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## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{1+\cos 3 x}{\sin ^{2} 7 x}$ | ## Solution
Substitution:
$x=y+\pi \Rightarrow y=x-\pi$
$x \rightarrow \pi \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \pi} \frac{1+\cos 3 x}{\sin ^{2} 7 x}=\lim _{y \rightarrow 0} \frac{1+\cos 3(y+\pi)}{\sin ^{2} 7(y+\pi)}= \\
& =\lim _{y \rightarrow 0} \frac{1+\cos (3 y+3 \pi)}... | \frac{9}{98} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,315 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 2} \frac{\ln (x-\sqrt[3]{2 x-3})}{\sin \left(\frac{\pi x}{2}\right)-\sin ((x-1) \pi)}
$$ | ## Solution
Substitution:
$x=y+2 \Rightarrow y=x-2$
$x \rightarrow 2 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{\ln (x-\sqrt[3]{2 x-3})}{\sin \left(\frac{\pi x}{2}\right)-\sin ((x-1) \pi)}=\lim _{y \rightarrow 0} \frac{\ln ((y+2)-\sqrt[3]{2(y+2)-3})}{\sin \left(\frac{\p... | \frac{2}{3\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,316 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{6^{2 x}-7^{-2 x}}{\sin 3 x-2 x}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{6^{2 x}-7^{-2 x}}{\sin 3 x-2 x}=\lim _{x \rightarrow 0} \frac{\left(36^{x}-1\right)-\left(49^{-x}-1\right)}{\sin 3 x-2 x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 36}\right)^{x}-1\right)-\left(\left(e^{\ln 49}\right)^{-x}-1\right)}{\sin 3 x-... | \ln(6^{2}\cdot7^{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,317 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{x^{3}+1}{\sin (x+1)}$ | ## Solution
Substitution:
$x=y-1 \Rightarrow y=x+1$
$x \rightarrow-1 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{3}+1}{\sin (x+1)}=\lim _{y \rightarrow 0} \frac{(y-1)^{3}+1}{\sin ((y-1)+1)}= \\
& =\lim _{y \rightarrow 0} \frac{y^{3}-3 y^{2}+3 y-1+1}{\sin y}=\lim _{y \... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,318 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{1+x \cdot 2^{x}}{1+x \cdot 3^{x}}\right)^{\frac{1}{x^{2}}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{1+x \cdot 2^{x}}{1+x \cdot 3^{x}}\right)^{\frac{1}{x^{2}}}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln \left(\left(1+x \cdot 2^{x}\right) /\left(1+x \cdot 3^{x}\right)\right)}\right)^{\frac{1}{x^{2}}}=$
$=\lim _{x \rightarrow 0} e^{\frac{1}{x^{2}} \ln \left(\left(1+x \cdot... | \frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,319 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(\frac{\sin 4 x}{x}\right)^{\frac{2}{x+2}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{\sin 4 x}{x}\right)^{\frac{2}{x+2}}=\left(\lim _{x \rightarrow 0} \frac{\sin 4 x}{x}\right)^{\lim _{x \rightarrow 0} \frac{2}{x+2}}= \\
& =\left(\lim _{x \rightarrow 0} \frac{\sin 4 x}{x}\right)^{\frac{2}{0+2}}=\left(\lim _{x \rightarrow 0} \frac{\sin... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,320 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\frac{1}{\sqrt[3]{x}-1}}
$$ | ## Solution
$\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\frac{1}{\sqrt[3]{x}-1}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{2 x-1}{x}\right)}\right)^{\frac{1}{\sqrt[3]{x}-1}}=$
$=\lim _{x \rightarrow 1} e^{\frac{1}{\sqrt[3]{x}-1} \cdot \ln \left(\frac{2 x-1}{x}\right)}=\exp \left\{\lim _{x \rightarrow ... | e^3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,321 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{4}}\left(\frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{1 /\left(x+\frac{\pi}{4}\right)}$ | ## Solution
$\lim _{x \rightarrow \frac{\pi}{4}}\left(\frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{1 /\left(x+\frac{\pi}{4}\right)}=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{\lim _{x \rightarrow \frac{\pi}{4}} 1 /\left(x+\frac{\pi}{4}... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,322 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{2 n-\sin n}{\sqrt{n}-\sqrt[3]{n^{3}-7}}$ | ## Solution
$$
\lim _{n \rightarrow \infty} \frac{2 n-\sin n}{\sqrt{n}-\sqrt[3]{n^{3}-7}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}(2 n-\sin n)}{\frac{1}{n}\left(\sqrt{n}-\sqrt[3]{n^{3}-7}\right)}=
$$
$=\lim _{n \rightarrow x} \frac{2-\frac{\sin n}{n}}{\sqrt{\frac{1}{n}}-\sqrt[3]{1-\frac{7}{n^{3}}}}=$
Since $\s... | -2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,323 |
## Task Condition
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{15 ;-20 ;-1\}$
$p=\{0 ; 2 ; 1\}$
$q=\{0 ; 1 ;-1\}$
$r=\{5 ;-3 ; 2\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\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... | -6p+q+3r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,324 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{-1 ; 2 ; 8\}$
$b=\{3 ; 7 ;-1\}$
$c_{1}=4 a-3 b$
$c_{2}=9 b-12 a$ | ## 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 $c_{2}=-3(4 a-3 b)=-3 c_{1}$ for any $a_{\text {and }} b$.
