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## Task Condition
Find the derivative.
$$
y=\frac{2}{3}\left(4 x^{2}-4 x+3\right) \sqrt{x^{2}-x}+(2 x-1)^{4} \cdot \arcsin \frac{1}{2 x-1}, 2 x-1>0
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2}{3}\left(4 x^{2}-4 x+3\right) \sqrt{x^{2}-x}+(2 x-1)^{4} \cdot \arcsin \frac{1}{2 x-1}\right)^{\prime}= \\
& =\frac{2}{3}\left((8 x-4) \sqrt{x^{2}-x}+\left(4 x^{2}-4 x+3\right) \cdot \frac{1}{2 \sqrt{x^{2}-x}} \cdot(2 x-1)\right)+ \\
& +\left(4(2 x-1) \cdot 2 \... | 8(2x-1)\cdot\arcsin\frac{1}{2x-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,046 |
## Task Condition
Find the derivative.
$y=\operatorname{arctg} \sqrt{x^{2}-1}-\frac{\ln x}{\sqrt{x^{2}-1}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg} \sqrt{x^{2}-1}-\frac{\ln x}{\sqrt{x^{2}-1}}\right)^{\prime}= \\
& =\frac{1}{1+\left(\sqrt{x^{2}-1}\right)^{2}} \cdot \frac{1}{2 \sqrt{x^{2}-1}} \cdot 2 x-\frac{\frac{1}{x} \cdot \sqrt{x^{2}-1}-\ln x \cdot \frac{1}{2 \sqrt{x^{2}-1}} \cdot 2 x}{x^{2}... | \frac{x\cdot\lnx}{\sqrt{(x^{2}-1)^{3}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,047 |
## Task Condition
Find the derivative.
$$
y=\frac{4^{x}(\ln 4 \cdot \sin 4 x-4 \cos 4 x)}{16+\ln ^{2} 4}
$$ | ## Solution
$y^{\prime}=\left(\frac{4^{x}(\ln 4 \cdot \sin 4 x-4 \cos 4 x)}{16+\ln ^{2} 4}\right)^{\prime}=$
$=\frac{1}{16+\ln ^{2} 4} \cdot\left(4^{x} \cdot \ln 4 \cdot(\ln 4 \cdot \sin 4 x-4 \cos 4 x)+4^{x}(4 \ln 4 \cdot \cos 4 x+16 \sin 4 x)\right)=$
$$
\begin{aligned}
& =\frac{4^{x}}{16+\ln ^{2} 4} \cdot\left(\l... | 4^{x}\cdot\sin4x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,048 |
## Task Condition
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=t \sqrt{t^{2}+1} \\
y=\ln \frac{1+\sqrt{1+t^{2}}}{t}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(t \sqrt{t^{2}+1}\right)^{\prime}=\sqrt{t^{2}+1}+t \cdot \frac{1}{2 \sqrt{t^{2}+1}} \cdot 2 t=\sqrt{t^{2}+1}+\frac{t^{2}}{\sqrt{t^{2}+1}}=$
$=\frac{t^{2}+1+t^{2}}{\sqrt{t^{2}+1}}=\frac{2 t^{2}+1}{\sqrt{t^{2}+1}}$
$y_{t}^{\prime}=\left(\ln \frac{1+\sqrt{1-t^{2}}}{t}\right)^{\prime}=\f... | -\frac{1}{(2^{2}+1)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,049 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{
\begin{array}{l}
x=t(1-\sin t) \\
y=t \cdot \cos t
\end{array}
\right.
\]
$t_{0}=0$ | ## Solution
Since $t_{0}=0$, then
$x_{0}=0 \cdot(1-\sin 0)=0$
$y_{0}=0 \cdot \cos 0=0$
Let's find the derivatives:
$x_{t}^{\prime}=(t(1-\sin t))^{\prime}=(1-\sin t)-t \cdot \cos t$
$y_{t}^{\prime}=(t \cdot \cos t)^{\prime}=t \cdot \cos t-\sin t$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{t \cdot... | 0,\;0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,050 |
## Problem Statement
Find the $n$-th order derivative.
$y=a^{2 x+3}$ | ## Solution
$y=a^{2 x+3}=a^{2 x} \cdot a^{3}=\left(e^{\ln a}\right)^{2 x} \cdot a^{3}=e^{2 x \cdot \ln a} \cdot a^{3}$
$y^{\prime}=\left(e^{2 x \cdot \ln a} \cdot a^{3}\right)^{\prime}=e^{2 x \cdot \ln a} \cdot 2 \ln a \cdot a^{3}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(e^{2 x \cdot \ln a} \cdot 2... | ^{2x+3}\cdot2^{n}\cdot\ln^{n} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,051 |
## Task Condition
Find the derivative of the specified order.
$$
y=\frac{\ln (2 x+5)}{2 x+5}, y^{\prime \prime \prime}=?
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\ln (2 x+5)}{2 x+5}\right)^{\prime}=\frac{\frac{1}{2 x+5} \cdot 2 \cdot(2 x+5)-\ln (2 x+5) \cdot 2}{(2 x+5)^{2}}= \\
& =\frac{2-2 \ln (2 x+5)}{(2 x+5)^{2}}=2 \cdot \frac{1-\ln (2 x+5)}{(2 x+5)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(2 \... | \frac{88-48\ln(2x+5)}{(2x+5)^{4}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,052 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\cos t \\
y=\ln (\sin t)
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\cos t)^{\prime}=-\sin t$
$y_{t}^{\prime}=(\ln (\sin t))^{\prime}=\frac{1}{\sin t} \cdot \cos t=\frac{\cos t}{\sin t}$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{\cos t}{\sin t}\right) /(-\sin t)=-\frac{\cos t}{\sin ^{2} t}$
$\left(y_{x}^{\prime}\right... | -\frac{1+\cos^{2}}{\sin^{4}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,053 |
## Problem Statement
Show that the function $y$ satisfies equation (1).
