IB Maths AA SL · Paper 1

1. A maximum at $x=0$ to the next minimum at $x=2\pi$ is half a cycle, so the period is $4\pi$.
2. The period of $\cos bx$ is $\frac{2\pi }{b}$. Set it equal to $4\pi$.

Another way. At the minimum $\cos (2\pi b)=-1$, and the first positive solution is $2\pi b=\pi$, so $b=\frac{1}{2}$.

1. The graph sits above the $x$-axis the whole way, so the area is just the definite integral from $0$ to $2\pi$.
2. Integrate: $3$ gives $3x$, and $\cos \frac{x}{2}$ gives $2\sin \frac{x}{2}$ (divide by the $\frac{1}{2}$).
3. Substitute the limits and subtract.

1. $(f\circ g)(x)$ means put $g(x)$ into $f$: replace every $x$ in $f$ with ${\mathrm{e}}^x-10$.
2. Expand and collect like terms.

1. Set the expression from (a) equal to $50$; the constants cancel.
2. Factor out ${\mathrm{e}}^x$ (think of it as a quadratic in $u={\mathrm{e}}^x$).
3. ${\mathrm{e}}^x=0$ has no solution, so only ${\mathrm{e}}^x=15$ is left. Take logs.

Another way. Write ${\mathrm{e}}^{2x}=15{\mathrm{e}}^x$, take logs of both sides to get $2x=\ln 15+x$, so $x=\ln 15$.

1. Simplify each bracket: $1-\frac{1}{n}=\frac{n-1}{n}$, so the terms are $\ln \frac{1}{2},\ln \frac{2}{3},\dots ,\ln \frac{7}{8}$.
2. A sum of logs is the log of the product, so multiply the fractions together.
3. Every numerator cancels with the denominator before it, leaving $\frac{1}{8}$.
4. Write $\ln \frac{1}{8}$ as $-\ln 8$ to match the required form.

Another way. Split each term as $\ln (n-1)-\ln n$; adding them up, everything cancels except $\ln 1-\ln 8$.

1. Write the root as a power: $(2x+5{)}^{\frac{1}{2}}$.
2. Bring the power down, reduce it by one, then multiply by the derivative of the inside, which is $2$.

1. Find the gradient of $L$: substitute $x=-1$ into $g^{\prime }(x)$.
2. The gradient of a line equals $\tan$ of the angle it makes with the $x$-axis.
3. Recognise $\tan \theta =\frac{1}{\sqrt{3}}$ as an exact value.

1. Vertical asymptote: where the denominator is zero, $2x-5=0$.
2. Horizontal asymptote: for large $x$ the function tends to the ratio of the $x$ coefficients, $\frac{4}{2}$.

1. Draw dashed lines for the asymptotes $x=2.5$ and $y=2$. They split the plane into four regions around the point $(2.5,2)$.
2. Left branch: passes through $(0,1.8)$ and $(2.25,0)$, approaches $y=2$ from below as $x\to -\infty$, and drops down towards $-\infty$ as $x$ nears $2.5$ from the left.
3. Right branch: comes down from $+\infty$ just right of $x=2.5$ and flattens towards $y=2$ from above as $x$ grows.
4. The two branches sit in opposite corners: bottom-left and top-right of where the asymptotes cross.

1. Write $y=\frac{4x-9}{2x-5}$ and swap $x$ and $y$.
2. Multiply out, gather the $y$ terms on one side and factor out $y$.
3. Divide to make $y$ the subject.
4. The inverse's domain is the original range: every value except the horizontal asymptote $y=2$.

1. Set $f(x)$ equal to $f^{-1}(x)$ and cross-multiply, or, more quickly, solve $f(x)=x$, since the graphs of a function and its inverse meet on $y=x$ here.
2. Rearrange into a quadratic and factorise.
3. Both roots avoid $x=2$ and $x=2.5$, so both count.

Another way. Cross-multiplying $\frac{4x-9}{2x-5}=\frac{5x-9}{2x-4}$ gives $(4x-9)(2x-4)=(5x-9)(2x-5)$, which simplifies to the same quadratic.

1. Use $\text{P}(A\mid W)=\frac{\text{P}(A\cap W)}{\text{P}(W)}$.
2. The top is the head-then-white route only; the bottom is the answer to (b).

1. Each round is white with probability $\frac{1}{5}$ and red with probability $\frac{4}{5}$, independently, because the ball goes back.
2. (i) A score of $9$ needs three whites.
3. (ii) A score of $5$ needs one white and two reds. The white can come in any of the three rounds, so multiply by $3$.

Another way. For (ii), the four probabilities must add to $1$: $1-\frac{64}{125}-\frac{12}{125}-\frac{1}{125}=\frac{48}{125}$.

1. Find $\text{E}(X)$, the expected score for one day of three rounds: $\sum x\text{P}(X=x)$.
2. Over $n$ days the expected total is $n\text{E}(X)$. Set it equal to $105$.

Another way. Expected whites per round is $\frac{1}{5}$, so the expected score per round is $1+2\times \frac{1}{5}=\frac{7}{5}$ and per day $3\times \frac{7}{5}=\frac{21}{5}$.

1. Split the trapezium into triangles $OAB$, $OBC$ and $OCD$. Each has two sides of length $2$ (radii), so use $\frac{1}{2}ab\sin C$.
2. With $\theta =\frac{\pi }{6}$, the outer triangles each have angle $\frac{\pi }{6}$ at $O$. The middle one gets what is left of the straight angle: $\pi -2\times \frac{\pi }{6}=\frac{2\pi }{3}$.
3. Use $\sin \frac{\pi }{6}=\frac{1}{2}$ and $\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2}$, then add.

Another way. Use the trapezium formula: height $2\sin \frac{\pi }{6}=1$, parallel sides $4$ and $2\times 2\cos \frac{\pi }{6}=2\sqrt{3}$, giving $\frac{1}{2}(4+2\sqrt{3})\times 1=2+\sqrt{3}$.

1. Repeat (a) with a general $\theta$: the two outer triangles have angle $\theta$ at $O$, and the middle one has $\pi -2\theta$.
2. Use $\sin (\pi -2\theta )=\sin 2\theta$.