That is,
$$
c_{1}=-\frac{1}{3} \cdot c_{2}, \text {... | c_{1}=-\frac{1}{3}\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,325 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(-2 ; 4 ;-6), B(0 ; 2 ;-4), C(-6 ; 8 ;-10)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-(-2) ; 2-4 ;-4-(-6))=(2 ;-2 ; 2)$
$\overrightarrow{A C}=(-6-(-2) ; 8-4 ;-10-(-6))=(-4 ; 4 ;-4)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\ov... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,326 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=3 p+2 q$
$b=2 p-q$
$|p|=4$
$|q|=3$
$(\widehat{p, q})=\frac{3 \pi}{4}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$$
\begin{aligned}
& a \times b=(3 p+2 q) \times(2 p-q)=3 \cdot 2 \cdot p \times p+3 \cdot(-1... | 42\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,327 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{7 ; 4 ; 6\}$
$b=\{2 ; 1 ; 1\}$
$c=\{19 ; 11 ; 17\}$ | ## 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 & 4 & 6 \\
2 & 1 & 1 \\
19 & 11 & 17
\end{array}\right|= \\
& =7 \cdot\left|\... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,328 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(1 ; -1 ; 2) \)
\( A_{2}(2 ; 1 ; 2) \)
\( A_{3}(1 ; 1 ; 4) \)
\( A_{4}(6 ; -3 ; 8) \) | ## Solution
From vertex $A_{1}$, we draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{2-1 ; 1-(-1) ; 2-2\}=\{1 ; 2 ; 0\} \\
& \overrightarrow{A_{1} A_{3}}=\{1-1 ; 1-(-1) ; 4-2\}=\{0 ; 2 ; 2\} \\
& \overrightarrow{A_{1} A_{4}}=\{6-1 ;-3-(-1) ; 8-2\}=\{5 ;-2 ; 6\}
\end{aligned}
$$
$$
V_{A_{1} A_{2} A_{... | 3\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,329 |
## problem statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ; 5 ;-7)$
$M_{2}(-3 ; 6 ; 3)$
$M_{3}(-2 ; 7 ; 3)$
$M_{0}(1 ;-1 ; 2)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$\left|\begin{array}{ccc}x-1 & y-5 & z-(-7) \\ -3-1 & 6-5 & 3-(-7) \\ -2-1 & 7-5 & 3-(-7)\end{array}\right|=0$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-5 & z+7 \\
-4 & 1 & 10 \\
-3 & ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,330 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(2 ; 5 ; -3)$
$B(7 ; 8 ; -1)$
$C(9 ; 7 ; 4)$ | ## Solution
Let's find the vector $\overrightarrow{B C}:$
$\overrightarrow{B C}=\{9-7 ; 7-8 ; 4-(-1)\}=\{2 ;-1 ; 5\}$
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:
$$
2 \cdot(x-2)-(y-5)+5 \cdot(z-(-3... | 2x-y+5z+16=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,331 |
## Task Condition
Find the angle between the planes
$x+2 y-2 z-7=0$
$x+y-35=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{1 ; 2 ;-2\}$
$\overrightarrow{n_{2}}=\{1 ; 1 ; 0\}$
The angle $\phi_{\text{between the planes is determined by the formula: }}$
$\cos \phi=\frac{\left(... | \frac{\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,332 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(x ; 0 ; 0)$
$B(-2 ;-4 ;-6)$
$C(-1 ;-2 ;-3)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(-2-x)^{2}+(-4-0)^{2}+(-6-0)^{2}}=\sqrt{4+4 x+x^{2}+16+36}=\sqrt{x^{2}+4 x+56} \\
& A C=\sqrt{(-1-x)^{2}+(-2-0)^{2}+(-3-0)^{2}}=\sqrt{1+2 x+x^{2}+4+9}=\sqrt{x^{2}+2 x+14}
\end{aligned}
$$
Since by the condition of the problem $A B=A... | A(-21;0;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,333 |
## Task Condition
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(0 ; 3 ;-1)$
$a: 2x - y + 3z - 1 = 0$
$k = 2$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 2 x-y+3 z-2=0$
Substitute the coordinates of point $A$ into the equati... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,334 |
## problem statement
Write the canonical equations of the line.
$$
\begin{aligned}
& 2 x-3 y-2 z+6=0 \\
& x-3 y+z+3=0
\end{aligned}
$$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$,
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vec... | \frac{x+3}{9}=\frac{y}{4}=\frac{z}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,335 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-7}{3}=\frac{y-3}{1}=\frac{z+1}{-2}$
$2 x+y+7 z-3=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-7}{3}=\frac{y-3}{1}=\frac{z+1}{-2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=7+3 t \\
y=3+t \\
z=-1-2 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$2(7+3 t)+(3+t)+7(-1-2 t)-3=0$
$14+6 t+... | (10;4;-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,336 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(-2 ; 0 ; 3)$
$2 x-2 y+10 z+1=0$ | ## Solution
Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$. Since the line is perpendicular to the given plane, we can use the normal vector of the plane as its direction vector:
$\vec{s}=\vec{n}=\{2 ;-2 ; 10\}$
Then the equation of the desired line is:
$\f... | M^{\}(-3;1;-2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,337 |
## Task Condition
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\sqrt{t-1} \\
y=\frac{t}{\sqrt{1-t}}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\sqrt{t-1})^{\prime}=\frac{1}{2 \sqrt{t-1}}$
$$
\begin{aligned}
& y_{t}^{\prime}=\left(\frac{t}{\sqrt{1-t}}\right)^{\prime}=\frac{1 \cdot \sqrt{1-t}-t \cdot \frac{1}{2 \sqrt{1-t}} \cdot(-1)}{1-t}= \\
& =\frac{2(1-t)+t}{2 \sqrt{(1-t)^{3}}}=\frac{2-2 t+t}{2 \sqrt{(1-t)^{3}}}=\frac{2-t}{2 \s... | \frac{2}{\sqrt{(1-)^{3}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,338 |
## problem statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{3 ;-3 ; 4\}$
$p=\{1 ; 0 ; 2\}$
$q=\{0 ; 1 ; 1\}$
$r=\{2 ;-1 ; 4\}$ | ## 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:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | p-2q+r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,339 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{3 ; 4 ;-1\}$
$b=\{2 ;-1 ; 1\}$
$c_{1}=6 a-3 b$
$c_{2}=b-2 a$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
It is not hard to notice that $c_{1}=-3(b-2 a)=-3 c_{2}$ for any $a$ and $b$.
That is, $c_{1}=-3 \cdot c_{2}$, which means the vectors $c_{1}... | c_{1}=-3\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,340 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(2, -4, 6), B(0, -2, 4), C(6, -8, 10)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(0-2 ;-2-(-4) ; 4-6)=(-2 ; 2 ;-2) \\
& \overrightarrow{A C}=(6-2 ;-8-(-4) ; 10-6)=(4 ;-4 ; 4)
\end{aligned}
$$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\ov... | \cos(\overrightarrow{AB,AC})=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,341 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=4 p+q$
$b=p-q$
$|p|=7$
$|q|=2$
$(\widehat{p, q})=\frac{\pi}{4}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$$
\begin{aligned}
& a \times b=(4 p+q) \times(p-q)=4 \cdot p \times p-4 \cdot p \times q+q \... | 35\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,342 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{4 ; 3 ; 1\}$
$b=\{6 ; 7 ; 4\}$
$c=\{2 ; 0 ;-1\}$ | ## 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}
4 & 3 & 1 \\
6 & 7 & 4 \\
2 & 0 & -1
\end{array}\right|= \\
& =4 \cdot\left|\be... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,343 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(2 ; -1 ; -2) \)
\( A_{2}(1 ; 2 ; 1) \)
\( A_{3}(5 ; 0 ; -6) \)
\( A_{4}(-10 ; 9 ; -7) \) | ## Solution
From vertex $A_{1}$, we will draw vectors:
$$
\begin{aligned}
& A_{1} A_{2}=\{1-2 ; 2-(-1) ; 1-(-2)\}=\{-1 ; 3 ; 3\} \\
& A_{1} A_{3}=\{5-2 ; 0-(-1) ;-6-(-2)\}=\{3 ; 1 ;-4\}
\end{aligned}
$$
$\overrightarrow{A_{1} A_{4}}=\{-10-2 ; 9-(-1) ;-7-(-2)\}=\{-12 ; 10 ;-5\}$
According to the geometric meaning of... | 46\frac{2}{3},4\sqrt{14} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,344 |
## problem statement
Find the distance from the point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$$
\begin{aligned}
& M_{1}(3 ; 10 ;-1) \\
& M_{2}(-2 ; 3 ;-5) \\
& M_{3}(-6 ; 0 ;-3) \\
& M_{0}(-6 ; 7 ;-10)
\end{aligned}
$$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-3 & y-10 & z-(-1) \\
-2-3 & 3-10 & -5-(-1) \\
-6-3 & 0-10 & -3-(-1)
\end{array}\right|=0
$$
Perform the transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-3 & y-10 & z-(-1) \\
-... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,345 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-2 ; 0 ;-5)$
$B(2 ; 7 ;-3)$
$C(1 ; 10 ;-1)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{1-2 ; 10-7 ;-1-(-3)\}=\{-1 ; 3 ; 2\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$-(x-(-2))+3 \cdot(y-0)+2 \cdot(z-(-... | -x+3y+2z+8=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,346 |
## Task Condition
Find the angle between the planes
$6 x+3 y-2 z=0$
$x+2 y+6 z-12=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{6 ; 3 ;-2\}$
$\overrightarrow{n_{2}}=\{1 ; 2 ; 6\}$
The angle $\phi_{\text{between the planes is determined by the formula: }}$
$\cos \phi=\frac{\left(... | \frac{\pi}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,347 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; 0 ; z)$
$B(10 ; 0 ;-2)$
$C(9 ;-2 ; 1)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$A B=\sqrt{(10-0)^{2}+(0-0)^{2}+(-2-z)^{2}}=\sqrt{100+0+4+4 z+z^{2}}=\sqrt{z^{2}+4 z+104}$
$A C=\sqrt{(9-0)^{2}+(-2-0)^{2}+(1-z)^{2}}=\sqrt{81+4+1-2 z+z^{2}}=\sqrt{z^{2}-2 z+86}$
Since by the condition of the problem $A B=A C$, then
$$
\begin{aligned}
& \sqrt{z... | A(0;0;-3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,348 |
