IB Maths AI HL · Paper 2

1. The quadratic has a positive $x^2$ coefficient, so its lowest point is the vertex. Find it with $x=-\frac{b}{2a}$ or from the GDC graph.
2. Substitute that $x$ back into $h$ and give a height, not a coordinate pair.

1. Each hour mark is $\frac{360^{\circ }}{12}=30^{\circ }$ apart. At eight o'clock the hands are four marks apart, so the angle between them is $120^{\circ }$.
2. You know two sides and the angle between them, so use the cosine rule.

Another way. The minute hand points at the 4, which is $30^{\circ }$ below the horizontal, so its tip is $15\sin 30^{\circ }=7.5$ cm below the centre.

1. In 20 minutes the hour hand turns $\frac{20}{60}\times 30^{\circ }=10^{\circ }$ past the 8. The 8 is $30^{\circ }$ below the horizontal, so the hour hand is now $20^{\circ }$ below the horizontal.
2. Height of the hour hand's tip above the bottom is $15-9\sin 20^{\circ }$.
3. Subtract the minute hand's height from part (c).

Another way. Measure the hour hand from 12: it sits at $250^{\circ }$, so its tip is at height $15+9\cos 250^{\circ }$, giving the same $11.92$ cm.

1. First stage: two branches, hit $0.34$ and miss $0.66$.
2. From the hit branch: hit $0.72$, miss $0.28$.
3. From the miss branch: hit $0.68$, miss $0.32$. Label every branch with the outcome and its probability.

Another way. Add the three branches directly: $0.34(0.28)+0.66(0.68)+0.66(0.32)=0.7552$.

Another way. The inverse binomial function on the GDC with area $0.25$ gives $92$ directly.

1. Distance from $O$ is $|\mathbf{r}|=\sqrt{x^2+y^2}$. Square each component.
2. The cross terms $\pm 24\sin 2t\cos 2t$ cancel.
3. Group what is left and use ${\sin }^{2}2t+{\cos }^{2}2t=1$.

1. Two non-zero vectors are perpendicular exactly when their scalar product is zero, so aim to show $\mathbf{r}\cdot \mathbf{v}=0$.
2. Multiply matching components and add. Write $s=\sin 2t$, $c=\cos 2t$ to keep it tidy.
3. Expand and watch every term cancel.

1. With $y=\frac{\mathrm{d}x}{\mathrm{d}t}$, the second derivative of $x$ is $\frac{\mathrm{d}y}{\mathrm{d}t}$.
2. Rearrange the equation to make $\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}$ the subject, then replace the derivatives.

Another way. You can also write the full vector form: $\left(\begin{array}{c}x\\ y\end{array}\right)=A{\mathrm{e}}^{-t}\left(\begin{array}{c}1\\ -1\end{array}\right)+B{\mathrm{e}}^{-2t}\left(\begin{array}{c}1\\ -2\end{array}\right)$.

1. Differentiate the general solution to get the velocity.
2. At $t=0$ the displacement is $20$ and the velocity is $-50$. That gives two simultaneous equations.
3. Solve them and substitute back.

1. Twenty minutes is two 10-minute periods, so double the mean.
2. Then find the probability of zero calls.

1. Independent Poisson variables add: the combined 10-minute mean is $1.25+0.82=2.07$.
2. Scale to 30 minutes by multiplying by $3$.
3. Fewer than 4 means at most 3.

1. The mean of 6 independent normal times is normal, with the same mean and variance divided by $6$.
2. Find the probability below $10$.

Another way. Work with the total instead: $\sum C\sim \mathrm{N}(72,54)$ and find $P(\sum C<60)$.

1. Turn the comparison into one variable: Allison longer than Jean means $A-J>0$.
2. Means subtract, but variances add because the times are independent.

1. Put the two fractions over the common denominator $x(2-x)$.
2. The numerator must equal $1$ for every $x$, so the $x$ terms cancel and the constant term fixes $A$.

1. Move everything involving $x$ to one side, and $\mathrm{d}t$ to the other.
2. Replace the left-hand integrand with the partial fractions from (a).
3. Integrate each term. The $\frac{1}{2-x}$ term gives $-\ln |2-x|$ because of the minus inside.
4. Add one constant on the right.

Another way. Combine the logs into one: $\frac{1}{2}\ln \left|\frac{x}{2-x}\right|=t+C$.

1. Drop the modulus and double both sides.
2. Exponentiate to remove the log.
3. Multiply out the denominator, gather the $x$ terms, factorise and divide.

Another way. Dividing top and bottom by ${\mathrm{e}}^{2t}$ gives the tidy logistic form $x=\frac{2}{1+{\mathrm{e}}^{-2t}}$.