$$
\begin{aligned}
& y=\sqrt[4]{\sqrt{x}+\sqrt{x+1}} \\
& 8 \cdot x \cdot y^{\prime}-y=\frac{-1}{y^{3} \sqrt{x+1}}
\end{aligned}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=(\sqrt[4]{\sqrt{x}+\sqrt{x+1}})^{\prime}=\frac{1}{4} \cdot \frac{1}{\sqrt[4]{(\sqrt{x}+\sqrt{x+1})^{3}}} \cdot\left(\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x+1}}\right)= \\
& =\frac{1}{4} \cdot \frac{1}{\sqrt[4]{(\sqrt{x}+\sqrt{x+1})^{3}}} \cdot \frac{\sqrt{x}+\sqrt{x+1}}{2 \s... | proof | Calculus | proof | Yes | Yes | olympiads | false | 47,054 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{23-4 n}{2-n}, a=4$ | ## Solution
By the definition of the limit:
$\forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{23-4 n-8+4 n}{2-n}\right| \\
& \left|\frac{15}{2-n}\right| \\
& \left|\frac{\frac{15}{n-2}}{n}\right| \\
& \frac{15}{n-2} \\
& n-2>\frac{15}... | N(\varepsilon)=3+[\frac{15}{\varepsilon}] | Calculus | proof | Yes | Yes | olympiads | false | 47,055 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}+(n-1)^{2}}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n-1)^{3}}{(n+1)^{2}+(n-1)^{2}}=\lim _{n \rightarrow \infty} \frac{n^{3}+3 n^{2}+3 n+1-n^{3}+3 n^{2}-3 n+1}{n^{2}+2 n+1+n^{2}-2 n+1}=$
$=\lim _{n \rightarrow \infty} \frac{6 n^{2}+2}{2 n^{2}+2}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(3 n^{2}+1\... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,056 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}\right)}{\frac{1}{n^{2}}(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}= \\
& =\lim _{n \rightarrow \infty}... | -4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,057 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}(n \sqrt{n}-\sqrt{n(n+1)(n+2)})
$$ | ## Solution
$\lim _{n \rightarrow \infty}(n \sqrt{n}-\sqrt{n(n+1)(n+2)})=$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty} \frac{(n \sqrt{n}-\sqrt{n(n+1)(n+2)})(n \sqrt{n}+\sqrt{n(n+1)(n+2)})}{n \sqrt{n}+\sqrt{n(n+1)(n+2)}}= \\
& =\lim _{n \rightarrow \infty} \frac{n^{3}-n(n+1)(n+2)}{n \sqrt{n}+\sqrt{n(n+1)(n+2)}}... | -\infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,058 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1-2+3-4+\ldots+(2 n-1)-2 n}{\sqrt[3]{n^{3}+2 n+2}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1-2+3-4+\ldots+(2 n-1)-2 n}{\sqrt[3]{n^{3}+2 n+2}}= \\
& =\{1-2=3-4=\ldots=(2 n-1)-2 n=-1\}= \\
& =\lim _{n \rightarrow \infty} \frac{-1 \cdot n}{\sqrt[3]{n^{3}+2 n+2}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n} \cdot(-1) \cdot n}{\frac{1}{n} \sqr... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,059 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{2 n-1}{2 n+1}\right)^{n+1}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{2 n-1}{2 n+1}\right)^{n+1}=\lim _{n \rightarrow \infty}\left(\frac{2 n+1}{2 n-1}\right)^{-n-1}= \\
& =\lim _{n \rightarrow \infty}\left(\frac{2 n-1+2}{2 n-1}\right)^{-n-1}=\lim _{n \rightarrow \infty}\left(1+\frac{2}{2 n-1}\right)^{-n-1}= \\
& =\... | \frac{1}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,060 |
## Task Condition
Prove that (find $\delta(\varepsilon)$ :
$$
\lim _{x \rightarrow 10} \frac{5 x^{2}-51 x+10}{x-10}=49
$$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). The number
=\frac{\varepsilon}{5} | Calculus | proof | Yes | Yes | olympiads | false | 47,061 |
## problem statement
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0 \text { (find }} \delta(\varepsilon)_{\text {): }}$ :
$f(x)=-3 x^{2}-9, x_{0}=3$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0$ :
$$
\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, such that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x... | proof | Calculus | proof | Yes | Yes | olympiads | false | 47,062 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-3} \frac{x^{2}+2 x-3}{x^{3}+4 x^{2}+3 x}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-3} \frac{x^{2}+2 x-3}{x^{3}+4 x^{2}+3 x}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-3} \frac{(x+3)(x-1)}{(x+3)\left(x^{2}+x\right)}= \\
& =\lim _{x \rightarrow-3} \frac{x-1}{x^{2}+x}=\frac{-3-1}{(-3)^{2}-3}=\frac{-4}{9-3}=-\frac{4}{6}=-\frac{2}{3}
\end{aligned... | -\frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,063 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 8} \frac{\sqrt{9+2 x}-5}{\sqrt[3]{x}-2}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 8} \frac{\sqrt{9+2 x}-5}{\sqrt[3]{x}-2}=\lim _{x \rightarrow 8} \frac{(\sqrt{9+2 x}-5)(\sqrt{9+2 x}+5)}{(\sqrt[3]{x}-2)(\sqrt{9+2 x}+5)}= \\
& =\lim _{x \rightarrow 8} \frac{9+2 x-25}{(\sqrt[3]{x}-2)(\sqrt{9+2 x}+5)}=\lim _{x \rightarrow 8} \frac{2 x-16}{(\sqrt[3]{... | \frac{12}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,064 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\ln (e-x)-1}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$\arcsin 2 x \sim 2 x$, as $x \rightarrow 0 (2 x \rightarrow 0)$
$\ln \left(1+\left(-\frac{x}{e}\right)\right) \sim -\frac{x}{e}$, as $x \rightarrow 0 \left(-\frac{x}{e} \rightarrow 0\right)$
We obtain:
$$
\begin{aligned}
& \lim _{x \rightarrow... | -2e | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,065 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2} \frac{\operatorname{arctg}\left(x^{2}-2 x\right)}{\sin 3 \pi x}$ | ## Solution
Substitution:
$x=y+2 \Rightarrow y=x-2$
$x \rightarrow 2 \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow 2} \frac{\operatorname{arctg}\left(x^{2}-2 x\right)}{\sin 3 \pi x}=\lim _{y \rightarrow 0} \frac{\operatorname{arctg}\left((y+2)^{2}-2(y+2)\right)}{\sin 3 \pi(y+2)}=$
