Del 1
Oppgave 1
[tex] a)\quad 1)\;\quad f^{\tiny\prime}\left( t \right) = 0.06{t^2} + 1.2t[/tex]
[tex] {\rm{ }}\quad 2)\quad g^{\tiny\prime}\left( x \right) = \frac{{\left( {{x^2} - 1} \right)^{\tiny\prime}}}{{2\sqrt {{x^2} - 1} }} = \frac{x}{{\sqrt {{x^2} - 1} }} [/tex]
[tex] {\rm{ }}\quad 3)\quad h^{\tiny\prime}\left( x \right) = 2x \cdot {e^{2x}} + {x^2} \cdot 2 \cdot {e^{2x}} = 2x{e^{2x}}\left( {x + 1} \right) [/tex]
[tex] b){\rm{ }}1)\quad P\left( 2 \right) = {2^3} - 4 \cdot {2^3} - 4 \cdot 2 + {2^4} = {2^3} - {2^4} - {2^3} + {2^4} = 0 [/tex]
[tex]\quad \;\;\;2)\quad P\left( x \right) = {x^2}\left( {x - 4} \right) - 4\left( {x - 4} \right) = \left( {{x^2} - 4} \right)\left( {x - 4} \right) = \left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 4} \right) [/tex]
[tex] \quad \;\;\;3)\quad 2 \le x \le 4 \wedge x \le - 2 \Leftrightarrow x \in \left( { - \infty ,\left. { - 2} \right]} \right. \cap \left[ {2,4} \right] [/tex]
[tex] c)\quad x = \frac{{\log \left( {y - a} \right)}}{{\log \left( b \right)}}\quad ,\quad Siden{\rm{ }}{{\rm{b}}^x} \ge 0\forall x.{\rm{ om x < 0 s{\aa} har vi }}{{\rm{b}}^{ - x}} \quad = \frac{1}{{{b^x}}}{\rm{ som er > 0}} [/tex]
[tex] d){\rm{ }}1)\quad \vec {AB} = \left[ {2,4} \right]\qquad \vec {AC} = \left[ {1,t} \right] [/tex]
[tex] \quad \;\;\;{\rm{ }}2)\quad \angle A = 90^\circ \Leftrightarrow \vec {AB} \bot \vec {AC} \Leftrightarrow \vec{AB} \cdot \vec {AC} = 0 \Leftrightarrow 2 + 4t = 0 \Leftrightarrow t = - 1/2 [/tex]
[tex] \quad \;\;\;{\rm{ }}3)\quad \left| {\vec{AB} } \right| = \sqrt {{2^2} + {4^2}} = 2\sqrt 5 \quad ,\quad Midtpunkt{\rm{ AB = }}\left( {2,2} \right) [/tex]
[tex] \qquad \quad \;\;{\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\frac{{2\sqrt 5 }}{2}} \right)^2} = 5 [/tex]
[tex] e){\rm{ }}1)\quad {\rm{synker }}x < - 1{\rm{ }}eller{\rm{ }}x > 3\quad ,\quad stiger{\rm{ }} - 1 < x < 1 [/tex]
[tex] \quad \;\;\;2)\quad f^{\tiny\prime\prime}\left( a \right) > 0{\rm{ gir bunn}}{\rm{, f^{\prime\prime}}}\left( a \right) < 0{\rm{ gir topp alts\aa}} [/tex]
[tex] \qquad \;\;\;\quad \;{\rm{bunn}}\left( {x = - 1} \right){\rm{ }}topp\left( {x = 1} \right) [/tex]
[tex] \quad \;\;\;3) [/tex]
[tex] f){\rm{ }}1)\quad = {\lim }\limits_{\Delta x \to 0 } \left( \frac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}} = \frac{{\left[ {{{\left( {x + \Delta x} \right)}^2} + 1} \right] - \left[ {{x^2} + 1} \right]}}{{\Delta x}} \right. [/tex]
