problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
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## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{4} e^{\sqrt{(4-x) /(4+x)}} \cdot \frac{d x}{(4+x) \sqrt{16-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{4} e^{\sqrt{(4-x) /(4+x)}} \cdot \frac{d x}{(4+x) \sqrt{16-x^{2}}}=
$$
Let's make the substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{4-x}{4+x}} \\
& d t=\frac{1}{2 \sqrt{\frac{4-x}{4+x}}} \cdot \frac{-4-x-4+x}{(4+x)^{2}} d x=-4 \sqrt{\frac{4+x}{4-x}} \cdot \frac{d x}{(4+x)^{2}}=-4 \frac{4+... | \frac{1}{4}\cdot(e-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,996 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1 / 8}^{1} \frac{15 \sqrt{x+3}}{(x+3)^{2} \sqrt{x}} d x
$$ | ## Solution
$$
\int_{1 / 8}^{1} \frac{15 \sqrt{x+3}}{(x+3)^{2} \sqrt{x}} d x=\int_{1 / 8}^{1} \frac{15}{(x+3) \sqrt{(x+3) x}} d x=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{x+3}{x}} \\
& d t=\frac{1}{2} \sqrt{\frac{x}{x+3}} \cdot \frac{1 \cdot x-(x+3) \cdot 1}{x^{2}} d x=\frac{1}{2} \sqrt{\frac{x}{x+3}} \c... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,997 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-5 / 3}^{1} \frac{\sqrt[3]{3 x+5}+2}{1+\sqrt[3]{3 x+5}} d x
$$ | ## Solution
$$
\int_{-5 / 3}^{1} \frac{\sqrt[3]{3 x+5}+2}{1+\sqrt[3]{3 x+5}} d x
$$
Perform the variable substitution:
$$
\begin{aligned}
& t=\sqrt[3]{3 x+5} \\
& x=\frac{t^{3}-5}{3} \\
& d x=t^{2} d t
\end{aligned}
$$
Recalculate the limits of integration:
$$
\begin{aligned}
& x_{1}=-\frac{5}{3} \Rightarrow t_{1}... | \frac{8}{3}+\ln3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,998 |
## Problem Statement
Calculate the definite integral:
$$
\int_{2}^{3} \sqrt{\frac{3-2 x}{2 x-7}} d x
$$ | ## Solution
$$
\int_{2}^{3} \sqrt{\frac{3-2 x}{2 x-7}} d x=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{3-2 x}{2 x-7}} \Rightarrow t^{2}=\frac{3-2 x}{2 x-7}=-\frac{2 x-7+4}{2 x-7}=-1-\frac{4}{2 x-7} \Rightarrow \\
& \Rightarrow 7-2 x=\frac{4}{t^{2}+1} \Rightarrow x=-\frac{2}{t^{2}+1}+\frac{7}{2} \\
& d x=\fr... | \frac{\pi}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,999 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{7} \frac{\sqrt{x+25}}{(x+25)^{2} \sqrt{x+1}} d x
$$ | ## Solution
$$
\int_{0}^{7} \frac{\sqrt{x+25}}{(x+25)^{2} \sqrt{x+1}} d x=\int_{0}^{7} \frac{1}{(x+25) \sqrt{(x+25)(x+1)}} d x=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{x+25}{x+1}} \\
& d t=\frac{1}{2} \sqrt{\frac{x+1}{x+25}} \cdot \frac{1 \cdot(x+1)-(x+25) \cdot 1}{(x+1)^{2}} d x=\frac{1}{2} \sqrt{\frac{... | \frac{1}{40} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,000 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2} \frac{(4 \sqrt{2-x}-\sqrt{3 x+2}) d x}{(\sqrt{3 x+2}+4 \sqrt{2-x})(3 x+2)^{2}}
$$ | ## Solution
Introduce the substitution:
$$
t=\sqrt{\frac{2-x}{3 x+2}}
$$
Then:
$$
\begin{aligned}
& x=\frac{2-2 t^{2}}{3 t^{2}+1}, \quad d x=-\frac{16 t}{\left(3 t^{2}+1\right)^{2}} d t \\
& \text { When } x=0, \quad t=1 \\
& \text { When } x=2, \quad t=0 \\
& 3 x+2=\frac{8}{3 t^{2}+1}
\end{aligned}
$$
We obtain:
... | \frac{1}{32}\ln5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,001 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2} e^{\sqrt{(2-x) /(2+x)}} \cdot \frac{d x}{(2+x) \sqrt{4-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{2} e^{\sqrt{(2-x) /(2+x)}} \cdot \frac{d x}{(2+x) \sqrt{4-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{2-x}{2+x}} \\
& d t=\frac{1}{2} \cdot \sqrt{\frac{2+x}{2-x}} \cdot\left(\frac{2-x}{2+x}\right)^{\prime} \cdot d x=\frac{1}{2} \cdot \sqrt{\frac{2+x}{2-x}} \cdot \frac{-4}{(2... | \frac{e-1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,002 |
## Problem Statement
Calculate the definite integral:
$$
\int_{3}^{5} \sqrt{\frac{2-x}{x-6}} d x
$$ | ## Solution
$$
\int_{3}^{5} \sqrt{\frac{2-x}{x-6}} d x=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{2-x}{x-6}} \Rightarrow t^{2}=\frac{2-x}{x-6}=-\frac{x-6+4}{x-6}=-1-\frac{4}{x-6} \Rightarrow \\
& \Rightarrow 6-x=\frac{4}{t^{2}+1} \Rightarrow x=-\frac{4}{t^{2}+1}+6 \\
& d x=\frac{4}{\left(t^{2}+1\right)^{2}... | \frac{2\pi}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,003 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1 / 24}^{1 / 3} \frac{5 \sqrt{x+1}}{(x+1)^{2} \sqrt{x}} d x
$$ | ## Solution
$$
\int_{1 / 24}^{1 / 3} \frac{5 \sqrt{x+1}}{(x+1)^{2} \sqrt{x}} d x=\int_{1 / 24}^{1 / 3} \frac{5}{(x+1) \sqrt{(x+1) x}} d x=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{x+1}{x}} \\
& d t=\frac{1}{2} \sqrt{\frac{x}{x+1}} \cdot \frac{1 \cdot x-(x+1) \cdot 1}{x^{2}} d x=\frac{1}{2} \sqrt{\frac{x}{... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,004 |
## Problem Statement
Calculate the definite integral:
$$
\int_{9}^{15} \sqrt{\frac{6-x}{x-18}} d x
$$ | ## Solution
$$
\int_{9}^{15} \sqrt{\frac{6-x}{x-18}} d x=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{6-x}{x-18}} \Rightarrow t^{2}=\frac{6-x}{x-18}=-\frac{x-18+12}{x-18}=-1-\frac{12}{x-18} \Rightarrow \\
& \Rightarrow 18-x=\frac{12}{t^{2}+1} \Rightarrow x=-\frac{12}{t^{2}+1}+18 \\
& d x=\frac{12}{\left(t^{2... | 2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,005 |
## Problem Statement
Calculate the definite integral:
$$
\int_{16 / 15}^{4 / 3} \frac{4 \sqrt{x}}{x^{2} \sqrt{x-1}} d x
$$ | ## Solution
## Problem 11.
## Calculate the definite integral:
$$
I=\int_{16 / 15}^{4 / 3} \frac{4 \sqrt{x}}{x^{2} \sqrt{x-1}} d x
$$
## Perform a variable substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{x}{x-1}}, \text { hence } t^{2}=\frac{x}{x-1}, \text { that is } x=1+\frac{1}{t^{2}-1}, \text { and } \\
& d x... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,008 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{6} \frac{e^{\sqrt{(6-x) /(6+x)}} \cdot d x}{(6+x) \sqrt{36-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{6} \frac{e^{\sqrt{(6-x) /(6+x)}} \cdot d x}{(6+x) \sqrt{36-x^{2}}}=\int_{0}^{6} \frac{e^{\sqrt{(6-x) /(6+x)}} \cdot d x}{(6+x)^{2} \sqrt{\frac{6-x}{6+x}}}=-\left.\frac{1}{6} \cdot e^{\sqrt{\frac{6-x}{6+x}}}\right|_{0} ^{6}=\frac{e-1}{6}
$$
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�... | \frac{e-1}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,009 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{64} \frac{6-\sqrt{x}+\sqrt[4]{x}}{\sqrt{x^{3}}-7 x-6 \sqrt[4]{x^{3}}} d x
$$ | ## Solution
$$
\int_{1}^{64} \frac{6-\sqrt{x}+\sqrt[4]{x}}{\sqrt{x^{3}}-7 x-6 \sqrt[4]{x^{3}}} d x=
$$
Substitution:
$$
\begin{aligned}
& x=t^{4}, d x=4 t^{3} d t \\
& x=1 \Rightarrow t=\sqrt[4]{1}=1 \\
& x=64 \Rightarrow t=\sqrt[4]{64}=2 \sqrt{2}
\end{aligned}
$$
We get:
$$
=\int_{1}^{2 \sqrt{2}} \frac{\left(6-t^... | 4\ln(\frac{2}{2\sqrt{2}+1}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,010 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} \frac{(4 \sqrt{1-x}-\sqrt{x+1}) d x}{(\sqrt{x+1}+4 \sqrt{1-x})(x+1)^{2}}
$$ | ## Solution
$$
\int_{0}^{1} \frac{(4 \sqrt{1-x}-\sqrt{x+1}) d x}{(\sqrt{x+1}+4 \sqrt{1-x})(x+1)^{2}}=\int_{0}^{1} \frac{\left(4 \sqrt{\frac{1-x}{x+1}}-1\right) d x}{\left(1+4 \sqrt{\frac{1-x}{x+1}}\right)(x+1)^{2}}=
$$
Introduce the substitution:
$$
t=\sqrt{\frac{1-x}{x+1}}
$$
Then
$$
\begin{aligned}
& t^{2}=\frac... | \frac{1}{8}\ln5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,011 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{3} \frac{e^{\sqrt{(3-x) /(3+x)}} d x}{(3+x) \sqrt{9-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{3} \frac{e^{\sqrt{(3-x) /(3+x)}} d x}{(3+x) \sqrt{9-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{3-x}{3+x}} \\
& d t=\frac{1}{2} \cdot \sqrt{\frac{3+x}{3-x}} \cdot\left(\frac{3-x}{3+x}\right)^{\prime} \cdot d x=\frac{1}{2} \cdot \sqrt{\frac{3+x}{3-x}} \cdot \frac{-6}{(3+x)^{2... | \frac{e-1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,012 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2} \frac{(4 \sqrt{2-x}-\sqrt{x+2}) d x}{(\sqrt{x+2}+4 \sqrt{2-x})(x+2)^{2}}
$$ | ## Solution
$$
\int_{0}^{2} \frac{(4 \sqrt{2-x}-\sqrt{x+2}) d x}{(\sqrt{x+2}+4 \sqrt{2-x})(x+2)^{2}}=\int_{0}^{2} \frac{\left(4 \sqrt{\frac{2-x}{x+2}}-1\right) d x}{\left(1+4 \sqrt{\frac{2-x}{x+2}}\right)(x+2)^{2}}=
$$
Introduce the substitution:
$$
t=\sqrt{\frac{2-x}{x+2}}
$$
Then
$$
\begin{aligned}
& t^{2}=\frac... | \frac{1}{4}\ln5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,013 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{2-3 n^{2}}{4+5 n^{2}}, a=-\frac{3}{5}$ | ## Solution
By the definition of the limit:
$$
\begin{aligned}
& \forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{10-15 n^{2}+12+15 n^{2}}{5\left(4+5 n^{2}\right)}\right| \\
& \left.\frac{22}{5\left(4+5 n^{2}\right)} \right\rvert\, \\... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,014 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(n+2)^{2}-(n-2)^{2}}{(n+3)^{2}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+2)^{2}-(n-2)^{2}}{(n+3)^{2}}=\lim _{n \rightarrow \infty} \frac{n^{2}+4 n+4-n^{2}+4 n-4}{n^{2}+6 n+9}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n} 8 n}{\frac{1}{n}\left(n^{2}+6 n+9\right)}=\lim _{n \rightarrow \infty} \frac{8}{n+6+\frac{9}... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,015 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+1}-\sqrt[3]{n^{3}+1}}{\sqrt[4]{n+1}-\sqrt[5]{n^{5}+1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+1}-\sqrt[3]{n^{3}+1}}{\sqrt[4]{n+1}-\sqrt[5]{n^{5}+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+1}-\sqrt[3]{n^{3}+1}\right)}{\frac{1}{n}\left(\sqrt[4]{n+1}-\sqrt[5]{n^{5}+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,016 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \sqrt{n(n+1)(n+2)}\left(\sqrt{n^{3}-3}-\sqrt{n^{3}-2}\right)$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \sqrt{n(n+1)(n+2)}\left(\sqrt{n^{3}-3}-\sqrt{n^{3}-2}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{n(n+1)(n+2)}\left(\sqrt{n^{3}-3}-\sqrt{n^{3}-2}\right)\left(\sqrt{n^{3}-3}+\sqrt{n^{3}-2}\right)}{\sqrt{n^{3}-3}+\sqrt{n^{3}-2}}= \\
& =\lim _{n \ri... | -\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,017 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{n+5}{n-7}\right)^{\frac{n}{6}+1}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n+5}{n-7}\right)^{\frac{n}{6}+1}=\lim _{n \rightarrow \infty}\left(\frac{n-7+12}{n-7}\right)^{\frac{n}{6}+1}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{12}{n-7}\right)^{\frac{n}{6}+1}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac... | e^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,019 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$$
\lim _{x \rightarrow-\frac{1}{5}} \frac{15 x^{2}-2 x-1}{x+\frac{1}{5}}=-8
$$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). The number
_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$f(x)=4 x^{2}+6, x_{0}=7$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$.
