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IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2023 TZ2 paper, which your school or the IB can provide.

Practice mode

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Section A · Short response

Questions 1–9 · 57 marks

Question 1

6 marks

A circular sector: perimeter, angle and area

Geometry and trigonometry

(a)

2 marks

The sector is bounded by two radii and the arc PQR.

perimeter = 4 + 4 + 10 = 18 cm

(b)

2 marks

Arc length $= r θ$ .

10 = 4 θ \Rightarrow θ = 2.5

(c)

2 marks

Area of a sector.

A = \frac{1}{2} r^{2} θ = \frac{1}{2} (16) (2.5) = 20 cm^{2}

Question 2

5 marks

f(x) = 1 - 1/(x - 2): asymptotes, intercepts and sketch

Functions

(a)

2 marks

The function is $f (x) = 1 - \frac{1}{x - 2}$ . The denominator is zero at $x = 2$ , and for large $∣ x ∣$ the fraction tends to 0.

x = 2, y = 1

(b)

2 marks

At $x = 0$ : $f (0) = 1 + \frac{1}{2}$ . At $f (x) = 0$ : $\frac{1}{x - 2} = 1$ .

(0, \frac{3}{2}), (3, 0)

(c)

1 mark

Sketch: two branches of a hyperbola, with the asymptotes $x = 2$ and $y = 1$ . The left branch passes through $(0, \frac{3}{2})$ and lies above $y = 1$ for $x < 2$ and rises without bound as $x \to 2^{-}$ . The right branch passes through $(3, 0)$ and goes to $- \infty$ as $x \to 2^{+}$ , approaching $y = 1$ from below for large $x$ .

Question 3

5 marks

Finding P(B) from conditional probability and the union

Number and algebra

5 marks

Addition rule: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .
Conditional probability: $P (A \cap B) = P (A ∣ B) P (B)$ .
Combine them.

0.55 = 0.4 + P (B) - P (A \cap B) P (A \cap B) = 0.25 P (B) 0.55 = 0.4 + P (B) - 0.25 P (B) \Rightarrow 0.75 P (B) = 0.15 \Rightarrow P (B) = 0.2

Question 4

6 marks

Area under y = x/(x^2 + 2) equal to ln 3

Number and algebra

6 marks

Set the area equal to $ln 3$ .
Combine the logs.

\int_{0}^{c} \frac{x}{x ^{2} + 2} d x = [\frac{1}{2} ln (x^{2} + 2)]_{0}^{c} = \frac{1}{2} ln (c^{2} + 2) - \frac{1}{2} ln 2 = ln 3 ln \frac{c ^{2} + 2}{2} = 2 ln 3 = ln 9 \Rightarrow \frac{c ^{2} + 2}{2} = 9 \Rightarrow c^{2} = 16 c = 4

Question 5

7 marks

Two linear functions f with (g after f) given

Number and algebra

7 marks

Substitute $f$ into $g$ .
Compare with $4 x^{2} - 14 x + 15$ .
So $a = \pm 2$ . Use the middle equation to find $b$ : for $a = 2$ , $4 b + 2 = - 14$ gives $b = - 4$ . For $a = - 2$ , $- 4 b - 2 = - 14$ gives $b = 3$ .
Check the constant: $16 - 4 + 3 = 15$ and $9 + 3 + 3 = 15$ .

(g \circ f) (x) = (a x + b)^{2} + (a x + b) + 3 = a^{2} x^{2} + (2 ab + a) x + b^{2} + b + 3 a^{2} = 4, 2 ab + a = - 14, b^{2} + b + 3 = 15 f (x) = 2 x - 4 or f (x) = - 2 x + 3

Question 6

5 marks

Mean and variance of a uniform density

Statistics and probability

(a)

1 mark

The mean of a flat density is the midpoint of the interval.

E (X) = \frac{2 a + 3 a}{2} = \frac{5 a}{2}

(b)

4 marks

Find $E (X^{2})$ by integration.

