IB Maths AA HL · Paper 1

Another way. From maximum to minimum is half a period, so the period is $4\pi$, and $\frac{2\pi }{b}=4\pi$ gives the same value.

1. The curve is above the $x$-axis throughout (its lowest value is $2$), so the area is just the integral of $f$ from $0$ to $2\pi$.
2. Integrate term by term. Integrating $\cos \frac{x}{2}$ gives $2\sin \frac{x}{2}$, because you divide by the inner coefficient $\frac{1}{2}$.
3. Substitute the limits; both sine terms vanish.

1. Build $f(g(x))$ by putting $e^x-10$ in place of every $x$ in $f$.
2. Let $w=e^x-10$ so the equation becomes a quadratic in $w$, and solve it.
3. Turn each $w$ back into $e^x$ and reject any value that is not positive.
4. Take logs to finish.

Another way. Expand fully to get $e^{2x}-15e^x=0$, factorise as $e^x(e^x-15)=0$, and note $e^x\ne 0$.

1. Since $x=-1$ is a root, $(x+1)$ is a factor. Divide it out (long division or matching coefficients).
2. Solve the quadratic factor with the formula or by completing the square.

1. Simplify each bracket: $1-\frac{1}{n}=\frac{n-1}{n}$.
2. Use the log law for a sum to combine everything into the log of one product.
3. In the product, each numerator cancels with the denominator before it, leaving only the first numerator and the last denominator.
4. Rewrite $\ln \frac{1}{8}$ with a minus sign to match the required form.

Another way. Write each term as $\ln (n-1)-\ln n$ and watch the middle terms cancel in pairs, leaving $\ln 1-\ln 8$.

1. Choose the teachers and the students separately, then multiply.
2. Teachers: any $3$ of $5$.
3. Students: all pairs from $4$, minus the single pair that is not allowed.
4. Multiply the two counts.

Another way. Count all committees, $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\left(\genfrac{}{}{0ex}{}{4}{2}\right)=60$, then subtract the $10$ that contain the forbidden pair.

1. Base case: check $n=1$.
2. Assume the result for $n=k$, i.e. $k^3+5k=6p$ for some integer $p$.
3. Expand the $n=k+1$ expression and pull out $k^3+5k$.
4. What is left is $3k(k+1)+6$. One of $k$ and $k+1$ is even, so $k(k+1)$ is even and $3k(k+1)$ is a multiple of $6$.
5. Write the concluding sentence linking the base case and the inductive step.

1. Differentiate every term with respect to $x$. Each $y$ term picks up a factor $\frac{dy}{dx}$.
2. Use the product rule on $3x{y}^2$: it gives $3y^2+6xy\frac{dy}{dx}$.
3. Collect the $\frac{dy}{dx}$ terms on one side and factorise.

1. Find $\frac{dy}{dx}$ at $P(2,-1)$ using (a).
2. Link the rates with the chain rule: $\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}$.
3. Substitute $\frac{dy}{dt}=-6$ and solve.

1. Separate the variables: all the $y$ terms with $dy$, all the $x$ terms with $dx$.
2. Integrate. The left side is of the form $\frac{f^{\prime }}{f}$ up to a factor $\frac{1}{2}$.
3. Use $y=0$ when $x=2$ to find the constant.
4. Combine the logs and remove them to get $y^2$ on its own.

Another way. Keep the constant as $\ln A$ from the start: $1+y^2=A{x}^2$, then $y=0,x=2$ gives $A=\frac{1}{4}$.

1. Split the trapezoid into three triangles meeting at $O$, each with two sides of length $2$.
2. The two outer triangles have angle $\theta =\frac{\pi }{6}$ at $O$; the middle one has what is left of $\pi$, namely $\frac{2\pi }{3}$.
3. Use $\frac{1}{2}ab\sin C$ for each and add.

1. Repeat (a) with a general $\theta$: the outer triangles each give $2\sin \theta$.
2. The middle triangle has angle $\pi -2\theta$, and $\sin (\pi -2\theta )=\sin 2\theta$.

1. Factorise the denominator as $(x+3)(x+1)$ and write $\frac{A}{x+3}+\frac{B}{x+1}$.
2. Multiply through, then substitute $x=-1$ and $x=-3$ to find $B$ and $A$.

1. Integrate the partial fractions from $0$ to $1$; each gives a log.
2. Substitute the limits and use log laws to combine into a single log.
3. The integral comes out negative because the curve lies below the $x$-axis on $[0,1]$. Area is positive, so take the modulus.

1. Write $\frac{2}{x+3}=\frac{2}{3}{\left(1+\frac{x}{3}\right)}^{-1}$ and $\frac{3}{x+1}=3(1+x{)}^{-1}$.
2. Expand each with $(1+u{)}^{-1}=1-u+u^2-\cdots$ up to the $x^2$ term.
3. Subtract and collect like terms.

1. Write $z^6=-64=64\mathrm{cis}(\pi +2k\pi )$.
2. Take the sixth root of the modulus and divide the argument by $6$.
3. List the six roots, one every $\frac{\pi }{3}$, starting from $\frac{\pi }{6}$.

1. All six roots lie on the circle of radius $2$ centred at the origin, equally spaced $60^{\circ }$ apart, forming a regular hexagon.
2. Two sit on the imaginary axis, at $2i$ and $-2i$; the others are at $\pm \frac{\pi }{6}$ and $\pm \frac{5\pi }{6}$, i.e. $\pm \sqrt{3}\pm i$.
3. A natural labelling goes anticlockwise: $z_1=\sqrt{3}+i$, $z_2=2i$, $z_3=-\sqrt{3}+i$, $z_4=-\sqrt{3}-i$, $z_5=-2i$, $z_6=\sqrt{3}-i$.

Another way. The roots of $z^2+4=0$ are $\pm 2i$, which are $z_2$ and $z_5$ from (b).

1. Pair each remaining root with its conjugate so the quadratics have real coefficients.
2. $\sqrt{3}\pm i$ are the roots of $Q(z)$ from (a), so one factor is already known.
3. The pair $-\sqrt{3}\pm i$ gives the same quadratic with the sign of the middle term flipped.

Another way. Divide to get $P(z)=(z^2+4)(z^4-4z^2+16)$, then write $z^4-4z^2+16=(z^2+4{)}^2-12z^2$ and use difference of two squares.

1. Use the hexagon from (c). Neighbouring vertices of a regular hexagon of radius 2 are 2 apart.
2. Vertices two steps away are $2\sqrt{3}$ apart, and the opposite vertex is a diameter, $4$, away.
3. From $z_2$ there are two neighbours, two vertices two steps away, and one opposite.