[tex] \left. \qquad \quad \; \; \; = \frac{{\left[ {\left( {{x^2} + 2x\Delta x + \Delta {x^2}} \right) + 1} \right] - \left[ {{x^2} + 1} \right]}}{{\Delta x}} = \frac{{\left[ {\left( {2x\Delta x + \Delta {x^2}} \right)} \right]}}{{\Delta x}} \right) = 2x [/tex]
[tex]g){\rm{ }}1)\;\quad \angle ADB = 30^\circ [/tex]
[tex] \quad \;\;\;2)\quad \angle DBE = 10^\circ \quad \quad [/tex]
[tex] \quad \; \; \;3)\quad \angle CDB = 180^\circl - ADB = 180^\circ - 30^\circ = 150^\circ [/tex]
[tex] \qquad \quad \angle ACB = 180^\circ - \angle CDB - \angle DBC = 180^\circ - 150^\circ - 10^\circ = 20^\circ [/tex]
Del 2
Oppgave 2
[tex] a)\qquad f\left( x \right) = {x^3} - 4{x^2} + 4x = x\left( {{x^2} - 2 \cdot 2x + {2^2}} \right) = x{\left( {x - 2} \right)^2} \to x = 0 \wedge x = 2{\rm{ husk at x}} \in \left[ { - 1,3} \right] [/tex]
[tex] b)\qquad {f^\prime }\left( x \right) = {\left( {x - 2} \right)^2} + 2x\left( {x - 2} \right) = \left( {x - 2} \right)\left( {\left( {x - 2} \right) + 2x} \right) = \left( {x - 2} \right)\left( {3x - 2} \right) [/tex]
[tex]\qquad \quad bunn\left( {2,0} \right){\rm{ topp}}\left( {\frac{2}{3},\frac{{32}}{{27}}} \right) [/tex]
[tex] c)\qquad [/tex] Drittkjedelig
[tex] {f^{\prime \prime }}\left( x \right) = 6x - 8 \Rightarrow 2\left( {3x - 4} \right).{\rm{ Vendepunkt }}\left( {\frac{4}{3},\frac{{16}}{{27}}} \right) [/tex]
[tex] d)\qquad y = {f^\prime }\left( a \right)\left( {x - a} \right) + f\left( a \right) = {f^\prime }\left( 1 \right)\left( {x - 1} \right) + f\left( 1 \right) = - 1\left( {x - 1} \right) + 1 = - x + 2 [/tex]
[tex] e)[/tex]
[tex] f)\qquad y = f\left( x \right) \Leftrightarrow - x + 2 = {x^3} - 4{x^2} + 4x \Rightarrow {x^3} - 4{x^2} + 5x - 2 = 0 [/tex]
[tex] \qquad \quad {x^3} - 4{x^2} + 5x - 2 = x\left( {{x^2} - 2x + 1} \right) - 2\left( {{x^2} - 2x + 1} \right) = x{\left( {x - 1} \right)^2} - 2{\left( {x - 1} \right)^2} = \left( {x - 2} \right){\left( {x - 1} \right)^2}[/tex]
[tex] Q\left( {2,0} \right) [/tex]
Oppgave 3
[tex]\qquad a) [/tex]
[tex] \qquad \quad 1){\rm{ }}\quad \;{\rm{er en likebenet trekant}}{\rm{. AH halverer vinkel A}}{\rm{, og er dermed h{\o}yden i trekant ADE}}{\rm{.}} [/tex]
[tex] \qquad \qquad \quad {\rm{Dette f{\o}rer til at }}AH{\rm{ st{\aa}r vinkelrett p{\aa} }}DE{\rm{ og }}DHA = 90^\circ .{\rm{ grunnet likhet s{\aa} er GHE = DHA}} [/tex]
[tex] \text{Vinkel FSD er 90 grader, grunnet rettvinklet trekant. Vinkel FED er en del av sirkelperiferien, og vil v{\ae}re halvparten av FSD.}[/tex]
