p-q \\
& b=p+2 q \\
& |p|=5 \\
& |q|=4 \\
& (\widehat{p, q})=\frac{\pi}{4}
\end{aligned}
\] | ## 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=(4 p-q) \times(p+2 q)=4 \cdot p \times p+4 \cdot 2 \cdot p \times q-q \times p-2 ... | 90\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,412 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(2 ; -1 ; 2) \)
\( A_{2}(1 ; 2 ; -1) \)
\( A_{3}(3 ; 2 ; 1) \)
\( A_{4}(-4 ; 2 ; 5) \) | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{1-2 ; 2-(-1) ;-1-2\}=\{-1 ; 3 ;-3\} \\
& \overrightarrow{A_{1} A_{3}}=\{3-2 ; 2-(-1) ; 1-2\}=\{1 ; 3 ;-1\} \\
& \overrightarrow{A_{1} A_{4}}=\{-4-2 ; 2-(-1) ; 5-2\}=\{-6 ; 3 ; 3\}
\end{aligned}
$$
Accordin... | 3\sqrt{\frac{11}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,413 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(-2; -1; -1)$
$M_{2}(0; 3; 2)$
$M_{3}(3; 1; -4)$
$M_{0}(-21; 20; -16)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-(-2) & y-(-1) & z-(-1) \\
0-(-2) & 3-(-1) & 2-(-1) \\
3-(-2) & 1-(-1) & -4-(-1)
\end{array}\right|=0
$$
Perform the transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+2 & y+1 & ... | \frac{1023}{\sqrt{1021}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,414 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(5, -1, 2)$
$B(2, -4, 3)$
$C(4, -1, 3)$ | ## Solution
Let's find the vector $\overrightarrow{BC}$:
$\overrightarrow{BC}=\{4-2 ;-1-(-4) ; 3-3\}=\{2 ; 3 ; 0\}$
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 of the form:
$2 \cdot(x-5)+3 \cdot(y+1)+... | 2x+3y-7=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,415 |
## Problem Statement
Find the angle between the planes:
\[
\begin{aligned}
& 3 x+y+z-4=0 \\
& y+z+5=0
\end{aligned}
\] | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$$
\begin{aligned}
& \overrightarrow{n_{1}}=\{3 ; 1 ; 1\} \\
& \overrightarrow{n_{2}}=\{0 ; 1 ; 1\}
\end{aligned}
$$
The angle $\phi_{\text{between the planes is determined by... | \arccos\sqrt{\frac{2}{11}}\approx6445^{\}38^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,416 |
## problem statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; y ; 0)$
$B(3 ; 0 ; 3)$
$C(0 ; 2 ; 4)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(3-0)^{2}+(0-y)^{2}+(3-0)^{2}}=\sqrt{9+y^{2}+9}=\sqrt{y^{2}+18} \\
& A C=\sqrt{(0-0)^{2}+(2-y)^{2}+(4-0)^{2}}=\sqrt{4-4 y+y^{2}+16}=\sqrt{y^{2}-4 y+20}
\end{aligned}
$$
Since according to the problem $A B=A C$, then
$\sqrt{y^{2}+18... | A(0;\frac{1}{2};0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,417 |
## 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 $a$?
$A\left(\frac{1}{4} ; \frac{1}{3} ; 1\right)$
a: $4 x-3 y+5 z-10=0$
$k=\frac{1}{2}$ | ## Solution
When transforming similarity with the center at the origin of the coordinate plane, the plane $a: A x + B y + C z + D = 0$ and the 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 $a$:
$a^{\prime}: 4 x - 3 y + 5 z - 5 = 0$
Substitute t... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,418 |
## Task Condition
Write the canonical equations of the line.
$3 x+3 y-2 z-1=0$
$2 x-3 y+z+6=0$ | ## 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}{-3}=\frac{y-\frac{4}{3}}{-7}=\frac{z}{-15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,419 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-1}{1}=\frac{y+1}{0}=\frac{z-1}{-1}$
$3 x-2 y-4 z-8=0$ | ## Solution
Let's write the parametric equations of the line.