$=\lim _{y \rightar... | \frac{2}{3\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,066 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln (\sin x)}{(2 x-\pi)^{2}}$ | ## Solution
Substitution:
$x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2}$
$x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}} \frac{\ln (\sin x)}{(2 x-\pi)^{2}}=\lim _{y \rightarrow 0} \frac{\ln \left(\sin \left(y+\frac{\pi}{2}\right)\right)}... | -\frac{1}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,067 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{x}-e^{-2 x}}{x+\sin x^{2}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{x}-e^{-2 x}}{x+\sin x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)}{x+\sin x^{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{-2 x}-1\right)\right)}{\frac{1}{x}\left(x+\sin x^{2}... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,068 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{3^{x+1}-3}{\ln \left(1+x \sqrt{1+x e^{x}}\right)}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{3^{x+1}-3}{\ln \left(1+x \sqrt{1+x e^{x}}\right)}=\lim _{x \rightarrow 0} \frac{3\left(3^{x}-1\right)}{\ln \left(1+x \sqrt{1+x e^{x}}\right)}= \\
& =\lim _{x \rightarrow 0} \frac{3\left(\left(e^{\ln 3}\right)^{x}-1\right)}{\ln \left(1+x \sqrt{1+x e^{x}}\ri... | 3\ln3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,069 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(\frac{1+\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right)^{\frac{1}{x^{3}}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{1+\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right)^{\frac{1}{x^{3}}}= \\
& =\lim _{x \rightarrow 0}\left(e^{\ln \left(\frac{1+\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right)}\right)^{\frac{1}{x^{3}}}= \\
& =\lim _{x \rightarrow 0} e^{\fra... | e^{\frac{21}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,070 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{x+6}}=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\lim _{x \rightarrow 0} \frac{1}{x+6}}=$
$=\left(\lim _{x \rightarrow 0} \frac{\sin 5 x^{2}}{\sin x}\right)^{\frac{1}{0+6}}=\left(\lim _{x \rightarrow 0} \frac... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,071 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 3}\left(\frac{\sin x}{\sin 3}\right)^{\frac{1}{x-3}}$ | ## Solution
$\lim _{x \rightarrow 3}\left(\frac{\sin x}{\sin 3}\right)^{\frac{1}{x-3}}=\lim _{x \rightarrow 3}\left(e^{\ln \left(\frac{\sin x}{\sin 3}\right)}\right)^{\frac{1}{x-3}}=$
$=\lim _{x \rightarrow 3} e^{\frac{1}{x-3} \cdot \ln \left(\frac{\sin x}{\sin 3}\right)}=\exp \left\{\lim _{x \rightarrow 3} \frac{1}{... | e^{\cot3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,072 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{1}{2}}(\arcsin x+\arccos x)^{\frac{1}{x}}$ | ## Solution
$\lim _{x \rightarrow \frac{1}{2}}(\arcsin x+\arccos x)^{\frac{1}{x}}=\left(\arcsin \frac{1}{2}+\arccos \frac{1}{2}\right)^{\frac{1}{\left(\frac{1}{2}\right)}}=\left(\frac{\pi}{6}+\frac{\pi}{3}\right)^{2}=\left(\frac{\pi}{2}\right)^{2}=\frac{\pi^{2}}{4}$
## Problem Kuznetsov Limits 20-26 | \frac{\pi^{2}}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,073 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 2} \sqrt[3]{\lg (x+2)+\sin \sqrt{4-x^{2}} \cdot \cos \frac{x+2}{x-2}}
$$ | ## Solution
Since $\cos \frac{x+2}{x-2}$ is bounded, and $\sin \sqrt{4-x^{2}} \rightarrow 0$, as $x \rightarrow 2$, then $\sin \sqrt{4-x^{2}} \cdot \cos \frac{x+2}{x-2} \rightarrow 0$, as $x \rightarrow 2$
Then:
$\lim _{x \rightarrow 2} \sqrt[3]{\lg (x+2)+\sin \sqrt{4-x^{2}} \cdot \cos \frac{x+2}{x-2}}=\sqrt[3]{\lg ... | \sqrt[3]{\lg4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,074 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{11 ; 5 ;-3\}$
$p=\{1 ; 0 ; 2\}$
$q=\{-1 ; 0 ; 1\}$
$r=\{2 ; 5 ;-3\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 3p-6q+r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,075 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{-1 ; 2 ;-1\}$
$b=\{2 ;-7 ; 1\}$
$c_{1}=6 a-2 b$
$c_{2}=b-3 a$ | ## 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 | 47,076 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(2, -8, -1), B(4, -6, 0), C(-2, -5, -1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(4-2 ;-6-(-8) ; 0-(-1))=(2 ; 2 ; 1)$
$\overrightarrow{A C}=(-2-2 ;-5-(-8) ;-1-(-1))=(-4 ; 3 ; 0)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos ... | -\frac{2}{15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,077 |
## Problem Statement
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
\[
\begin{aligned}
& a=2 p+3 q \\
& b=p-2 q \\
& |p|=2 \\
& |q|=3 \\
& (\widehat{p, q})=\frac{\pi}{4}
\end{aligned}
\] | ## 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:
$$
\begin{aligned}
& a \times b=(2 p+3 q) \times(p-2 q)=2 \cdot p \times p+2 \cdot(-2) \c... | 21\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,078 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{5 ; 3 ; 4\}$
$b=\{-1 ; 0 ;-1\}$
$c=\{4 ; 2 ; 4\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
5 & 3 & 4 \\
-1 & 0 & -1 \\
4 & 2 & 4
\end{array}\right|= \\
& =5 \cdot\left|\b... | 2\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,079 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(1 ; 1 ;-1) \)
\( A_{2}(2 ; 3 ; 1) \)
\( A_{3}(3 ; 2 ; 1) \)
\( A_{4}(5 ; 9 ;-8) \) | ## Solution
From vertex $A_{1}$, we draw vectors:
$\overrightarrow{A_{1} A_{2}}=\{2-1 ; 3-1 ; 1-(-1)\}=\{1 ; 2 ; 2\}$
$\overrightarrow{A_{1} A_{3}}=\{3-1 ; 2-1 ; 1-(-1)\}=\{2 ; 1 ; 2\}$
$\vec{A}_{1} A_{4}=\{5-1 ; 9-1 ;-8-(-1)\}=\{4 ; 8 ;-7\}$
According to the geometric meaning of the scalar triple product, we have... | 7.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,080 |