[tex]\text{ Dermed er } FED=GED=45^{\circ}. \text{ Ved \aa bruke at vinkelsummen i en trekant alltid er } 180^{\circ} \text{f{\aa}r vi at } HGE=45^{\circ}[/tex]
Oppgave 4
[tex] a = {x^2} + {y^2} + 6x + 4y - 12 = 0 \Leftrightarrow \left( {{x^2} + 6x + 9} \right) + \left( {{y^2} + 4y + 4} \right) = 25 \Leftrightarrow {\left( {x + 3} \right)^2} + {\left( {y + 2} \right)^2} = {5^2} [/tex]
[tex] b = {x^2} + {y^2} - 6x - 12y + 20 = 0 \Leftrightarrow \left( {{x^2} - 6x + 9} \right) + \left( {{y^2} - 2 \cdot 6y + {6^2}} \right) = 25 \Leftrightarrow {\left( {x - 3} \right)^2} + {\left( {y - 6} \right)^2} = {5^2} [/tex]
[tex]P{\rm{ midtpunkt mellom sentrene P}}\left( {0,2} \right)\left[ {maple:{\rm{ solve}}\left( {\{ a,b\} ,\{ x,y\} } \right) \Rightarrow x = 0,y = 2} \right] [/tex]
[tex]Siden{\rm{ A + }}\left( {3,4} \right) = P{\rm{ og }}B = P + \left( {3,4} \right).{\rm{ Evnt bruke at det finnes en t}}{\rm{, slik at }}t \cdot \vec{AP} = \vec{AB} [/tex]
[tex] c)\qquad l:\left[ { - 3 + 3t, - 2 + 4t} \right] = \left[ {3t,2 + 4t} \right] [/tex]
[tex] d)\qquad {\left( {x - 3} \right)^2} + {\left( {y - 6} \right)^2} = {5^2} \Rightarrow {\left( {\left[ {3t} \right] - 3} \right)^2} + {\left( {\left[ {2 + 4t} \right] - 6} \right)^2} = {5^2} \Rightarrow 9{\left( {t - 1} \right)^2} + 16{\left( {t - 1} \right)^2} - {5^2} = 0 [/tex]
[tex] \qquad \qquad {\left( {t - 1} \right)^2}\left[ {9 + 16} \right] - {5^2} = 0 \Rightarrow 25\left[ {{{\left( {t - 1} \right)}^2} - {1^2}} \right] = 0 \Rightarrow \left( t \right)\left( {t - 2} \right) = 0 [/tex]
[tex] \qquad \qquad C = \left[ {3\left( 2 \right),2 + 4\left( 2 \right)} \right] = \left[ {6,10} \right] [/tex]
Oppgave 5
[tex] a)\qquad P\left( B \right) = \frac{{80+60}}{{120 + 80}} = \frac{8+6}{{12 + 8}} = \frac{4+3}{{6 + 4}} = \frac{7}{10} [/tex]
[tex] b)\qquad P\left( {B\mid J} \right) = \frac{{60}}{{120}} = \frac{6}{{12}} = \frac{1}{2} [/tex]
[tex] c)\qquad P\left( {J\mid B} \right) = \frac{{P\left( J \right)P\left( {B\mid J} \right)}}{{P\left( B \right)}} = \frac{{\frac{{120}}{{120 + 80}} \cdot \frac{1}{2}}}{{\frac{{80 + 60}}{{120 + 80}}}} = \frac{{\frac{3}{{3 + 2}} \cdot \frac{1}{2}}}{{\frac{{4 + 3}}{{6 + 4}}}} = \frac{3}{{10}}:\frac{7}{{10}} = \frac{3}{7} [/tex]
Oppgave 6
[tex] a)\quad 28 \: \S \: 1\,,\,2\,,\,4\,,\,7\,,\,14 \Rightarrow 1 + 2 + 4 + 7 + 14 = 28,\qquad \left( {6\,,\,28\,,\,496\,,\,8128} \right) [/tex]
[tex] b)\quad 284 \, = \, 4 \cdot 71 \: \S \: 1\,,\,2\,,\,4\,,\,71\,,\,142 \Rightarrow 1 + 2 + 4 + 71 + 142 = 220 [/tex]