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IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 110
Duration: 2 hours
Calculator: 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 TZ1 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 · 55 marks

Question 1

5 marks

Three cities: the sine rule with a bearing

Geometry and trigonometry

(a)

2 marks

The angle at J is $5 6^{\circ}$ and the angle at L is $2 5^{\circ}$ . The angles of triangle JLD add to $18 0^{\circ}$ .

J D L = 180 - 56 - 25 = 9 9^{\circ}

(b)

3 marks

Use the sine rule, with LD opposite the angle at J.

\frac{L D}{sin 5 6 ^{\circ}} = \frac{500}{sin 9 9 ^{\circ}} \Rightarrow L D \approx 420 km

Question 2

5 marks

A composite function of tan and a cubic: expression and sketch

Functions

(a)

2 marks

Substitute $g (x) = tan x$ into $f$ .

(f \circ g) (x) = 2 tan x - tan^{3} x

(b)

3 marks

The graph is an odd function through the origin.
The turning points on $- 1 \leq x \leq 1$ are
Sketch: starting at $(- 1, 0.66)$ the curve falls to the local minimum $(- 0.685, - 1.09)$ , rises through the origin to the local maximum $(0.685, 1.09)$ and then falls to $(1, - 0.66)$ .

local maximum (0.685, 1.09), local minimum (- 0.685, - 1.09)

Question 3

7 marks

Children in a park and ice creams: a model and a regression of x on y

Calculus
Statistics and probability

(a)

2 marks

Substitute $T = 25$ into the model.

y = - 0.6 (25)^{2} + 23 (25) + 110 = 310

(b)

3 marks

Regress $x$ on $y$ using the five data points.

x = 0.0935 y + 7.43

(c)

2 marks

Use the 310 children from part (a).
About 36 ice creams.

x = 0.0935 (310) + 7.43 \approx 36.4

Question 4

5 marks

Tube diameters: finding sigma from the interquartile range

Number and algebra

5 marks

The quartiles are the 25th and 75th percentiles. For $N (32, s^{2})$ they are $32 \pm 0.6745 s$ .

IQR = 2 (0.6745) s = 0.28 s = 0.208

Question 5

7 marks

Finding a and b from two binomial coefficients

Number and algebra

7 marks

The term in $x^{6}$ is $(2 n) (a x^{3})^{2} b^{n - 2}$ .
Divide the second by the first.
Then $a^{2} = \frac{16}{64} = \frac{1}{4}$ .

(2 8) a^{2} b^{6} = 28 a^{2} b^{6} = 448 \Rightarrow a^{2} b^{6} = 16 (2 10) a^{2} b^{8} = 45 a^{2} b^{8} = 2880 \Rightarrow a^{2} b^{8} = 64 b^{2} = 4 \Rightarrow b = 2 (b > 0) a = \frac{1}{2}, b = 2

Question 6

7 marks

Powers of a complex number on the unit circle

Number and algebra

(a)

4 marks

$z^{n} = cos \frac{11 nπ}{18} + i sin \frac{11 nπ}{18}$ . The number $- i$ has argument $\frac{3 π}{2}$ , up to multiples of $2 π$ .
The smallest positive integer solution (try $k = 0, 1, 2$ ) is

\frac{11 nπ}{18} = \frac{3 π}{2} + 2 k π \Rightarrow 11 n = 27 + 36 k k = 2 : 11 n = 99 \Rightarrow n = 9

(b)

3 marks

From (a), $z^{9} = - i$ , so $z^{10} = z^{9} \cdot z = - i z$ .
Multiplying by $- i$ rotates a point by $\frac{π}{2}$ clockwise about the origin. The transformation is a rotation of $9 0^{\circ}$ clockwise about the origin.

Question 7

6 marks

A gene test: tree diagram and estimating the proportion

Statistics and probability

(a)

2 marks

The branches are $G$ (probability $p$ ) and $G^{'}$ (probability $1 - p$ ). From $G$ : positive 0.95, negative 0.05. From $G^{'}$ : positive 0.02, negative 0.98.