$$
\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, such that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$ when $\left|x-x_{0}\righ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,021 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-3} \frac{x^{3}+7 x^{2}+15 x+9}{x^{3}+8 x^{2}+21 x+18}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-3} \frac{x^{3}+7 x^{2}+15 x+9}{x^{3}+8 x^{2}+21 x+18}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-3} \frac{(x+3)\left(x^{2}+4 x+3\right)}{(x+3)\left(x^{2}+5 x+6\right)}= \\
& =\lim _{x \rightarrow-3} \frac{x^{2}+4 x+3}{x^{2}+5 x+6}=\left\{\frac{0}{0}\right\}=\l... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,022 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow-8} \frac{10-x-6 \sqrt{1-x}}{2+\sqrt[3]{x}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-8} \frac{10-x-6 \sqrt{1-x}}{2+\sqrt[3]{x}}=\lim _{x \rightarrow-8} \frac{(10-x-6 \sqrt{1-x})(10-x+6 \sqrt{1-x})}{(2+\sqrt[3]{x})}= \\
& =\lim _{x \rightarrow-8} \frac{(10-x+6 \sqrt{1-x})}{(2+\sqrt[3]{x})(10-x+6 \sqrt{1-x})}=\lim _{x \rightarrow-8} \frac{100-20 x+x^... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,023 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{2\left(e^{\pi x}-1\right)}{3(\sqrt[3]{1+x}-1)}
$$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$e^{\pi x}-1 \sim \pi x$, as $x \rightarrow 0 (\pi x \rightarrow 0)$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{2\left(e^{\pi x}-1\right)}{3(\sqrt[3]{1+x}-1)}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{2 \pi x\left(... | 2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,024 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\sin 5 x}{\tan 3 x}$ | ## Solution
Substitution:
$x=y+\pi \Rightarrow y=x-\pi$
$x \rightarrow \pi \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow \pi} \frac{\sin 5 x}{\tan 3 x}=\lim _{y \rightarrow 0} \frac{\sin 5(y+\pi)}{\tan 3(y+\pi)}=$
$=\lim _{y \rightarrow 0} \frac{\sin (5 y+5 \pi)}{\tan(3 y+3 \pi)}=\lim _{y \rightarrow... | -\frac{5}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,025 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\operatorname{tg}\left(3^{\pi / x}-3\right)}{3^{\cos (3 x / 2)}-1}$ | ## Solution
Substitution:
$x=y+\pi \Rightarrow y=x-\pi$
$x \rightarrow \pi \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow \pi} \frac{\operatorname{tg}\left(3^{\pi / x}-3\right)}{3^{\cos (3 x / 2)}-1}=\lim _{y \rightarrow 0} \frac{\operatorname{tg}\left(3^{\pi /(y+\pi)}-3\right)}{3^{\cos (3(y+\pi) / 2)}... | -\frac{2}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,026 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{2^{3 x}-3^{2 x}}{x+\arcsin x^{3}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{2^{3 x}-3^{2 x}}{x+\arcsin x^{3}}=\lim _{x \rightarrow 0} \frac{\left(8^{x}-1\right)-\left(9^{x}-1\right)}{x+\arcsin x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 8}\right)^{x}-1\right)-\left(\left(e^{\ln 9}\right)^{x}-1\right)}{x+\arcsin ... | \ln\frac{2^{3}}{3^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,027 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 3} \frac{\log _{3} x-1}{\tan \pi x}$ | ## Solution
$\lim _{x \rightarrow 3} \frac{\log _{3} x-1}{\tan \pi x}=\lim _{x \rightarrow 3} \frac{\log _{3} x-\log _{3} 3}{\tan \pi x}=$
$=\lim _{x \rightarrow 3} \frac{\log _{3} \frac{x}{3}}{\tan \pi x}=\lim _{x \rightarrow 3} \frac{\ln \frac{x}{3} / \ln 3}{\tan \pi x}=$
$=\lim _{x \rightarrow 3} \frac{\ln \frac{x... | \frac{1}{3\pi\ln3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,028 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(1-\sin ^{2} \frac{x}{2}\right)^{\frac{1}{\ln \left(1+\operatorname{tg}^{2} 3 x\right)}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(1-\sin ^{2} \frac{x}{2}\right)^{\frac{1}{\ln \left(1+\operatorname{tg}^{2} 3 x\right)}}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln \left(1-\sin ^{2} \frac{x}{2}\right)}\right)^{\frac{1}{\ln \left(1+\operatorname{tg}^{2} 3 x\right)}}=$
$=\lim _{x \rightarrow 0} e^{\frac{\ln \lef... | e^{-\frac{1}{36}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,029 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{\arcsin ^{2} x}{\arcsin ^{2} 4 x}\right)^{2 x+1}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{\arcsin ^{2} x}{\arcsin ^{2} 4 x}\right)^{2 x+1}=\left(\lim _{x \rightarrow 0} \frac{\arcsin ^{2} x}{\arcsin ^{2} 4 x}\right)^{\lim _{x \rightarrow 0} 2 x+1}= \\
& =\left(\lim _{x \rightarrow 0} \frac{\arcsin ^{2} x}{\arcsin ^{2} 4 x}\right)^{2 \cdot ... | \frac{1}{16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,030 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi}\left(\operatorname{ctg}\left(\frac{x}{4}\right)\right)^{1 / \cos \left(\frac{x}{2}\right)}$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=4 y+\pi \Rightarrow y=\frac{1}{4}(x-\pi) \\
& x \rightarrow \pi \Rightarrow y \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \pi}\left(\operatorname{ctg}\left(\frac{4 y+\pi}{4}\right)\right)^{1 / \cos \left(\frac{4 y+\pi}{2}\right)}=... | e | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,031 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}\left(\frac{1+\cos \pi x}{\tan^{2} \pi x}\right)^{x^{2}}$ | ## Solution
$\lim _{x \rightarrow 1}\left(\frac{1+\cos \pi x}{\tan^{2} \pi x}\right)^{x^{2}}=\lim _{x \rightarrow 1}\left(\frac{1+1-2 \sin ^{2}\left(\frac{\pi x}{2}\right)}{\tan^{2} \pi x}\right)^{x^{2}}=$
$$
\begin{aligned}
& =\lim _{x \rightarrow 1}\left(\frac{2-2 \sin ^{2}\left(\frac{\pi x}{2}\right)}{\tan^{2} \pi... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,032 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{\sin x+\sin (\pi x) \cdot \operatorname{arctg} \frac{1+x}{1-x}}{1+\cos x}$ | ## Solution
Since $\operatorname{arctg} \frac{1+x}{1-x}_{\text {- is bounded, and }}$
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \sin (\pi x)=\sin (\pi \cdot 1)=\sin \pi=0 \\
& \sin (\pi x) \cdot \operatorname{arctg} \frac{1+x}{1-x} \rightarrow 0, \text { then } \\
& \text {, as } x \rightarrow 0
\end{aligned}
$$
... | \frac{\sin1}{1+\cos1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,033 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{3 ; 1 ; 8\}$
$p=\{0 ; 1 ; 3\}$
$q=\{1 ; 2 ;-1\}$
$r=\{2 ; 0 ;-1\}$ | ## 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... | 3p-q+2r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,034 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{5 ; 0 ;-2\}$
$b=\{6 ; 4 ; 3\}$
$c_{1}=5 a-3 b$
$c_{2}=6 b-10 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}=-2(5 a-3 b)=-2 c_{1}$ for any $a$ and $b$.