$\frac{x-1}{1}=\frac{y+1}{0}=\frac{z-1}{-1}=t \Rightarrow$
$\left\{\begin{array}{l}x=1+t \\ y=-1 \\ z=1-t\end{array}\right.$
Substitute into the equation of the plane:
$3(1+t)-2(-1)-4(1-t)-8=0$
$3+3 t+2-4+4 t-8=0$
$7 t-7=0$
$t=1$
Find the coordinate... | (2,-1,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,420 |
## Task Condition
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$$
\begin{aligned}
& M(-1 ; 2 ; 0) \\
& \frac{x+0.5}{1}=\frac{y+0.7}{-0.2}=\frac{z-2}{2}
\end{aligned}
$$ | ## Solution
We find the equation of the plane that is perpendicular to the given line and passes through point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as its normal vector:
$\vec{n}=\vec{s}=\{1 ;-0.2 ; 2\}$
Then the equation of the desired plane is:
$1 \c... | M^{\}(-2;-3;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,421 |
## Task Condition
Prove that (find $\delta(\varepsilon)$ :
$\lim _{x \rightarrow-\frac{1}{2}} \frac{6 x^{2}+x-1}{x+\frac{1}{2}}=-5$ | ## 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
 \cos 2 x \, d x
$$ | ## Solution
$$
\int_{-2}^{0}\left(x^{2}+5 x+6\right) \cos 2 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x^{2}+5 x+6 ; d u=(2 x+5) d x \\
& d v=\cos 2 x d x ; v=\frac{1}{2} \sin 2 x
\end{aligned}
$$
Using the integration by parts formula \(\int u d v=u v-\int v d u\), we get:
$$
\begin{aligned}
& =\left.\left(x^... | \frac{5-\cos4-\sin4}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,423 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-2}^{0}\left(x^{2}-4\right) \cos 3 x \, d x
$$ | ## Solution
$$
\int_{-2}^{0}\left(x^{2}-4\right) \cos 3 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x^{2}-4 ; d u=2 x \cdot d x \\
& d v=\cos 3 x d x ; v=\frac{1}{3} \sin 3 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:
$$
\begin{aligned}
& =\left.\left(x^{2}-4\ri... | \frac{12\cos6-2\sin6}{27} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,424 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-1}^{0}\left(x^{2}+4 x+3\right) \cos x d x
$$ | ## Solution
$$
\int_{-1}^{0}\left(x^{2}+4 x+3\right) \cos x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x^{2}+4 x+3 ; d u=(2 x+4) d x \\
& d v=\cos x d x ; v=\sin x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:
$$
\begin{aligned}
& =\left.\left(x^{2}+4 x+3\right) \si... | 4-2\cos1-2\sin1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,425 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-2}^{0}(x+2)^{2} \cos 3 x \, d x
$$ | ## Solution
$$
\int_{-2}^{0}(x+2)^{2} \cos 3 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=(x+2)^{2} ; d u=2(x+2) \cdot d x \\
& d v=\cos 3 x d x ; v=\frac{1}{3} \sin 3 x
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.(x+2)^{2} \cdot ... | \frac{12-2\sin6}{27} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,426 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-4}^{0}\left(x^{2}+7 x+12\right) \cos x d x
$$ | ## Solution
$$
\int_{-4}^{0}\left(x^{2}+7 x+12\right) \cos x d x=
$$
Let:
$$
\begin{aligned}
& u=x^{2}+7 x+12 ; d u=(2 x+7) d x \\
& d v=\cos x d x ; v=\sin x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.\left(x^{2}+7 x+12\right) \sin x\ri... | 7+\cos4-2\sin4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,427 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi}\left(2 x^{2}+4 x+7\right) \cos 2 x \, d x
$$ | ## Solution
$$
\int_{0}^{\pi}\left(2 x^{2}+4 x+7\right) \cos 2 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=2 x^{2}+4 x+7 ; d u=(4 x+4) d x \\
& d v=\cos 2 x d x ; v=\frac{1}{2} \sin 2 x
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left... | \pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,428 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi}\left(9 x^{2}+9 x+11\right) \cos 3 x d x
$$ | ## Solution
$$
\int_{0}^{\pi}\left(9 x^{2}+9 x+11\right) \cos 3 x d x=
$$
Let:
$$
\begin{aligned}
& u=9 x^{2}+9 x+11 ; d u=(18 x+9) d x \\
& d v=\cos 3 x d x ; v=\frac{1}{3} \sin 3 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.\left(9 x^{... | -2\pi-2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,429 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi}\left(8 x^{2}+16 x+17\right) \cos 4 x \, dx
$$ | ## Solution
$$
\int_{0}^{\pi}\left(8 x^{2}+16 x+17\right) \cos 4 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=8 x^{2}+16 x+17 ; d u=(16 x+16) d x \\
