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Added grade for dis hw1
1 files changed, 11 insertions, 8 deletions
| -rw-r--r-- | Year 1/Calculus/Homeworks/Homework1.tex | 19 |
diff --git a/Year 1/Calculus/Homeworks/Homework1.tex b/Year 1/Calculus/Homeworks/Homework1.tex index 351cd0a..9cf6aee 100644 --- a/Year 1/Calculus/Homeworks/Homework1.tex +++ b/Year 1/Calculus/Homeworks/Homework1.tex @@ -27,10 +27,13 @@ \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 +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) @@ -49,7 +52,7 @@ & \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 \\ + & \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} \\ @@ -57,14 +60,14 @@ & \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}{$f^{-1}: \mathbb{R} \rightarrow \left( -\infty, 1 \right) \cup \left( 1, +\infty \right) \qquad f^{-1}\left( y \right) = 1 \pm \sqrt{5^y}$} + & \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 +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] @@ -76,8 +79,8 @@ & \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 \left[ -\frac{\pi}{4}, \frac{\pi}{4} \right] \implies \sin \left( \beta \right) = +\sqrt{1 - \cos ^ 2 \left( \beta \right)} = \sqrt{1 - x ^ 2} \\ - & \omit\hfill \\ + & \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*} @@ -86,7 +89,7 @@ \paragraph{\colorbox{yellow}{Задача 3.}} -Пресметнете границата на редицата +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}} @@ -107,7 +110,7 @@ \paragraph{\colorbox{yellow}{Задача 4.}} -Пресметнете границата на редицата. +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} |