IB Maths AI HL · Paper 1

1. Add the left-handed row: $16+6=22$ left-handed students in total.
2. Given left-handed, the group shrinks to those $22$ students, and $16$ of them prefer geometry.

1. The complement of at least one left-handed is that neither student is.
2. There are $178$ right-handed students, and the second pick comes from $199$ people because nobody is chosen twice.

Another way. Add the three cases directly: left then right, right then left, and left then left, each built from fractions over $200$ and $199$.

1. Inverse variation means $c=\frac{k}{n}$.
2. Use the case with four people to find $k$.

1. Convert to litres: $1{\text{ mm}}^3=0.001$ ml and $1000$ ml make one litre.
2. The stated $20$ litres is the exact value, so it goes in the denominator.

Another way. Another valid reason: the dimensions are outside measurements, while the capacity depends on the inside, since the walls take up space.

1. Check the net: find $h(9)$ and compare with $2.24$.
2. Check the landing: solve $h(x)=0$ for the positive root, or find $h(18)$, and compare with the end line.
3. State a conclusion that uses both checks.

Another way. Instead of the root, show $h(18)=-0.26<0$: the ball is already below ground level at the end line, so it has landed in the court.

Another way. The GDC's numerical derivative at $x=6$ gives $-1.5$ directly.

1. Draw a straight line through $A(6,5)$ that crosses the $y$-axis at $(0,1)$.
2. Another easy point to plot: $(3,3)$. The line should cut the curve at right angles at $A$.

1. Solve $f(x)=\frac{2}{3}x+1$ on the GDC, graphically or with the polynomial solver.
2. One root is $x=6$ (point $A$); keep the other two and find their $y$-values.

1. Inside the bracket, $3x$ squashes horizontally by a factor of $\frac{1}{3}$, so $x=6$ moves to $x=2$.
2. Outside, the minus sign reflects in the $x$-axis and the $+4$ shifts up: $y\mapsto -y+4$.

1. $AB=AC=5$ km, so triangle $ABC$ is isosceles with apex angle $6^{\circ }$ at $A$.
2. The two base angles are equal.

Another way. Split the isosceles triangle down the middle: $CB=2\times 5\sin 3^{\circ }$.

1. This month's vector is $T$ times last month's, so undo it with $T^{-1}$.
2. $\det T=0.56-0.06=0.5$.

Another way. Write $0.8r+0.3d=0.65$ and $0.2r+0.7d=0.35$ and solve simultaneously.

1. A steady state $\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)$ is unchanged by $T$, and its entries are proportions adding to $1$.
2. The first row gives $0.8x_1+0.3x_2=x_1$, so $0.3x_2=0.2x_1$, that is $2x_1=3x_2$.
3. Combine with $x_1+x_2=1$; the pair of linear equations has exactly one solution.

Another way. Diagonalise $T$: its eigenvalues are $1$ and $0.5$, so $T^n$ tends to a matrix with both columns equal to $\left(\begin{array}{c}0.6\\ 0.4\end{array}\right)$.

1. Displacement is the area under the velocity graph. Four equal strips from $0$ to $8$ give $h=2$.
2. Ends once, middle values twice.

1. Start at $(0,1.5)$ and follow the slope segments both ways.
2. The curve rises to a maximum near $x=\frac{\pi }{2}\approx 1.57$ with height about $4.1$, then falls; to the left of $0$ it drops towards a minimum near $x=-1.57$ of about $0.55$.

1. Separate the variables and integrate both sides.
2. Remove the logarithm and write the constant as $A$.
3. Use $(0,1.5)$ to find $A$.

1. First solve $E(C)=2$ for $C$ on the GDC.
2. Then set $C(t)$ equal to that value and solve for $t$, keeping the positive root.

Another way. Build the composite $E(C(t))$ and solve $E(C(t))=2$ in one go by graphing it against $y=2$.

1. Make a new list of $\ln N$ values.
2. Run linear regression of $\ln N$ on $t$ with the GDC.

1. Exponentiate both sides of the line from (a).
2. Split the exponent: ${\mathrm{e}}^{at+b}={\mathrm{e}}^b{\mathrm{e}}^{at}$.

1. Differentiating $a\sin (x^n)+b$ gives $an{x}^{n-1}\cos (x^n)$. Match this with $6x^2\cos (x^3)$.
2. So $n=3$ and $3a=6$, giving $a=2$.
3. The $y$-intercept gives $b$ because $\sin 0=0$.

1. The curve meets $y=3$ when $\sin (x^3)=1$, so $x^3=\frac{\pi }{2}$ and $x=1.16244\dots$ is the upper limit.
2. (i) The region lies between $y=g(x)$ below and $y=3$ above, so the solid is a cylinder with the curve's solid removed.
3. (ii) Evaluate on the GDC.

1. $r$ is the modulus, so square both parts and add.
2. Expand the bracket carefully: $(\sqrt{6}-\sqrt{2}{)}^2=6-2\sqrt{12}+2$.

1. $z$ is in the second quadrant (real part negative, imaginary part positive).
2. Find the reference angle, then subtract it from $\pi$.

1. $w=z^3=r^3\text{cis}3\theta$, compared with $z=r\text{cis}\theta$.
2. The length is multiplied by $r^2$ and the angle grows by $2\theta$.

Another way. The same rotation can be described as clockwise by $2\pi -5.78\approx 0.507$ radians.

1. Water volume = large cone of radius $r$, height $x+4$ minus the removed cone.
2. Replace $x+4$ with $2r$ from (a) and compare with the given form.

1. Substitute $\frac{\mathrm{d}V}{\mathrm{d}t}=5$ and $\frac{\mathrm{d}r}{\mathrm{d}t}=0.06$ into the given rate equation.
2. Solve for $r$ and keep the positive value.
3. Use $x=2r-4$ to get the height.

1. Write the system as a matrix and solve $\det (M-\lambda I)=0$.
2. The characteristic equation is a quadratic with negative discriminant.

1. Complex eigenvalues mean the trajectory rotates: a spiral or an ellipse.
2. The real part $0.5$ is positive, so it moves away from the origin: a spiral outwards.
3. At $(1,0)$: $\dot{x}=2>0$ and $\dot{y}=1>0$, so on the positive $x$-axis the point moves upwards, which is anticlockwise.
4. Tick: anticlockwise spiral outwards from the origin.