## 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 $a$?
$A(1; -2; 1)$
$a: 5x + y - z + 6 = 0$
$k = \frac{2}{3}$ | ## 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 $k$, it transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 5 x+y-z+4=0$
Substitute the coordinates of point $A$ into the equatio... | 6\neq0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,349 |
## Problem Statement
Find the point of intersection of the line and the plane.
$$
\begin{aligned}
& \frac{x-1}{2}=\frac{y-2}{0}=\frac{z-4}{1} \\
& x-2 y+4 z-19=0
\end{aligned}
$$ | ## Solution
Let's write the parametric equations of the line.
$$
\frac{x-1}{2}=\frac{y-2}{0}=\frac{z-4}{1}=t \Rightarrow
$$
$$
\left\{\begin{array}{l}
x=1+2 t \\
y=2 \\
z=4+t
\end{array}\right.
$$
Substitute into the equation of the plane:
$(1+2 t)-2 \cdot 2+4(4+t)-19=0$
$1+2 t-4+16+4 t-19=0$
$6 t-6=0$
$t=1$
F... | (3;2;5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,351 |
## Task Condition
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$$
\begin{aligned}
& M(-1 ; 0 ;-1) \\
& \frac{x}{-1}=\frac{y-1.5}{0}=\frac{z-2}{1}
\end{aligned}
$$ | ## Solution
Find the equation of the plane that is perpendicular to the given line and passes through the point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as the normal vector of the plane:
$\vec{n}=\vec{s}=\{-1 ; 0 ; 1\}$
Then the equation of the desired pla... | M^{\}(3;3;3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,352 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\sqrt{1+\ln \left(1+3 x^{2} \cos \frac{2}{x}\right)}-1, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} ... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,353 |
## Task Condition
Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=\frac{x^{2}-3 x+3}{3}, x_{0}=3$ | ## Solution
Let's find $y^{\prime}:$
$y^{\prime}=\left(\frac{x^{2}-3 x+3}{3}\right)^{\prime}=\frac{2 x-3}{3}$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{2 \cdot 3-3}{3}=\frac{3}{3}=1$
Since the function $y^{\prime}$ at the point $x_{0}$ has a finite derivative, the equation of the tangent line is:
$... | x-2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,354 |
## Task Condition
Find the differential $d y$.
$$
y=\ln \left|2 x+2 \sqrt{x^{2}+x}+1\right|
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\ln \left|2 x+2 \sqrt{x^{2}+x}+1\right|\right)^{\prime} d x= \\
& =\frac{1}{2 x+2 \sqrt{x^{2}+x}+1} \cdot\left(2+\frac{2}{2 \sqrt{x^{2}+x}} \cdot(2 x+1)\right) \cdot d x= \\
& =\frac{1}{2 x+2 \sqrt{x^{2}+x}+1} \cdot\left(\frac{2 \sqrt{x^{2}+x}+2 x+1}{\sqr... | \frac{}{\sqrt{x^{2}+x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,355 |
## Problem Statement
Approximately calculate using the differential.
$y=\sqrt{4 x-3}, x=1.78$ | ## 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:
$x_{0} = 1.75$
Then:
$\Delta x = 0.03$
Calculate:
$y(1.75) = \sqrt{4 \cdot 1.75 - 3} = ... | 2.03 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,356 |
## Task Condition
Find the derivative.
$y=2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right)^{\prime}=2 \cdot \frac{1}{2} \cdot \sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}} \cdot\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)^{\prime}= \\
& =\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}} \cdot \frac{-\frac{1}{2 \sqrt{x}} \cdot(1+\sqrt{x}... | -\frac{1}{\sqrt{x(1-x)}\cdot(1+\sqrt{x})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,357 |
## Task Condition
Find the derivative.
$$
y=3 e^{\sqrt[3]{x}}\left(\sqrt[3]{x^{2}}-2 \sqrt[3]{x}+2\right)
$$ | ## Solution
$y^{\prime}=\left(3 e^{\sqrt[3]{x}}\left(\sqrt[3]{x^{2}}-2 \sqrt[3]{x}+2\right)\right)^{\prime}=$
$=3 e^{\sqrt[3]{x}} \cdot \frac{1}{3} \cdot \frac{1}{\sqrt[3]{x^{2}}} \cdot\left(\sqrt[3]{x^{2}}-2 \sqrt[3]{x}+2\right)+3 e^{\sqrt[3]{x}}\left(\frac{2}{3} \cdot \frac{1}{\sqrt[3]{x}}-2 \cdot \frac{1}{3} \cdot... | e^{\sqrt[3]{x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,358 |
## Task Condition
Find the derivative.