E (X^{2}) = \int_{2 a}^{3 a} \frac{x ^{2}}{a} d x = \frac{1}{3 a} [x^{3}]_{2 a}^{3 a} = \frac{19 a ^{2}}{3} Var (X) = \frac{19 a ^{2}}{3} - \frac{25 a ^{2}}{4} = \frac{a ^{2}}{12}

Question 7

7 marks

Proof by induction of a sum involving factorials

Number and algebra

7 marks

Claim: $r = 1 \sum n \frac{r}{( r + 1 )!} = 1 - \frac{1}{( n + 1 )!}$ for $n \geq 1$ .
Base case $n = 1$ : the left side is $\frac{1}{2 !} = \frac{1}{2}$ and the right side is $1 - \frac{1}{2 !} = \frac{1}{2}$ . It holds.
Assume it holds for $n = k$ .
Then for $n = k + 1$ add the next term.
This is the formula for $n = k + 1$ .
The statement is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By mathematical induction it is true for all integers $n \geq 1$ .

r = 1 \sum k \frac{r}{( r + 1 )!} = 1 - \frac{1}{( k + 1 )!} r = 1 \sum k + 1 \frac{r}{( r + 1 )!} = 1 - \frac{1}{( k + 1 )!} + \frac{k + 1}{( k + 2 )!} = 1 - \frac{k + 2}{( k + 2 )!} + \frac{k + 1}{( k + 2 )!} = 1 - \frac{1}{( k + 2 )!}

Question 8

7 marks

Cosine and tangent curves meeting at right angles

Functions
Calculus

(a)

1 mark

At the intersection $cos k = tan k$ .

cos k = \frac{sin k}{cos k} \Rightarrow cos^{2} k = sin k

(b)

3 marks

The gradients at $P$ are $f^{'} (k) = - sin k$ and $g^{'} (k) = sec^{2} k = \frac{1}{cos ^{2} k} = \frac{1}{sin k}$ (using part (a)).
The product is $- 1$ , so the tangents are perpendicular.

f^{'} (k) g^{'} (k) = - sin k \cdot \frac{1}{sin k} = - 1

(c)

3 marks

Use $cos^{2} k = 1 - sin^{2} k$ in part (a).
The other root is negative and so is rejected, since $0 < k < \frac{π}{2}$ .

1 - sin^{2} k = sin k \Rightarrow sin^{2} k + sin k - 1 = 0 sin k = \frac{- 1 + 5}{2}

Question 9

9 marks

A parallelogram and a point M with OM perpendicular to MC

Geometry and trigonometry

(a)

2 marks

$M$ is on $A B$ , which is parallel to $O C$ , with $A M = k A B = k c$ .

O M = a + k c, M C = (1 - k) c - a

(b)

3 marks

$O M \cdot M C = 0$ . Expand.
Use $∣ c ∣ = 2∣ a ∣$ and $a \cdot c = 2∣ a ∣^{2} cos θ$ .

(a + k c) \cdot ((1 - k) c - a) = (1 - k) a \cdot c - ∣ a ∣^{2} + k (1 - k) ∣ c ∣^{2} - k a \cdot c ∣ a ∣^{2} [2 (1 - 2 k) cos θ - 1 + 4 k (1 - k)] = ∣ a ∣^{2} [2 (1 - 2 k) cos θ - (1 - 2 k)^{2}] ∣ a ∣^{2} (1 - 2 k) (2 cos θ - (1 - 2 k)) = 0

(c)

4 marks

Since $∣ a ∣^{2} > 0$ , either $k = \frac{1}{2}$ or $cos θ = \frac{1 - 2 k}{2}$ .
For $0 \leq k \leq 1$ , the value $1 - 2 k$ runs from $- 1$ to 1, so $- \frac{1}{2} \leq cos θ \leq \frac{1}{2}$ .
Two positions need a second value of $k$ besides $\frac{1}{2}$ . That happens for $cos θ = \frac{1 - 2 k}{2}$ with $k \neq = \frac{1}{2}$ , so $cos θ \neq = 0$ (when $θ = \frac{π}{2}$ both factors give the same $k$ ).

