IB Maths AA HL · Paper 3

1. The terms $1,2,3,\dots ,n$ form an arithmetic sequence with first term $1$ and common difference $1$.
2. Use the arithmetic series formula from the booklet, then expand.

Another way. Use the other booklet form, $\frac{n}{2}(u_1+u_n)$, with $u_1=1$ and $u_n=n$.

1. Each rectangle has width $1$, so the total area of the first $n$ rectangles is $1^2+2^2+\cdots +n^2=S_2(n)$.
2. The curve passes through the top right corners and is increasing, so on each strip it stays below the rectangle. The rectangles therefore have more area than the region under the curve from $0$ to $n$.
3. Work out that area with an integral.

1. Now use a curve through the top left corners, $(k-1,k^2)$. Shifting $y=x^2$ one unit left gives $y=(x+1{)}^2$.
2. This curve lies above every rectangle, so the rectangles have less area than the region under it from $0$ to $n$.
3. Integrate and expand.

Another way. Integrate $(x+1{)}^2$ directly as ${\left[\frac{(x+1{)}^3}{3}\right]}_0^n=\frac{(n+1{)}^3-1}{3}$, which expands to the same thing.

1. Write the sum out term by term using $u_r=r^3-(r-1{)}^3$.
2. Each positive cube is cancelled by the negative cube in the next term (a telescoping sum).
3. Only $n^3$ and $-0^3$ survive.

1. Replace the left side with $n^3$ from (f), $S_1(n)$ with (b) and $S_0(n)$ with (a)(ii).
2. Rearrange to make $S_2(n)$ the subject, then divide by $3$.

1. Lower bound: the extra terms $\frac{n^2}{2}+\frac{n}{6}$ are positive for positive $n$, so $S_2(n)$ is bigger than $\frac{n^3}{3}$.
2. Upper bound: compare term by term, since $\frac{n^2}{2}<n^2$ and $\frac{n}{6}<n$.

1. Every point is four units from the origin, so the curve is a circle centred at $0$ with radius $4$.
2. Mark where it crosses the axes: $\pm 4$ on the real axis and $\pm 4\mathrm{i}$ on the imaginary axis.

1. The maximum happens when $\sin \theta =1$, which in $0\le \theta \le \pi$ means $\theta =\frac{\pi }{2}$.
2. Build $z$ from $r=4$ and that argument.

1. Multiply both sides of $r=4\sin \theta$ by $r$, so the right side becomes $4r\sin \theta =4y$.
2. Replace $r^2$ with $x^2+y^2$.
3. Complete the square in $y$.

Another way. Write $x=4\sin \theta \cos \theta =2\sin 2\theta$ and $y=4{\sin }^2\theta =2-2\cos 2\theta$, then $x^2+(y-2{)}^2=4{\sin }^{2}2\theta +4{\cos }^{2}2\theta =4$.

1. Differentiate both sides with respect to $x$, using the chain rule on the $y$ term.
2. Make $\frac{\mathrm{d}y}{\mathrm{d}x}$ the subject.

1. The gradient is undefined where the denominator is zero, so $y=2$.
2. Put $y=2$ back into the circle to find $x$.

1. From (d), the curve is a circle with centre $2\mathrm{i}$ (the point $(0,2)$) and radius $2$.
2. It touches the real axis only at the origin and reaches $4\mathrm{i}$ at the top of the imaginary axis.

1. With $r=4+4\sin \theta$, write $x$ and $y$ as functions of $\theta$.
2. Differentiate each with respect to $\theta$ using the product rule.
3. Tidy both, using ${\cos }^2\theta =1-{\sin }^2\theta$ in $\frac{\mathrm{d}x}{\mathrm{d}\theta }$.
4. Divide: $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}\theta }÷\frac{\mathrm{d}x}{\mathrm{d}\theta }$, and cancel the factor $4$.

1. A horizontal tangent means the gradient is zero, so the numerator is zero: $\cos \theta =0$ or $\sin \theta =-\frac{1}{2}$.
2. $A$ has a negative real part and, from the diagram, a negative imaginary part, so it is in the third quadrant. That rules out $\cos \theta =0$ and the fourth-quadrant root.