& d v=\cos 4 x d x ; v=\frac{1}{4} \sin 4 x
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& ... | \pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,430 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi}\left(3 x^{2}+5\right) \cos 2 x \, d x
$$ | ## Solution
$$
\int_{0}^{2 \pi}\left(3 x^{2}+5\right) \cos 2 x d x=
$$
Let:
$$
\begin{aligned}
& u=3 x^{2}+5 ; d u=6 x \cdot d x \\
& d v=\cos 2 x d x ; v=\frac{1}{2} \sin 2 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.\left(3 x^{2}+5\ri... | 3\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,431 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi}\left(2 x^{2}-15\right) \cos 3 x \, d x
$$ | ## Solution
$$
\int_{0}^{2 \pi}\left(2 x^{2}-15\right) \cos 3 x d x=
$$
Let:
$$
\begin{aligned}
& u=2 x^{2}-15 ; d u=4 x \cdot d x \\
& d v=\cos 3 x d x ; v=\frac{1}{3} \sin 3 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.\left(2 x^{2}-15... | \frac{8\pi}{9} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,432 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi}\left(3-7 x^{2}\right) \cos 2 x \, d x
$$ | ## Solution
$$
\int_{0}^{2 \pi}\left(3-7 x^{2}\right) \cos 2 x d x=
$$
Let:
$$
\begin{aligned}
& u=3-7 x^{2} ; d u=-14 x \cdot d x \\
& d v=\cos 2 x d x ; v=\frac{1}{2} \sin 2 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.\left(3-7 x^{2}\... | -7\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,433 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \pi}\left(1-8 x^{2}\right) \cos 4 x \, d x
$$ | ## Solution
$$
\int_{0}^{2 \pi}\left(1-8 x^{2}\right) \cos 4 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=1-8 x^{2} ; d u=-16 x \cdot d x \\
& d v=\cos 4 x d x ; v=\frac{1}{4} \sin 4 x
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.\... | -2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,434 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-1}^{0}\left(x^{2}+2 x+1\right) \sin 3 x d x
$$ | ## Solution
$$
\int_{-1}^{0}\left(x^{2}+2 x+1\right) \sin 3 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x^{2}+2 x+1 ; d u=(2 x+2) d x=2(x+1) d x \\
& d v=\sin 3 x d x ; v=-\frac{1}{3} \cos 3 x
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
&... | -\frac{7+2\cos3}{27} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,435 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{3}\left(x^{2}-3 x\right) \sin 2 x \, d x
$$ | ## Solution
$$
\int_{0}^{3}\left(x^{2}-3 x\right) \sin 2 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x^{2}-3 x ; d u=(2 x-3) d x \\
& d v=\sin 2 x d x ; v=-\frac{1}{2} \cos 2 x
\end{aligned}
$$
Using the integration by parts formula \(\int u d v=u v-\int v d u\), we get:
$$
\begin{aligned}
& =\left.\left(x^{2}-... | \frac{3\sin6+\cos6-1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,436 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\pi}\left(x^{2}-3 x+2\right) \sin x d x
$$ | ## Solution
$$
\int_{0}^{\pi}\left(x^{2}-3 x+2\right) \sin x d x=
$$
Let:
$$
\begin{aligned}
& u=x^{2}-3 x+2 ; d u=(2 x-3) d x \\
& d v=\sin x d x ; v=-\cos x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.\left(x^{2}-3 x+2\right) \cdot(-\co... | \pi^{2}-3\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,437 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}}\left(x^{2}-5 x+6\right) \sin 3 x \, dx
$$ | ## Solution
$$
\int_{0}^{\frac{\pi}{2}}\left(x^{2}-5 x+6\right) \sin 3 x d x=
$$
Let:
$$
\begin{aligned}
& u=x^{2}-5 x+6 ; d u=(2 x-5) d x \\
& d v=\sin 3 x d x ; v=-\frac{1}{3} \cos 3 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.\left(x... | \frac{67-3\pi}{27} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,438 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-3}^{0}\left(x^{2}+6 x+9\right) \sin 2 x \, d x
$$ | ## Solution
$$
\int_{-3}^{0}\left(x^{2}+6 x+9\right) \sin 2 x d x=
$$
Let:
$$
\begin{aligned}
& u=x^{2}+6 x+9 ; d u=(2 x+6) d x \\
& d v=\sin 2 x d x ; v=-\frac{1}{2} \cos 2 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.\left(x^{2}+6 x+9\... | -\frac{17+\cos6}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,439 |
## Condition of the problem
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{4}}\left(x^{2}+17.5\right) \sin 2 x d x
$$ | ## Solution
$$
\int_{0}^{\frac{\pi}{4}}\left(x^{2}+17.5\right) \sin 2 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x^{2}+17.5 ; d u=2 x \cdot d x \\
