IB Maths AA HL · Paper 2

1. Enter leg length in one list and race time in another.
2. Run linear regression with $L$ as the independent variable and $T$ as the dependent one.

1. Make $h$ the subject of both equations and set them equal.
2. Collect the $x$ terms and solve.

Another way. Use the sine rule in the triangle formed by $A$, $B$ and the cliff top (angles $15^{\circ }$, $140^{\circ }$, $25^{\circ }$) to find the slant distance from $B$, then multiply by $\sin 40^{\circ }$.

1. Write one equation for the large sector's area and one for its perimeter (two radii plus the arc).
2. From the perimeter, $R=\frac{40}{2+\theta }$. Substitute this into the area equation.
3. Solve for $\theta$ and keep the root that is acute.

1. Displacement has a turning point where $v(t)=0$. Graph $v$ on $0\le t\le 6$ and find its zeros: $t=1.96227\dots$ and $t=5.39302\dots$ (and $t=0$).
2. Between $1.96$ and $5.39$ the velocity is positive, so the particle moves forwards and displacement climbs to a peak at $t=5.39$.
3. Integrate $v$ from $0$ to that time, since the particle starts at $O$.

1. Acceleration is $v^{\prime }(t)$, so it is negative wherever the velocity graph slopes downwards.
2. Find the turning points of $v$ on the GDC: $t=0.791904\dots$ (a minimum) and $t=3.50201\dots$ (a maximum).
3. The graph of $v$ falls before the minimum and after the maximum.

1. Rotating about the $y$-axis needs $\pi \int x^2\mathrm{d}y$, so rearrange for $x^2$ in terms of $y$.
2. The $y$ limits run from the $x$-axis ($y=0$) up to $y=a=3$.
3. Integrate on the GDC or by hand.

1. Join the centre of the hemisphere to the edge of the water surface: that line is a radius, $15$.
2. The vertical side from the centre down to the water surface is $15-h$, and the horizontal side is $r$. Apply Pythagoras.
3. Expand, simplify, and take the positive root since $r>0$.

1. Chain rule: $\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{\mathrm{d}r}{\mathrm{d}h}\times \frac{\mathrm{d}h}{\mathrm{d}V}\times \frac{\mathrm{d}V}{\mathrm{d}t}$.
2. Differentiate $V$ with respect to $h$ and evaluate at $h=5$.
3. Differentiate $r=(30h-h^2{)}^{1/2}$ with respect to $h$ and evaluate at $h=5$.
4. Combine with $\frac{\mathrm{d}V}{\mathrm{d}t}=20$.

1. Seat the girls first: $7!$ orders.
2. The girls create $8$ gaps (six between them and one at each end). Put each boy in a different gap, in order: $8\times 7\times 6$ ways.

Another way. Complement with inclusion-exclusion: $10!-6\times 9!+3!\times 8!$, subtracting arrangements with a pair of boys together and adding back the triple block counted twice.

1. Both ends must be boys: $3\times 2$ ways to choose which boys sit there.
2. Seat the girls in a line between them: $7!$ ways.
3. The third boy cannot touch an end boy, so he goes in one of the $6$ gaps strictly between two girls.

1. With $n$ appointments, $Y\sim \mathrm{B}(n,0.826)$. Every attender is seen when $Y\le 40$.
2. The chance of $Y\le 40$ falls as $n$ grows, so tabulate $\mathrm{P}(Y\le 40)$ for $n=41,42,43$.
3. Pick the largest $n$ that still gives at least $0.99$.

1. The oblique asymptote is the quotient when the numerator is divided by $x-3$. So the numerator equals $(x+7)(x-3)$ plus a constant.
2. Expand and compare the $x$ coefficients.

1. Write $f(x)=x+7+\frac{A}{x-3}$ where $A=c+21$.
2. Differentiate and set $f^{\prime }(-3)=0$.
3. Solve for $A$, then for $c$.

Another way. Differentiate $\frac{x^2+4x+c}{x-3}$ with the quotient rule and set the numerator to zero at $x=-3$.

1. Two branches either side of the vertical asymptote $x=3$, both hugging the oblique asymptote $y=x+7$ as $x\to \pm \infty$.
2. Left branch: rises from the bottom left along $y=x+7$, reaches a local maximum at $(-3,-2)$, crosses the $y$-axis at $(0,-5)$, then plunges down towards $x=3$.
3. Right branch: comes down from the top next to $x=3$, reaches a local minimum at $(9,22)$, then rises alongside $y=x+7$.
4. There are no $x$-intercepts: $x^2+4x+15$ has a negative discriminant.

1. Graph $|f(x)|$: the left branch is reflected above the $x$-axis, so it has a local minimum of height $2$ at $x=-3$ and goes to infinity on both sides.
2. A horizontal line $y=k$ meets the reflected left branch twice for any $k>2$, and meets the right branch twice only when $k>22$.
3. Four solutions need both, so $k>22$.

1. The translation gives $g(x)=f(x-2)+3$.
2. Vertical asymptotes of $\frac{1}{g(x)}$ are where $g(x)=0$. Simplify $g$ and solve.
3. The asymptote of $g$ at $x=5$ is not one for $\frac{1}{g}$: there $\frac{1}{g}\to 0$.

Another way. Solve $f(x)=-3$ to get $x=-6$ and $x=-1$, then shift both $2$ to the right.

1. Use $y_{n+1}=y_n+0.01\left(y_n^3-\frac{y_n}{x_n}\right)$ with $x_0=1$, $y_0=2$.
2. Five steps take $x$ from $1$ to $1.05$. A GDC sequence or spreadsheet saves time.

1. Differentiate $z=-\frac{1}{2}{y}^{-2}$ with respect to $x$ using the chain rule.
2. Substitute the expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$.
3. Replace $\frac{1}{y^2}$ with $-2z$ and rearrange.

1. The equation from (b) is linear. The integrating factor is $e^{\int -2/x\mathrm{d}x}=x^{-2}$ (valid as $x>0$).
2. Multiply through so the left side becomes $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{z}{x^2}\right)$, then integrate.
3. Find the constant: $y=2$ gives $z=-\frac{1}{8}$ at $x=1$.
4. Swap back to $y$ and make $y$ the subject, taking the positive root as $y>0$.

Another way. Put $z=vx$ to turn the equation into the separable form $x\frac{\mathrm{d}v}{\mathrm{d}x}=1+v$, then integrate both sides.

1. Domain: the expression under the root must be strictly positive (it is also a denominator), and $x>0$ is given.
2. Range: $y$ is smallest where $8x-7x^2$ is largest, at its vertex $x=\frac{4}{7}$. As $x$ approaches either end of the domain, $y\to \infty$.