That is, $c_{1}=-2 \cdot c_{2}$, which means the vectors $c_{... | c_{1}=-2\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,035 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(0 ; 2 ; -4), B(8 ; 2 ; 2), C(6 ; 2 ; 4)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(8-0 ; 2-2 ; 2-(-4))=(8 ; 0 ; 6)$
$\overrightarrow{A C}=(6-0 ; 2-2 ; 4-(-4))=(6 ; 0 ; 8)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrightarr... | 0.96 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,036 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=5 p+q$
$b=p-3 q$
$|p|=1$
$|q|=2$
$(\widehat{p, q})=\frac{\pi}{3}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(5 p+q) \times(p-3 q)=5 \cdot p \times p+5 \cdot(-3) \cdot p \times q+q \times p-... | 16\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,037 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{-2 ;-4 ;-3\}$
$b=\{4 ; 3 ; 1\}$
$c=\{6 ; 7 ; 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}
-2 & -4 & -3 \\
4 & 3 & 1 \\
6 & 7 & 4
\end{array}\right|= \\
& =-2 \cdot\left|... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,038 |
## Task Condition
Calculate the volume of the tetrahedron with vertices at points $A_{1}, A_{2}, A_{3}, A_{4_{\text {and }}}$ its height dropped from vertex $A_{4 \text { to the face }} A_{1} A_{2} A_{3}$.
$A_{1}(-3 ; 4 ;-7)$
$A_{2}(1 ; 5 ;-4)$
$A_{3}(-5 ;-2 ; 0)$
$A_{4}(2 ; 5 ; 4)$ | ## Solution
From vertex $A_{1}$, we will draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{1-(-3) ; 5-4 ;-4-(-7)\}=\{4 ; 1 ; 3\} \\
& \overrightarrow{A_{1} A_{3}}=\{-5-(-3) ;-2-4 ; 0-(-7)\}=\{-2 ;-6 ; 7\} \\
& \overrightarrow{A_{1} A_{4}}=\{2-(-3) ; 5-4 ; 4-(-7)\}=\{5 ; 1 ; 11\}
\end{aligned}
$$
Acco... | 25\frac{1}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,039 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(-4 ; 2 ; 6)$
$M_{2}(2 ;-3 ; 0)$
$M_{3}(-10 ; 5 ; 8)$
$M_{0}(-12 ; 1 ; 8)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-(-4) & y-2 & z-6 \\
2-(-4) & -3-2 & 0-6 \\
-10-(-4) & 5-2 & 8-6
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+4 & y-2 & z-6 \\
6 & -5 & -6 \... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,040 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(3, -3, -6)$
$B(1, 9, -5)$
$C(6, 6, -4)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{6-1 ; 6-9 ;-4-(-5)\}=\{5 ;-3 ; 1\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$5 \cdot(x-3)-3 \cdot(y-(-3))+(z-(-6))... | 5x-3y+z-18=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,041 |
## Task Condition
Find the angle between the planes:
$x-3 y-2 z-8=0$
$x+y-z+3=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{1 ;-3 ;-2\}$
$\overrightarrow{n_{2}}=\{1 ; 1 ;-1\}$
The angle $\phi$ between the planes is determined by the formula:
$\cos \phi=\frac{\left(\overr... | \frac{\pi}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,042 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; y ; 0)$
$B(0 ;-4 ; 1)$
$C(1 ;-3 ; 5)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(0-0)^{2}+(-4-y)^{2}+(1-0)^{2}}=\sqrt{0+16+8 y+y^{2}+1}=\sqrt{y^{2}+8 y+17} \\
& A C=\sqrt{(1-0)^{2}+(-3-y)^{2}+(5-0)^{2}}=\sqrt{1+9+6 y+y^{2}+25}=\sqrt{y^{2}+6 y+35}
\end{aligned}
$$
Since by the condition of the problem $A B=A C$,... | A(0;9;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,043 |
## problem statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point
$A$ belongs to the image of plane $a$?
$A(1 ; 1 ; 1)$
$a: 7 x-6 y+z-5=0$
$k=-2$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0$ and the coefficient $k$, the plane transitions to
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 7 x-6 y+z+10=0$
Substitute the coordinates of point $A$ into the equatio... | 12\neq0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,044 |
## Task Condition
Write the canonical equations of the line.
$$
\begin{aligned}
& 2 x-3 y+z+6=0 \\
& x-3 y-2 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}
$$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction ... | \frac{x+3}{9}=\frac{y}{5}=\frac{z}{-3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,045 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x+3}{2}=\frac{y-1}{3}=\frac{z-1}{5}$
$2 x+3 y+7 z-52=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x+3}{2}=\frac{y-1}{3}=\frac{z-1}{5}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-3+2 t \\
y=1+3 t \\
z=1+5 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$$
\begin{aligned}
& 2(-3+2 t)+3(1+3 t... | (-1;4;6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,046 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$$
M(-1 ; 0 ;-1)
$$
$2 x+6 y-2 z+11=0$ | ## Solution
Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$.
Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:
$\vec{s}=\vec{n}=\{2 ; 6 ;-2\}$
Then the equation of the desired line is:
$\... | M^{\}(-2;-3;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,047 |
## Task Condition
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{6 ; 12 ;-1\}$
$p=\{1 ; 3 ; 0\}$
$q=\{2 ;-1 ; 1\}$
$r=\{0 ;-1 ; 2\}$ | ## 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:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 4p+q-r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,048 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{1 ; 0 ; 1\}$
$b=\{-2 ; 3 ; 5\}$
$c_{1}=a+2 b$
$c_{2}=3 a-b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
\begin{aligned}
& c_{1}=a+2 b=\{1+2 \cdot(-2) ; 0+2 \cdot 3 ; 1+2 \cdot 5\}=\{-3 ; 6 ; 11\} \\
& c_{2}=3 a-b=\{3 \cdot 1-(-2) ; 3... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,049 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$$
A(0 ;-3 ; 6), B(-12 ;-3 ;-3), C(-9 ;-3 ;-6)
$$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(-12-0 ;-3-(-3) ;-3-6)=(-12 ; 0 ;-9) \\
& \overrightarrow{A C}=(-9-0 ;-3-(-3) ;-6-6)=(-9 ; 0 ;-12)
\end{aligned}
$$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and... | 0.96 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,050 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=3 p+q$
$b=p-2 q$
$|p|=4$
$|q|=1$
$\widehat{(\widehat{p}, q})=\frac{\pi}{4}$ | ## 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:
$$
\begin{aligned}
& a \times b=(3 p+q) \times(p-2 q)=3 \cdot p \times p+3 \cdot(-2) \cdot p ... | 14\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,051 |
## problem statement
Are the vectors $a, b$ and $c$ coplanar?
$a=\{3 ; 2 ; 1\}$
$b=\{2 ; 3 ; 4\}$
$c=\{3 ; 1 ;-1\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
3 & 2 & 1 \\
2 & 3 & 4 \\
3 & 1 & -1
\end{array}\right|= \\
& =3 \cdot\left|\be... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,052 |
## problem statement
Calculate the volume of the tetrahedron with vertices at points $A_{1}, A_{2}, A_{3}, A_{4_{\text {and }}}$ its height dropped from vertex $A_{4 \text { to the face }} A_{1} A_{2} A_{3}$.
