IB Maths AI SL · Paper 1

1. The path is a circle centred on the Earth's centre, so its radius is the Earth's radius plus the height.
2. Find one circumference with $C=2\pi r$, then multiply by $6$ for six orbits.
3. Show the unrounded total before rounding to $3$ s.f.

1. Percentage error compares the difference with the true value, so divide by the real distance of $231718$ km.
2. Use your unrounded total from part (a).

Another way. Halve backwards from $48$: $24,12,6,3$, so the first term is $3$.

1. Enter the nine observed counts as a $3\times 3$ matrix and run the chi-squared two-way test on your GDC.
2. The degrees of freedom are $(3-1)(3-1)=4$.

1. Compare the $p$-value with the $5\%$ level. It is smaller, so reject $H_0$.
2. Say what that means for these students in words.

1. Let $X$ be the number of heads. There are $12$ independent coins with the same chance, so $X\sim \mathrm{B}(12,0.73)$.
2. Use the binomial pdf at $8$.

1. With $12$ coins, more heads than tails means at least $7$ heads.
2. Use the cumulative function: $1-\mathrm{P}(X\le 6)$.

Another way. Count tails instead: $T\sim \mathrm{B}(12,0.27)$ and find $\mathrm{P}(T\le 5)$.

1. Cut the L-shape into two rectangles with one straight line.
2. One piece is $6x$ by $x+4$. The other is $2x$ by the leftover length $(2x+10)-(x+4)=x+6$.
3. Expand each area and add.

Another way. Take the full $6x$ by $(2x+10)$ rectangle and subtract the missing $4x$ by $(x+6)$ corner.

1. Set the area equal to $72$ and solve the quadratic with the GDC's polynomial solver.
2. A length cannot be negative, so reject the negative root.

1. The fence runs along the four outer sides that do not touch the house: $6x$, $2x+10$, $x+4$ and $2x$.
2. Add them and substitute $x=1.5$.

1. Fill in by subtraction from the totals: not lost $=80-5$, not heavy $=80-24$.
2. Heavy and not lost $=24-2$; not heavy and lost $=5-2$; the last cell is $56-3$.

1. Let $W\sim \mathrm{N}(16,3^2)$. Heavy means $W>18$, so find that probability with the normal cdf.
2. Multiply by the chance of being lost when heavy, from (b)(ii).

1. The exact area is the integral of the curve from $0$ to $3$. Evaluate it on the GDC.
2. Compare with $53.4$ and say which is bigger.

Another way. Integrate by hand: $\frac{1.37}{4}(81)-\frac{6.07}{3}(27)+\frac{7.87}{2}(9)+45=53.5275$.

1. Each roll is $0.6$ m wide, so divide the exact area by $0.6$ to get the total length of turf needed.
2. Divide by $1.4$ m per roll and round up, since part rolls cannot be bought.
3. Cost is the rolls times $12.50$ plus the $28$ delivery charge.

Another way. Divide the area by one roll's area, $0.6\times 1.4=0.84{\text{ m}}^2$; you get the same $63.72\dots$.

1. Each bisector passes through the midpoint and is perpendicular to the segment (negative reciprocal gradient).
2. $[AD]$: midpoint $(2.5,2)$, gradient of $AD$ is $-1$, so the bisector has gradient $1$. It passes through $(0.5,0)$ and $(2.5,2)$.
3. $[BD]$: midpoint $(3.5,2)$, gradient of $BD$ is $1$, so the bisector has gradient $-1$. It passes through $(3.5,2)$ and $(5.5,0)$.
4. $[CD]$ is horizontal with midpoint $(3.5,1.5)$, so its bisector is the vertical line $x=3.5$.

1. D's cell is bounded by the three new bisectors. They meet the old edges at two new vertices: $(3,2.5)$ (equidistant from A, B and D) and $(3.5,2)$ (equidistant from B, C and D).
2. Keep: $x=3$ above $(3,2.5)$ between A and B, and $y=2$ to the right of $(3.5,2)$ between B and C.
3. Add: $y=x-0.5$ from $(3,2.5)$ down to $(0.5,0)$; $y=-x+5.5$ from $(3,2.5)$ to $(3.5,2)$; and $x=3.5$ from $(3.5,2)$ down to the $x$-axis.
4. Remove the old parts of the A, B and C edges that now fall inside D's cell.

1. Integrate term by term and include the constant $c$.
2. Use $P(5)=254$ to find $c$.

1. The population is gone when $P(t)=0$. Solve with the GDC's polynomial solver or a graph.
2. Only a positive time makes sense.

Another way. Graph both sides on the GDC and find the intersection.

1. Let $d$ be the horizontal distance to the helicopter. Its height above each person is $d\tan (\text{angle})$, and the two heights differ by $15$ m.
2. Solve for $d$, then find the height of the helicopter above Benito.
3. Carlos's eyes are at the helicopter's height, so divide that height by $3.75$ to count the floors above Benito.
4. Add to $22$ and round to a whole floor.

Another way. Use triangle Alberta, Benito, helicopter: its angles are $44^{\circ }$, $122^{\circ }$ and $14^{\circ }$, so the sine rule gives the distance from Alberta as $52.58\dots$ m, and $52.58\dots \times \sin 46^{\circ }=37.82\dots$ m above Alberta, which is $18+10.09\dots \approx 28$.