IB Maths AI HL · Paper 2

The height is three times the length of the base.

The volume is $l\times l\times d=3l^3$.

Two square faces and four rectangles.

The volume of the cylinder is $\pi r^{2}h=375$.

The total external area is two discs plus the curved surface, $A=2\pi r^2+2\pi rh$.

Differentiate $A=2\pi r^2+750r^{-1}$.

(i) Set the derivative to zero.

(ii) Substitute $r=3.908$.

(i) Differentiate again.

(ii) For $r>0$ both terms are positive, so $\frac{d^{2}A}{d{r}^2}>0$ and the graph of $A$ is concave up.

The tide repeats every 12 hours, so the period in degrees is 360 over $b$.

Substitute $t=5$, with the angle in degrees.

(i) The amplitude is 2.5 m.

(ii) The principal axis.

With strips of width 1.9 m.

The volume per second is the area times the speed.

The deepest point of the river is 2.81 m, and the boat's lowest part is 1.6 m below the surface.

The maximum of $y=-0.15x^2+1.14x+0.9$ is at the vertex $x=\frac{1.14}{0.3}=3.8$.

The matrix has 1 in row $A$, column $E$ (route $A\to E$) but 0 in row $E$, column $A$.

(i) Compute $M^3$.

$M^3$ contains zeros, so some pairs of ports cannot be reached in exactly three trips.

(ii) Add the matrices for one, two and three trips, $M+M^2+M^3$.

This matrix has no zeros, so yes: every port can be reached from any other port in three trips or fewer.

From $A$ the cheapest port is $B$ (3). From $B$ the cheapest unvisited is $C$ (4). From $C$ the cheapest unvisited is $E$ (10) (since $D$ costs 13). Then $D$ (3) and back to $A$ (8).

At $t=0$ the position is $(2,1,0)$.

(i) $P_1$ moves with velocity $(-4,4,7)$, so after 5 seconds it is at $B=(2-20,1+20,35)=(-18,21,35)$.

(ii) The length of the vector.

The directions are $\overrightarrow{AB}$, parallel to $(-4,4,7)$ with length 9, and $P_2$'s direction $(-1,2,0)$ with length $\sqrt{5}$.

At $t=5$, $P_2$ has travelled $12\times 5=60$ m from $A$. Triangle $ABD$ has $AB=45$, $AD=60$ and the angle $\widehat{DAB}=53.4^{\circ }$ between them. Use the cosine rule.

(i) At least two customers in the queue.

(ii) The expected number.

The mean service time is 115 seconds.

Three minutes is 180 seconds.

The total for two customers is $T_1+T_2\sim N\left(230,2\times 28^2\right)$, with standard deviation $28\sqrt{2}=39.6$.

(i) A new customer waits more than three minutes in each case, depending on the number $c$ ahead: for 0, never; for 1, probability 0.0101; for 2, probability 0.8966; for 3 or more, probability 1.

(ii) Since $0.425>0.4$, Daphne will decide to employ more staff.

The first branch is rotated $\frac{\pi }{12}$ clockwise from the vertical segment (argument $\frac{\pi }{2}$) so its argument is $\frac{\pi }{2}-\frac{\pi }{12}$.

(i) After stage 1 there are two branches of length 5. After stage 2 there are four more, each 70% of the previous length, $0.7\times 5=3.5$.

(ii) Stage $n$ adds $2^n$ branches of length $5(0.7{)}^{n-1}$, a total length of $5\times 0.7^{n-1}\times 2^n=10(1.4{)}^{n-1}$. This is a geometric series with first term 10 and common ratio 1.4.

(iii) The common ratio $1.4>1$, so the length added at each stage keeps growing and the series diverges. There is no limit to the total length.

(i) Branch $B$ is made of three branches in order. The first has length 5 and argument $\frac{5\pi }{12}$. The second is $0.7\times 5=3.5$ long and is rotated $\frac{\pi }{12}$ clockwise, with argument $\frac{4\pi }{12}=\frac{\pi }{3}$. The third has length $0.7\times 3.5=2.45$ and is rotated $\frac{\pi }{12}$ anticlockwise, with argument $\frac{5\pi }{12}$.

(ii) Convert to Cartesian form.