R1 2009 vår LØSNING

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)

1)<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>2x^2+10x+12 = 0 \\ x= \frac {-10 \pm \sqrt{100-96}}{4} = \frac{-10 \pm 2}{4} \\ x = -3 \vee x= -2 \\ f(x)= 2x^3+8x^2+2x-12 = (x-1)(x+2)(x+3)</tex>

2)

<tex>f(x) \leq 0 \\ x \in <\leftarrow, -3] \cup [-2,1]</tex>

f)

<tex> lg(\frac{1}{a^2}) + 3lga = lg1 - lga^2+3lga = -2lga+3lga = lga</tex>

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>

Del 2

b)

<tex>(lnx)^2+lnx^2 =3 \\ (lnx)^2+2lnx-3=0 \\ u = lnx \\ u^2+2u-3 =0 \\ u = \frac{-2 \pm \sqrt{4+12}}{2} \\ u = -3 \quad \vee\quad u = 1 \\ lnx=-3 \quad\vee\quad lnx = 1 \\x = e^{-3} \quad\vee\quad x = e</tex>

c)

1)

<tex>P(F) = P(A)\cdot P(F|A) + P(B) \cdot P(F|B) = 0,7 \cdot 0,05 + 0,3 \cdot 0,1 = 0,065</tex>

2)

<tex>P(A|F) = \frac{P(A\cap B)}{P(B)} = \frac {0,7 \cdot 0,05}{0,065} = 0,54</tex>

a)

<tex>f(x) = -x^3+ax^2+bx-11 \\ f'(x) = -3x^2+2ax+b \\ f'(x)=0 \Rightarrow -3-2a+b = 0 \Rightarrow b = 2a+3 \\ f(-1)= -16 \Rightarrow 1+a-b-11 = - 16 \Rightarrow a-b =-6 \\ a- (2a+3) = -6 \\ a =3 \\ b= 2a+3 = 2 \cdot 3 +3 =9 </tex>

b)

<tex>f(x)= -x^3+3x^2+9x-11 \\ f'(x) = -3x^2+6x+9 \\ f'(x)=0 \Rightarrow x=-1 \quad \vee \quad x=3 </tex>