(b)

4 marks

Let the sample have $S$ parrots. The true positives are $0.95 pS = 132$ and the false positives are $0.02 (1 - p) S = 18$ .

\frac{0.95 p}{0.02 ( 1 - p )} = \frac{132}{18} 0.95 p \cdot 18 = 132 \cdot 0.02 (1 - p) \Rightarrow 17.1 p = 2.64 - 2.64 p \Rightarrow 19.74 p = 2.64 p \approx 0.134

Question 8

7 marks

The angle between a line and a plane

Number and algebra

7 marks

The direction vector of the line is $d = 32 - 1$ and the normal of the plane is $n = 4 cos α sin α$ .
The angle between the line and the plane is $α$ , so the angle between $d$ and $n$ is $\frac{π}{2} - α$ .
Solve on a graph for $0 < α < \frac{π}{2}$ .

d \cdot n = 12 + 2 cos α - sin α, ∣ d ∣ = 14, ∣ n ∣ = 16 + cos^{2} α + sin^{2} α = 17 sin α = \frac{12 + 2 cos α - sin α}{14 17} α \approx 0.932

Question 9

6 marks

Proof by contradiction that p^2 - 8q - 11 is never zero

Number and algebra

6 marks

Assume $p^{2} - 8 q - 11 = 0$ for integers $p$ and $q$ , so $p^{2} = 8 q + 11$ .
The right side is odd, so $p^{2}$ is odd and so $p$ is odd. Write $p = 2 m + 1$ for an integer $m$ .
The left side $2 (m^{2} + m - 2 q)$ is even, but the right side 5 is odd. This is a contradiction.
So the assumption is false and $p^{2} - 8 q - 11 \neq = 0$ for all integers $p, q$ .

(2 m + 1)^{2} = 8 q + 11 \Rightarrow 4 m^{2} + 4 m + 1 = 8 q + 11 \Rightarrow 2 m^{2} + 2 m - 4 q = 5

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

15 marks

A rain gutter: angle, cross-section area and overflow

Geometry and trigonometry
Calculus

(a)

3 marks

The width of the sheet is the two vertical sides plus the arc AB.

45 = 10 + 10 + 12 θ \Rightarrow θ = \frac{25}{12} = 2.08

(b)

7 marks

Segment AHB: sector minus triangle.
The width of the gutter is the chord AB.
Add the rectangle $10 \times w$ below the arc.

\frac{1}{2} (12)^{2} (2.08) - \frac{1}{2} (12)^{2} sin (2.08) = 149.76 - 62.87 = 86.89 w = 2 (12) sin (1.04) = 20.70 area = 86.89 + 10 (20.70) \approx 294 cm^{2}

(c)

5 marks

The gutter holds $294 \times 600 \approx 176000 cm^{3}$ .
In any 60 seconds the rainwater is
The cosine term gives 0 since 60 seconds is exactly 12 periods of 5 seconds, so every 60-second period brings 180 000 cm $^{3}$ .
Since $180000 > 176000$ , the gutter overflowed.

\int_{0}^{60} (50 cos \frac{2 π t}{5} + 3000) d t = [\frac{250}{2 π} sin \frac{2 π t}{5} + 3000 t]_{0}^{60} = 180000 cm^{3}

Question 11

19 marks

A density with a square root: mean, probability and consecutive successes

Calculus
Statistics and probability

(a)

5 marks

$f (x) = \frac{6}{π 16 - x ^{2}}$ , so $E (X) = \int_{0}^{2} \frac{6 x}{π 16 - x ^{2}} d x$ . Use $u = 16 - x^{2}$ .

E (X) = \frac{6}{π} [- 16 - x^{2}]_{0}^{2} = \frac{6}{π} (4 - 12) = \frac{24 - 12 3}{π}

(b)

2 marks

Integrate the density from 0 to 0.5.