\frac{π}{3} \leq θ \leq \frac{2 π}{3}, θ \neq = \frac{π}{2}

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

14 marks

The largest triangle inscribed in a circle

Geometry and trigonometry
Calculus

(a)

1 mark

Q lies on the circle $x^{2} + y^{2} = 9$ and in the first quadrant.

y = 9 - x^{2}

(b)

3 marks

The base QR has length $2 y$ and the height from P is $x + 3$ .

A = \frac{1}{2} (2 y) (x + 3) = (x + 3) 9 - x^{2}

(c)

4 marks

Product rule.

\frac{d A}{d x} = 9 - x^{2} - \frac{x ( x + 3 )}{9 - x ^{2}} = \frac{9 - x ^{2} - x ^{2} - 3 x}{9 - x ^{2}} \frac{d A}{d x} = \frac{9 - 3 x - 2 x ^{2}}{9 - x ^{2}}

(d)

6 marks

Set $\frac{d A}{d x} = 0$ .
Reject $x = - 3$ . For $x = \frac{3}{2}$ , $y = 9 - \frac{9}{4} = \frac{3 3}{2}$ , and $R$ is $(x, - y)$ .

2 x^{2} + 3 x - 9 = 0 \Rightarrow (2 x - 3) (x + 3) = 0 \Rightarrow x = \frac{3}{2} y_{R} = - \frac{3 3}{2} \approx - 2.60

Question 11

22 marks

Complex numbers: powers, roots of a cubic and z squared equals 2 z-conjugate

Number and algebra

(a)

3 marks

$∣ u ∣ = 1 + 3 = 2$ . The point $(- 1, 3)$ is in the second quadrant with reference angle $\frac{π}{3}$ .

ar g u = π - \frac{π}{3} = \frac{2 π}{3} \Rightarrow u = 2 e^{2 π i /3}

(b)

5 marks

$u^{n} = 2^{n} e^{2 nπ i /3}$ is real when $sin \frac{2 nπ}{3} = 0$ , that is when $\frac{2 nπ}{3}$ is a multiple of $π$ , so $n$ is a multiple of 3.

n = 3 u^{3} = 2^{3} e^{2 π i} = 8

(c)

9 marks

By the conjugate root theorem $- 1 - 3 i$ is also a root. Let the real root be $c$ . The sum of roots is $- 5$ .
The roots of the cubic are $- 1 \pm 3 i$ and $- 3$ .
For $1 + 5 w + 10 w^{2} + 12 w^{3} = 0$ , divide by $w^{3}$ and let $z = \frac{1}{w}$ .

(- 1 + 3 i) + (- 1 - 3 i) + c = - 5 \Rightarrow c = - 3 z^{3} + 5 z^{2} + 10 z + 12 = 0 \Rightarrow w = \frac{1}{z} w = - \frac{1}{3}, w = \frac{1}{- 1 \pm 3 i} = \frac{- 1 \mp 3 i}{4}

(d)

5 marks

Write $z = a + bi$ , so $z^{2} = a^{2} - b^{2} + 2 abi$ and $2 \overset{z}{ˉ} = 2 a - 2 bi$ . Compare real and imaginary parts.
From the second equation $b = 0$ or $a = - 1$ . If $b = 0$ then $a^{2} = 2 a$ , and $z \neq = 0$ gives $a = 2$ . If $a = - 1$ then $1 - b^{2} = - 2$ , so $b = \pm 3$ .

a^{2} - b^{2} = 2 a, 2 ab = - 2 b z = 2, z = - 1 \pm 3 i

Question 12

17 marks

Integrating cos(sqrt x) and the areas between the zeros

Calculus
Functions

(a)

6 marks

Let $t = x$ , so $x = t^{2}$ and $d x = 2 t d t$ .
Integrate by parts with $u = t$ and $d v = cos t d t$ .