## Problem Statement
Find the distance from the point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ; -1 ; 2)$
$M_{2}(2 ; 1 ; 2)$
$M_{3}(1 ; 1 ; 4)$
$M_{0}(-3 ; 2 ; 7)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{lll}
x-1 & y-(-1) & z-2 \\
2-1 & 1-(-1) & 2-2 \\
1-1 & 1-(-1) & 4-2
\end{array}\right|=0
$$
Perform transformations:
$$
\left|\begin{array}{ccc}
x-1 & y+1 & z-2 \\
1 & 2 & 0 \\
0 & 2 & 2
\end{array}... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,081 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(1; -1; 5)$
$B(0; 7; 8)$
$C(-1; 3; 8)$ | ## Solution
Let's find the vector $\overrightarrow{BC}$:
$\overrightarrow{BC}=\{-1-0 ; 3-7 ; 8-8\}=\{-1 ;-4 ; 0\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$-(x-1)-4 \cdot(y-(-1))+0 \cdot(z-5)=0$... | x+4y+3=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,082 |
## Task Condition
Find the angle between the planes:
$x+2 y+2 z-3=0$
$2 x-y+2 z+5=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{1 ; 2 ; 2\}$
$\overrightarrow{n_{2}}=\{2 ;-1 ; 2\}$
The angle $\phi$ between the planes is determined by the formula:
$$
\begin{aligned}
& \cos \ph... | \arccos\frac{4}{9}\approx63^{0}36^{\}44^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,083 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; y ; 0)$
$B(-2 ;-4 ; 6)$
$C(7 ; 2 ; 5)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(-2-0)^{2}+(-4-y)^{2}+(6-0)^{2}}=\sqrt{4+16+8 y+y^{2}+36}=\sqrt{y^{2}+8 y+56} \\
& A C=\sqrt{(7-0)^{2}+(2-y)^{2}+(5-0)^{2}}=\sqrt{49+4-4 y+y^{2}+25}=\sqrt{y^{2}-4 y+78}
\end{aligned}
$$
Since by the condition of the problem $A B=A C... | A(0;1\frac{5}{6};0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,084 |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-2; -1; 1)$
$a: x-2y+6z-10=0$
$k=\frac{3}{5}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: x-2 y+6 z-6=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,085 |
## Task Condition
Write the canonical equations of the line.
$6 x-5 y-4 z+8=0$
$6 x+5 y+3 z+4=0$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- are coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction v... | \frac{x+1}{5}=\frac{y-\frac{2}{5}}{-42}=\frac{z}{60} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,086 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-2}{2}=\frac{y-2}{-1}=\frac{z-4}{3}$
$x+3 y+5 z-42=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-2}{2}=\frac{y-2}{-1}=\frac{z-4}{3}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=2+2 t \\
y=2-t \\
z=4+3 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$(2+2 t)+3(2-t)+5(4+3 t)-42=0$
$2+2 t+6... | (4;1;7) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,087 |
## Task Condition
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$M(0; -3; -2)$
$$
\frac{x-0.5}{0}=\frac{y+1.5}{-1}=\frac{z-1.5}{1}
$$ | ## Solution
We 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 its normal vector:
$\vec{n}=\vec{s}=\{0 ;-1 ; 1\}$
Then the equation of the desired plane is:
$0 ... | M^{\}(1;2;3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,088 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{4 n^{2}+1}{3 n^{2}+2}, a=\frac{4}{3}$ | ## Solution
By the definition of the limit:
$\forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{12 n^{2}+3-12 n^{2}-8}{3\left(3 n^{2}+2\right)}\right| \\
& \left.\frac{-5}{3\left(3 n^{2}+2\right)} \right\rvert\, \\
& \frac{5}{3\left(3 n... | proof | Calculus | proof | Yes | Yes | olympiads | false | 47,089 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n+1)^{2}}{(n-1)^{3}-(n+1)^{3}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{3}-(n+1)^{2}}{(n-1)^{3}-(n+1)^{3}}=\lim _{n \rightarrow \infty} \frac{(n+1)^{2} \cdot((n+1)-1)}{n^{3}-3 n^{2}+3 n-1-n^{3}-3 n^{2}-3 n-1}= \\
& =\lim _{n \rightarrow \infty} \frac{n(n+1)^{2}}{-6 n^{2}-2}=\lim _{n \rightarrow \infty} \frac{n^{3}\... | -\infty | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,090 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}}{(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}}{(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[5]{n}-\sqrt[3]{27 n^{6}+n^{2}}\right)}{\frac{1}{n^{2}}(n+\sqrt[4]{n}) \sqrt{9+n^{2}}}= \\
& =\lim _{n \rightarrow \... | -3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,091 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}-3 n+2}-n\right)
$$ | ## Solution
$\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}-3 n+2}-n\right)=\lim _{n \rightarrow \infty} \frac{\left(\sqrt{n^{2}-3 n+2}-n\right)\left(\sqrt{n^{2}-3 n+2}+n\right)}{\sqrt{n^{2}-3 n+2}+n}=$
$=\lim _{n \rightarrow \infty} \frac{n^{2}-3 n+2-n^{2}}{\sqrt{n^{2}-3 n+2}+n}=\lim _{n \rightarrow \infty} \frac{-3 n... | -\frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,092 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{1+2+3+\ldots+n} \cdot \frac{(1+(2 n-1)) n}{2}= \\
& =\lim _{n \rightarrow \infty} \frac{1}{1+2+3+\ldots+n} \cdot n^{2}=\lim _{n \rightarrow \infty} \frac{1}{\frac{(1+n)... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,093 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-6 n+7}{3 n^{2}+20 n-1}\right)^{-n+1}
$$ | ## Solution
$\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-6 n+7}{3 n^{2}+20 n-1}\right)^{-n+1}=\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}+20 n-1}{3 n^{2}-6 n+7}\right)^{n-1}=$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-6 n+7+26 n-8}{3 n^{2}-6 n+7}\right)^{n-1}= \\
& =\lim _{n \right... | e^{\frac{26}{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,094 |