$A_{1}(-4 ; 2 ; 6)$
$A_{2}(2 ;-3 ; 0)$
$A_{3}(-10 ; 5 ; 8)$
$A_{4}(-5 ; 2 ;-4)$ | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{2-(-4) ;-3-2 ; 0-6\}=\{6 ;-5 ;-6\} \\
& \vec{A}_{1} A_{3}=\{-10-(-4) ; 5-2 ; 8-6\}=\{-6 ; 3 ; 2\} \\
& \overrightarrow{A_{1} A_{4}}=\{-5-(-4) ; 2-2 ;-4-6\}=\{-1 ; 0 ;-10\}
\end{aligned}
$$
According to the... | 18\frac{2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,053 |
## problem statement
Find the distance from the point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(-1 ; 2 ;-3)$
$M_{2}(4 ;-1 ; 0)$
$M_{3}(2 ; 1 ;-2)$
$M_{0}(1 ;-6 ;-5)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-(-1) & y-2 & z-(-3) \\
4-(-1) & -1-2 & 0-(-3) \\
2-(-1) & 1-2 & -2-(-3)
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+1 & y-2 & z+3 \\
5 & -... | 5\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,054 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-1 ; 3 ; 4)$
$B(-1 ; 5 ; 0)$
$C(2 ; 6 ; 1)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{2-(-1) ; 6-5 ; 1-0\}=\{3 ; 1 ; 1\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$3 \cdot(x-(-1))+(y-3)+(z-4)=0$
$3 x+3... | 3x+y+z-4=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,055 |
## Problem Statement
Find the angle between the planes:
$x-3 y+z-1=0$
$x+z-1=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{1 ;-3 ; 1\}$
$\overrightarrow{n_{2}}=\{1 ; 0 ; 1\}$
The angle $\phi_{\text {between the planes is determined by }}$ the formula:
$$
\begin{aligned}... | \arccos\sqrt{\frac{2}{11}}\approx6445^{\}38^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,056 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; 0 ; z)$
$B(3 ; 3 ; 1)$
$C(4 ; 1 ; 2)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(3-0)^{2}+(3-0)^{2}+(1-z)^{2}}=\sqrt{9+9+1-2 z+z^{2}}=\sqrt{z^{2}-2 z+19} \\
& A C=\sqrt{(4-0)^{2}+(1-0)^{2}+(2-z)^{2}}=\sqrt{16+1+4-4 z+z^{2}}=\sqrt{z^{2}-4 z+21}
\end{aligned}
$$
Since according to the problem's condition $A B=A C... | A(0;0;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,057 |
## 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 $a?$
$A(2 ; 1 ; 2)$
$a: x-2 y+z+1=0$
$k=-2$ | ## 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 $a$:
$a^{\prime}: x-2 y+z-2=0$
Substitute the coordinates of point $A$ into the equatio... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,058 |
## Task Condition
Write the canonical equations of the line.
$x-3 y+2 z+2=0$
$x+3 y+z+14=0$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vecto... | \frac{x+8}{-9}=\frac{y+2}{1}=\frac{z}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,059 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x+1}{3}=\frac{y-3}{-4}=\frac{z+1}{5}$
$x+2 y-5 z+20=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x+1}{3}=\frac{y-3}{-4}=\frac{z+1}{5}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-1+3 t \\
y=3-4 t \\
z=-1+5 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$(-1+3 t)+2(3-4 t)-5(-1+5 t)+20=0$
... | (2,-1,4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,060 |
## Task Condition
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$M(2, -1, 1)$
$$
\frac{x-4.5}{1}=\frac{y+3}{-0.5}=\frac{z-2}{1}
$$ | ## Solution
We find the equation of the plane that is perpendicular to the given line and passes through point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as its normal vector:
$\vec{n}=\vec{s}=\{1 ;-0.5 ; 1\}$
Then the equation of the desired plane is:
$1 \c... | M^{\}(3;-3;-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,061 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\sin \left(x \sin \frac{3}{x}\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$$
f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}
$$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} ... | f^{\}(0) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,062 |
## Task Condition
Derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=x+\sqrt{x^{3}}, x_{0}=1$ | ## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=\left(x+\sqrt{x^{3}}\right)^{\prime}=\left(x+x^{\frac{3}{2}}\right)^{\prime}=1+\frac{3}{2} \cdot x^{\frac{1}{2}}=1+\frac{3}{2} \sqrt{x}
$$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=1+\frac{3}{2} \sqrt{x_{0}}=1+\frac{3}{2} \sqrt{1}=1+\frac{3}{2}=\frac{5}{2}... | -\frac{2x}{5}+2\frac{2}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,063 |
## Condition of the problem
Find the differential $d y$.
$$
y=\arccos \left(\frac{1}{\sqrt{1+2 x^{2}}}\right), x>0
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\arccos \left(\frac{1}{\sqrt{1+2 x^{2}}}\right)\right)^{\prime} d x= \\
& =\left(\frac{-1}{\sqrt{1-\left(\frac{1}{\sqrt{1+2 x^{2}}}\right)^{2}}}\right) \cdot\left(\frac{1}{\sqrt{1+2 x^{2}}}\right)^{\prime} d x= \\
& =\left(\frac{-1}{\sqrt{1-\frac{1}{1+2 x... | \frac{\sqrt{2}\cdotx}{1+2x^{2}}\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,064 |
## Task Condition
Approximately calculate using the differential.
$y=\arcsin x, x=0.08$ | ## Solution
If the increment $\Delta x = x - x_{0 \text{ of the argument }} x$ 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:
$x_{0} = 0$
Then:
$\Delta x = 0.08$
Calculate:
$y(0) = \arcsin 0 = 0$
$y^{\pr... | 0.08 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,065 |
## Task Condition
Find the derivative.
$$
y=\frac{\left(1+x^{8}\right) \sqrt{1+x^{8}}}{12 x^{12}}
$$ | ## Solution
## Method 1
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\left(1+x^{8}\right) \sqrt{1+x^{8}}}{12 x^{12}}\right)^{\prime}= \\
& =\frac{\left(8 x^{7} \cdot \sqrt{1+x^{8}}+\left(1+x^{8}\right) \cdot \frac{1}{2 \sqrt{1+x^{8}}} \cdot 8 x^{7}\right) \cdot x^{12}-\left(1+x^{8}\right) \sqrt{1+x^{8}} \cdot 12 x^{11... | \sqrt{e^{x}+1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,066 |
## Task Condition
Find the derivative.
$$
y=\ln (\sqrt{x}+\sqrt{x+1})
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=(\ln (\sqrt{x}+\sqrt{x+1}))^{\prime}=\frac{1}{\sqrt{x}+\sqrt{x+1}} \cdot\left(\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x+1}}\right)= \\
& =\frac{1}{\sqrt{x}+\sqrt{x+1}} \cdot \frac{\sqrt{x+1}+\sqrt{x}}{2 \sqrt{x} \cdot \sqrt{x+1}}=\frac{1}{2 \sqrt{x^{2}+x}}
\end{aligned}
$$
##... | \frac{1}{2\sqrt{x^{2}+x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,067 |
## Task Condition
Find the derivative.