$y=\ln \arccos \sqrt{1-e^{4 x}}$ | Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \arccos \sqrt{1-e^{4 x}}\right)^{\prime}=\frac{1}{\arccos \sqrt{1-e^{4 x}}} \cdot \frac{-1}{\sqrt{1-\left(\sqrt{1-e^{4 x}}\right)^{2}}} \cdot \frac{1}{2 \sqrt{1-e^{4 x}}} \cdot\left(-4 e^{4 x}\right)= \\
& =\frac{1}{\arccos \sqrt{1-e^{4 x}}} \cdot \frac{1}{\sqrt{e^{4 ... | \frac{2e^{2x}}{\sqrt{1-e^{4x}}\cdot\arccos\sqrt{1-e^{4x}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,359 |
Condition of the problem
Find the derivative.
$$
y=\cos (\ln 13)-\frac{1}{44} \cdot \frac{\cos ^{2} 22 x}{\sin 44 x}
$$ | ## Solution
$y^{\prime}=\left(\cos (\ln 13)-\frac{1}{44} \cdot \frac{\cos ^{2} 22 x}{\sin 44 x}\right)^{\prime}=0-\frac{1}{44} \cdot\left(\frac{\cos ^{2} 22 x}{\sin 44 x}\right)^{\prime}=$
$=-\frac{1}{44} \cdot\left(\frac{\cos ^{2} 22 x}{2 \sin 22 x \cdot \cos 22 x}\right)^{\prime}=-\frac{1}{88} \cdot\left(\frac{\cos... | \frac{1}{4\sin^{2}22x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,360 |
## Problem Statement
Find the derivative.
$$
y=\arcsin \frac{x-2}{(x-1) \sqrt{2}}
$$ | Solution
$y^{\prime}=\left(\arcsin \frac{x-2}{(x-1) \sqrt{2}}\right)^{\prime}=\frac{1}{\sqrt{1-\left(\frac{x-2}{(x-1) \sqrt{2}}\right)^{2}}} \cdot\left(\frac{x-2}{(x-1) \sqrt{2}}\right)^{\prime}=$
$$
\begin{aligned}
& =\frac{(x-1) \sqrt{2}}{\sqrt{\left.(x-1)^{2} \cdot 2-(x-2)\right)^{2}}} \cdot \frac{1}{\sqrt{2}} \cd... | \frac{1}{(x-1)\sqrt{x^{2}-2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,361 |
## Task Condition
Find the derivative.
$y=\frac{1-8 \cosh^{2} x}{4 \cosh^{4} x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1-8 \cosh^{2} x}{4 \cosh^{4} x}\right)^{\prime}=\frac{-16 \cosh x \cdot \sinh x \cdot \cosh^{4} x-\left(1-8 \cosh^{2} x\right) \cdot 4 \cosh^{3} x \cdot \sinh x}{4 \cosh^{8} x}= \\
& =\frac{-4 \cosh^{2} x \cdot \sinh x-\left(1-8 \cosh^{2} x\right) \cdot \sinh x}{... | \frac{4\tanh^{3}x}{\cosh^{2}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,362 |
## Problem Statement
Find the derivative.
$y=\frac{2 x-1}{4 x^{2}-4 x+3}+\frac{1}{\sqrt{2}} \cdot \operatorname{arctg} \frac{2 x-1}{\sqrt{2}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2 x-1}{4 x^{2}-4 x+3}+\frac{1}{\sqrt{2}} \cdot \operatorname{arctg} \frac{2 x-1}{\sqrt{2}}\right)^{\prime}= \\
& =\frac{2 \cdot\left(4 x^{2}-4 x+3\right)-(2 x-1) \cdot(8 x-4)}{\left(4 x^{2}-4 x+3\right)^{2}}+\frac{1}{\sqrt{2}} \cdot \frac{1}{1+\left(\frac{2 x-1}{... | \frac{8}{(4x^{2}-4x+3)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,364 |
## Task Condition
Find the derivative.
$$
y=3 \arcsin \frac{3}{x+2}+\sqrt{x^{2}+4 x-5}
$$ | ## Solution
$y^{\prime}=\left(3 \arcsin \frac{3}{x+2}+\sqrt{x^{2}+4 x-5}\right)^{\prime}=$
$=3 \cdot \frac{1}{\sqrt{1-\left(\frac{3}{x+2}\right)^{2}}} \cdot\left(-\frac{3}{(x+2)^{2}}\right)+\frac{1}{2 \sqrt{x^{2}+4 x-5}} \cdot(2 x+4)=$
$=-\frac{x+2}{\sqrt{(x+2)^{2}-3^{2}}} \cdot \frac{9}{(x+2)^{2}}+\frac{x+2}{\sqrt{... | \frac{\sqrt{x^{2}+4x-5}}{x+2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,365 |
## Task Condition
Find the derivative.
$$
y=\frac{\cos x}{\sin ^{2} x}-2 \cos x-3 \ln \left(\tan \frac{x}{2}\right)
$$ | ## Solution
$y^{\prime}=\left(\frac{\cos x}{\sin ^{2} x}-2 \cos x-3 \ln \left(\operatorname{tg} \frac{x}{2}\right)\right)^{\prime}=$
$=\frac{-\sin x \cdot \sin ^{2} x-\cos x \cdot 2 \sin x \cdot \cos x}{\sin ^{4} x}+2 \sin x-3 \cdot \frac{1}{\operatorname{tg} \frac{x}{2}} \cdot \frac{1}{\cos ^{2} \frac{x}{2}}=$
$=\f... | -\frac{2+3\sin^{2}x}{\sin^{3}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,366 |
## Condition of the problem
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\operatorname{arctg} t \\
y=\ln \frac{\sqrt{1+t^{2}}}{t+1}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\operatorname{arctg} t)^{\prime}=\frac{1}{1+t^{2}}$
$y_{t}^{\prime}=\left(\ln \frac{\sqrt{1+t^{2}}}{t+1}\right)^{\prime}=\frac{t+1}{\sqrt{1+t^{2}}} \cdot \frac{\frac{1}{2 \sqrt{1+t^{2}}} \cdot 2 t \cdot(t+1)-\sqrt{1+t^{2}} \cdot 1}{(t+1)^{2}}=$
$=\frac{1}{\sqrt{1+t^{2}}} \cdot \frac{t^{2... | \frac{-1}{+1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,367 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{
\begin{array}{l}
x=\frac{1+t^{3}}{t^{2}-1} \\
y=\frac{t}{t^{2}-1}
\end{array}
\right.
\]
$t_{0}=2$ | ## Solution
Since $t_{0}=2$, then
$x_{0}=\frac{1+2^{3}}{2^{2}-1}=\frac{9}{3}=3$
$y_{0}=\frac{2}{2^{2}-1}=\frac{2}{3}$
Let's find the derivatives:
$x_{t}^{\prime}=\left(\frac{1+t^{3}}{t^{2}-1}\right)^{\prime}=\frac{3 t^{2} \cdot\left(t^{2}-1\right)-\left(1+t^{3}\right) \cdot 2 t}{\left(t^{2}-1\right)^{2}}=\frac{3 t... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,368 |
## problem statement
Find the $n$-th order derivative.
$y=\sin (3 x+1)+\cos 5 x$ | ## Solution
$y^{\prime}=(\sin (3 x+1)+\cos 5 x)^{\prime}=3 \cos (3 x+1)-5 \sin 5 x$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=(3 \cos (3 x+1)-5 \sin 5 x)^{\prime}=-3^{2} \cdot \sin (3 x+1)-5^{2} \cdot \cos 5 x$
$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-3^{2} \cdot \sin (3 x+1... | y^{(n)}=3^{n}\cdot\sin(\frac{3\pi}{2}\cdotn+3x+1)+5^{n}\cdot\cos(\frac{3\pi}{2}\cdotn+5x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,369 |
Condition of the problem
Find the derivative of the specified order.