P (X < 0.5) = \int_{0}^{0.5} \frac{6}{π 16 - x ^{2}} d x = 0.239

(c)

3 marks

The chance of at least one success in $n$ trials is $1 - (1 - 0.2394)^{n}$ .
The least number of trials is 17.

1 - 0.760 6^{n} \geq 0.99 \Rightarrow n \geq \frac{ln 0.01}{ln 0.7606} = 16.8

(d)

2 marks

The number of successes is $B (10, 0.2394)$ .

P (X = 3) = (3 10) (0.2394)^{3} (0.7606)^{7} \approx 0.242

(e)

1 mark

A block of three in a row can start at positions 1 to 8.

8

(f)

6 marks

For $n$ trials there are $n - 2$ blocks of three consecutive trials, and all choices of three trials number $(3 n)$ .
The probability that three given successes are consecutive is $\frac{n - 2}{( 3 n )}$ . Evaluate for increasing $n$ .
The greatest $n$ with the probability above 0.05 is

(i) n - 2 n = 10 : 0.0667, n = 11 : 0.0545, n = 12 : 0.0455 n = 11

Question 12

21 marks

Euler's method, Maclaurin series and exact solution of a differential equation

Calculus

(a)

4 marks

The slope is $\frac{d y}{d x} = \frac{x ^{2} y - y}{x ^{2} + 1}$ . Start from $x_{0} = 0$ , $y_{0} = 3$ and take five steps of 0.03.
At $x = 0.15$ :

y_{1} = 2.91, y_{2} = 2.82286, y_{3} = 2.73878, y_{4} = 2.65794, y_{5} = 2.58046 y \approx 2.58046

(b)

5 marks

At $x = 0$ , with $y = 3$ :
Differentiate $(x^{2} + 1) \frac{d y}{d x} = x^{2} y - y$ with the product rule and implicit differentiation.
At $x = 0$ , $y = 3$ , $\frac{d y}{d x} = - 3$ :

\frac{d y}{d x} = \frac{0 - 3}{1} = - 3 2 x \frac{d y}{d x} + (x^{2} + 1) \frac{d ^{2} y}{d x ^{2}} = 2 x y + x^{2} \frac{d y}{d x} - \frac{d y}{d x} \frac{d ^{2} y}{d x ^{2}} = 0 + 0 + 3 = 3

(c)

3 marks

Maclaurin series $y = y (0) + y^{'} (0) x + \frac{y ^{''} ( 0 )}{2 !} x^{2} + \frac{y ^{'''} ( 0 )}{3 !} x^{3} + \dots$
At $x = 0.15$ :

y = 3 - 3 x + \frac{3}{2} x^{2} + \frac{3}{2} x^{3} y \approx 2.58881

(d)

7 marks

Separate the variables, using the given identity $\frac{x ^{2} - 1}{x ^{2} + 1} = 1 - \frac{2}{x ^{2} + 1}$ .
Use $y = 3$ when $x = 0$ .
At $x = 0.15$ :

\int \frac{d y}{y} = \int (1 - \frac{2}{x ^{2} + 1}) d x \Rightarrow ln y = x - 2 arctan x + c ln 3 = c y = 3 e^{x - 2 a r c t a n x} y = 3 e^{0.15 - 2 a r c t a n 0.15} \approx 2.58786

(e)

2 marks

Differentiate $\frac{y ^{'}}{y} = 1 - \frac{2}{x ^{2} + 1}$ .
For $y > 0$ and $x \geq 0$ the bracket is positive, so $y^{''} > 0$ : the graph is concave up. Each Euler step follows a tangent, and a tangent to a concave-up curve lies below it, so every Euler estimate is too small.