## Condition of the problem
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 \( M \). A number \( A \in \mathbb{R} \) is called the limit of the function \( f \) as \( x \) approaches \( a (x \rightarrow... | \delta(\varepsilon)=\frac{\varepsilon}{6} | Calculus | proof | Yes | Yes | olympiads | false | 47,095 |
## problem statement
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0 \text { (find }} \delta(\varepsilon)_{\text {) }}$ :
$f(x)=-3 x^{2}-6, x_{0}=1$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0$ :
$$
\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, such that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$ when $\left|x-x_{0}\righ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 47,096 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)^{2}}{x^{4}+2 x+1}$ | Solution
$\lim _{x \rightarrow-1} \frac{\left(x^{3}-2 x-1\right)^{2}}{x^{4}+2 x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-1\right)^{2}(x+1)^{2}}{\left(x^{3}-x^{2}+x+1\right)(x+1)}=$
$=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-1\right)^{2}(x+1)}{x^{3}-x^{2}+x+1}=\frac{\left((-1)^{2}-(-... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,097 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt{x}-4}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt{x}-4}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{(\sqrt[4]{x}-2)(\sqrt[4]{x}+2)}= \\
& =\lim _{x \rightarrow 16} \frac{1}{\sqrt[4]{x}+2}=\frac{1}{\sqrt[4]{16}+2}=\frac{1}{2+2}=\frac{1}{4}
\end{aligned}
$$ | \frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,098 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{2 x}{\tan\left(2 \pi\left(x+\frac{1}{2}\right)\right)}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$\operatorname{tg} 2 \pi x \sim 2 \pi x$, as $x \rightarrow 0 (2 \pi x \rightarrow 0)$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{2 x}{\operatorname{tg}\left(2 \pi\left(x+\frac{1}{2}\right)\right)}=\lim _{x \rightarrow 0} \frac{2... | \frac{1}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,099 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 3x}{\tan x}$ | ## Solution
Substitution:
$x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2}$
$x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 3 x}{\tan x}=\lim _{y \rightarrow 0} \frac{\tan 3\left(y+\frac{\pi}{2}\right)}{\tan \left(y+\frac{\pi}{2}... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,100 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow \frac{\pi}{6}} \frac{\ln (\sin 3 x)}{(6 x-\pi)^{2}}
$$ | ## Solution
Substitution:
$x=y+\frac{\pi}{6} \Rightarrow y=x-\frac{\pi}{6}$
$x \rightarrow \frac{\pi}{6} \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{6}} \frac{\ln (\sin 3 x)}{(6 x-\pi)^{2}}=\lim _{y \rightarrow 0} \frac{\ln \left(\sin 3\left(y+\frac{\pi}{6}\right)\righ... | -\frac{1}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,101 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{3 x}}{\operatorname{arctg} x-x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)}{\operatorname{arctg} x-x^{2}}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\frac{1}{x}\left... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,102 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}=\lim _{x \rightarrow 0} \frac{\left(e^{\alpha x}-1\right)-\left(e^{\beta x}-1\right)}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}=$
$=\lim _{x \rightarrow 0} \frac{e^{\alpha x}-1}{2 \sin \frac{x(\alpha... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,103 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(5-\frac{4}{\cos x}\right)^{\frac{1}{\sin ^{2} 3 x}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(5-\frac{4}{\cos x}\right)^{\frac{1}{\sin ^{2} 3 x}}=\lim _{x \rightarrow 0}\left(e^{\ln \left(5-\frac{4}{\cos x}\right)}\right)^{\frac{1}{\sin ^{2} 3 x}}=$
$=\lim _{x \rightarrow 0} e^{\frac{\ln \left(5-\frac{4}{\cos x}\right)}{\sin ^{2} 3 x}}=\exp \left\{\lim _{x \rightarrow ... | e^{-\frac{2}{9}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,104 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{x^{2}+4}{x+2}\right)^{x^{2}+3}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{x^{2}+4}{x+2}\right)^{x^{2}+3}=\left(\frac{0^{2}+4}{0+2}\right)^{0^{2}+3}=$
$=\left(\frac{4}{2}\right)^{3}=2^{3}=8$
## Problem Kuznetsov Limits 18-6 | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,105 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{4}}(\tan x)^{1 / \cos \left(\frac{3 \pi}{4}-x\right)}$ | ## Solution
$\lim _{x \rightarrow \frac{\pi}{4}}(\operatorname{tg} x)^{1 / \cos \left(\frac{3 \pi}{4}-x\right)}=\lim _{x \rightarrow \frac{\pi}{4}}\left(e^{\ln (\operatorname{tg} x)}\right)^{1 / \cos \left(\frac{3 \pi}{4}-x\right)}=$
$=\lim _{x \rightarrow \frac{\pi}{4}} e^{\ln (\operatorname{tg} x) / \cos \left(\fra... | e^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,106 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{6}}(\sin x)^{\frac{6 x}{\pi}}$ | ## Solution
$\lim _{x \rightarrow \frac{\pi}{6}}(\sin x)^{\frac{6 x}{\pi}}=\left(\sin \frac{\pi}{6}\right)^{\frac{6}{\pi} \cdot \frac{\pi}{6}}=\left(\frac{1}{2}\right)^{1}=\frac{1}{2}$
## Problem Kuznetsov Limits 20-6 | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,107 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt[4]{2+n^{5}}-\sqrt{2 n^{3}+3}}{(n+\sin n) \sqrt{7 n}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt[4]{2+n^{5}}-\sqrt{2 n^{3}+3}}{(n+\sin n) \sqrt{7 n}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n \sqrt{n}}\left(\sqrt[4]{2+n^{5}}-\sqrt{2 n^{3}+3}\right)}{\frac{1}{n \sqrt{n}}(n+\sin n) \sqrt{7 n}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sq... | -\frac{\sqrt{2}}{\sqrt{7}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,108 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{3 n-1}{5 n+1}, a=\frac{3}{5}$ | ## Solution
By the definition of the limit:
$\forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left.\frac{-8}{5(5 n+1)} \right\rvert\, \\
& \left|\frac{8}{5(5 n+1)}\right| \\
& \frac{8}{5(5 n+1)}
\end{aligned}
$$
$5 n+1>\frac{8}{5 \varepsilon} ;... | N(\varepsilon)=[\frac{8+20\varepsilon}{25\varepsilon}] | Calculus | proof | Yes | Yes | olympiads | false | 47,109 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}-(2 n+3)^{3}}{(2 n+1)^{2}+(2 n+3)^{2}}$ | ## Solution
$\lim _{n \rightarrow \infty} \frac{(2 n+1)^{3}-(2 n+3)^{3}}{(2 n+1)^{2}+(2 n+3)^{2}}=\lim _{n \rightarrow \infty} \frac{8 n^{3}+3 \cdot 4 n^{2}+3 \cdot 2 n+1-8 n^{3}-3 \cdot 3 \cdot 4 n^{2}-3 \cdot 3^{2} \cdot 2 n-3^{3}}{(2 n+1)^{2}+(2 n+3)^{2}}=$
$=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,110 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}}{(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}}{(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}\right)}{\frac{1}{n^{2}}(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}= \\
& =\lim _{n \rightarrow \inft... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,111 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}-\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}\right)
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}-\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}+2\right)}-\sqrt{\left(n^{2}-1\right)\left(n^{2}-2\right)}\right... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,112 |
## Task Condition
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{1+2+\ldots+n}{n-n^{2}+3}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1+2+\ldots+n}{n-n^{2}+3}=\lim _{n \rightarrow \infty} \frac{\frac{(1+n) n}{2}}{n-n^{2}+3}= \\
& =\lim _{n \rightarrow \infty} \frac{n+n^{2}}{2\left(n-n^{2}+3\right)}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n+n^{2}\right)}{\frac{1}{n^{2... | -\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,113 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n}{3 n^{2}-5 n+7}\right)^{n+1}$ | ## Solution
$\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n}{3 n^{2}-5 n+7}\right)^{n+1}=\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n+7}{3 n^{2}-5 n}\right)^{-n-1}=$
$=\lim _{n \rightarrow \infty}\left(1+\frac{7}{3 n^{2}-5 n}\right)^{-n-1}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{3 n^{2}-5... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,114 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$$
\lim _{x \rightarrow-7} \frac{2 x^{2}+15 x+7}{x+7}=-13
$$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). The number
=\frac{\varepsilon}{2} | Calculus | proof | Yes | Yes | olympiads | false | 47,115 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$f(x)=-2 x^{2}-4, x_{0}=3$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$. $\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x-x_{0}\r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 47,116 |
Condition of the problem
Calculate the limit of the function:
$$
\lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x^{2}+2 x+1}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x^{2}+2 x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+1)\left(x^{2}-x-2\right)}{(x+1)^{2}}= \\
& =\lim _{x \rightarrow-1} \frac{x^{2}-x-2}{x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+1)(x-2)}{x+1}= \\
& ... | -3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,117 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{1}{4}} \frac{\sqrt[3]{\frac{x}{16}}-\frac{1}{4}}{\sqrt{\frac{1}{4}+x}-\sqrt{2 x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{1}{4}} \frac{\sqrt[3]{\frac{x}{16}}-\frac{1}{4}}{\sqrt{\frac{1}{4}+x}-\sqrt{2 x}}=\lim _{x \rightarrow \frac{1}{4}} \frac{\left(\sqrt[3]{\frac{x}{16}}-\frac{1}{4}\right)\left(\sqrt[3]{\left(\frac{x}{16}\right)^{2}}+\sqrt[3]{\frac{x}{16}} \cdot \frac{1}{4}+\le... | -\frac{2}{3}\sqrt{\frac{1}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,118 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{x \cdot \sin x}$ | ## Solution
Let's use the substitution of equivalent infinitesimals:
$1-\cos x \sim \frac{x^{2}}{2}$, as $x \rightarrow 0$
$\sin x \sim x$, as $x \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{x \cdot \sin x}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{(1-\s... | \frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,119 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 2} \frac{1-2^{4-x^{2}}}{2\left(\sqrt{2 x}-\sqrt{3 x^{2}-5 x+2}\right)}
$$ | ## Solution
Substitution:
$x=y+2 \Rightarrow y=x-2$
$x \rightarrow 2 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{1-2^{4-x^{2}}}{2\left(\sqrt{2 x}-\sqrt{3 x^{2}-5 x+2}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{1-2^{4-(y+2)^{2}}}{2\left(\sqrt{2(y+2)}-\sqrt{3(y+2)^{2}-5(... | -\frac{8\ln2}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,120 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1} \frac{\sqrt{2^{x}+7}-\sqrt{2^{x+1}+5}}{x^{3}-1}
$$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=y+1 \Rightarrow y=x-1 \\
& x \rightarrow 1 \Rightarrow y \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\sqrt{2^{x}+7}-\sqrt{2^{x+1}+5}}{x^{3}-1}=\lim _{y \rightarrow 0} \frac{\sqrt{2^{y+1}+7}-\sqrt{2^{(y+1)+1}+5}}{(y+1)^{3}... | -\frac{\ln2}{9} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,121 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{4^{5 x}-9^{-2 x}}{\sin x-\tan x^{3}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{4^{5 x}-9^{-2 x}}{\sin x-\operatorname{tg} x^{3}}=\lim _{x \rightarrow 0} \frac{\left(1024^{x}-1\right)-\left(81^{-x}-1\right)}{\sin x-\operatorname{tg} x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 1024}\right)^{x}-1\right)-\left(\left(e^... | \ln(2^{10}\cdot9^{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,122 |
