(a) Find the modulus | 𝑧 | . [1]

(a) |z|=12+(3)2=1+3=2. (b) arg⁡z=arctan(31)=π3, so k=3. (c) z6=26[cos⁡(6⋅π3)+isin⁡(6⋅π3)]=64(cos⁡2π+isin⁡2π)=64+0i, so a=64,b=0.

Given that cos ⁡ 2 𝜃 = 7 25 and 0 < 2 𝜃 < 𝜋 2 , find the exact value of:

(a) sin2⁡2θ=1−49625=576625; since 2θ acute, sin⁡2θ=2425. (b) cos⁡2θ=2cos2⁡θ−1⇒2cos2⁡θ=1+725=3225⇒cos2⁡θ=1625; θ acute, so cos⁡θ=45. (c) sin2⁡θ=1−1625=925⇒sin⁡θ=35; hence tan⁡θ=34.

Let 𝑓 ( 𝑥 ) = 𝑥 2 𝑒 − 𝑥 for 𝑥 ∈ 𝑅 .

(a) Product rule: f′(x)=2xe−x+x2(−e−x)=e−x(2x−x2)=xe−x(2−x). Coefficient of x in (2−x) is −1. (b) f′(x)=0⇒x=0 or x=2 (since e−x≠0). (c) Sign of f′(x): just to the left of x=2 (e.g. x=1.5), f′(1.5)=1.5e−1.5(0.5)>0; just to the right (e.g. x=2.5), f′(2.5)=2.5e−2.5(−0.5)<0. Sign change +→−, so x=2 is a local maximum.

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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Progress: 0 / 19 correct

Question 1

6 marks

Complex numbers — modulus, argument, De Moivre

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

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

| z | =

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

k =

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

a =

b =

(a)

| z | = \sqrt{1^{2} + (\sqrt{3})^{2}} = \sqrt{1 + 3} = 2

.
(b)

\arg z = \arctan (\frac{\sqrt{3}}{1}) = \frac{π}{3}

, so

k = 3

.
(c)

z^{6} = 2^{6} [\cos (6 \cdot \frac{π}{3}) + i \sin (6 \cdot \frac{π}{3})] = 64 (\cos 2 π + i \sin 2 π) = 64 + 0 i

, so

a = 64, b = 0

Question 2

6 marks

Sequences & series — nth term from S_n

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

(a) Find $u_{1}$ . [1]

u_{1} =

(b) Find $u_{n}$ in the form $a n + b$ . Type the two integers. [3]

a =

b =

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

d =

(a)

u_{1} = S_{1} = 2 (1)^{2} + 3 (1) = 5

.
(b) For

n \geq 2

u_{n} = S_{n} - S_{n - 1} = (2 n^{2} + 3 n) - (2 (n - 1)^{2} + 3 (n - 1)) = (2 n^{2} + 3 n) - (2 n^{2} - 4 n + 2 + 3 n - 3) = 4 n + 1

. Check

u_{1} = 4 (1) + 1 = 5

✓, so

u_{n} = 4 n + 1

.
(c)

u_{n + 1} - u_{n} = 4 (n + 1) + 1 - (4 n + 1) = 4

Question 3

6 marks

Trigonometry — double angle identities

Given that $\cos 2 θ = \frac{7}{25}$ and $0 < 2 θ < \frac{π}{2}$ , find the exact value of:

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

\sin 2 θ =

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

\cos θ =

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

\tan θ =

(a)

\sin^{2} 2 θ = 1 - \frac{49}{625} = \frac{576}{625}

; since

2 θ

acute,

\sin 2 θ = \frac{24}{25}

.
(b)

\cos 2 θ = 2 \cos^{2} θ - 1 \Rightarrow 2 \cos^{2} θ = 1 + \frac{7}{25} = \frac{32}{25} \Rightarrow \cos^{2} θ = \frac{16}{25}

;

θ

acute, so

\cos θ = \frac{4}{5}

.
(c)

\sin^{2} θ = 1 - \frac{16}{25} = \frac{9}{25} \Rightarrow \sin θ = \frac{3}{5}

; hence

\tan θ = \frac{3}{4}

Question 4

6 marks

Differentiation — product rule & stationary points

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

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

coefficient

=

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

smaller

x =

larger

x =

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

type =

(a) Product rule:

f^{'} (x) = 2 x e^{- x} + x^{2} (- e^{- x}) = e^{- x} (2 x - x^{2}) = {xe}^{- x} (2 - x)

. Coefficient of

x

(2 - x)

- 1

.
(b)

f^{'} (x) = 0 \Rightarrow x = 0

x = 2

(since

e^{- x} \neq 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 $\int_{1}^{e} \frac{1}{x} d x$ . Give the exact value. [1]

value

=

(b) Evaluate $\int_{0}^{π / 2} \sin x d x$ . Give the exact value. [1]

value

=

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

a =

b =

(a)

\int_{1}^{e} \frac{1}{x} d x = [\ln x]_{1}^{e} = \ln e - \ln 1 = 1

.
(b)

\int_{0}^{π / 2} \sin x d x = [- \cos x]_{0}^{π / 2} = 0 - (- 1) = 1

.
(c) Let

u = x, d v = \cos x d x \Rightarrow d u = d x, v = \sin x

. Then

\int x \cos x d x = x \sin x - \int \sin x d x = x \sin x + \cos x

. Evaluated from

0

\frac{π}{2}

[\frac{π}{2} (1) + 0] - [0 + 1] = \frac{π}{2} - 1

. So

a = 2, b = 1

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