\frac{y ^{''}}{y} - (\frac{y ^{'}}{y})^{2} = \frac{4 x}{( x ^{2} + 1 ) ^{2}} \Rightarrow y^{''} = y [(\frac{y ^{'}}{y})^{2} + \frac{4 x}{( x ^{2} + 1 ) ^{2}}]

Our analysis

Where the marks are

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

Calculus35 marks · 32%
Q3, Q10, Q11, Q12
Number and algebra32 marks · 29%
Q4, Q5, Q6, Q8, Q9
Statistics and probability23 marks · 21%
Q3, Q7, Q11
Geometry and trigonometry15 marks · 14%
Q1, Q10
Functions5 marks · 5%
Q2

Mentor insights

Traps to watch for

01
Regress x on y · Q3
The vendor wants to predict ice creams ( $x$ ) from the number of children ( $y$ ), so use the regression line of $x$ on $y$ .
02
Pick the right term · Q5
In $(a x^{3} + b)^{n}$ the $x^{6}$ term needs $(a x^{3})^{2}$ , which is the term with $r = 2$ .
03
The sheet width fixes the arc · Q10
The 45 cm sheet is two vertical sides of 10 cm plus the curved arc, so the arc is 25 cm.
04
Check the density first · Q11
$\frac{6}{π 16 - x ^{2}}$ integrates to 1 on $[0, 2]$ because $arcsin \frac{1}{2} = \frac{π}{6}$ .
05
Euler uses the old values · Q12
Each step is $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ with $h = 0.03$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 110
Duration: 2 hours
Calculator: 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 TZ1 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 · 55 marks

Question 1

5 marks

Three cities: the sine rule with a bearing

Geometry and trigonometry

(a)

2 marks

The angle at J is $5 6^{\circ}$ and the angle at L is $2 5^{\circ}$ . The angles of triangle JLD add to $18 0^{\circ}$ .

J D L = 180 - 56 - 25 = 9 9^{\circ}

(b)

3 marks

Use the sine rule, with LD opposite the angle at J.

\frac{L D}{sin 5 6 ^{\circ}} = \frac{500}{sin 9 9 ^{\circ}} \Rightarrow L D \approx 420 km

Question 2

5 marks

A composite function of tan and a cubic: expression and sketch

Functions

(a)

2 marks

Substitute $g (x) = tan x$ into $f$ .

(f \circ g) (x) = 2 tan x - tan^{3} x

(b)

3 marks

The graph is an odd function through the origin.
The turning points on $- 1 \leq x \leq 1$ are
Sketch: starting at $(- 1, 0.66)$ the curve falls to the local minimum $(- 0.685, - 1.09)$ , rises through the origin to the local maximum $(0.685, 1.09)$ and then falls to $(1, - 0.66)$ .

local maximum (0.685, 1.09), local minimum (- 0.685, - 1.09)

Question 3

7 marks

Children in a park and ice creams: a model and a regression of x on y

Calculus
Statistics and probability

(a)

2 marks

Substitute $T = 25$ into the model.

y = - 0.6 (25)^{2} + 23 (25) + 110 = 310

(b)

3 marks

Regress $x$ on $y$ using the five data points.

x = 0.0935 y + 7.43

(c)

2 marks

Use the 310 children from part (a).
About 36 ice creams.

x = 0.0935 (310) + 7.43 \approx 36.4

Question 4

5 marks

Tube diameters: finding sigma from the interquartile range

Number and algebra

5 marks

The quartiles are the 25th and 75th percentiles. For $N (32, s^{2})$ they are $32 \pm 0.6745 s$ .

IQR = 2 (0.6745) s = 0.28 s = 0.208

Question 5

7 marks

Finding a and b from two binomial coefficients

Number and algebra

7 marks

The term in $x^{6}$ is $(2 n) (a x^{3})^{2} b^{n - 2}$ .
Divide the second by the first.
Then $a^{2} = \frac{16}{64} = \frac{1}{4}$ .