$y=e^{\frac{x}{2}} \cdot \sin 2 x, y^{IV}=?$ | ## Solution
$y^{\prime}=\left(e^{\frac{x}{2}} \cdot \sin 2 x\right)^{\prime}=\frac{1}{2} \cdot e^{\frac{x}{2}} \cdot \sin 2 x+2 e^{\frac{x}{2}} \cdot \cos 2 x$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{2} \cdot e^{\frac{x}{2}} \cdot \sin 2 x+2 e^{\frac{x}{2}} \cdot \cos 2 x\right)^{\prime}=$
... | \frac{161}{16}\cdote^{\frac{x}{2}}\cdot\sin2x-15\cdote^{\frac{x}{2}}\cdot\cos2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,370 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\cos t+t \cdot \sin t \\
y=\sin t-t \cdot \cos t
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\cos t+t \cdot \sin t)^{\prime}=-\sin t+\sin t+t \cdot \cos t=t \cdot \cos t$
$y_{t}^{\prime}=(\sin t-t \cdot \cos t)^{\prime}=\cos t-\cos t+t \cdot \sin t=t \cdot \sin t$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{t \cdot \sin t}{t \cdot \cos t}=\frac{\sin t... | \frac{1}{\cdot\cos^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,371 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{0 ; -8 ; 9\}$
$p=\{0 ; -2 ; 1\}$
$q=\{3 ; 1 ; -1\}$
$r=\{4 ; 0 ; 1\}$ | ## 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:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 2p-4q+3r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,373 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{-1 ; 3 ; 4\}$
$b=\{2 ;-1 ; 0\}$
$c_{1}=6 a-2 b$
$c_{2}=b-3 a$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
It is not hard to notice that $c_{1}=-2(b-3 a)=-2 c_{2}$ for any $a$ and $b$.
That is, $c_{1}=-2 \cdot c_{2}$, which means the vectors $c_{1}... | c_{1}=-2\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,374 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(1 ; 4 ;-1), B(-2 ; 4 ;-5), C(8 ; 4 ; 0)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(-2-1 ; 4-4 ;-5-(-1))=(-3 ; 0 ;-4)$
$\overrightarrow{A C}=(8-1 ; 4-4 ; 0-(-1))=(7 ; 0 ; 1)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overright... | -\frac{1}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,375 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=6 p-q$
$b=5 q+p$
$|p|=\frac{1}{2}$
$|q|=4$
$(\widehat{p, q})=\frac{5 \pi}{6}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$$
\begin{aligned}
& a \times b=(6 p-q) \times(5 q+p)=6 \cdot 5 \cdot p \times q+6 \cdot p \t... | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,376 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{4 ; 1 ; 1\}$
$b=\{-9 ;-4 ;-9\}$
$c=\{6 ; 2 ; 6\}$ | ## 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 & 1 & 1 \\
-9 & -4 & -9 \\
6 & 2 & 6
\end{array}\right|= \\
& =4 \cdot\left|\... | -18\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,377 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(-2 ; -1 ; -1) \)
\( A_{2}(0 ; 3 ; 2) \)
\( A_{3}(3 ; 1 ; -4) \)
\( A_{4}(-4 ; 7 ; 3) \) | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$$
\begin{aligned}
& A_{1} \overrightarrow{A_{2}}=\{0-(-2) ; 3-(-1) ; 2-(-1)\}=\{2 ; 4 ; 3\} \\
& A_{1} \overrightarrow{A_{3}}=\{3-(-2) ; 1-(-1) ;-4-(-1)\}=\{5 ; 2 ;-3\} \\
& \vec{A}_{1} A_{4}=\{-4-(-2) ; 7-(-1) ; 3-(-1)\}=\{-2 ; 8 ; 4\}
\end{aligned}
$$
Ac... | 23\frac{1}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,378 |
## problem statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ; 1 ; 2)$
$M_{2}(-1 ; 1 ; 3)$
$M_{3}(2 ;-2 ; 4)$
$M_{0}(2 ; 3 ; 8)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$\left|\begin{array}{ccc}x-1 & y-1 & z-2 \\ -1-1 & 1-1 & 3-2 \\ 2-1 & -2-1 & 4-2\end{array}\right|=0$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-1 & z-2 \\
-2 & 0 & 1 \\
1 & -3 & 2
\end... | 7\sqrt{\frac{7}{10}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,379 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-1 ; 2 ; -2)$
$B(13 ; 14 ; 1)$
$C(14 ; 15 ; 2)$ | ## Solution
Let's find the vector $\overrightarrow{B C}:$
$\overrightarrow{B C}=\{14-13 ; 15-14 ; 2-1\}=\{1 ; 1 ; 1\}$
Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$$
\begin{aligned}
& (x-(-1))+(y-2... | x+y+z+1=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,380 |
## Task Condition
Find the angle between the planes:
$x-y+7 z-1=0$
$2 x-2 y-5=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{1 ;-1 ; 7\}$
$\overrightarrow{n_{2}}=\{2 ;-2 ; 0\}$
The angle $\phi$ between the planes is determined by the formula:
$$
\begin{aligned}
& \cos \ph... | 7834^{\}42^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,381 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(x ; 0 ; 0)$
$B(1 ; 5 ; 9)$
$C(3 ; 7 ; 11)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$A B=\sqrt{(1-x)^{2}+(5-0)^{2}+(9-0)^{2}}=\sqrt{1-2 x+x^{2}+25+81}=\sqrt{x^{2}-2 x+107}$
$A C=\sqrt{(3-x)^{2}+(7-0)^{2}+(11-0)^{2}}=\sqrt{9-6 x+x^{2}+49+121}=\sqrt{x^{2}-6 x+179}$
Since by the condition of the problem $A B=A C$, then
$$
\sqrt{x^{2}-2 x+107}=\sq... | A(18;0;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,382 |
## 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 $a$?
$A(1 ; 2 ; 2)$
$a: 3x - z + 5 = 0$
$k = -\frac{1}{5}$ | ## Solution
When transforming similarity with the center at the origin of the coordinate 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 $a$:
$a^{\prime}: 3 x-z-1=0$
Substitute the coordinates of point $A$ into the eq... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,383 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-1}{2}=\frac{y+2}{-5}=\frac{z-3}{-2}$
$x+2 y-5 z+16=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-1}{2}=\frac{y+2}{-5}=\frac{z-3}{-2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+2 t \\
y=-2-5 t \\
z=3-2 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$(1+2 t)+2(-2-5 t)-5(3-2 t)+16=0$ $1... | (3,-7,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,385 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(2; -2; -3)$
$y+z+2=0$ | ## 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:
$\vec{s}=\vec{n}=\{0 ; 1 ; 1\}$
Then the equation of the desired line is:
$\f... | M^{\}(2;1;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,386 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\frac{\ln (\cos x)}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | -\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,387 |
## Problem Statement
Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=\frac{x^{5}+1}{x^{4}+1}, x_{0}=1$ | ## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x^{5}+1}{x^{4}+1}\right)^{\prime}=\frac{\left(x^{5}+1\right)^{\prime}\left(x^{4}+1\right)-\left(x^{5}+1\right)\left(x^{4}+1\right)^{\prime}}{\left(x^{4}+1\right)^{2}}= \\
& =\frac{5 x^{4}\left(x^{4}+1\right)-\left(x^{5}+1\right) \cdot 4 ... | \frac{x}{2}+\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,388 |
## Task Condition
Find the differential $d y$.
$$
y=\ln \left|\frac{x+\sqrt{x^{2}+1}}{2 x}\right|
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\ln \left|\frac{x+\sqrt{x^{2}+1}}{2 x}\right|\right)^{\prime} d x=\frac{2 x}{x+\sqrt{x^{2}+1}} \cdot\left(\frac{x+\sqrt{x^{2}+1}}{2 x}\right)^{\prime} d x= \\
& =\frac{2 x}{x+\sqrt{x^{2}+1}} \cdot \frac{\left(1+\frac{1}{2 \sqrt{x^{2}+1}} \cdot 2 x\right) ... | -\frac{}{x(x+\sqrt{x^{2}+1})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,389 |
## Task Condition
Approximately calculate using the differential.
