HL AA · Practice Paper 1 · Interactive

IB Maths Analysis & Approaches HL — Interactive Practice Paper 1

HL AA Paper 1 is no calculator — proof-flavour, complex numbers, deep calculus and exact-value trigonometry. Two modes: per-part feedback, or full timed exam with a countdown.

30 marks · 5 questions · No GDC · Exact answers
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Question 1

6 marks
Complex numbers — modulus, argument, De Moivre

Let $z = 1 + i\sqrt{3}$.

(a) Find the modulus $|z|$. [1]

(b) Find $\arg(z)$ in the form $\dfrac{\pi}{k}$ where $k$ is a positive integer. Type just $k$. [2]

(c) Use De Moivre's theorem to write $z^{6}$ as $a + bi$. Type both integers. [3]

(a) $|z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = \mathbf{2}$.
(b) $\arg z = \arctan\!\left(\tfrac{\sqrt{3}}{1}\right) = \tfrac{\pi}{3}$, so $k = \mathbf{3}$.
(c) $z^6 = 2^6\!\left[\cos(6\cdot\tfrac{\pi}{3}) + i\sin(6\cdot\tfrac{\pi}{3})\right] = 64(\cos 2\pi + i\sin 2\pi) = 64 + 0i$, so $\mathbf{a = 64,\;b = 0}$.

Question 2

6 marks
Sequences & series — nth term from Sn

The sum of the first $n$ terms of a sequence is $S_n = 2n^2 + 3n$.

(a) Find $u_1$. [1]

(b) Find $u_n$ in the form $an + b$. Type the two integers. [3]

(c) Hence, state the common difference of the sequence. [2]

(a) $u_1 = S_1 = 2(1)^2 + 3(1) = \mathbf{5}$.
(b) For $n \geq 2$, $u_n = S_n - S_{n-1} = (2n^2+3n) - \bigl(2(n-1)^2 + 3(n-1)\bigr) = (2n^2+3n) - (2n^2 - 4n + 2 + 3n - 3) = 4n + 1$. Check $u_1 = 4(1)+1 = 5$ ✓, so $\mathbf{u_n = 4n+1}$.
(c) $u_{n+1} - u_n = 4(n+1)+1 - (4n+1) = \mathbf{4}$.

Question 3

6 marks
Trigonometry — double angle identities

Given that $\cos 2\theta = \tfrac{7}{25}$ and $0 < 2\theta < \tfrac{\pi}{2}$, find the exact value of:

(a) $\sin 2\theta$. Type in the form $p/q$ (or a decimal). [2]

(b) $\cos\theta$. Type in the form $p/q$. [2]

(c) $\tan\theta$. Type in the form $p/q$. [2]

(a) $\sin^2 2\theta = 1 - \tfrac{49}{625} = \tfrac{576}{625}$; since $2\theta$ acute, $\sin 2\theta = \mathbf{\tfrac{24}{25}}$.
(b) $\cos 2\theta = 2\cos^2\theta - 1 \Rightarrow 2\cos^2\theta = 1 + \tfrac{7}{25} = \tfrac{32}{25} \Rightarrow \cos^2\theta = \tfrac{16}{25}$; $\theta$ acute, so $\cos\theta = \mathbf{\tfrac{4}{5}}$.
(c) $\sin^2\theta = 1 - \tfrac{16}{25} = \tfrac{9}{25} \Rightarrow \sin\theta = \tfrac{3}{5}$; hence $\tan\theta = \mathbf{\tfrac{3}{4}}$.

Question 4

6 marks
Differentiation — product rule & stationary points

Let $f(x) = x^{2} e^{-x}$ for $x \in \mathbb{R}$.

(a) Show that $f'(x) = xe^{-x}(2 - x)$. Then type the coefficient of $x$ inside the bracket, i.e. the coefficient of $x$ in $(2-x)$. [3]

(b) Find the $x$-coordinates of the two stationary points. Type the smaller value first. [2]

(c) Classify the stationary point at $x = 2$. Type max or min. [1]

(a) Product rule: $f'(x) = 2xe^{-x} + x^2(-e^{-x}) = e^{-x}(2x - x^2) = \mathbf{xe^{-x}(2-x)}$. Coefficient of $x$ in $(2-x)$ is $\mathbf{-1}$.
(b) $f'(x) = 0 \Rightarrow x = 0$ or $x = 2$ (since $e^{-x} \ne 0$).
(c) Sign of $f'(x)$: just to the left of $x=2$ (e.g. $x=1.5$), $f'(1.5) = 1.5 e^{-1.5}(0.5) > 0$; just to the right (e.g. $x=2.5$), $f'(2.5) = 2.5 e^{-2.5}(-0.5) < 0$. Sign change $+ \to -$, so $x=2$ is a local maximum.

Question 5

6 marks
Integration — parts & definite integral

Consider the following definite integrals.

(a) Evaluate $\displaystyle\int_{1}^{e} \dfrac{1}{x}\,dx$. Give the exact value. [1]

(b) Evaluate $\displaystyle\int_{0}^{\pi/2} \sin x\,dx$. Give the exact value. [1]

(c) Use integration by parts to show that $\displaystyle\int_{0}^{\pi/2} x\cos x\,dx = \dfrac{\pi}{a} - b$. Type each integer. [4]

(a) $\int_1^e \tfrac{1}{x}\,dx = [\ln x]_1^e = \ln e - \ln 1 = \mathbf{1}$.
(b) $\int_0^{\pi/2} \sin x\,dx = [-\cos x]_0^{\pi/2} = 0 - (-1) = \mathbf{1}$.
(c) Let $u = x,\;dv = \cos x\,dx \Rightarrow du = dx,\;v = \sin x$. Then $\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x$. Evaluated from $0$ to $\tfrac{\pi}{2}$: $\bigl[\tfrac{\pi}{2}(1) + 0\bigr] - \bigl[0 + 1\bigr] = \tfrac{\pi}{2} - 1$. So $\mathbf{a = 2,\;b = 1}$.

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