(2 8) a^{2} b^{6} = 28 a^{2} b^{6} = 448 \Rightarrow a^{2} b^{6} = 16 (2 10) a^{2} b^{8} = 45 a^{2} b^{8} = 2880 \Rightarrow a^{2} b^{8} = 64 b^{2} = 4 \Rightarrow b = 2 (b > 0) a = \frac{1}{2}, b = 2

Question 6

7 marks

Powers of a complex number on the unit circle

Number and algebra

(a)

4 marks

$z^{n} = cos \frac{11 nπ}{18} + i sin \frac{11 nπ}{18}$ . The number $- i$ has argument $\frac{3 π}{2}$ , up to multiples of $2 π$ .
The smallest positive integer solution (try $k = 0, 1, 2$ ) is

\frac{11 nπ}{18} = \frac{3 π}{2} + 2 k π \Rightarrow 11 n = 27 + 36 k k = 2 : 11 n = 99 \Rightarrow n = 9

(b)

3 marks

From (a), $z^{9} = - i$ , so $z^{10} = z^{9} \cdot z = - i z$ .
Multiplying by $- i$ rotates a point by $\frac{π}{2}$ clockwise about the origin. The transformation is a rotation of $9 0^{\circ}$ clockwise about the origin.

Question 7

6 marks

A gene test: tree diagram and estimating the proportion

Statistics and probability

(a)

2 marks

The branches are $G$ (probability $p$ ) and $G^{'}$ (probability $1 - p$ ). From $G$ : positive 0.95, negative 0.05. From $G^{'}$ : positive 0.02, negative 0.98.

(b)

4 marks

Let the sample have $S$ parrots. The true positives are $0.95 pS = 132$ and the false positives are $0.02 (1 - p) S = 18$ .

\frac{0.95 p}{0.02 ( 1 - p )} = \frac{132}{18} 0.95 p \cdot 18 = 132 \cdot 0.02 (1 - p) \Rightarrow 17.1 p = 2.64 - 2.64 p \Rightarrow 19.74 p = 2.64 p \approx 0.134

Question 8

7 marks

The angle between a line and a plane

Number and algebra

7 marks

The direction vector of the line is $d = 32 - 1$ and the normal of the plane is $n = 4 cos α sin α$ .
The angle between the line and the plane is $α$ , so the angle between $d$ and $n$ is $\frac{π}{2} - α$ .
Solve on a graph for $0 < α < \frac{π}{2}$ .

d \cdot n = 12 + 2 cos α - sin α, ∣ d ∣ = 14, ∣ n ∣ = 16 + cos^{2} α + sin^{2} α = 17 sin α = \frac{12 + 2 cos α - sin α}{14 17} α \approx 0.932

Question 9

6 marks

Proof by contradiction that p^2 - 8q - 11 is never zero

Number and algebra

6 marks

Assume $p^{2} - 8 q - 11 = 0$ for integers $p$ and $q$ , so $p^{2} = 8 q + 11$ .
The right side is odd, so $p^{2}$ is odd and so $p$ is odd. Write $p = 2 m + 1$ for an integer $m$ .
The left side $2 (m^{2} + m - 2 q)$ is even, but the right side 5 is odd. This is a contradiction.
So the assumption is false and $p^{2} - 8 q - 11 \neq = 0$ for all integers $p, q$ .

(2 m + 1)^{2} = 8 q + 11 \Rightarrow 4 m^{2} + 4 m + 1 = 8 q + 11 \Rightarrow 2 m^{2} + 2 m - 4 q = 5

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

15 marks

A rain gutter: angle, cross-section area and overflow

Geometry and trigonometry
Calculus

(a)

3 marks

The width of the sheet is the two vertical sides plus the arc AB.

45 = 10 + 10 + 12 θ \Rightarrow θ = \frac{25}{12} = 2.08

(b)

7 marks

Segment AHB: sector minus triangle.
The width of the gutter is the chord AB.
Add the rectangle $10 \times w$ below the arc.