$y=\sqrt{4 x-1}, x=2.56$ | ## Solution
If the increment $\Delta x = x - x_{0}$ of the argument $x$ is small in absolute value, then
$f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x$
Choose:
$x_{0} = 2.5$
Then:
$\Delta x = 0.06$
Calculate:
$y(2.5) = \sqrt{4 \cdot 2.5 - 1} = 3$... | 3.04 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,390 |
## Task Condition
Find the derivative.
$y=\frac{\sqrt{2 x+3}(x-2)}{x^{2}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\sqrt{2 x+3}(x-2)}{x^{2}}\right)^{\prime}=\frac{\left(\frac{1}{2 \sqrt{2 x+3}} \cdot 2 \cdot(x-2)+\sqrt{2 x+3}\right) x^{2}-\sqrt{2 x+3}(x-2) \cdot 2 x}{x^{4}}= \\
& =\frac{((x-2)+(2 x+3)) x-(2 x+3)(x-2) \cdot 2}{x^{3} \sqrt{2 x+3}}= \\
& =\frac{(3 x+1) x-2\left(... | \frac{-x^{2}+3x+12}{x^{3}\sqrt{2x+3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,391 |
## Task Condition
Find the derivative.
$y=\ln \left(e^{x}+\sqrt{e^{2 x}-1}\right)+\arcsin e^{-x}$ | ## Solution
$y^{\prime}=\left(\ln \left(e^{x}+\sqrt{e^{2 x}-1}\right)+\arcsin e^{-x}\right)^{\prime}=$
$=\frac{1}{e^{x}+\sqrt{e^{2 x}-1}} \cdot\left(e^{x}+\frac{1}{2 \sqrt{e^{2 x}-1}} \cdot 2 e^{2 x}\right)+\frac{1}{\sqrt{1-e^{-2 x}}} \cdot\left(-e^{-x}\right)=$
$$
\begin{aligned}
& =\frac{e^{x}}{e^{x}+\sqrt{e^{2 x}... | \sqrt{\frac{e^{x}-1}{e^{x}+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,392 |
## Task Condition
Find the derivative.
$$
y=\ln \cos \frac{2 x+3}{x+1}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \cos \frac{2 x+3}{x+1}\right)^{\prime}=\frac{1}{\cos \frac{2 x+3}{x+1}} \cdot\left(\cos \frac{2 x+3}{x+1}\right)^{\prime}= \\
& =\frac{1}{\cos \frac{2 x+3}{x+1}} \cdot\left(-\sin \frac{2 x+3}{x+1}\right) \cdot\left(\frac{2 x+3}{x+1}\right)^{\prime}= \\
& =-\tan \fr... | \frac{\tan\frac{2x+3}{x+1}}{(x+1)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,393 |
## Task Condition
Find the derivative.
$y=\ln \cos \frac{2 x+3}{2 x+1}$ | ## Solution
$y^{\prime}=\left(\ln \cos \frac{2 x+3}{2 x+1}\right)^{\prime}=\frac{1}{\cos \frac{2 x+3}{2 x+1}} \cdot\left(\cos \frac{2 x+3}{2 x+1}\right)^{\prime}=$
$=\frac{1}{\cos \frac{2 x+3}{2 x+1}} \cdot\left(-\sin \frac{2 x+3}{2 x+1}\right) \cdot\left(\frac{2 x+3}{2 x+1}\right)^{\prime}=$
$=-\operatorname{tg} \f... | \frac{4\operatorname{tg}\frac{2x+3}{2x+1}}{(2x+1)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,394 |
## Condition of the problem
Find the derivative.
$$
y=\frac{\operatorname{ctg}\left(\sin \frac{1}{3}\right) \cdot \sin ^{2} 17 x}{17 \cos 34 x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\operatorname{ctg}\left(\sin \frac{1}{3}\right) \cdot \sin ^{2} 17 x}{17 \cos 34 x}\right)^{\prime}= \\
& =\frac{\operatorname{ctg}\left(\sin \frac{1}{3}\right)}{17} \cdot \frac{2 \sin 17 x \cdot \cos 17 x \cdot 17 \cdot \cos 34 x-\sin ^{2} 17 x \cdot(-\sin 34 x)... | \frac{\operatorname{ctg}(\sin\frac{1}{3})\cdot\operatorname{tg}34x}{\cos34x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,395 |
## Task Condition
Find the derivative.
$$
y=\frac{x-3}{2} \sqrt{6 x-x^{2}-8}+\arcsin \sqrt{\frac{x}{2}-1}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x-3}{2} \sqrt{6 x-x^{2}-8}+\arcsin \sqrt{\frac{x}{2}-1}\right)^{\prime}= \\
& =\frac{1}{2} \cdot \sqrt{6 x-x^{2}-8}+\frac{x-3}{2} \cdot \frac{1}{2 \sqrt{6 x-x^{2}-8}} \cdot(6-2 x)+\frac{1}{\sqrt{1-\left(\sqrt{\frac{x}{2}-1}\right)^{2}}} \cdot \frac{1}{2 \sqrt{\fr... | \sqrt{6x-x^{2}-8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,396 |
## Task Condition
Find the derivative.
$$
y=-\frac{\operatorname{sh} x}{2 \operatorname{ch}^{2} x}+\frac{3}{2} \cdot \arcsin (\operatorname{th} x)
$$ | ## Solution
$y^{\prime}=\left(-\frac{\operatorname{sh} x}{2 \operatorname{ch}^{2} x}+\frac{3}{2} \cdot \arcsin (\operatorname{th} x)\right)^{\prime}=$
$=-\frac{\operatorname{ch} x \cdot \operatorname{ch}^{2} x-\operatorname{sh} x \cdot 2 \operatorname{ch} x \cdot \operatorname{sh} x}{2 \operatorname{ch}^{4} x}+\frac{... | \frac{\operatorname{ch}2x}{\operatorname{ch}^{3}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,397 |
## Task Condition
Find the derivative.
$y=(\sin x)^{\frac{\varsigma_{x}}{2}}$ | ## Solution
$y=(\sin x)^{\frac{5 x}{2}}$
$\ln y=\ln \left((\sin x)^{\frac{5 x}{2}}\right)=\frac{5 x}{2} \cdot \ln (\sin x)$
$\frac{y^{\prime}}{y}=\left(\frac{5 x}{2} \cdot \ln (\sin x)\right)^{\prime}=\frac{5}{2} \cdot \ln (\sin x)+\frac{5 x}{2} \cdot \frac{1}{\sin x} \cdot \cos x=\frac{5}{2} \cdot(\ln (\sin x)+x \c... | \frac{5}{2}\cdot(\sinx)^{\frac{5x}{2}}\cdot(\ln(\sinx)+x\cdot\operatorname{ctg}x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,398 |
## Condition of the problem
Find the derivative.