& d v=\sin 2 x d x ; v=-\frac{1}{2} \cos 2 x
\end{aligned}
$$
Using the integration by parts formula \(\int u d v=u v-\int v d u\), we get:
$$
\begin{aligned}
& =\l... | \frac{17}{2}+\frac{\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,440 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}}\left(1-5 x^{2}\right) \sin x \, dx
$$ | ## Solution
$$
\int_{0}^{\frac{\pi}{2}}\left(1-5 x^{2}\right) \sin x d x=
$$
Let:
$$
\begin{aligned}
& u=1-5 x^{2} ; d u=-10 x \cdot d x \\
& d v=\sin x d x ; v=-\cos x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.\left(1-5 x^{2}\right) \c... | 11-5\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,441 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\frac{\pi}{4}}^{3}\left(3 x-x^{2}\right) \sin 2 x \, d x
$$ | ## Solution
$$
\int_{\frac{\pi}{4}}^{3}\left(3 x-x^{2}\right) \sin 2 x d x=
$$
Let's denote:
$$
\begin{aligned}
u=3 x-x^{2} ; d u & =(3-2 x) d x \\
d v=\sin 2 x d x ; v & =-\frac{1}{2} \cos 2 x
\end{aligned}
$$
Using the integration by parts formula \(\int u d v=u v-\int v d u\), we get:
$$
\begin{aligned}
& =\lef... | \frac{\pi-6+2\cos6-6\sin6}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,442 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{2} x \cdot \ln ^{2} x d x
$$ | ## Solution
$$
\int_{1}^{2} x \cdot \ln ^{2} x d x=
$$
Let:
$$
\begin{aligned}
& u=\ln ^{2} x ; d u=2 \ln x \cdot \frac{1}{x} \cdot d x \\
& d v=x \cdot d x ; v=\frac{x^{2}}{2}
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.\ln ^{2} x \cdot ... | 2\ln^{2}2-2\ln2+\frac{3}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,443 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{e^{2}} \frac{\ln ^{2} x}{\sqrt{x}} d x
$$ | ## Solution
$$
\int_{1}^{\epsilon^{2}} \frac{\ln ^{2} x}{\sqrt{x}} d x=
$$
Let:
$$
\begin{aligned}
& u=\ln ^{2} x ; d u=2 \ln x \cdot \frac{1}{x} \cdot d x \\
& d v=\frac{1}{\sqrt{x}} \cdot d x ; v=2 \sqrt{x}
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{ali... | 8e-16 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,444 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{8} \frac{\ln ^{2} x}{\sqrt[3]{x^{2}}} d x
$$ | ## Solution
$$
\int_{1}^{8} \frac{\ln ^{2} x}{\sqrt[3]{x^{2}}} d x=
$$
Let:
$$
\begin{aligned}
& u=\ln ^{2} x ; d u=2 \ln x \cdot \frac{1}{x} \cdot d x \\
& d v=\frac{1}{\sqrt[3]{x^{2}}} \cdot d x ; v=3 \sqrt[3]{x}
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\beg... | 6\ln^{2}8-36\ln8+54 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,445 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1}(x+1) \cdot \ln ^{2}(x+1) d x
$$ | ## Solution
$$
\int_{0}^{1}(x+1) \cdot \ln ^{2}(x+1) d x=
$$
Let's denote:
$$
\begin{aligned}
& u=\ln ^{2}(x+1) ; d u=2 \ln (x+1) \cdot \frac{1}{x+1} \cdot d x \\
& d v=(x+1) \cdot d x ; v=\frac{(x+1)^{2}}{2}
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\beg... | 2\ln^{2}2-2\ln2+\frac{3}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,446 |
## Problem Statement
Calculate the definite integral:
$$
\int_{2}^{3}(x-1)^{3} \cdot \ln ^{2}(x-1) d x
$$ | ## Solution
$$
\int_{2}^{3}(x-1)^{3} \cdot \ln ^{2}(x-1) d x=
$$
Let's denote:
$$
\begin{aligned}
& u=\ln ^{2}(x-1) ; d u=2 \ln (x-1) \cdot \frac{1}{x-1} \cdot d x \\
& d v=(x-1)^{3} \cdot d x ; v=\frac{(x-1)^{4}}{4}
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:
$$
\b... | 4\ln^{2}2-2\ln2+\frac{15}{32} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,447 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-1}^{0}(x+2)^{3} \cdot \ln ^{2}(x+2) d x
$$ | ## Solution
$$
\int_{-1}^{0}(x+2)^{3} \cdot \ln ^{2}(x+2) d x=
$$
Let's denote:
$$
\begin{aligned}
& u=\ln ^{2}(x+2) ; d u=2 \ln (x+2) \cdot \frac{1}{x+2} \cdot d x \\
& d v=(x+2)^{3} \cdot d x ; v=\frac{(x+2)^{4}}{4}
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:
$$
\... | 4\ln^{2}2-2\ln2+\frac{15}{32} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,448 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2}(x+1)^{2} \cdot \ln ^{2}(x+1) d x
$$ | ## Solution
$$
\int_{0}^{2}(x+1)^{2} \cdot \ln ^{2}(x+1) d x=
$$
Let's denote:
$$
\begin{aligned}
& u=\ln ^{2}(x+1) ; d u=2 \ln (x+1) \cdot \frac{1}{x+1} \cdot d x \\
& d v=(x+1)^{2} \cdot d x ; v=\frac{(x+1)^{3}}{3}
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:
$$
