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% Document MUST be built with XeLaTeX!
\documentclass[11pt,a4paper]{article}
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\usepackage{amsmath} % Математически символи
\usepackage{amsfonts} % Още математически символи
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\usepackage{cancel} % Зачертаване
\usepackage[dvipsnames]{xcolor} % Цветни кутийки + допълнителни цветове
\title{\vspace{-2cm}Домашна работа 1 по ДИС 1, за спец. КН 1 курс, 2 поток. Вариант 2 за студентите с четна последна
цифра от факултетния номер.}
\author{Камен Димитров Младенов, Група 6}
\date{10 Ноември 2021}
\begin{document}
\maketitle
\noindent
\colorbox{YellowGreen}{\parbox{\textwidth}{\; Отличен 5,55/6 \qquad [Коментарите са под определените места]}}
\paragraph{\colorbox{yellow}{Задача 1.}}
Изследвайте функцията за обратимост. Намерете максимален интервал
на обратимост, както и нейната обратна \hfill \colorbox{Dandelion}{0,8/1,0 т.}
\begin{equation*}
f(x) = \log_{5} \left( x^2 - 2x + 1 \right)
\end{equation*}
\paragraph{Решение:}
\begin{align*}
& f(x) = \log_{5} \left( x^2 - 2x + 1 \right) = \log_{5} \left( \left(x - 1\right) ^ 2 \right) \quad \text{Д.С. } x \neq 1 \text{ (аргумент на логаритъм)} \\
& \omit\hfill \\
& \text{Разглеждаме за инективност. Нека } f\left( x_1 \right) = f\left( x_2 \right) \qquad x_1, x_2 \in \mathbb{R} \setminus \{-1\} \\
& \log_{5} \left( \left(x_1 - 1\right) ^ 2 \right) = \log_{5} \left( \left(x_2 - 1\right) ^ 2 \right) \iff \left(x_1 - 1\right) ^ 2 = \left(x_2 - 1\right) ^ 2 \iff |x_1 - 1| = |x_2 - 1| \\
& \textbf{I. Случай: } x_1 - 1 > 0 \text{ и } x_2 - 1 > 0 \;,\; x_1 > 1 \quad x_2 > 1 \;,\; x_1,x_2 \in \left( 1, +\infty \right) \\
& \implies |x_1 - 1| = x_1 - 1 \text{ и } |x_2 - 1| = x_2 - 1 \\
& \text{Тоест } |x_1 - 1| = |x_2 - 1| \iff x_1 - 1 = x_2 - 1 \implies x_1 = x_2 \; \colorbox{yellow}{има инекция} \\
& \textbf{II. Случай: } x_1 - 1 < 0 \text{ и } x_2 - 1 < 0 \;,\; x_1 < 1 \quad x_2 < 1 \;,\; x_1,x_2 \in \left( -\infty, 1 \right) \\
& \implies |x_1 - 1| = 1 - x_1\text{ и } |x_2 - 1| = 1 - x_2 \\
& \text{Тоест } |x_1 - 1| = |x_2 - 1| \iff 1 - x_1 = 1 - x_2 \implies x_1 = x_2 \; \colorbox{yellow}{има инекция} \\
& \omit\hfill \colorbox{RedOrange}{\text{А ако са в различни интервали?}} \\
& \text{Нека } y \in \mathbb{R} \text{. Тогава търсим x, такова че: } f(x) = y \\
& y = f(x) \quad y = \log_{5} \left( \left(x - 1\right) ^ 2 \right) \quad 5^y = \left(x - 1\right) ^ 2 \quad \sqrt{5^y} = |x - 1| \\
& \textbf{I. Случай: } x \in \left( 1, +\infty \right) \implies \sqrt{5^y} = x - 1 \quad x = 1 + \sqrt{5^y} \\
& x > 1 \quad 1 + \sqrt{5^y} > 1 \quad \sqrt{5^y} > 0 \quad \text{изпълнено за всяко y} \\