$y=\frac{1}{\sqrt{2}} \operatorname{arctg} \frac{x-1}{\sqrt{2}}+\frac{x-1}{x^{2}-2 x+3}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{\sqrt{2}} \operatorname{arctg} \frac{x-1}{\sqrt{2}}+\frac{x-1}{x^{2}-2 x+3}\right)^{\prime}= \\
& =\frac{1}{\sqrt{2}} \cdot \frac{1}{1+\left(\frac{x-1}{\sqrt{2}}\right)^{2}} \cdot \frac{1}{\sqrt{2}}+\frac{1 \cdot\left(x^{2}-2 x+3\right)-(x-1) \cdot(2 x-2)}{\le... | \frac{4}{(x^{2}-2x+3)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,399 |
## Task Condition
Find the derivative.
$$
y=\frac{1}{3}(x-2) \sqrt{x+1}+\ln (\sqrt{x+1}+1)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{3}(x-2) \sqrt{x+1}+\ln (\sqrt{x+1}+1)\right)^{\prime}= \\
& =\frac{1}{3} \cdot \sqrt{x+1}+\frac{1}{3} \cdot(x-2) \cdot \frac{1}{2 \sqrt{x+1}}+\frac{1}{\sqrt{x+1}+1} \cdot \frac{1}{2 \sqrt{x+1}}= \\
& =\frac{2(x+1)}{6 \sqrt{x+1}}+\frac{x-2}{6 \sqrt{x+1}}+\frac{... | \frac{3x\sqrt{x+1}+3x-\sqrt{x+1}+2}{6\sqrt{x+1}\cdot(\sqrt{x+1}+1)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,400 |
## Task Condition
Find the derivative.
$$
y=\operatorname{arctg} \frac{2 \sin x}{\sqrt{9 \cos ^{2} x-4}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg} \frac{2 \sin x}{\sqrt{9 \cos ^{2} x-4}}\right)^{\prime}= \\
& =\frac{1}{1+\left(\frac{2 \sin x}{\sqrt{9 \cos ^{2} x-4}}\right)^{2}} \cdot 2 \cdot \frac{\cos x \cdot \sqrt{9 \cos ^{2} x-4}-\sin x \cdot \frac{1}{2 \sqrt{9 \cos ^{2} x-4}} \cdot 18 \co... | \frac{2}{\cosx\sqrt{9\cos^{2}x-4}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,401 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
\[
\left\{\begin{array}{l}
x=\arccos \frac{1}{t} \\
y=\sqrt{t^{2}-1}+\arcsin \frac{1}{t}
\end{array}\right.
\] | ## Solution
$x_{t}^{\prime}=\left(\arccos \frac{1}{t}\right)^{\prime}=\frac{-1}{\sqrt{1-\left(\frac{1}{t}\right)^{2}}} \cdot\left(-\frac{1}{t^{2}}\right)=\frac{1}{t \cdot \sqrt{t^{2}-1}}$
$y_{t}^{\prime}=\left(\sqrt{t^{2}-1}+\arcsin \frac{1}{t}\right)^{\prime}=\frac{1}{2 \sqrt{t^{2}-1}} \cdot 2 t+\frac{1}{\sqrt{1-\le... | ^2-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,402 |
## Problem Statement
Derive the equations of the tangent and normal to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{\begin{array}{l}
x=a(t \cdot \sin t+\cos t) \\
y=a(\sin t-t \cdot \cos t)
\end{array}\right.
\]
$t_{0}=\frac{\pi}{4}$ | ## Solution
Since $t_{0}=\frac{\pi}{4}$, then
$x_{0}=a\left(\frac{\pi}{4} \cdot \sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right)=a\left(\frac{\pi}{4} \cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2} \cdot a}{2} \cdot \frac{\pi+4}{4}$
$y_{0}=a\left(\sin \frac{\pi}{4}-\frac{\pi}{4} \cdot \cos \frac{\pi}{4... | x+\frac{\sqrt{2}\cdot\cdot\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,403 |
## Task Condition
Find the $n$-th order derivative.
$y=\frac{x}{9(4 x+9)}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x}{9(4 x+9)}\right)^{\prime}=\frac{1 \cdot(4 x+9)-x \cdot 4}{9(4 x+9)^{2}}=\frac{4 x+9-4 x}{9(4 x+9)^{2}}=\frac{1}{(4 x+9)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{(4 x+9)^{2}}\right)^{\prime}=\frac{-2}{(4 x+9)^{3}} \cdot 4=\fra... | y^{(n)}=\frac{(-1)^{n-1}\cdotn!\cdot4^{n-1}}{(4x+9)^{n+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,404 |
## Task Condition
Find the derivative of the specified order.
$$
y=\left(1-x-x^{2}\right) e^{\frac{x-1}{2}}, y^{IV}=?
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\left(1-x-x^{2}\right) e^{\frac{x-1}{2}}\right)^{\prime}=(-1-2 x) e^{\frac{x-1}{2}}+\frac{1}{2} \cdot\left(1-x-x^{2}\right) e^{\frac{x-1}{2}}= \\
& =\frac{1}{2} \cdot\left(-2-4 x+1-x-x^{2}\right) e^{\frac{x-1}{2}}=-\frac{1}{2} \cdot\left(1+5 x+x^{2}\right) e^{\frac{x-1... | -\frac{1}{16}\cdot(55+17x+x^{2})e^{\frac{x-1}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,405 |
## Task Condition
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\sqrt{t-3} \\
y=\ln (t-2)
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\sqrt{t-1})^{\prime}=\frac{1}{2 \sqrt{t-1}}$
$y_{t}^{\prime}=(\ln (t-2))^{\prime}=\frac{1}{t-2}$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{1}{t-2}\right) /\left(\frac{1}{2 \sqrt{t-1}}\right)=\frac{2 \sqrt{t-1}}{t-2} \\
& \left(y_{x}... | -\frac{2}{(-2)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,406 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
\[
\begin{aligned}
& y=-\sqrt{\frac{2}{x^{2}}-1} \\
& 1+y^{2}+x \cdot y \cdot y^{\prime}=0
\end{aligned}
\] | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(-\sqrt{\frac{2}{x^{2}}-1}\right)^{\prime}=-\frac{1}{2 \sqrt{\frac{2}{x^{2}}-1}} \cdot\left(-\frac{4}{x^{3}}\right)= \\
& =\frac{2}{x^{3} \cdot \sqrt{\frac{2}{x^{2}}-1}}
\end{aligned}
$$
$1+\left(-\sqrt{\frac{2}{x^{2}}-1}\right)^{2}+x \cdot\left(-\sqrt{\frac{2}{x^{2}}-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 46,407 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\arcsin \left(x^{2} \cos \left(\frac{1}{9 x}\right)\right)+\frac{2}{3} x, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | \frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,408 |
## Condition of the problem
To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=2 x^{2}+3 x-1, x_{0}=-2$ | ## Solution
Let's find $y^{\prime}:$
$y^{\prime}=\left(2 x^{2}+3 x-1\right)^{\prime}=4 x+3$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=4 x_{0}+3=4 \cdot(-2)+3=-8+3=-5$
Since $y^{\prime}\left(x_{0}\right) \neq 0$, the equation of the normal line is:
$y-y_{0}=-\frac{1}{y_{0}^{\prime}} \cdot\left(x-x_{0}\rig... | \frac{x}{5}+1\frac{2}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,409 |
Condition of the problem
Find the differential $d y$
$y=\operatorname{tg}\left(2 \arccos \sqrt{1-2 x^{2}}\right), x>0$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\operatorname{tg}\left(2 \arccos \sqrt{1-2 x^{2}}\right)\right)^{\prime} d x= \\
& =\frac{1}{\cos ^{2}\left(2 \arccos \sqrt{1-2 x^{2}}\right)} \cdot\left(2 \arccos \sqrt{1-2 x^{2}}\right)^{\prime} d x= \\
& =\frac{1}{\left(2 \cos ^{2}\left(\arccos \sqrt{1... | \frac{1}{(1-4x^{2})^{2}\cdot\sqrt{1-2x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,410 |
Condition of the problem
Calculate approximately using the differential.