\b... | 9\ln^{2}3-6\ln3+1\frac{25}{27} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,449 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{e} \sqrt{x} \cdot \ln ^{2} x d x
$$ | ## Solution
$$
\int_{1}^{e} \sqrt{x} \cdot \ln ^{2} x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=\ln ^{2} x ; d u=2 \ln x \cdot \frac{1}{x} \cdot d x \\
& d v=\sqrt{x} \cdot d x ; v=\frac{\sqrt{x^{3}}}{\left(\frac{3}{2}\right)}=\frac{2 \sqrt{x^{3}}}{3}
\end{aligned}
$$
Using the integration by parts formula $\int... | \frac{10e\sqrt{e}-16}{27} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,450 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-1}^{1} x^{2} \cdot e^{-\frac{x}{2}} d x
$$ | ## Solution
$$
\int_{-1}^{1} x^{2} \cdot e^{-\frac{x}{2}} d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x^{2} ; d u=2 x \cdot d x \\
& d v=e^{-\frac{x}{2}} \cdot d x ; v=-2 e^{-\frac{x}{2}}
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left... | -\frac{26}{\sqrt{e}}+10\sqrt{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,451 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} x^{2} \cdot e^{3 x} d x
$$ | ## Solution
$$
\int_{0}^{1} x^{2} \cdot e^{3 x} d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x^{2} ; d u=2 x \cdot d x \\
& d v=e^{3 x} \cdot d x ; v=\frac{e^{3 x}}{3}
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\left.x^{2} \cdot \frac{e^... | \frac{5e^{3}-2}{27} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,452 |
## Task Condition
Based on the definition of the derivative, find $f^{\prime}(0)$ :
$$
f(x)=\left\{\begin{array}{c}
x^{2} \cos ^{2} \frac{11}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,454 |
## Condition of the problem
To derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$$
y=2 x+\frac{1}{x}, x_{\overline{0}}=1
$$ | ## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=\left(2 x+\frac{1}{x}\right)^{\prime}=2-\frac{1}{x^{2}}
$$
Then:
$$
y_{\overline{0}}^{\prime}=y^{\prime}\left(x_{\overline{0}}\right)=2-\frac{1}{x_{\overline{0}}^{2}}=2-\frac{1}{1^{2}}=2-1=1
$$
Since the function $y^{\prime}$ at the point $x_{0}$ has a finite deri... | x+2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,455 |
## Condition of the problem
Find the differential $d y$
$$
y=2 x+\ln |\sin x+2 \cos x|
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=(2 x-\ln |\sin x+2 \cos x|)^{\prime} d x=\left(2+\frac{1}{\sin x+2 \cos x} \cdot(\cos x-2 \sin x)\right) d x= \\
& =\left(\frac{2 \sin x+1 \cos x+\cos x-2 \sin x}{\sin x+2 \cos x}\right) d x=\frac{5 \cos x}{\sin x+2 \cos x} \cdot d x
\end{aligned}
$$ | \frac{5\cosx}{\sinx+2\cosx}\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,456 |
## Task Condition
Find the derivative.
$y=\frac{\sqrt{\left(1+x^{2}\right)^{3}}}{3 x^{3}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\sqrt{\left(1+x^{2}\right)^{3}}}{3 x^{3}}\right)^{\prime}=\frac{\frac{3}{2} \sqrt{1+x^{2}} \cdot 2 x \cdot x^{5}-\sqrt{\left(1+x^{2}\right)^{5}} \cdot 3 x^{2}}{3 x^{6}}= \\
& =\frac{\left(1+x^{2}\right) \cdot x^{-2}-\left(1+x^{2}\right)^{2}}{x^{2} \sqrt{1+x^{2}}}... | -\frac{\sqrt{1+x^{2}}}{x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,458 |
## Task Condition
Find the derivative.
$y=\log _{4} \log _{2} \operatorname{tg} x$ | ## Solution
$y^{\prime}=\left(\log _{4} \log _{2} \operatorname{tg} x\right)^{\prime}=\frac{1}{\log _{2} \operatorname{tg} x \cdot \ln 4} \cdot\left(\log _{2} \operatorname{tg} x\right)^{\prime}=$
$=\frac{1}{\log _{2} \operatorname{tg} x \cdot \ln 4} \cdot \frac{1}{\operatorname{tg} x \cdot \ln 2} \cdot \frac{1}{\cos... | \frac{1}{\sin2x\cdot\ln2\cdot\ln\operatorname{tg}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,460 |
## Task Condition
Find the derivative.
$$
y=\frac{\cos \left(\tan \frac{1}{3}\right) \cdot \sin ^{2} 15 x}{15 \cos 30 x}
$$ | ## Solution
$y^{\prime}=\left(\frac{\cos \left(\operatorname{tg} \frac{1}{3}\right) \cdot \sin ^{2} 15 x}{15 \cos 30 x}\right)^{\prime}=$
$=\frac{\cos \left(\operatorname{tg}_{\frac{1}{3}}\right)}{15} \cdot \frac{2 \sin 15 x \cdot \cos 15 x \cdot 15 \cdot \cos 30 x-\sin ^{2} 15 x \cdot(-\sin 30 x) \cdot 30}{\cos ^{2}... | \frac{\cos(\operatorname{tg}\frac{1}{3})\cdot\operatorname{tg}30x}{\cos30x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,461 |
## Problem Statement
Find the derivative.
