IB Maths AI HL · Paper 2

(i) New Zealand and Australia both score 7.2, which would be ranks 3 and 4, so they share the rank 3.5.

(ii) Spain's score of 6.5 is the 8th highest.

(iii) Australia has the other tied 7.2.

(i) Use the calculator on the ranks 1 to 10 of the countries and the happiness ranks.

(ii) France's score would change from 6.7 to 6.9, but it would still be below 7.0 (Ireland) and above 6.6 (UAE), so its rank stays at 6 and the ranks do not change.

Jose's conclusion is not appropriate, because $r_s\approx 0.164$ is close to zero. It shows only a weak correlation between the ranks, so we cannot say that happy countries are favourite expatriate countries.

Use five intervals of width $h=1$.

Integrate the model.

The volume of a cylinder is $\pi r^{2}h$.

(i) Differentiate $A=2\pi r^2+1200r^{-1}$.

(ii) Set the derivative to zero.

The widths are geometric with $u_1=3$ and $r=\frac{2}{3}$.

Sum the first five widths.

The total width can never exceed the sum to infinity.

Each area is $\frac{4}{9}$ of the previous one, since both height and width are multiplied by $\frac{2}{3}$.

The first area is $4.5\times 3=13.5$ cm${}^2$.

Triangle $AMT$ is right-angled at $M$, with $AM=2$ and $AT=3.4$.

(i) The base is six equilateral triangles of side 2.

(ii) The volume of a pyramid.

In the right triangle $AMT$.

The velocity of the car is $(50,-33,0)$.

The velocities are $\mathbf{u}=(50,-33,0)$ and $\mathbf{v}=(-15,-20,0)$.

The drone starts at $(200,-100,0.02)$.

(i) Equate the $x$-components.

(ii) Use the $y$-component.

(i) Subtract the car's position from the drone's.

(ii) The distance between them.

Find the minimum on the calculator. It happens at $t\approx 3.09$ hours.

The matrix is a rotation matrix with $\theta =15^{\circ }$.

$A^n=I$ when the total rotation is a whole number of full turns. The smallest is one full turn.

(i) An enlargement with scale factor 1.05 about the origin.

(ii) $B^{24}$ is an enlargement with scale factor $1.05^{24}$.

It is an enlargement, scale factor $1.05^{24}\approx 3.23$, centre $(0,0)$.

Multiply.

The limit point is a fixed point, so $T\left(\begin{array}{c}p\\ q\end{array}\right)=\left(\begin{array}{c}p\\ q\end{array}\right)$.

(i) The general solution.

Use $x(0)=5$ and $y(0)=\frac{dx}{dt}=2$.

Substitute $A=5-B$.

(ii) Sketch: the curve starts at $x=5$, rises very slightly at first (since the initial speed is positive), then decreases smoothly towards the asymptote $x=0$ without crossing it.

(i) For $a=18$, $b=85$ the eigenvalues are $-9\pm 2i$, so the phase portrait is a spiral going into the origin. It starts at $(5,2)$. At that point $\frac{dy}{dt}=-85(5)-18(2)=-461<0$, and $\frac{dx}{dt}=2>0$, so the motion is clockwise.

(ii) The graph of $x$ against $t$ starts at $x=5$ and oscillates about 0 with decreasing amplitude.

Spring 1 might be better because $x$ returns smoothly to equilibrium without oscillating, so it gives a smoother ride, whereas Spring 2 keeps bouncing.