$y=\sqrt[3]{x^{3}+7 x}, x=1,012$ | ## 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:
$\Delta x = 0.012$
Calculate:
$y(1) = \sqrt[3]{1^{3} + 7 \cdot 1} = ... | 2.01 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,411 |
## Task Condition
Find the derivative.
$$
y=\frac{\left(2 x^{2}-1\right) \sqrt{1+x^{2}}}{3 x^{3}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\left(2 x^{2}-1\right) \sqrt{1+x^{2}}}{3 x^{3}}\right)^{\prime}=\frac{\left(\left(2 x^{2}-1\right) \sqrt{1+x^{2}}\right)^{\prime} x^{3}-\left(2 x^{2}-1\right) \sqrt{1+x^{2}}\left(x^{3}\right)^{\prime}}{3 x^{6}}= \\
& =\frac{\left(4 x \sqrt{1+x^{2}}+\left(2 x^{2}-... | \frac{1}{x^{4}\sqrt{1+x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,412 |
## Task Condition
Find the derivative.
$y=\frac{e^{2 x}(2-\sin 2 x-\cos 2 x)}{8}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{e^{2 x}(2-\sin 2 x-\cos 2 x)}{8}\right)^{\prime}= \\
& =\frac{1}{8} \cdot\left(2 e^{2 x}(2-\sin 2 x-\cos 2 x)+e^{2 x}(-\cos 2 x \cdot 2+\sin 2 x \cdot 2)\right)= \\
& =\frac{1}{4} \cdot e^{2 x} \cdot(2-\sin 2 x-\cos 2 x-\cos 2 x+\sin 2 x)= \\
& =\frac{1}{2} \cdot... | e^{2x}\cdot\sin^{2}x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,413 |
## Task Condition
Find the derivative.
$y=\ln \left(x+\sqrt{a^{2}+x^{2}}\right)$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \left(x+\sqrt{a^{2}+x^{2}}\right)\right)^{\prime}=\frac{1}{x+\sqrt{a^{2}+x^{2}}} \cdot\left(1+\frac{1}{2 \sqrt{a^{2}+x^{2}}} \cdot 2 x\right)= \\
& =\frac{1}{x+\sqrt{a^{2}+x^{2}}} \cdot \frac{\sqrt{a^{2}+x^{2}}+x}{\sqrt{a^{2}+x^{2}}}=\frac{1}{\sqrt{a^{2}+x^{2}}}
\e... | \frac{1}{\sqrt{^{2}+x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,414 |
Condition of the problem
Find the derivative.
$y=\cos \ln 2-\frac{1}{3} \cdot \frac{\cos ^{2} 3 x}{\sin 6 x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\cos \ln 2-\frac{1}{3} \cdot \frac{\cos ^{2} 3 x}{\sin 6 x}\right)^{\prime}=0-\frac{1}{3} \cdot\left(\frac{\cos ^{2} 3 x}{\sin 6 x}\right)^{\prime}= \\
& =-\frac{1}{3} \cdot\left(\frac{\cos ^{2} 3 x}{2 \sin 3 x \cdot \cos 3 x}\right)^{\prime}=-\frac{1}{6} \cdot\left(\f... | \frac{1}{2\sin^{2}3x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,415 |
## Problem Statement
Find the derivative.
$y=\arcsin \frac{\sqrt{x}-2}{\sqrt{5 x}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\arcsin \frac{\sqrt{x}-2}{\sqrt{5 x}}\right)^{\prime}=\frac{1}{\sqrt{1-\left(\frac{\sqrt{x}-2}{\sqrt{5 x}}\right)^{2}}} \cdot\left(\frac{\sqrt{x}-2}{\sqrt{5 x}}\right)^{\prime}= \\
& =\frac{\sqrt{5 x}}{\sqrt{5 x-(\sqrt{x}-2)^{2}}} \cdot \frac{(\sqrt{x}-2)^{\prime} \cdo... | \frac{1}{2x\sqrt{x+\sqrt{x}-1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,416 |
## Task Condition
Find the derivative.
$y=\frac{\operatorname{sh} x}{4 \operatorname{ch}^{4} x}+\frac{3 \operatorname{sh} x}{8 \operatorname{ch}^{2} x}+\frac{3}{8} \operatorname{arctg}(\operatorname{sh} x)$ | Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\operatorname{sh} x}{4 \operatorname{ch}^{4} x}+\frac{3 \operatorname{sh} x}{8 \operatorname{ch}^{2} x}+\frac{3}{8} \operatorname{arctg}(\operatorname{sh} x)\right)^{\prime}= \\
& =\frac{(\operatorname{sh} x)^{\prime} \cdot \operatorname{ch}^{4} x-\operatorname{sh} ... | \frac{1-3\operatorname{ch}^{5}x}{4\operatorname{ch}^{5}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,417 |
## Task Condition
Find the derivative.
$y=(\sin \sqrt{x})^{\ln (\sin \sqrt{x})}$ | ## Solution
$y=(\sin \sqrt{x})^{\ln (\sin \sqrt{x})}$
$\ln y=\ln \left((\sin \sqrt{x})^{\ln (\sin \sqrt{x})}\right)=\ln (\sin \sqrt{x}) \cdot \ln (\sin \sqrt{x})=\ln ^{2}(\sin \sqrt{x})$
$\frac{y^{\prime}}{y}=\left(\ln ^{2}(\sin \sqrt{x})\right)^{\prime}=2 \ln (\sin \sqrt{x}) \cdot \frac{1}{\sin \sqrt{x}} \cdot \cos... | (\sin\sqrt{x})^{\ln(\sin\sqrt{x})}\cdot\frac{\ln(\sin\sqrt{x})\cdot\operatorname{ctg}\sqrt{x}}{\sqrt{x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,418 |
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