$$
y=\frac{1}{2} \cdot \sqrt{\frac{1}{x^{2}}-1}-\frac{\arccos x}{2 x^{2}}
$$ | ## Solution
$y^{\prime}=\left(\frac{1}{2} \cdot \sqrt{\frac{1}{x^{2}}-1}-\frac{\arccos x}{2 x^{2}}\right)^{\prime}=\left(\frac{1}{2} \cdot \frac{\sqrt{x^{2}-1}}{x}-\frac{\arccos x}{2 x^{2}}\right)^{\prime}=$
$$
\begin{aligned}
& =\frac{1}{2} \cdot \frac{\frac{1}{2 \sqrt{x^{2}-1}} \cdot 2 x \cdot x-\sqrt{x^{2}-1} \cdo... | \frac{x+\sqrt{1-x^{2}}\cdot\arccosx}{x^{3}\sqrt{1-x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,462 |
## Condition of the problem
Find the derivative.
$$
y=\frac{1+8 \operatorname{ch}^{2} x \cdot \ln (\operatorname{ch} x)}{2 \operatorname{ch}^{2} x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1+8 \cosh^{2} x \cdot \ln (\cosh x)}{2 \cosh^{2} x}\right)^{\prime}=\left(\frac{1}{2 \cosh^{2} x}+4 \ln (\cosh x)\right)^{\prime}= \\
& =-2 \cdot \frac{1}{2 \cosh^{3} x} \cdot \sinh x+4 \cdot \frac{1}{\cosh x} \cdot \sinh x=-\frac{\sinh x}{\cosh^{3} x}+\frac{4 \s... | \frac{\sinhx\cdot(4\cosh^{2}x-1)}{\cosh^{3}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,463 |
## Task Condition
Find the derivative.
$y=\left(x^{2}-1\right)^{\sin x}$ | ## Solution
$y=\left(x^{2}-1\right)^{\sin x}$
$\ln y=\ln \left(x^{2}-1\right)^{\sinh x}=\sinh x \cdot \ln \left(x^{2}-1\right)$
$(\ln y)^{\prime}=\frac{y^{\prime}}{y}=\cosh x \cdot \ln \left(x^{2}-1\right)+\sinh x \cdot \frac{1}{x^{2}-1} \cdot 2 x=$
$=\cosh x \cdot \ln \left(x^{2}-1\right)+\frac{2 x \cdot \sinh x}{... | (x^{2}-1)^{\sinhx}\cdot(\coshx\cdot\ln(x^{2}-1)+\frac{2x\cdot\sinhx}{x^{2}-1}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,464 |
## Problem Statement
Find the derivative.
$$
y=\ln \frac{1+\sqrt{-3+4 x-x^{2}}}{2-x}+\frac{2}{2-x} \sqrt{-3+4 x-x^{2}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \frac{1+\sqrt{-3+4 x-x^{2}}}{2-x}+\frac{2}{2-x} \sqrt{-3+4 x-x^{2}}\right)^{\prime}= \\
& =\frac{2-x}{1+\sqrt{-3+4 x-x^{2}}} \cdot \frac{\frac{1}{2 \sqrt{-3+4 x-x^{2}}} \cdot(4-2 x) \cdot(2-x)-\left(1+\sqrt{-3+4 x-x^{2}}\right) \cdot(-1)}{(2-x)^{2}}+ \\
& +\frac{2 ... | \frac{4-x}{(2-x)^{2}\cdot\sqrt{-3+4x-x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,465 |
## Task Condition
Find the derivative.
$y=2 \arcsin \frac{2}{3 x+1}+\sqrt{9 x^{2}+6 x-3}, 3 x+1>0$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(2 \arcsin \frac{2}{3 x+1}+\sqrt{9 x^{2}+6 x-3}\right)^{\prime}= \\
& =2 \cdot \frac{1}{\sqrt{1-\left(\frac{2}{3 x+1}\right)^{2}}} \cdot\left(-\frac{2}{(3 x+1)^{2}} \cdot 3\right)+\frac{1}{2 \sqrt{9 x^{2}+6 x-3}} \cdot(18 x+6)= \\
& =-\frac{2(3 x+1)}{\sqrt{(3 x+1)^{2}-2... | \frac{3\sqrt{9x^{2}+6x-3}}{3x+1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,466 |
## Task Condition
Find the derivative.