\int cos x d x = 2 \int t cos t d t 2 (t sin t - \int sin t d t) = 2 t sin t + 2 cos t + C \int cos x d x = 2 x sin x + 2 cos x + C

(b)

1 mark

Replace $n$ by $n + 1$ .

x_{n + 1} = \frac{( 2 n + 1 ) ^{2} π ^{2}}{4}

(c)

7 marks

The area of $R_{n}$ is $\int_{x_{n}}^{x_{n + 1}} cos x d x$ . Here $x_{n} = \frac{( 2 n - 1 ) π}{2}$ and $x_{n + 1} = \frac{( 2 n + 1 ) π}{2}$ .
At each endpoint the cosine is 0 and the sine is $\pm 1$ .
The area is positive.

F (x_{n + 1}) = (2 n + 1) π (- 1)^{n}, F (x_{n}) = (2 n - 1) π (- 1)^{n - 1} F (x_{n + 1}) - F (x_{n}) = (- 1)^{n} π [(2 n + 1) + (2 n - 1)] = (- 1)^{n} 4 nπ Area (R_{n}) = 4 nπ

(d)

3 marks

Find the difference between consecutive areas.
The difference is constant, so the areas form an arithmetic sequence with common difference $4 π$ .

Area (R_{n + 1}) - Area (R_{n}) = 4 (n + 1) π - 4 nπ = 4 π

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra47 marks · 43%
Q3, Q4, Q5, Q7, Q11
Calculus29 marks · 26%
Q8, Q10, Q12
Geometry and trigonometry19 marks · 17%
Q1, Q9, Q10
Functions10 marks · 9%
Q2, Q8, Q12
Statistics and probability5 marks · 5%
Q6

Mentor insights

Traps to watch for

01
Compare coefficients · Q5
Expand $(g \circ f) (x) = (a x + b)^{2} + (a x + b) + 3$ and match each power of $x$ .
02
Use the scalar product · Q9
Perpendicular vectors have scalar product 0. Expand it with $a \cdot c = ∣ a ∣∣ c ∣ cos θ$ .
03
Express the area in one variable · Q10
Use $y = 9 - x^{2}$ so that $A$ depends on $x$ only.
04
Conjugate root · Q11
The cubic has real coefficients, so $\overset{u}{ˉ}$ is also a root. The third root is real.
05
Substitute t = sqrt x · Q12
Then $d x = 2 t d t$ and the integral becomes $2 \int t cos t d t$ , which is by parts.

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Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2023 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–9 · 57 marks

Question 1

6 marks

A circular sector: perimeter, angle and area

Geometry and trigonometry

(a)

2 marks

The sector is bounded by two radii and the arc PQR.

perimeter = 4 + 4 + 10 = 18 cm

(b)

2 marks

Arc length $= r θ$ .

10 = 4 θ \Rightarrow θ = 2.5

(c)

2 marks

Area of a sector.

A = \frac{1}{2} r^{2} θ = \frac{1}{2} (16) (2.5) = 20 cm^{2}

Question 2

5 marks

f(x) = 1 - 1/(x - 2): asymptotes, intercepts and sketch

Functions

(a)

2 marks

The function is $f (x) = 1 - \frac{1}{x - 2}$ . The denominator is zero at $x = 2$ , and for large $∣ x ∣$ the fraction tends to 0.

x = 2, y = 1

(b)

2 marks

At $x = 0$ : $f (0) = 1 + \frac{1}{2}$ . At $f (x) = 0$ : $\frac{1}{x - 2} = 1$ .

(0, \frac{3}{2}), (3, 0)

(c)

1 mark

Question 3

5 marks

Finding P(B) from conditional probability and the union

Number and algebra

5 marks

Addition rule: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .
Conditional probability: $P (A \cap B) = P (A ∣ B) P (B)$ .
Combine them.

0.55 = 0.4 + P (B) - P (A \cap B) P (A \cap B) = 0.25 P (B) 0.55 = 0.4 + P (B) - 0.25 P (B) \Rightarrow 0.75 P (B) = 0.15 \Rightarrow P (B) = 0.2

Question 4

6 marks

Area under y = x/(x^2 + 2) equal to ln 3

Number and algebra

6 marks

Set the area equal to $ln 3$ .
Combine the logs.

