R2 2014 vår LØSNING: Forskjell mellom sideversjoner
Linje 60: | Linje 60: | ||
a) $\displaystyle |k| < 1 \Rightarrow |1 - x| < 1 \Rightarrow 0 < x < 2$ | a) $\displaystyle |k| < 1 \Rightarrow |1 - x| < 1 \Rightarrow 0 < x < 2$ | ||
b) $\displaystyle \begin{align*} s(x) & = 3 \\ | b) | ||
$\displaystyle \begin{align*} s(x) & = 3 \\ | |||
1 + \left(1 - x\right) + \left(1 - x\right)^2 + \left(1 - x\right)^3 + ... & = 3 \\ | 1 + \left(1 - x\right) + \left(1 - x\right)^2 + \left(1 - x\right)^3 + ... & = 3 \\ | ||
\frac{1}{1 - \left(1 - x\right)} & = 3 \\ | \frac{1}{1 - \left(1 - x\right)} & = 3 \\ |
Sideversjonen fra 19. mai 2014 kl. 21:52
DEL 1
Oppgave 1
a) $\displaystyle f(x) = \sin(3x)$
$\displaystyle f'(x) = 3\cos(3x)$
b) $\displaystyle g(x) = e^{2x} \cdot \cos x$
$\displaystyle g'(x) = 2e^{2x} \cdot \cos x + e^{2x} \cdot (-\sin x) = e^{2x} (2\cos x - \sin x)$
Oppgave 2
a) $\displaystyle \int 2x \cdot \sin (x^2) \, \mathrm{d}x$
La $\displaystyle u = x^2$
$\displaystyle \begin{align*} & \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = 2x \\ & \Rightarrow \mathrm{d}u = 2x \space \mathrm{d}x \end{align*}$
$\displaystyle \int 2x \cdot \sin (x^2) \, \mathrm{d}x = \int \sin u \, \mathrm{d}u = -\cos u + C = -\cos (x^2) + C$
b) $\displaystyle \int_1^{e} x \cdot \ln x \, \mathrm{d}x$
La $\displaystyle u = \ln x$ og $\displaystyle v' = x$:
$\displaystyle \begin{align*} \int_1^{e} x \cdot \ln x \, \mathrm{d}x & = \left[ \ln x \cdot \frac{1}{2} x^2 - \int \frac{1}{x} \cdot \frac{1}{2} x^2 \right]_1^{e} \\ & = \left[ \frac{1}{2} x^2 \cdot \ln x - \frac{1}{2} \int x \, \mathrm{d}x \right]_1^{e} \\ & = \left[ \frac{1}{2} x^2 \cdot \ln x - \frac{1}{2} \cdot \frac{1}{2} x^2 \right]_1^{e} \\ & = \frac{1}{2} \left[ x^2 \cdot \ln x - \frac{1}{2} x^2 \right]_1^{e} \\ & = \frac{1}{2} \left( (e^2 \cdot \ln e - \frac{1}{2} \cdot e^2) - (1^2 \cdot \ln1 - \frac{1}{2} \cdot 1^2) \right) \\ & = \frac{1}{2} \left( (e^2 - \frac{1}{2} \cdot e^2) - (0 - \frac{1}{2}) \right) \\ & = \frac{1}{2} \left( \frac{e^2}{2} + \frac{1}{2} \right) \\ & = \frac{1}{2} \cdot \frac{e^2 + 1}{2} \\ & = \frac{e^2 + 1}{4} \end{align*}$
Oppgave 3
$\displaystyle f(x) = e^{2x} - 4e^x \space , \space D_f = \R$
$\displaystyle f'(x) = 2e^{2x} - 4e^x$
$\displaystyle f ' ' (x) = 4e^{2x} - 4e^x$
$\displaystyle \begin{align*} f ' ' (x) & = 0 \\ 4e^{2x} - 4e^x & = 0 \\ 4\left(e^x\right)^2 - 4e^x & = 0 \\ 4e^x\left(e^x - 1\right) & = 0 \\ e^x - 1 & = 0 \\ e^x & = 1 \\ x & = 0 \end{align*}$
Vendepunkt: $\displaystyle \left( 0 \space , \space f(0)\right) = \left( 0 \space , \space e^{2 \cdot 0} - 4e^0\right) = \left( 0 \space , \space 1 - 3 \right) = \left( 0 \space , \space -3\right)$
Oppgave 4
$\displaystyle s(x) = 1 + \left(1 - x\right) + \left(1 - x\right)^2 + \left(1 - x\right)^3 + ...$
a) $\displaystyle |k| < 1 \Rightarrow |1 - x| < 1 \Rightarrow 0 < x < 2$
b)
$\displaystyle \begin{align*} s(x) & = 3 \\ 1 + \left(1 - x\right) + \left(1 - x\right)^2 + \left(1 - x\right)^3 + ... & = 3 \\ \frac{1}{1 - \left(1 - x\right)} & = 3 \\ \frac{1}{x} & = 3 \\ 1 & = 3x \\ x & = \frac{1}{3}\end{align*}$
$\displaystyle \begin{align*} s(x) & = \frac{1}{3} \\ 1 + \left(1 - x\right) + \left(1 - x\right)^2 + \left(1 - x\right)^3 + ... & = \frac{1}{3} \\ \frac{1}{x} & = \frac{1}{3} \end{align*}$
$\displaystyle x ≠ 3$ ettersom denne verdien ligger utenfor rekkens konvergensområde. Likningen har ingen løsning.
Oppgave 5
$\displaystyle \alpha$: $\displaystyle 2x + y - 2z + 3 = 0$
a) Punktet $\displaystyle P(3,4,2)$ ligger ikke i planet $\displaystyle \alpha$ kun dersom punktets koordinater ikke tilfredstiller likningen til planet.
$\displaystyle 2\left(3\right) + \left(4\right) - 2\left(2\right) + 3 = 6 + 4 - 4 = 6 ≠ 0 \Leftrightarrow$ punktet $\displaystyle P(3,4,2)$ ligger ikke i planet $\displaystyle\alpha$.
Hvilket skulle vises.
b) $\displaystyle l \perp \alpha \Leftrightarrow \vec{r}_{l} = \vec{n}_{\alpha}$
$\displaystyle\vec{r}_{l} = [2,1,-2]$
$\displaystyle \Rightarrow l$: $\displaystyle \begin{align*} x & = 3 + 2t \\ y & = 4 + t \\ z & = 2 - 2t\end{align*}$
c)
$\displaystyle \begin{align*} 2\left( 3 + 2t \right) + \left(4 + t\right) - 2\left( 2 - 2t\right) + 3 & = 0 \\ 6 + 4t + 4 + t - 4 + 4t + 3 & = 0 \\ 9 + 9t & = 0 \\ 9t & = -9 \\ t & = -1\end{align*}$
Skjæringspunkt $\displaystyle = \left( 3 + 2\left( -1\right), 4 + \left( -1\right), 2 - 2\left(-1\right)\right) = \left(1,3,4\right)$
d) $\displaystyle D = \frac{|2\cdot 3 + 1 \cdot 4 - 2 \cdot 2 + 3|}{\sqrt{2^2 + 1^2 + \left(-2\right)^2}} = \frac{|6 + 4 - 4 + 3|}{\sqrt{4 + 1 + 4}} = \frac{9}{\sqrt{9}} = \frac{9}{3} = 3$