$$
y=\frac{6^{x}(\sin 4 x \cdot \ln 6-4 \cos 4 x)}{16+\ln ^{2} 6}
$$ | ## Solution
$y^{\prime}=\left(\frac{6^{x}(\sin 4 x \cdot \ln 6-4 \cos 4 x)}{16+\ln ^{2} 6}\right)^{\prime}=\left(\frac{6^{x} \sin 4 x \cdot \ln 6-4 \cdot 6^{x} \cos 4 x}{16+\ln ^{2} 6}\right)^{\prime}=$
$$
=\frac{\ln 6\left(6^{x} \sin 4 x\right)^{\prime}-4\left(6^{x} \cos 4 x\right)^{\prime}}{16+\ln ^{2} 6}=
$$
$$
\... | 6^{x}\sin4x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,467 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\left(1+\cos ^{2} t\right)^{2} \\
y=\frac{\cos t}{\sin ^{2} t}
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\left(1+\cos ^{2} t\right)^{2}\right)^{\prime}=2\left(1+\cos ^{2} t\right) \cdot 2 \cos t \cdot(-\sin t)= \\
& =-4\left(1+\cos ^{2} t\right) \cdot \cos t \cdot \sin t \\
& y_{t}^{\prime}=\left(\frac{\cos t}{\sin ^{2} t}\right)^{\prime}=\frac{-\sin t \cdot \sin ^{2}... | \frac{1}{4\sin^{4}\cdot\cos} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,468 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\frac{1+t}{t^{2}} \\
y=\frac{3}{2 t^{2}}+\frac{2}{t}
\end{array}\right. \\
& t_{0}=2
\end{aligned}
$$ | ## Solution
Since $t_{0}=2$, then
$x_{0}=\frac{1+2}{2^{2}}=\frac{3}{4}$
$y_{0}=\frac{3}{2 \cdot 2^{2}}+\frac{2}{2}=\frac{3}{8}+1=\frac{11}{3}$
Let's find the derivatives:
$$
\begin{aligned}
x_{t}^{\prime} & =\left(\frac{1+t}{t^{2}}\right)^{\prime}=\frac{1 \cdot t^{2}-(1+t) \cdot 2 t}{t^{4}}=\frac{t^{2}-2 t-2 t^{2}}... | \frac{7x}{4}+\frac{113}{48} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,469 |
## Task Condition
Find the $n$-th order derivative.
$y=\lg (3 x+1)$ | ## Solution
$$
\begin{aligned}
& y=\lg (3 x+1) \\
& y^{\prime}=(\lg (3 x+1))^{\prime}=\frac{3}{(3 x+1) \ln 10}=\frac{3}{\ln 10} \cdot(3 x+1)^{-1} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{3}{\ln 10} \cdot(3 x+1)^{-1}\right)^{\prime}=-\frac{3^{2}}{\ln 10} \cdot(3 x+1)^{-2} \\
& y^{\prime \prim... | y^{(n)}=\frac{(-1)^{n-1}\cdot(n-1)!\cdot3^{n}}{\ln10\cdot(3x+1)^{n}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,470 |
## Task Condition
Find the derivative of the specified order.
$$
y=\left(2 x^{3}+1\right) \cos x, y^{V}=?
$$ | ## Solution
$y^{\prime}=\left(\left(2 x^{3}+1\right) \cos x\right)^{\prime}=6 x^{2} \cdot \cos x-\left(2 x^{3}+1\right) \sin x_{.}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(6 x^{2} \cdot \cos x-\left(2 x^{3}+1\right) \sin x\right)^{\prime}=$
$=12 x \cdot \cos x-6 x^{2} \cdot \sin x-6 x^{2} \cdot \si... | (30x^{2}-120)\cosx-(2x^{3}-120x+1)\sinx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,471 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$\left\{\begin{array}{l}x=\sqrt{t-1} \\ y=\frac{1}{\sqrt{t}}\end{array}\right.$ | ## Solution
$x_{t}^{\prime}=(\sqrt{t-1})^{\prime}=\frac{1}{2 \sqrt{t-1}}$
$y_{t}^{\prime}=\left(\frac{1}{\sqrt{t}}\right)^{\prime}=\left(t^{-\frac{1}{2}}\right)^{\prime}=-\frac{1}{2} \cdot t^{-\frac{3}{2}}=-\frac{1}{2 t \sqrt{t}}$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=... | \frac{(2-3)\sqrt{}}{^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,472 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
\[
\begin{aligned}
& y=\sqrt{\ln \left(\frac{1+e^{x}}{2}\right)^{2}+1} \\
& \left(1+e^{x}\right) \cdot y \cdot y^{\prime}=e^{x}
\end{aligned}
\] | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sqrt{\ln \left(\frac{1+e^{x}}{2}\right)^{2}+1}\right)^{\prime}= \\
& =\frac{1}{2 \sqrt{\ln \left(\frac{1+e^{x}}{2}\right)^{2}+1}} \cdot \frac{1}{\left(\frac{1+e^{x}}{2}\right)^{2}} \cdot 2\left(\frac{1+e^{x}}{2}\right) \cdot \frac{e^{x}}{2}= \\
& =\frac{1}{2 \sqrt{\ln... | proof | Calculus | proof | Yes | Yes | olympiads | false | 47,473 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{3-n^{2}}{4+2 n^{2}}, a=-\frac{1}{2}$ | ## Solution
By the definition of the limit:
$$
\begin{aligned}
& \forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{6-2 n^{2}+4+2 n^{2}}{2\left(4+2 n^{2}\right)}\right| \\
& \left.\frac{10}{2\left(4+2 n^{2}\right)} \right| \\