\int_{0}^{c} \frac{x}{x ^{2} + 2} d x = [\frac{1}{2} ln (x^{2} + 2)]_{0}^{c} = \frac{1}{2} ln (c^{2} + 2) - \frac{1}{2} ln 2 = ln 3 ln \frac{c ^{2} + 2}{2} = 2 ln 3 = ln 9 \Rightarrow \frac{c ^{2} + 2}{2} = 9 \Rightarrow c^{2} = 16 c = 4

Question 5

7 marks

Two linear functions f with (g after f) given

Number and algebra

7 marks

Substitute $f$ into $g$ .
Compare with $4 x^{2} - 14 x + 15$ .
So $a = \pm 2$ . Use the middle equation to find $b$ : for $a = 2$ , $4 b + 2 = - 14$ gives $b = - 4$ . For $a = - 2$ , $- 4 b - 2 = - 14$ gives $b = 3$ .
Check the constant: $16 - 4 + 3 = 15$ and $9 + 3 + 3 = 15$ .

(g \circ f) (x) = (a x + b)^{2} + (a x + b) + 3 = a^{2} x^{2} + (2 ab + a) x + b^{2} + b + 3 a^{2} = 4, 2 ab + a = - 14, b^{2} + b + 3 = 15 f (x) = 2 x - 4 or f (x) = - 2 x + 3

Question 6

5 marks

Mean and variance of a uniform density

Statistics and probability

(a)

1 mark

The mean of a flat density is the midpoint of the interval.

E (X) = \frac{2 a + 3 a}{2} = \frac{5 a}{2}

(b)

4 marks

Find $E (X^{2})$ by integration.

E (X^{2}) = \int_{2 a}^{3 a} \frac{x ^{2}}{a} d x = \frac{1}{3 a} [x^{3}]_{2 a}^{3 a} = \frac{19 a ^{2}}{3} Var (X) = \frac{19 a ^{2}}{3} - \frac{25 a ^{2}}{4} = \frac{a ^{2}}{12}

Question 7

7 marks

Proof by induction of a sum involving factorials

Number and algebra

7 marks

Claim: $r = 1 \sum n \frac{r}{( r + 1 )!} = 1 - \frac{1}{( n + 1 )!}$ for $n \geq 1$ .
Base case $n = 1$ : the left side is $\frac{1}{2 !} = \frac{1}{2}$ and the right side is $1 - \frac{1}{2 !} = \frac{1}{2}$ . It holds.
Assume it holds for $n = k$ .
Then for $n = k + 1$ add the next term.
This is the formula for $n = k + 1$ .
The statement is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By mathematical induction it is true for all integers $n \geq 1$ .

r = 1 \sum k \frac{r}{( r + 1 )!} = 1 - \frac{1}{( k + 1 )!} r = 1 \sum k + 1 \frac{r}{( r + 1 )!} = 1 - \frac{1}{( k + 1 )!} + \frac{k + 1}{( k + 2 )!} = 1 - \frac{k + 2}{( k + 2 )!} + \frac{k + 1}{( k + 2 )!} = 1 - \frac{1}{( k + 2 )!}

Question 8

7 marks

Cosine and tangent curves meeting at right angles

Functions
Calculus

(a)

1 mark

At the intersection $cos k = tan k$ .

cos k = \frac{sin k}{cos k} \Rightarrow cos^{2} k = sin k

(b)

3 marks

The gradients at $P$ are $f^{'} (k) = - sin k$ and $g^{'} (k) = sec^{2} k = \frac{1}{cos ^{2} k} = \frac{1}{sin k}$ (using part (a)).
The product is $- 1$ , so the tangents are perpendicular.

f^{'} (k) g^{'} (k) = - sin k \cdot \frac{1}{sin k} = - 1

(c)

3 marks

Use $cos^{2} k = 1 - sin^{2} k$ in part (a).
The other root is negative and so is rejected, since $0 < k < \frac{π}{2}$ .