& \textbf{II. Случай: } x \in \left( -\infty, 1 \right) \implies \sqrt{5^y} = 1 - x \quad x = 1 - \sqrt{5^y} \\
& x < 1 \quad 1 - \sqrt{5^y} < 1 \quad - \sqrt{5^y} < 0 \quad \sqrt{5^y} > 0 \quad \text{изпълнено за всяко y} \\
& \omit\hfill \\
& \colorbox{yellow}{$f^{-1}: \mathbb{R} \rightarrow \left( -\infty, 1 \right) \cup \left( 1, +\infty \right) \qquad$} \colorbox{RedOrange}{$f^{-1}\left( y \right) = 1 \pm \sqrt{5^y}$}
\end{align*}
\hfill
\paragraph{\colorbox{yellow}{Задача 2.}}
Докажете тъждеството \hfill \colorbox{Dandelion}{0,75/1,0 т.}
\begin{equation*}
\arcsin \left( 2x \sqrt{1-x^2} \right) = \pi - 2 \arccos \left( x \right) \text{, където } x \in \left[ \frac{-\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right]
\end{equation*}
\paragraph{Решение:}
\begin{align*}
& \arcsin \left( 2x \sqrt{1-x^2} \right) = \alpha \iff \sin \left( \alpha \right) = 2x \sqrt{1-x^2} \qquad \arccos \left( x \right) = \beta \iff \cos \left( \beta \right) = x \\
& \text{За } x = \frac{-\sqrt{2}}{2} \;,\quad \sin(\alpha) = \cancel{2} \cdot \frac{-\sqrt{2}}{\cancel{2}} \cdot \sqrt{1 - \left( \frac{-\sqrt{2}}{2} \right)^2} = -\sqrt{2} \cdot \sqrt{\frac{1}{2}} = -1 \;,\; \cos \left( \beta \right) = -\frac{\sqrt{2}}{2} \\
& \text{Аналогично за } x = \frac{\sqrt{2}}{2} \;,\quad \sin \left( \alpha \right) = 1 \;,\quad \cos \left( \beta \right) = \frac{\sqrt{2}}{2} \\
& \implies \alpha \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \;,\; \beta \in \colorbox{RedOrange}{$\left[ -\frac{\pi}{4}, \frac{\pi}{4} \right]$} \implies \sin \left( \beta \right) = +\sqrt{1 - \cos ^ 2 \left( \beta \right)} = \sqrt{1 - x ^ 2} \\
& \qquad\qquad\qquad\qquad\qquad \colorbox{RedOrange}{Трябва да е $\left[ \frac{\pi}{4}, \frac{3\pi}{4} \right]$} \hfill \\
& \sin \left( \alpha \right) = 2x\sqrt{1 - x^2} = 2\cos \left( \beta \right)\sin \left( \beta \right) = \sin \left( 2\beta \right) = \sin \left( \pi - 2\beta \right) \qquad\quad 2\beta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \\
& \implies \colorbox{yellow}{$\alpha = \pi - 2\beta$}
\end{align*}
\hfill
\paragraph{\colorbox{yellow}{Задача 3.}}
Пресметнете границата на редицата \hfill \colorbox{YellowGreen}{1,0/1,0 т.}
\begin{equation*}
\lim_{n \to \infty} \frac{\sqrt[n-5]{n-5} + 2^{2n} + n^{12}}{10^n-n^3+(-1)^{3n + 5}}
\end{equation*}
\paragraph{Решение:}
\begin{align*}
& \lim_{n \to \infty} \frac{\sqrt[n-5]{n-5} + 2^{2n} + n^{12}}{10^n-n^3+(-1)^{3n + 5}} = \lim_{n \to \infty} \frac{2^{2n} \cdot \left( \frac{\sqrt[n-5]{n-5}}{2^{2n}} + 1 + \frac{n^{12}}{2^{2n}} \right)}{10^n \cdot \left(1 - \frac{n^3}{10^n} + \frac{(-1)^{3n + 5}}{10^n} \right)} = a_n \\
& \omit\hfill \\
