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Solved 1st exercise
2 files changed, 23 insertions, 2 deletions
| -rw-r--r-- | Year 1/Mathematical analysis/Homework1.pdf | bin | 34958 -> 40673 bytes | |||
| -rw-r--r-- | Year 1/Mathematical analysis/Homework1.tex | 25 |
diff --git a/Year 1/Mathematical analysis/Homework1.pdf b/Year 1/Mathematical analysis/Homework1.pdf Binary files differindex 90bbe9a..e1a3345 100644 --- a/Year 1/Mathematical analysis/Homework1.pdf +++ b/Year 1/Mathematical analysis/Homework1.pdf diff --git a/Year 1/Mathematical analysis/Homework1.tex b/Year 1/Mathematical analysis/Homework1.tex index cb9c7cb..4514496 100644 --- a/Year 1/Mathematical analysis/Homework1.tex +++ b/Year 1/Mathematical analysis/Homework1.tex @@ -13,6 +13,7 @@ \usepackage[bulgarian,english]{babel} \usepackage{amsmath} % Математически символи +\usepackage{amsfonts} \usepackage{hyperref} % Линкове \usepackage{cancel} % Зачертаване \usepackage[dvipsnames]{xcolor} % Цветни кутийки + допълнителни цветове @@ -31,12 +32,32 @@ на обратимост, както и нейната обратна \begin{equation*} - f(x) = \log_{5} (x^2 - 2x + 1) + f(x) = \log_{5} \left( x^2 - 2x + 1 \right) \end{equation*} \paragraph{Решение:} -[TODO] +\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 \\ + & \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}{$f^{-1}: \mathbb{R} \rightarrow \left( -\infty, 1 \right) \cup \left( 1, +\infty \right) \qquad f^{-1}\left( y \right) = 1 \pm \sqrt{5^y}$} +\end{align*} \hfill |