1 - sin^{2} k = sin k \Rightarrow sin^{2} k + sin k - 1 = 0 sin k = \frac{- 1 + 5}{2}

Question 9

9 marks

A parallelogram and a point M with OM perpendicular to MC

Geometry and trigonometry

(a)

2 marks

$M$ is on $A B$ , which is parallel to $O C$ , with $A M = k A B = k c$ .

O M = a + k c, M C = (1 - k) c - a

(b)

3 marks

$O M \cdot M C = 0$ . Expand.
Use $∣ c ∣ = 2∣ a ∣$ and $a \cdot c = 2∣ a ∣^{2} cos θ$ .

(a + k c) \cdot ((1 - k) c - a) = (1 - k) a \cdot c - ∣ a ∣^{2} + k (1 - k) ∣ c ∣^{2} - k a \cdot c ∣ a ∣^{2} [2 (1 - 2 k) cos θ - 1 + 4 k (1 - k)] = ∣ a ∣^{2} [2 (1 - 2 k) cos θ - (1 - 2 k)^{2}] ∣ a ∣^{2} (1 - 2 k) (2 cos θ - (1 - 2 k)) = 0

(c)

4 marks

Since $∣ a ∣^{2} > 0$ , either $k = \frac{1}{2}$ or $cos θ = \frac{1 - 2 k}{2}$ .
For $0 \leq k \leq 1$ , the value $1 - 2 k$ runs from $- 1$ to 1, so $- \frac{1}{2} \leq cos θ \leq \frac{1}{2}$ .
Two positions need a second value of $k$ besides $\frac{1}{2}$ . That happens for $cos θ = \frac{1 - 2 k}{2}$ with $k \neq = \frac{1}{2}$ , so $cos θ \neq = 0$ (when $θ = \frac{π}{2}$ both factors give the same $k$ ).

\frac{π}{3} \leq θ \leq \frac{2 π}{3}, θ \neq = \frac{π}{2}

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

14 marks

The largest triangle inscribed in a circle

Geometry and trigonometry
Calculus

(a)

1 mark

Q lies on the circle $x^{2} + y^{2} = 9$ and in the first quadrant.

y = 9 - x^{2}

(b)

3 marks

The base QR has length $2 y$ and the height from P is $x + 3$ .

A = \frac{1}{2} (2 y) (x + 3) = (x + 3) 9 - x^{2}

(c)

4 marks

Product rule.

\frac{d A}{d x} = 9 - x^{2} - \frac{x ( x + 3 )}{9 - x ^{2}} = \frac{9 - x ^{2} - x ^{2} - 3 x}{9 - x ^{2}} \frac{d A}{d x} = \frac{9 - 3 x - 2 x ^{2}}{9 - x ^{2}}

(d)

6 marks

Set $\frac{d A}{d x} = 0$ .
Reject $x = - 3$ . For $x = \frac{3}{2}$ , $y = 9 - \frac{9}{4} = \frac{3 3}{2}$ , and $R$ is $(x, - y)$ .

2 x^{2} + 3 x - 9 = 0 \Rightarrow (2 x - 3) (x + 3) = 0 \Rightarrow x = \frac{3}{2} y_{R} = - \frac{3 3}{2} \approx - 2.60

Question 11

22 marks

Complex numbers: powers, roots of a cubic and z squared equals 2 z-conjugate

Number and algebra

(a)

3 marks

$∣ u ∣ = 1 + 3 = 2$ . The point $(- 1, 3)$ is in the second quadrant with reference angle $\frac{π}{3}$ .

ar g u = π - \frac{π}{3} = \frac{2 π}{3} \Rightarrow u = 2 e^{2 π i /3}

(b)

5 marks

$u^{n} = 2^{n} e^{2 nπ i /3}$ is real when $sin \frac{2 nπ}{3} = 0$ , that is when $\frac{2 nπ}{3}$ is a multiple of $π$ , so $n$ is a multiple of 3.

n = 3 u^{3} = 2^{3} e^{2 π i} = 8

(c)

9 marks

By the conjugate root theorem $- 1 - 3 i$ is also a root. Let the real root be $c$ . The sum of roots is $- 5$ .
The roots of the cubic are $- 1 \pm 3 i$ and $- 3$ .
For $1 + 5 w + 10 w^{2} + 12 w^{3} = 0$ , divide by $w^{3}$ and let $z = \frac{1}{w}$ .

