Matura (maj 2015, poziom podstawowy, (nowa formuła))

$A$A

$B$B

$C$C

$D$D

$-9$-9

$\displaystyle - \frac{1}{3}$\displaystyle - \frac{1}{3}

$\displaystyle \frac{1}{3}$\displaystyle \frac{1}{3}

$3$3

$\displaystyle 1000 \cdot \left(1- \frac{81}{100} \cdot \frac{4}{100} \right)$\displaystyle 1000 \cdot \left(1- \frac{81}{100} \cdot \frac{4}{100} \right)

$\displaystyle 1000 \cdot \left(1+ \frac{19}{100} \cdot \frac{4}{100} \right)$\displaystyle 1000 \cdot \left(1+ \frac{19}{100} \cdot \frac{4}{100} \right)

$\displaystyle 1000 \cdot \left(1+ \frac{81}{100} \cdot \frac{4}{100} \right)$\displaystyle 1000 \cdot \left(1+ \frac{81}{100} \cdot \frac{4}{100} \right)

$\displaystyle 1000 \cdot \left(1- \frac{19}{100} \cdot \frac{4}{100} \right)$\displaystyle 1000 \cdot \left(1- \frac{19}{100} \cdot \frac{4}{100} \right)

$m=5$m=5

$m=4$m=4

$m=1$m=1

$m=-5$m=-5

zbiór pusty

dokładnie jeden punkt

dokładnie dwa różne punkty

zbiór nieskończony

$-1$-1

$21$21

$1$1

$-21$-21

ma dokładnie jedno rozwiązanie: $x=1$x=1.

ma dokładnie jedno rozwiązanie: $x=0$x=0.

ma dokładnie jedno rozwiązanie: $x=-1$x=-1.

ma dokładnie dwa rozwiązania: $x=0,x=1$x=0,x=1.

$\left(-2,2\right)$\left(-2,2\right)

$[-2,2)$[-2,2)

$[-2,2]$[-2,2]

$(-2,2]$(-2,2]

$m=-1$m=-1

$m=0$m=0

$m=1$m=1

$m=2$m=2

$b=4$b=4

$\displaystyle b=- \frac{3}{2}$\displaystyle b=- \frac{3}{2}

$\displaystyle b=- \frac{8}{3}$\displaystyle b=- \frac{8}{3}

$\displaystyle b= \frac{4}{3}$\displaystyle b= \frac{4}{3}

$f(1)=-6$f(1)=-6

$f(1)=0$f(1)=0

$f(1)=6$f(1)=6

$f(1)=18$f(1)=18

$14$14

$15$15

$16$16

$17$17

$\displaystyle q= \frac{1}{3}$\displaystyle q= \frac{1}{3}

$\displaystyle q= \frac{1}{\sqrt[3]{3}}$\displaystyle q= \frac{1}{\sqrt[3]{3}}

$q=\sqrt[3]{3}$q=\sqrt[3]{3}

$q=3$q=3

$\displaystyle - \frac{\sqrt{3}}{3}$\displaystyle - \frac{\sqrt{3}}{3}

$\displaystyle - \frac{4}{5}$\displaystyle - \frac{4}{5}

$-1$-1

$\displaystyle - \frac{5}{4}$\displaystyle - \frac{5}{4}

$\displaystyle \cos\alpha= \frac{1}{2}$\displaystyle \cos\alpha= \frac{1}{2}

$\displaystyle \cos\alpha= \frac{\sqrt{2}}{2}$\displaystyle \cos\alpha= \frac{\sqrt{2}}{2}

$\displaystyle \cos\alpha= \frac{\sqrt{3}}{2}$\displaystyle \cos\alpha= \frac{\sqrt{3}}{2}

$\cos\alpha=1$\cos\alpha=1

$5^\circ$5^\circ

$10^\circ$10^\circ

$20^\circ$20^\circ

$30^\circ$30^\circ

$14^\circ<\alpha<15^\circ$14^\circ<\alpha<15^\circ

$29^\circ<\alpha<30^\circ$29^\circ<\alpha<30^\circ

$60^\circ<\alpha<61^\circ$60^\circ<\alpha<61^\circ

$75^\circ<\alpha<76^\circ$75^\circ<\alpha<76^\circ

$m=2$m=2

$m=-2$m=-2

$m=-2-2\sqrt{2}$m=-2-2\sqrt{2}

$m=2+2\sqrt{2}$m=2+2\sqrt{2}

$\displaystyle m=- \frac{1}{2}$\displaystyle m=- \frac{1}{2}

$\displaystyle m= \frac{1}{2}$\displaystyle m= \frac{1}{2}

$m=1$m=1

$m=2$m=2

$\displaystyle K'=\left(2,- \frac{3}{2} \right)$\displaystyle K'=\left(2,- \frac{3}{2} \right)

$\displaystyle K'=\left(2, \frac{3}{2} \right)$\displaystyle K'=\left(2, \frac{3}{2} \right)

$\displaystyle K'=\left(\frac{3}{2},2 \right)$\displaystyle K'=\left(\frac{3}{2},2 \right)

$\displaystyle K'=\left(\frac{3}{2},-2 \right)$\displaystyle K'=\left(\frac{3}{2},-2 \right)

$\measuredangle HOL$\measuredangle HOL

$\measuredangle OGL$\measuredangle OGL

$\measuredangle HLO$\measuredangle HLO

$\measuredangle OHL$\measuredangle OHL

$27\pi\sqrt{3}$27\pi\sqrt{3}

$9\pi\sqrt{3}$9\pi\sqrt{3}

$18\pi$18\pi

$6\pi$6\pi

$\displaystyle \frac{8^2}{3}\left( \frac{\sqrt{3}}{2}+3 \right)$\displaystyle \frac{8^2}{3}\left( \frac{\sqrt{3}}{2}+3 \right)

$8^2 \cdot \sqrt{3}$8^2 \cdot \sqrt{3}

$\displaystyle \frac{8^2\sqrt{6}}{3}$\displaystyle \frac{8^2\sqrt{6}}{3}

$\displaystyle 8^2\left( \frac{\sqrt{3}}{2}+3 \right)$\displaystyle 8^2\left( \frac{\sqrt{3}}{2}+3 \right)

$x=0$x=0

$x=3$x=3

$x=5$x=5

$x=6$x=6

$\displaystyle p= \frac{1}{4}$\displaystyle p= \frac{1}{4}

$\displaystyle p= \frac{3}{8}$\displaystyle p= \frac{3}{8}

$\displaystyle p= \frac{1}{2}$\displaystyle p= \frac{1}{2}

$\displaystyle p= \frac{2}{3}$\displaystyle p= \frac{2}{3}