$y=\arccos \frac{x^{2}-4}{\sqrt{x^{4}+16}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\arccos \frac{x^{2}-4}{\sqrt{x^{4}+16}}\right)^{\prime}=\frac{-1}{\sqrt{1-\left(\frac{x^{2}-4}{\sqrt{x^{4}+16}}\right)^{2}}} \cdot\left(\frac{x^{2}-4}{\sqrt{x^{4}+16}}\right)^{\prime}= \\
& =\frac{-\sqrt{x^{4}+16}}{\sqrt{x^{4}+16-\left(x^{2}-4\right)^{2}}} \cdot \frac{... | -\frac{2\sqrt{2}(4+x^{2})}{x^{4}+16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,068 |
## Problem Statement
Find the derivative.
$y=\frac{1}{2} \operatorname{th} x+\frac{1}{4 \sqrt{2}} \ln \frac{1+\sqrt{2} \operatorname{th} x}{1-\sqrt{2} \operatorname{th} x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2} \tanh x+\frac{1}{4 \sqrt{2}} \ln \frac{1+\sqrt{2} \tanh x}{1-\sqrt{2} \tanh x}\right)^{\prime}= \\
& =\frac{1}{2} \cdot \frac{1}{\cosh^{2} x}+\frac{1}{4 \sqrt{2}} \cdot \frac{1-\sqrt{2} \tanh x}{1+\sqrt{2} \tanh x} \cdot\left(\frac{1+\sqrt{2} \tanh x}{1-\sq... | \frac{1}{\cosh^{2}x(1-\sinh^{2}x)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,069 |
Condition of the problem
Find the derivative.
$y=(\ln x)^{3^{x}}$ | ## Solution
$y=(\ln x)^{3^{x}}$
$\ln y=3^{x} \cdot \ln (\ln x)$
$\frac{y^{\prime}}{y}=\left(3^{x} \cdot \ln (\ln x)\right)^{\prime}=3^{x} \cdot \ln 3 \cdot \ln (\ln x)+3^{x} \cdot \frac{1}{\ln x} \cdot \frac{1}{x}=$
$=3^{x} \cdot\left(\ln 3 \cdot \ln (\ln x)+\frac{1}{x \cdot \ln x}\right)$
$y^{\prime}=y \cdot 3^{x... | (\lnx)^{3^{x}}\cdot3^{x}\cdot(\ln3\cdot\ln(\lnx)+\frac{1}{x\cdot\lnx}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,070 |
## Problem Statement
Find the derivative.
$$
y=\frac{2}{x-1} \cdot \sqrt{2 x-x^{2}}+\ln \frac{1+\sqrt{2 x-x^{2}}}{x-1}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2}{x-1} \cdot \sqrt{2 x-x^{2}}+\ln \frac{1+\sqrt{2 x-x^{2}}}{x-1}\right)^{\prime}= \\
& =\frac{\frac{2}{2 \sqrt{2 x-x^{2}}} \cdot(2-2 x)-2 \sqrt{2 x-x^{2}} \cdot 1}{(x-1)^{2}}+\frac{x-1}{1+\sqrt{2 x-x^{2}}} \cdot\left(\frac{1+\sqrt{2 x-x^{2}}}{x-1}\right)^{\prime... | \frac{2x^{2}-7x+3}{(x-1)^{2}\sqrt{2x-x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,071 |
## Task Condition
Find the derivative.
$$
y=3 \arcsin \frac{3}{4 x+1}+2 \sqrt{4 x^{2}+2 x-2}, 4 x+1>0
$$ | ## Solution
$$
y^{\prime}=\left(3 \arcsin \frac{3}{4 x+1}+2 \sqrt{4 x^{2}+2 x-2}\right)^{\prime}=
$$
$$
\begin{aligned}
& =\frac{3}{\sqrt{1-\left(\frac{3}{4 x+1}\right)^{2}}}+\frac{2}{2 \sqrt{4 x^{2}+2 x-2}} \cdot(8 x+2)= \\
& =\frac{3(4 x+1)}{\sqrt{(4 x+1)^{2}-3^{2}}}+\frac{8 x+2}{\sqrt{4 x^{2}+2 x-2}}=\frac{3(4 x+1... | \frac{7\cdot(4x+1)}{2\sqrt{4x^{2}+2x-2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,072 |
## Problem Statement
Find the derivative.
$$
y=3 \frac{\sin x}{\cos ^{2} x}+2 \frac{\sin x}{\cos ^{4} x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=3 \frac{\cos ^{3} x-\sin x \cdot 2 \cos x \cdot(-\sin x)}{\cos ^{4} x}+2 \frac{\cos ^{5} x-\sin x \cdot 4 \cos ^{3} x \cdot(-\sin x)}{\cos ^{8} x}= \\
& =3 \frac{\cos ^{2} x+2 \sin ^{2} x}{\cos ^{3} x}+2 \frac{\cos ^{2} x+4 \sin ^{2} x}{\cos ^{5} x}=3 \frac{1-\sin ^{2} x+2 \... | \frac{3+3\sin^{2}x}{\cos^{3}x}+\frac{2-6\sin^{2}x}{\cos^{5}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,073 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\ln \left(t+\sqrt{t^{2}+1}\right) \\
y=t \sqrt{t^{2}+1}
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\ln \left(t+\sqrt{t^{2}+1}\right)\right)^{\prime}=\frac{1}{t+\sqrt{t^{2}+1}} \cdot\left(1+\frac{1}{2 \sqrt{t^{2}+1}} \cdot 2 t\right)= \\
& =\frac{1}{t+\sqrt{t^{2}+1}} \cdot\left(1+\frac{t}{\sqrt{t^{2}+1}}\right)=\frac{1}{t+\sqrt{t^{2}+1}} \cdot \frac{\sqrt{t^{2}+1... | 2^{2}+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,074 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{\begin{array}{l}
x=\frac{2 t+t^{2}}{1+t^{3}} \\
y=\frac{2 t-t^{2}}{1+t^{3}}
\end{array}\right.
\]
$t_{0}=1$ | ## Solution
Since $t_{0}=1$, then
$x_{0}=\frac{2 \cdot 1+1^{2}}{1+1^{3}}=\frac{3}{2}$
$y_{0}=\frac{2 \cdot 1-1^{2}}{1+1^{3}}=\frac{1}{2}$
Let's find the derivatives:
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\frac{2 t+t^{2}}{1+t^{3}}\right)^{\prime}=\frac{(2+2 t) \cdot\left(1+t^{3}\right)-\left(2 t+t^{2}\right) \c... | 3x-4-\frac{x}{3}+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,075 |
## Task Condition
Find the $n$-th order derivative.
$$
y=\lg (5 x+2)
$$ | ## Solution
$y^{\prime}=\frac{1}{(5 x+2) \cdot \ln 10} \cdot 5=\frac{5}{(5 x+2) \cdot \ln 10}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{5}{(5 x+2) \cdot \ln 10}\right)^{\prime}=\frac{5}{\ln 10} \cdot\left(-\frac{1}{(5 x+2)^{2}} \cdot 5\right)=$
$=-\frac{5^{2}}{\ln 10} \cdot \frac{1}{(5 x+2)^{2... | y^{(n)}=\frac{(-1)^{n-1}\cdot(n-1)!\cdot5^{n}}{\ln10\cdot(5x+2)^{n}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,076 |
## Task Condition
Find the derivative of the specified order.
$$
y=\frac{\log _{2} x}{x^{3}}, y^{\prime \prime \prime}=?
