IB Mathematics: Analysis and Approaches question bank

$d=3$ A2

$u_{20}=24+19\times 3$ M1

$=81$ seats A1

$S_{20}=\frac{20}{2}(24+81)$ M1A1

$=1050$ seats A1

$m=\frac{1-7}{5-(-3)}$ (M1)

$=-\frac{6}{8}=-\frac{3}{4}$ A1

$y-1=-\frac{3}{4}(x-5)$ (or $y-7=-\frac{3}{4}(x+3)$) (M1)

$3x+4y-19=0$ A1

Note: Accept any integer multiple, e.g. $-3x-4y+19=0$. Award (M1)A1FT for their gradient from part (a).

attempt to find at least one intercept (M1)

$x$-intercept $\frac{19}{3}$ and $y$-intercept $\frac{19}{4}$ A1

area $=\frac{1}{2}\times \frac{19}{3}\times \frac{19}{4}=\frac{361}{24}$ A1

Using ${\cos }^{2}x=1-{\sin }^{2}x$: $2-2{\sin }^{2}x+\sin x=1$ M1

$2{\sin }^{2}x-\sin x-1=0$ A1

$(2\sin x+1)(\sin x-1)=0$, so $\sin x=-\frac{1}{2}$ or $\sin x=1$ M1A1

$x=\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}$ A1A1

Award A1 for $\frac{\pi }{2}$ and A1 for both other values.

$P(M>200)$ (M1)

$=0.0478$ A1

$P(M>k)=0.15$ (M1)

$z=1.0364$ (A1)

$k=192$ g (192.437…) A1

$f^{\mathrm{\prime }}(x)=3x^2-12x+9$ A1A1

Award A1 for each two correct terms.

Setting $f^{\mathrm{\prime }}(x)=0$: $3(x-1)(x-3)=0$ M1

$x=1$ or $x=3$ A1

$(1,5)$ and $(3,1)$ A1A1

$f^{\mathrm{\prime }}(4)=48-48+9=9$ A1

$f(4)=64-96+36+1=5$ A1

$y-5=9(x-4)$, so $y=9x-31$ A1