S1 2009 høst LØSNING

Løsning laget av Svein Arneson: pdf ,word

1)

$5a^2+(a-2)(a+2)-(2a+1)=5a^2+a^2-4-2a-1=6a^2-2a-5$

2)

$\frac{2(x-2)}{3x}+\frac{1}{2}=\frac{2}{2}\cdot \frac{2x-4}{3x}+\frac{3x}{6x}=\frac{7x-8}{6x}=\frac{7}{6}-\frac{4}{3x}$

3)

$\frac{a^4\cdot 2b^2}{(2a{)}^3}=\frac{a^4\cdot 2b^2}{2^3\cdot a^3}=\frac{a{b}^2}{4}$

4)

$lg(a^{2}b)-lg(\frac{1}{ab})=2lga+lgb-(lg1-lga-lgb)=2lga+lgb-lg1+lga+lgb=3lga+2lgb=lg(a^3{b}^2)$

1)

$\frac{x}{4}-\frac{1}{6}=\frac{x}{6}-(\frac{x}{2}-1)=>\frac{x}{4}-\frac{x}{6}+\frac{x}{2}=1+\frac{1}{6}$

$x(\frac{1}{4}-\frac{1}{6}+\frac{1}{2})=x\cdot \frac{7}{12}=\frac{7}{6}=>x=2$

2)

$2x^2-2(x+2)=20=>2x^2-2x-24=0$

$2(x^2-x-12)=0=>2(x-4)(x+3)=0=>x=4\vee x=-3$

c)

masse til skrue målt i gram: s. masse til mutter målt i gram:m

$3s+2m=57$

$s+3m=33|\cdot 3=>3s+9m=99$

$3s+9m-(3s+2m)=99-57=42=>7m=42=>m=6=>s=33-3\cdot 6=15$

En skrue veier 15 gram og en mutter veier 6 gram.

d)

$x^2-2x\le 3=>x^2-2x-3=(x+1)(x-3)\le 0$

$x\in [-1,3]$

a)

$f(x)=x^3-3x^2+2$

$f(-1)=-1-3+2=-2$

$f(0)=2$

$f(1)=1-3+2=0$

$f(2)=8-12+2=-2$

$f(3)=27-27+2=2$

b)

$f(x)=x^3-3x^2+2=>f^{\prime }(x)=3x^2-6x=3x(x-2)$

Ekstremalpunkter:

$f^{\prime }(x)=3x(x-2)=0=>x=0\vee x=2$

Toppunkt: $(0,f(0))=(0,2)$

Bunnpunkt: $(2,f(2))=(2,-2)$

c)

d)

Ved regning:

$\frac{f(3)-f(2)}{3-2}=2-(-2)=4$

Grafisk:

a)

${10}^{2x}-{10}^x-6=0=>({10}^x{)}^2-{10}^x-6=0$

Bruker ABC-formelen.

${10}^x=-2\vee {10}^x=3=>x=lg3$

$({10}^x=-2$ har ingen løsning.)

b)

$lg(2x-2)-lg(x+5)=0=>lg(2x-2)=lg(x+5)$

$2x-2=x+5=>x=7$