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\log _{2} x}{x^{3}}\right)^{\prime}=\left(\frac{\ln x}{\ln 2 \cdot x^{3}}\right)^{\prime}=\frac{\frac{1}{x} \cdot x^{3}-\ln x \cdot 3 x^{2}}{\ln 2 \cdot x^{6}}= \\
& =\frac{1-3 \ln x}{\ln 2 \cdot x^{4}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\lef... | \frac{47-60\lnx}{\ln2\cdotx^{6}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,077 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=t+\sin t \\
y=2-\cos t
\end{array}\right.
$$ | ## Solution
$y_{x}^{\prime}=\frac{y^{\prime}(t)}{x^{\prime}(t)}$
$y_{x^{2}}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x^{\prime}(t)}$
$y^{\prime}(t)=\sin t$.
$x^{\prime}(t)=1+\cos t$.
$y_{x}^{\prime}=\frac{\sin t}{1+\cos t}$
$y_{x^{2}}^{\prime \prime}=\left(\frac{\sin t}{1+\cos t}\right)^{\p... | \frac{1}{(1+\cos)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,078 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
$y=x \cdot \sqrt{1-x^{2}}$
$y \cdot y^{\prime}=x-2 x^{3}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(x \cdot \sqrt{1-x^{2}}\right)^{\prime}=\sqrt{1-x^{2}}+x \cdot \frac{1}{2 \sqrt{1-x^{2}}} \cdot(-2 x)= \\
& =\sqrt{1-x^{2}}-\frac{x^{2}}{\sqrt{1-x^{2}}}
\end{aligned}
$$
Substitute into equation (1):
$$
x \cdot \sqrt{1-x^{2}} \cdot\left(\sqrt{1-x^{2}}-\frac{x^{2}}{\sq... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,079 |
## Task Condition
Based on the definition of the derivative, find $f^{\prime}(0)$ :
$$
f(x)=\left\{\begin{array}{c}
\sin x \cdot \cos \frac{5}{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... | f^{\}(0) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,080 |
## Task Condition
Compose the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=\frac{x^{2}-3 x+6}{x^{2}}, x_{0}=3$ | ## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=\left(\frac{x^{2}-3 x+6}{x^{2}}\right)^{\prime}=\left(1-\frac{3}{x}+\frac{6}{x^{2}}\right)^{\prime}=\frac{3}{x^{2}}-2 \cdot \frac{6}{x^{3}}=\frac{3}{x^{2}}-\frac{12}{x^{3}}
$$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{3}{x_{0}^{2}}-\frac{12}{x_{0}^{3... | 9x-26\frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,081 |
## Condition of the problem
Find the differential $d y$.
$$
y=\ln \left(x+\sqrt{1+x^{2}}\right)-\sqrt{1+x^{2}} \operatorname{arctg} x
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\ln \left(x+\sqrt{1+x^{2}}\right)-\sqrt{1+x^{2}} \operatorname{arctg} x\right)^{\prime} d x= \\
& =\left(\frac{1}{x+\sqrt{1+x^{2}}} \cdot\left(1+\frac{1}{2 \sqrt{1+x^{2}}} \cdot 2 x\right)-\left(\frac{1}{2 \sqrt{1+x^{2}}} \cdot 2 x \cdot \operatorname{arc... | -\frac{1}{\sqrt{1+x^{2}}}\cdot\operatorname{arctg}x\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,082 |
## Task Condition
Approximately calculate using the differential.
$$
y=\sqrt[3]{x}, x=1,21
$$ | ## Solution
If the increment $\Delta x = x - x_{\text{argument}}: x$ is small in absolute value, then
$$
f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x
$$
Choose:
$x_{0} = 1$
Then:
$\Delta x = 0.21$
## Calculation:
$$
y(1) = \sqrt[3]{1} = 1
$$
$y... | 1.07 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,083 |
## Task Condition
Find the derivative.
$y=\ln \operatorname{tg}\left(\frac{\pi}{4}+\frac{x}{2}\right)$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)^{\prime}=\frac{1}{\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot \frac{1}{\cos ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot \frac{1}{2}= \\
& =\frac{1}{2 \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \cdot \co... | \frac{1}{\cosx} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,086 |
## Task Condition
Find the derivative.
$$
y=\sqrt[3]{\operatorname{ctg} 2}-\frac{1}{20} \cdot \frac{\cos ^{2} 10 x}{\sin 20 x}
$$ | ## Solution
$y^{\prime}=\left(\sqrt[3]{\operatorname{ctg} 2}-\frac{1}{20} \cdot \frac{\cos ^{2} 10 x}{\sin 20 x}\right)^{\prime}=0-\frac{1}{20} \cdot\left(\frac{\cos ^{2} 10 x}{\sin 20 x}\right)^{\prime}=$
$=-\frac{1}{20} \cdot\left(\frac{\cos ^{2} 10 x}{2 \sin 10 x \cdot \cos 10 x}\right)^{\prime}=-\frac{1}{40} \cdo... | \frac{1}{4\sin^{2}10x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,087 |
## Problem Statement
Find the derivative.
$$
y=\frac{x^{3}}{3} \cdot \arccos x-\frac{2+x^{2}}{9} \cdot \sqrt{1-x^{2}}
$$ | ## Solution
$y^{\prime}=\left(\frac{x^{3}}{3} \cdot \arccos x-\frac{2+x^{2}}{9} \cdot \sqrt{1-x^{2}}\right)^{\prime}=$
$=x^{2} \cdot \arccos x+\frac{x^{3}}{3} \cdot \frac{-1}{\sqrt{1-x^{2}}}-\frac{2 x}{9} \cdot \sqrt{1-x^{2}}-\frac{2+x^{2}}{9} \cdot \frac{1}{2 \sqrt{1-x^{2}}} \cdot(-2 x)=$ $=x^{2} \cdot \arccos x-\fra... | x^{2}\cdot\arccosx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,088 |
## problem statement
Find the derivative.
$$
y=\frac{1}{6} \ln \frac{1-\operatorname{sh} 2 x}{2+\operatorname{sh} 2 x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{6} \ln \frac{1-\operatorname{sh} 2 x}{2+\operatorname{sh} 2 x}\right)^{\prime}= \\
& =\frac{1}{6} \cdot \frac{2+\operatorname{sh} 2 x}{1-\operatorname{sh} 2 x} \cdot \frac{-\operatorname{ch} 2 x \cdot 2 \cdot(2+\operatorname{sh} 2 x)-(1-\operatorname{sh} 2 x) ... | \frac{\operatorname{ch}2x}{\operatorname{sh}^{2}2x+\operatorname{sh}2x-2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,089 |
## Task Condition
Find the derivative.
$$
y=(\cos 5 x)^{e^{x}}
$$ | ## Solution
$$
\begin{aligned}
& y=(\cos 5 x')^{e^{e^{x}}} \\
& \ln y=e^{x} \cdot \ln (\cos 5 x) \\
& \frac{y'}{y}=e^{x} \cdot \ln (\cos 5 x)+e^{x} \cdot \frac{1}{\cos 5 x} \cdot(-\sin 5 x)= \\
& =e^{x} \cdot \ln (\cos 5 x)-e^{x} \cdot \tan 5 x=e^{x} \cdot(\ln (\cos 5 x)-\tan 5 x) \\
& y'=y \cdot e^{x} \cdot(\ln (\cos... | (\cos5x)^{e^{x}}\cdote^{x}\cdot(\ln(\cos5x)-\tan5x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,090 |
## Problem Statement
Find the derivative.