## Problem Statement
Calculate the limit of the function:
$\lim _{h \rightarrow 0} \frac{a^{x+h}+a^{x-h}-2 a^{x}}{h}$ | Solution
$\lim _{h \rightarrow 0} \frac{a^{x+h}+a^{x-h}-2 a^{x}}{h}=\lim _{h \rightarrow 0} \frac{a^{x+h}-a^{x}+a^{x-h}-a^{x}}{h}=$
$=\lim _{h \rightarrow 0} \frac{a^{x+h}-a^{x}}{h}+\lim _{h \rightarrow 0} \frac{a^{x-h}-a^{x}}{h}=$
$=\lim _{h \rightarrow 0} \frac{a^{x}\left(a^{h}-1\right)}{h}+\lim _{h \rightarrow 0}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,123 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{ctg}^{2} x}$ | ## Solution
$\lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{ctg}^{2} x}=\lim _{x \rightarrow 0}\left(e^{\ln \left(6-\frac{5}{\cos x}\right)}\right)^{\frac{1}{\operatorname{tg}^{2} x}}=$
$$
=\lim _{x \rightarrow 0} e^{\ln \left(6-\frac{5}{\cos x}\right) / \operatorname{tg}^{2} x}=\exp \left\{\li... | e^{-\frac{5}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,124 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{x+8}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{x+8}}=\left(\lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\lim _{x \rightarrow 0} \frac{3}{x+8}}=$
$=\left(\lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{x^{2}}\right)^{\frac{3}{0+8}}=\le... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,125 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2}\left(2 e^{x-2}-1\right)^{\frac{3 x+2}{x-2}}$ | ## Solution
$\lim _{x \rightarrow 2}\left(2 e^{x-2}-1\right)^{\frac{3 x+2}{x-2}}=\lim _{x \rightarrow 2}\left(e^{\ln \left(2 e^{x-2}-1\right)}\right)^{\frac{3 x+2}{x-2}}=$
$=\lim _{x \rightarrow 2} e^{\frac{3 x+2}{x-2} \cdot \ln \left(2 e^{x-2}-1\right)}=\exp \left\{\lim _{x \rightarrow 2} \frac{3 x+2}{x-2} \cdot \ln ... | e^{16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,126 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}\left(\ln ^{2} e x\right)^{\frac{1}{x^{2}+1}}
$$ | ## Solution
$$
\lim _{x \rightarrow 1}\left(\ln ^{2} e x\right)^{\frac{1}{x^{2}+1}}=\left(\ln ^{2}(e \cdot 1)\right)^{\frac{1}{1^{2}+1}}=\left(1^{2}\right)^{\frac{1}{2}}=1
$$
## Problem Kuznetsov Limits 20-21 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,127 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \ln \left(\left(e^{x^{2}}-\cos x\right) \cos \left(\frac{1}{x}\right)+\tan\left(x+\frac{\pi}{3}\right)\right)
$$ | ## Solution
Since $\cos \left(\frac{1}{x}\right)_{\text {- is bounded, and }}$
$$
\begin{aligned}
& \lim _{x \rightarrow 0} e^{x^{2}}-\cos x=e^{0^{2}}-\cos 0=e^{0}-1=1-1=0 \\
& \left(e^{x^{2}}-\cos x\right) \cos \left(\frac{1}{x}\right) \rightarrow 0, \text { then } \\
& \text {, as } x \rightarrow 0
\end{aligned}
$$... | \ln\sqrt{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,128 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{-5 ; 9 ;-13\}$
$p=\{0 ; 1 ;-2\}$
$q=\{3 ;-1 ; 1\}$
$r=\{4 ; 1 ; 0\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\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... | 5p-3q+r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,129 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{1 ;-2 ; 4\}$
$b=\{7 ; 3 ; 5\}$
$c_{1}=6 a-3 b$
$c_{2}=b-2 a$ | ## 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 | 47,130 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(1, -1, 0), B(-2, -1, 4), C(8, -1, -1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(-2-1 ;-1-(-1) ; 4-0)=(-3 ; 0 ; 4) \\
& \overrightarrow{A C}=(8-1 ;-1-(-1) ;-1-0)=(7 ; 0 ;-1)
\end{aligned}
$$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\ov... | -\frac{1}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,131 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=10 p+q$
$b=3 p-2 q$
$|p|=4$
$|q|=1$
$(\widehat{p, q})=\frac{\pi}{6}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(10 p+q) \times(3 p-2 q)=10 \cdot 3 \cdot p \times p+10 \cdot(-2) \cdot p \times ... | 46 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,132 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{4 ; 1 ; 2\}$
$b=\{9 ; 2 ; 5\}$
$c=\{1 ; 1 ;-1\}$ | ## 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 & 1 & 2 \\
9 & 2 & 5 \\
1 & 1 & -1
\end{array}\right|= \\
& =4 \cdot\left|\be... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,133 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(1 ; -1 ; 1) \)
\( A_{2}(-2 ; 0 ; 3) \)
\( A_{3}(2 ; 1 ; -1) \)
\( A_{4}(2 ; -2 ; -4) \) | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$$
\begin{aligned}
& \vec{A}_{1} \vec{A}_{2}=\{-2-1 ; 0-(-1) ; 3-1\}=\{-3 ; 1 ; 2\} \\
& A_{1} A_{3}=\{2-1 ; 1-(-1) ;-1-1\}=\{1 ; 2 ;-2\} \\
& \vec{A}_{1} \overrightarrow{A_{4}}=\{2-1 ;-2-(-1) ;-4-1\}=\{1 ;-1 ;-5\}
\end{aligned}
$$
According to the geometri... | 5.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,134 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(-1 ; -5 ; 2)$
$M_{2}(-6 ; 0 ; -3)$
$M_{3}(3 ; 6 ; -3)$
$M_{0}(10 ; -8 ; -7)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-(-1) & y-(-5) & z-2 \\
-6-(-1) & 0-(-5) & -3-2 \\
3-(-1) & 6-(-5) & -3-2
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+1 & y+5 & z-2 \\
-5 &... | 2\sqrt{38} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,135 |
## problem statement
Write the equation of the plane passing through point $A$ and perpendicular to the vector $\overrightarrow{B C}$.