(- 1 + 3 i) + (- 1 - 3 i) + c = - 5 \Rightarrow c = - 3 z^{3} + 5 z^{2} + 10 z + 12 = 0 \Rightarrow w = \frac{1}{z} w = - \frac{1}{3}, w = \frac{1}{- 1 \pm 3 i} = \frac{- 1 \mp 3 i}{4}

(d)

5 marks

Write $z = a + bi$ , so $z^{2} = a^{2} - b^{2} + 2 abi$ and $2 \overset{z}{ˉ} = 2 a - 2 bi$ . Compare real and imaginary parts.
From the second equation $b = 0$ or $a = - 1$ . If $b = 0$ then $a^{2} = 2 a$ , and $z \neq = 0$ gives $a = 2$ . If $a = - 1$ then $1 - b^{2} = - 2$ , so $b = \pm 3$ .

a^{2} - b^{2} = 2 a, 2 ab = - 2 b z = 2, z = - 1 \pm 3 i

Question 12

17 marks

Integrating cos(sqrt x) and the areas between the zeros

Calculus
Functions

(a)

6 marks

Let $t = x$ , so $x = t^{2}$ and $d x = 2 t d t$ .
Integrate by parts with $u = t$ and $d v = cos t d t$ .

\int cos x d x = 2 \int t cos t d t 2 (t sin t - \int sin t d t) = 2 t sin t + 2 cos t + C \int cos x d x = 2 x sin x + 2 cos x + C

(b)

1 mark

Replace $n$ by $n + 1$ .

x_{n + 1} = \frac{( 2 n + 1 ) ^{2} π ^{2}}{4}

(c)

7 marks

The area of $R_{n}$ is $\int_{x_{n}}^{x_{n + 1}} cos x d x$ . Here $x_{n} = \frac{( 2 n - 1 ) π}{2}$ and $x_{n + 1} = \frac{( 2 n + 1 ) π}{2}$ .
At each endpoint the cosine is 0 and the sine is $\pm 1$ .
The area is positive.

F (x_{n + 1}) = (2 n + 1) π (- 1)^{n}, F (x_{n}) = (2 n - 1) π (- 1)^{n - 1} F (x_{n + 1}) - F (x_{n}) = (- 1)^{n} π [(2 n + 1) + (2 n - 1)] = (- 1)^{n} 4 nπ Area (R_{n}) = 4 nπ

(d)

3 marks

Find the difference between consecutive areas.
The difference is constant, so the areas form an arithmetic sequence with common difference $4 π$ .

Area (R_{n + 1}) - Area (R_{n}) = 4 (n + 1) π - 4 nπ = 4 π

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra47 marks · 43%
Q3, Q4, Q5, Q7, Q11
Calculus29 marks · 26%
Q8, Q10, Q12
Geometry and trigonometry19 marks · 17%
Q1, Q9, Q10
Functions10 marks · 9%
Q2, Q8, Q12
Statistics and probability5 marks · 5%
Q6

Mentor insights

Traps to watch for

01
Compare coefficients · Q5
Expand $(g \circ f) (x) = (a x + b)^{2} + (a x + b) + 3$ and match each power of $x$ .
02
Use the scalar product · Q9
Perpendicular vectors have scalar product 0. Expand it with $a \cdot c = ∣ a ∣∣ c ∣ cos θ$ .
03
Express the area in one variable · Q10
Use $y = 9 - x^{2}$ so that $A$ depends on $x$ only.
04
Conjugate root · Q11
The cubic has real coefficients, so $\overset{u}{ˉ}$ is also a root. The third root is real.
05
Substitute t = sqrt x · Q12
Then $d x = 2 t d t$ and the integral becomes $2 \int t cos t d t$ , which is by parts.

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