$$
y=\ln \frac{1+2 \sqrt{-x-x^{2}}}{2 x+1}+\frac{4}{2 x+1} \sqrt{-x-x^{2}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \frac{1+2 \sqrt{-x-x^{2}}}{2 x+1}+\frac{4}{2 x+1} \sqrt{-x-x^{2}}\right)^{\prime}= \\
& =\frac{2 x+1}{1+2 \sqrt{-x-x^{2}}} \cdot \frac{\frac{1}{\sqrt{-x-x^{2}}} \cdot(-1-2 x) \cdot(2 x+1)-\left(1+2 \sqrt{-x-x^{2}}\right) \cdot 2}{(2 x+1)^{2}}+ \\
& +\frac{4 \cdot \... | -\frac{2x+3}{\sqrt{-x-x^{2}}\cdot(2x+1)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,091 |
## Problem Statement
Find the derivative.
$$
y=\sqrt{1-3 x-2 x^{2}}+\frac{3}{2 \sqrt{2}} \arcsin \frac{4 x+3}{\sqrt{17}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sqrt{1-3 x-2 x^{2}}+\frac{3}{2 \sqrt{2}} \arcsin \frac{4 x+3}{\sqrt{17}}\right)^{\prime}= \\
& =\frac{1}{2 \sqrt{1-3 x-2 x^{2}}} \cdot(-3-4 x)+\frac{3}{2 \sqrt{2}} \cdot \frac{1}{\sqrt{1-\left(\frac{4 x+3}{\sqrt{17}}\right)^{2}}} \cdot \frac{4}{\sqrt{17}}= \\
& =\frac... | -\frac{2x}{\sqrt{1-3x-2x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,092 |
## Task Condition
Find the derivative.
$$
y=-\frac{1}{3 \sin ^{3} x}-\frac{1}{\sin x}+\frac{1}{2} \ln \frac{1+\sin x}{1-\sin x}
$$ | ## Solution
$y^{\prime}=\left(-\frac{1}{3 \sin ^{3} x}-\frac{1}{\sin x}+\frac{1}{2} \ln \frac{1+\sin x}{1-\sin x}\right)^{\prime}=$
$=-\frac{-3}{3 \sin ^{4} x} \cdot \cos x-\frac{-1}{\sin ^{2} x} \cdot \cos x+$
$+\frac{1}{2} \cdot \frac{1-\sin x}{1+\sin x} \cdot \frac{\cos x \cdot(1-\sin x)-(1+\sin x) \cdot(-\cos x)... | \frac{1}{\cosx\cdot\sin^{4}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,093 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\ln \sqrt{\frac{1-t}{1+t}} \\
y=\sqrt{1-t^{2}}
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\ln \sqrt{\frac{1-t}{1+t}}\right)^{\prime}=\sqrt{\frac{1+t}{1-t}} \cdot \frac{1}{2 \sqrt{\frac{1-t}{1+t}}} \cdot \frac{-1 \cdot(1+t)-(1-t) \cdot 1}{(1+t)^{2}}= \\
& =\sqrt{\frac{1+t}{1-t}} \cdot \frac{1}{2} \cdot \sqrt{\frac{1+t}{1-t}} \cdot \frac{-2}{(1+t)^{2}}=-\... | \cdot\sqrt{1-^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,094 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{\begin{array}{l}
x=\frac{1}{2} \cdot t^{2}-\frac{1}{4} \cdot t^{4} \\
y=\frac{1}{2} \cdot t^{2}+\frac{1}{3} \cdot t^{3}
\end{array}\right.
\]
$t_{0}=0$ | ## Solution
Since $t_{0}=0$, then
$$
\begin{aligned}
& x_{0}=\frac{1}{2} \cdot 0^{2}-\frac{1}{4} \cdot 0^{4}=0 \\
& y_{0}=\frac{1}{2} \cdot 0^{2}+\frac{1}{3} \cdot 0^{3}=0
\end{aligned}
$$
Find the derivatives:
$x_{t}^{\prime}=\left(\frac{1}{2} \cdot t^{2}-\frac{1}{4} \cdot t^{4}\right)^{\prime}=t-t^{3}$
$y_{t}^{\p... | x-x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,095 |
Condition of the problem
Find the $n$-th order derivative.
$y=\frac{2 x+5}{13(3 x+1)}$ | ## Solution
$y=\frac{2 x+5}{13(3 x+1)}$
$y^{\prime}=\left(\frac{2 x+5}{13(3 x+1)}\right)^{\prime}=\frac{2 \cdot(3 x+1)-(2 x+5) \cdot 3}{13(3 x+1)^{2}}=$
$=\frac{6 x+2-6 x-15}{13(3 x+1)^{2}}=-\frac{13}{13(3 x+1)^{2}}=-\frac{1}{(3 x+1)^{2}}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-\frac{1}{(3 x+1)^... | y^{(n)}=\frac{(-1)^{n}\cdotn!\cdot3^{n-1}}{(3x+1)^{n+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,096 |
Condition of the problem
Find the derivative of the specified order.
$y=\left(1+x^{2}\right) \operatorname{arctg} x, y^{\prime \prime \prime}=?$ | ## Solution
$y^{\prime}=\left(\left(1+x^{2}\right) \operatorname{arctg} x\right)^{\prime}=2 x \cdot \operatorname{arctg} x+\left(1+x^{2}\right) \cdot \frac{1}{1+x^{2}}=$
$=2 x \cdot \operatorname{arctg} x+1$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=(2 x \cdot \operatorname{arctg} x+1)^{\prime}=2 \operator... | \frac{4}{(1+x^{2})^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,097 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\sqrt{t-1} \\
y=\frac{t}{\sqrt{1-t}}
\end{array}\right.
$$ | ## 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 | 46,098 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
\[
\begin{aligned}
& y=x(c-\ln x) \\
& (x-y) d x+x \cdot d y=0
\end{aligned}
\] | ## Solution
$y^{\prime}=(x(c-\ln x))^{\prime}=(c-\ln x)+x\left(-\frac{1}{x}\right)=c-\ln x-1$
Equation (1):
$(x-y) d x+x \cdot d y=0$
$x \cdot d y=(y-x) d x$
$x \cdot \frac{d y}{d x}=y-x$
$x \cdot y^{\prime}=y-x$
Substitute $y_{\text {and }} y^{\prime}$ :
$x \cdot(c-\ln x-1)=x(c-\ln x)-x$
Simplify:
$x \cdot(c-... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,099 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{1-2 n^{2}}{2+4 n^{2}}, a=-\frac{1}{2}$ | ## Solution
By the definition of the limit:
$$
\begin{aligned}
& \forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{2-4 n^{2}+2+4 n^{2}}{2\left(2+4 n^{2}\right)}\right| \\
& \left|\frac{4}{2\left(2+4 n^{2}\right)}\right| \\
& \left|\fra... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,100 |
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