& \left.\f... | N(\varepsilon)=[\sqrt{|\frac{5}{2\varepsilon}-2|}]+1 | Calculus | proof | Yes | Yes | olympiads | false | 47,474 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}+(3 n+2)^{3}}{(2 n+3)^{3}-(n-7)^{3}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}+(3 n+2)^{3}}{(2 n+3)^{3}-(n-7)^{3}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((2 n+1)^{3}+(3 n+2)^{3}\right)}{\frac{1}{n^{3}}\left((2 n+3)^{3}-(n-7)^{3}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(2+\frac{1}{n}\ri... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,475 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{4 n^{2}-\sqrt[4]{n^{3}}}{\sqrt[3]{n^{6}+n^{3}+1}-5 n}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{4 n^{2}-\sqrt[4]{n^{3}}}{\sqrt[3]{n^{6}+n^{3}+1}-5 n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(4 n^{2}-\sqrt[4]{n^{3}}\right)}{\frac{1}{n^{2}}\left(\sqrt[3]{n^{6}+n^{3}+1}-5 n\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{4-\sqrt[4]... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,476 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \sqrt{n^{3}+8}\left(\sqrt{n^{3}+2}-\sqrt{n^{3}-1}\right)$ | ## Solution
$\lim _{n \rightarrow \infty} \sqrt{n^{3}+8}\left(\sqrt{n^{3}+2}-\sqrt{n^{3}-1}\right)=$
$=\lim _{n \rightarrow \infty} \frac{\sqrt{n^{3}+8}\left(\sqrt{n^{3}+2}-\sqrt{n^{3}-1}\right)\left(\sqrt{n^{3}+2}+\sqrt{n^{3}-1}\right)}{\sqrt{n^{3}+2}+\sqrt{n^{3}-1}}=$
$=\lim _{n \rightarrow \infty} \frac{\sqrt{n^{... | \frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,477 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{2-5+4-7+\ldots+2 n-(2 n+3)}{n+3}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{2-5+4-7+\ldots+2 n-(2 n+3)}{n+3}=$
$=\{2-5=4-7=\ldots=2 n-(2 n+3)=-3\}=$
$=\lim _{n \rightarrow \infty} \frac{-3 \cdot n}{n+3}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n} \cdot(-3) \cdot n}{\frac{1}{n}(n+3)}=$
$=\lim _{n \rightarrow \infty} \frac{-3}{1+\frac{3}{n}}=\f... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,478 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+21 n-7}{2 n^{2}+18 n+9}\right)^{2 n+1}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+21 n-7}{2 n^{2}+18 n+9}\right)^{2 n+1}=\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+18 n+9+3 n-16}{2 n^{2}+18 n+9}\right)^{2 n+1}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{3 n-16}{2 n^{2}+18 n+9}\right)^{2 n+1}=\lim _{n \rightar... | e^3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,479 |
## Task Condition
Prove that (find $\delta(\varepsilon)$ :
$$
\lim _{x \rightarrow 11} \frac{2 x^{2}-21 x-11}{x-11}=23
$$ | ## 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 \( M \). A number \( A \in \mathbb{R} \) is called the limit of the function \( f \) as \( x \) approaches \( a (x \rightarrow... | \delta(\varepsilon)=\frac{\varepsilon}{2} | Calculus | proof | Yes | Yes | olympiads | false | 47,480 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0 \text { (find }} \delta(\varepsilon)$ ):
$f(x)=3 x^{2}-2, x_{0}=5$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0$ :
$$
\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, such that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x... | proof | Calculus | proof | Yes | Yes | olympiads | false | 47,481 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{\left(x^{2}-x-2\right)^{2}}$ | ## Solution
$\lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{\left(x^{2}-x-2\right)^{2}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+1)\left(x^{2}-x-2\right)}{(x+1)(x-2)\left(x^{2}-x-2\right)}=$
$=\lim _{x \rightarrow-1} \frac{1}{x-2}=\frac{1}{-1-2}=-\frac{1}{3}$
## Problem Kuznetsov Limits 10-19 | -\frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,482 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{1}{2}} \frac{\sqrt[3]{\frac{x}{4}}-\frac{1}{2}}{\sqrt{\frac{1}{2}+x}-\sqrt{2 x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{1}{2}} \frac{\sqrt[3]{\frac{x}{4}}-\frac{1}{2}}{\sqrt{\frac{1}{2}+x}-\sqrt{2 x}}=\lim _{x \rightarrow \frac{1}{2}} \frac{\left(\sqrt[3]{\frac{x}{4}}-\frac{1}{2}\right)\left(\sqrt[3]{\left(\frac{x}{4}\right)^{2}}+\sqrt[3]{\frac{x}{4}} \cdot \frac{1}{2}+\left(\... | -\frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,483 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{\sin (\pi(x+2))}$ | Solution
Let's use the substitution of equivalent infinitesimals:
$\sin \pi x \sim \pi x$, as $x \rightarrow 0(\pi x \rightarrow 0)$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{\sin (\pi(x+2))}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{\sin (\pi x+2 \pi)}= \... | \frac{1}{2\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,484 |
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