R1 2009 vår LØSNING: Forskjell mellom sideversjoner
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<p></p> <tex>\quad \quad 2x^3+8x^2+2x-12: (x-1) = 2x^2+10x+12 \\ -(2x^3 -2x^2) \\ \quad \quad\quad\quad \quad\quad \quad 10x^2+2x \\ \quad\quad \quad \quad -(10x^2-10x) \\ \quad \quad\quad \quad\quad \quad\quad \quad\quad\quad\quad\quad\quad 12x-12\\ \quad \quad\quad \quad\quad \quad\quad \quad\quad\quad\quad -(12x-12) \\\quad \quad\quad \quad \quad \quad \quad\quad \quad \quad \quad\quad \quad\quad \quad\quad \quad\quad\quad \quad \quad 0 </tex> | <p></p> <tex>\quad \quad 2x^3+8x^2+2x-12: (x-1) = 2x^2+10x+12 \\ -(2x^3 -2x^2) \\ \quad \quad\quad\quad \quad\quad \quad 10x^2+2x \\ \quad\quad \quad \quad -(10x^2-10x) \\ \quad \quad\quad \quad\quad \quad\quad \quad\quad\quad\quad\quad\quad 12x-12\\ \quad \quad\quad \quad\quad \quad\quad \quad\quad\quad\quad -(12x-12) \\\quad \quad\quad \quad \quad \quad \quad\quad \quad \quad \quad\quad \quad\quad \quad\quad \quad\quad\quad \quad \quad 0 </tex> | ||
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<tex>x^2-4x+3 = 0 \\ x= \frac {4 \pm \sqrt{16-12}}{2} = \frac{4 \pm 2}{2} \\ x = 1 \vee x=3 \\ f(x)= x^3-3x^2-x+3 = (x+1)(x-1)(x-3)</tex><p></p> | <tex>x^2-4x+3 = 0 \\ x= \frac {4 \pm \sqrt{16-12}}{2} = \frac{4 \pm 2}{2} \\ x = 1 \vee x=3 \\ f(x)= x^3-3x^2-x+3 = (x+1)(x-1)(x-3)</tex><p></p> | ||
Sideversjonen fra 25. sep. 2012 kl. 05:16
Del 1
Oppgave 1
a)
1)
<tex>f(x) = (x^2+1)^4 \\ f'(x)= 4(x^2+1)^3 \cdot 2x = 8x(x^2+1)^3</tex>
(kjerneregelen)
2)
<tex>g(x) = xe^{2x} \\ g'(x)= e^{2x}+xe^{2x} \cdot 2 = e^{2x}(1+2x)</tex>
(produktregelen)
b)
<tex>\lim_{x \to 2} \frac{x^2-2x}{x-2} =\lim_{x \to 2} \frac{x(x-2)}{x-2}=\lim_{x \to 2} x=2</tex>
c)
<tex> \frac{x-2}{x^2+2x}- \frac{x+2}{x^2-2x}-\frac{4x}{x^2-4} = \\ \frac{x-2}{x(x+2)}- \frac{x+2}{x(x-2)} - \frac{4x}{(x+2)(x-2)} = \\ \frac{(x-2)(x-2)-(x+2)(x+2)- 4x^2}{x(x+2)(x-2)} = \\ \frac{x^2-4x+4-(x^2+4x+4)- 4x^2}{x(x+2)(x-2)} = \\ \frac{- 4x}{(x+2)(x-2)} </tex>
d)
<tex> \vec{AB} = [5-(-2), 4-(-1)] = [7,5] \\ \vec{AC} = [4-(-2), 7-(-1)]= [6,8] \\ \vec{BC} = [4-5, 7-4] =[-1,3] </tex>
Dersom to vektorer står vinkelrett på hverandre er skalarproduktet lik null.
Det er ikke tilfelle her.
e)
<tex>f(x)= 2x^3+8x^2+2x-12 \\
f(-1) = 2(1)^3 +8 (1)^2 +2(1)-12 = 2+8+2-12 = 0 \quad</tex>
dvs.f(x) er delelig med (x-1)
<tex>\quad \quad 2x^3+8x^2+2x-12: (x-1) = 2x^2+10x+12 \\ -(2x^3 -2x^2) \\ \quad \quad\quad\quad \quad\quad \quad 10x^2+2x \\ \quad\quad \quad \quad -(10x^2-10x) \\ \quad \quad\quad \quad\quad \quad\quad \quad\quad\quad\quad\quad\quad 12x-12\\ \quad \quad\quad \quad\quad \quad\quad \quad\quad\quad\quad -(12x-12) \\\quad \quad\quad \quad \quad \quad \quad\quad \quad \quad \quad\quad \quad\quad \quad\quad \quad\quad\quad \quad \quad 0 </tex>
<tex>x^2-4x+3 = 0 \\ x= \frac {4 \pm \sqrt{16-12}}{2} = \frac{4 \pm 2}{2} \\ x = 1 \vee x=3 \\ f(x)= x^3-3x^2-x+3 = (x+1)(x-1)(x-3)</tex>
f)
<tex> lg(\frac{1}{a^2}) + 3lga = lg1 - lga^2+3lga = -2lga+3lga = lga</tex>
Oppgave 2
a)
<tex>\bigtriangleup ABC \sim \bigtriangleup ADC </tex>
Fordi vinkel A er den samme i begge trekanter og vinkel C (i ABC) er lik vinkel D (i ADC).
<tex>\bigtriangleup ABC \sim \bigtriangleup BCD </tex>
Fordi vinkel B er den samme i begge trekanter og vinkel C (i ABC) er lik vinkel D (i BCD).
b)
<tex> \frac{AC}{AB} = \frac{AD}{AC} \\ (AC)^2 = AD \cdot AB </tex>
<tex> \frac{BC}{AB} = \frac{BD}{BC} \\ (BC)^2 = BD \cdot AB</tex>
c)
<tex> (AC)^2 = AD \cdot AB \\ (BC)^2 = BD \cdot AB \\ \text{legger sammen likningene} \\ (AC)^2 + (BC)^2 = AB \cdot (AD + DB) \\ (AC)^2 + (BC)^2 = (AB)^2</tex>