& \lim_{n \to \infty} \frac{\sqrt[n-5]{n-5}}{2^{2n}} = \frac{\lim\limits_{n \to \infty} \sqrt[n-5]{n-5}}{ \lim\limits_{n \to \infty} 4^n} = \frac{1}{\infty} = 0 \;,\quad \lim_{n \to \infty} \frac{n^{12}}{2^{2n}} = 0 \;,\quad \lim_{n \to \infty} \frac{n^3}{10^n} = 0 \;,\; \\
& \lim_{n \to \infty} \frac{(-1)^{3n + 5}}{10^n} = \frac{\lim\limits_{n \to \infty} (-1)^{3n + 5}}{\lim\limits_{n \to \infty} 10^n} = \frac{\lim\limits_{n \to \infty} (-1)^{3n + 5}}{\infty} = 0 \quad \text{($(-1)^{3n + 5}$ е разходяща)} \\
& \omit\hfill \\
& \implies a_n = \lim_{n \to \infty} \frac{2^{2n}}{10^n} = \lim_{n \to \infty} \frac{4^n}{10^n} = \lim_{n \to \infty} \left( \frac{2}{5} \right) ^ n \qquad \frac{2}{5} < 1 \implies \lim_{n \to \infty} \left( \frac{2}{5} \right) ^ n = \colorbox{yellow}{0}
\end{align*}
\hfill
\paragraph{\colorbox{yellow}{Задача 4.}}
Пресметнете границата на редицата. \hfill \colorbox{YellowGreen}{1,0/1,0 т.}
\begin{equation*}
\lim_{n \to \infty} \left( \frac{n^2 - 5n + 14}{n^2 + 1} \right) ^ \frac{n^2-n+2}{n+5}
\end{equation*}
\paragraph{Решение:}
\begin{align*}
& \lim_{n \to \infty} \left( \frac{n^2 - 5n + 14}{n^2 + 1} \right) ^ \frac{n^2-n+2}{n+5} = \lim_{n \to \infty} \left( \frac{n^2 + 1 - 5n + 13}{n^2 + 1} \right) ^ \frac{n^2-n+2}{n+5} = \\
& = \lim_{n \to \infty} \left( 1 + \frac{13 - 5n}{n^2 + 1} \right) ^ \frac{n^2-n+2}{n+5} = \lim_{n \to \infty} \left( 1 + \frac{1}{\frac{n^2 + 1}{13 - 5n}} \right) ^ { \frac{n^2+1}{13-5n} \cdot \frac{13-5n}{n^2+1} \cdot \frac{n^2-n+2}{n+5} }
\end{align*}
От $\lim\limits_{n \to \infty} \left( 1 + \frac{1}{p_n} \right) ^ {p_n b_n} = e ^ {\lim_{n \to \infty} b_n}$, задачата се свежда до $e ^ {\lim_{n \to \infty} b_n} = \; ? \;$, където
$ \; b_n = \frac{13-5n}{n^2+1} \cdot \frac{n^2-n+2}{n+5}$
\begin{align*}
\lim_{n \to \infty} b_n &= \lim_{n \to \infty} \frac{13-5n}{n^2+1} \cdot \frac{n^2-n+2}{n+5} = \lim_{n \to \infty} \frac{13n^2 - 13n + 26 - 5n^3 + 5n^2 - 10n}{n^3 + 5n^2 + n + 5} = \\
& = \lim_{n \to \infty} \frac{-5n^3 + 18n^2 - 23n + 26}{n^3 + 5n^2 + n + 5} = \lim_{n \to \infty} \frac{\cancel{n^3} \cdot \left( -5 + \frac{18}{n} - \frac{23}{n^2} + \frac{26}{n^3} \right)}{\cancel{n^3} \cdot \left( 1 + \frac{5}{n} + \frac{1}{n^2} + \frac{5}{n^3} \right)} = \\
& = \frac{\lim\limits_{n \to \infty} -5 + \lim\limits_{n \to \infty} \frac{18}{n} - \lim\limits_{n \to \infty} \frac{23}{n^2} + \lim\limits_{n \to \infty} \frac{26}{n^3}}{\lim\limits_{n \to \infty} 1 + \lim\limits_{n \to \infty} \frac{5}{n} + \lim\limits_{n \to \infty} \frac{1}{n^2} + \lim\limits_{n \to \infty} \frac{5}{n^3}} = \frac{-5+0+0+0}{1+0+0+0} = -5 \\
& \implies e ^ {\lim_{n \to \infty} b_n} = e ^ {-5} = \colorbox{yellow}{$\frac{1}{e ^ 5}$}
\end{align*}
\end{document}
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