$A(1 ; 0 ;-6)$
$B(-7 ; 2 ; 1)$
$C(-9 ; 6 ; 1)$ | ## Solution
Let's find the vector $\overrightarrow{B C}$:
$\overrightarrow{B C}=\{-9-(-7) ; 6-2 ; 1-1\}=\{-2 ; 4 ; 0\}$
Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$$
\begin{aligned}
& -2 \cdot(x-1... | -x+2y+1=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,136 |
## Task Condition
Find the angle between the planes:
$x-2 y+2 z+17=0$
$x-2 y-1=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{1 ;-2 ; 2\}$
$\overrightarrow{n_{2}}=\{1 ;-2 ; 0\}$
The angle $\phi$ between the planes is determined by the formula:
$$
\begin{aligned}
& \cos \ph... | 4148^{\}37^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,137 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; y ; 0)$
$B(7 ; 3 ;-4)$
$C(1 ; 5 ; 7)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(7-0)^{2}+(3-y)^{2}+(-4-0)^{2}}=\sqrt{49+9-6 y+y^{2}+16}=\sqrt{y^{2}-6 y+74} \\
& A C=\sqrt{(1-0)^{2}+(5-y)^{2}+(7-0)^{2}}=\sqrt{1+25-10 y+y^{2}+49}=\sqrt{y^{2}-10 y+75}
\end{aligned}
$$
Since by the condition of the problem $A B=A ... | A(0;\frac{1}{4};0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,138 |
## Task Condition
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(-1 ; 2 ; 3)$
$a: x-3 y+z+2=0$
$k=2.5$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: x-3 y+z+5=0$
Substitute the coordinates of point $A$ into the equatio... | 1\neq0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,139 |
## Task Condition
Write the canonical equations of the line.
$2 x+y-3 z-2=0$
$2 x-y+z+6=0$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- are coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction ... | \frac{x+1}{-2}=\frac{y-4}{-8}=\frac{z}{-4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,140 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-1}{8}=\frac{y-8}{-5}=\frac{z+5}{12}$
$x-2 y-3 z+18=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-1}{8}=\frac{y-8}{-5}=\frac{z+5}{12}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+8 t \\
y=8-5 t \\
z=-5+12 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$$
\begin{aligned}
& (1+8 t)-2(8-5 ... | (9;3;7) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,141 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(-1 ; 2 ; 0)$
$4 x-5 y-z-7=0$ | ## 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:
$\vec{s}=\vec{n}=\{4 ;-5 ;-1\}$
Then the equation of the desired line is:
$\f... | M^{\}(3;-3;-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,142 |
## problem statement
Based on the definition of the derivative, find $f^{\prime}(0)$ :
$$
f(x)=\left\{\begin{array}{c}
\sqrt[3]{1-2 x^{3} \sin \frac{5}{x}}-1+x, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$$
f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}
$$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,143 |
## problem statement
Find the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=3 \sqrt[4]{x}-\sqrt{x}, x_{0}=1$ | ## Solution
Let's find $y^{\prime}$:
$y^{\prime}=(3 \sqrt[4]{x}-\sqrt{x})^{\prime}=\left(3 x^{\frac{1}{4}}-\sqrt{x}\right)^{\prime}=3 \cdot \frac{1}{4} \cdot x^{-\frac{3}{4}}-\frac{1}{2 \sqrt{x}}=\frac{3}{4 \sqrt[4]{x^{3}}}-\frac{1}{2 \sqrt{x}}$
Then
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{3}{4 \sqrt[4]{... | \frac{x}{4}+\frac{7}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,144 |
## Condition of the problem
Find the differential $d y$.
$$
y=\cos x \cdot \ln (\operatorname{tg} x)-\ln \left(\operatorname{tg} \frac{x}{2}\right)
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime}(x) \cdot \Delta x=y^{\prime}(x) \cdot d x \\
& y^{\prime}=(\cos x)^{\prime} \cdot \ln \operatorname{tg} x+\cos x \cdot(\ln \operatorname{tg} x)^{\prime}-\left(\operatorname{lntg} \frac{x}{2}\right)^{\prime}= \\
& =-\sin x \cdot \ln \operatorname{tg} x+\cos x \cdot \frac{... | -\sinx\cdot\ln\operatorname{tg}x\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,145 |
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