\frac{1}{2} (12)^{2} (2.08) - \frac{1}{2} (12)^{2} sin (2.08) = 149.76 - 62.87 = 86.89 w = 2 (12) sin (1.04) = 20.70 area = 86.89 + 10 (20.70) \approx 294 cm^{2}

(c)

5 marks

The gutter holds $294 \times 600 \approx 176000 cm^{3}$ .
In any 60 seconds the rainwater is
The cosine term gives 0 since 60 seconds is exactly 12 periods of 5 seconds, so every 60-second period brings 180 000 cm $^{3}$ .
Since $180000 > 176000$ , the gutter overflowed.

\int_{0}^{60} (50 cos \frac{2 π t}{5} + 3000) d t = [\frac{250}{2 π} sin \frac{2 π t}{5} + 3000 t]_{0}^{60} = 180000 cm^{3}

Question 11

19 marks

A density with a square root: mean, probability and consecutive successes

Calculus
Statistics and probability

(a)

5 marks

$f (x) = \frac{6}{π 16 - x ^{2}}$ , so $E (X) = \int_{0}^{2} \frac{6 x}{π 16 - x ^{2}} d x$ . Use $u = 16 - x^{2}$ .

E (X) = \frac{6}{π} [- 16 - x^{2}]_{0}^{2} = \frac{6}{π} (4 - 12) = \frac{24 - 12 3}{π}

(b)

2 marks

Integrate the density from 0 to 0.5.

P (X < 0.5) = \int_{0}^{0.5} \frac{6}{π 16 - x ^{2}} d x = 0.239

(c)

3 marks

The chance of at least one success in $n$ trials is $1 - (1 - 0.2394)^{n}$ .
The least number of trials is 17.

1 - 0.760 6^{n} \geq 0.99 \Rightarrow n \geq \frac{ln 0.01}{ln 0.7606} = 16.8

(d)

2 marks

The number of successes is $B (10, 0.2394)$ .

P (X = 3) = (3 10) (0.2394)^{3} (0.7606)^{7} \approx 0.242

(e)

1 mark

A block of three in a row can start at positions 1 to 8.

8

(f)

6 marks

For $n$ trials there are $n - 2$ blocks of three consecutive trials, and all choices of three trials number $(3 n)$ .
The probability that three given successes are consecutive is $\frac{n - 2}{( 3 n )}$ . Evaluate for increasing $n$ .
The greatest $n$ with the probability above 0.05 is

(i) n - 2 n = 10 : 0.0667, n = 11 : 0.0545, n = 12 : 0.0455 n = 11

Question 12

21 marks

Euler's method, Maclaurin series and exact solution of a differential equation

Calculus

(a)

4 marks

The slope is $\frac{d y}{d x} = \frac{x ^{2} y - y}{x ^{2} + 1}$ . Start from $x_{0} = 0$ , $y_{0} = 3$ and take five steps of 0.03.
At $x = 0.15$ :

y_{1} = 2.91, y_{2} = 2.82286, y_{3} = 2.73878, y_{4} = 2.65794, y_{5} = 2.58046 y \approx 2.58046

(b)

5 marks

At $x = 0$ , with $y = 3$ :
Differentiate $(x^{2} + 1) \frac{d y}{d x} = x^{2} y - y$ with the product rule and implicit differentiation.
At $x = 0$ , $y = 3$ , $\frac{d y}{d x} = - 3$ :

\frac{d y}{d x} = \frac{0 - 3}{1} = - 3 2 x \frac{d y}{d x} + (x^{2} + 1) \frac{d ^{2} y}{d x ^{2}} = 2 x y + x^{2} \frac{d y}{d x} - \frac{d y}{d x} \frac{d ^{2} y}{d x ^{2}} = 0 + 0 + 3 = 3

(c)

3 marks

Maclaurin series $y = y (0) + y^{'} (0) x + \frac{y ^{''} ( 0 )}{2 !} x^{2} + \frac{y ^{'''} ( 0 )}{3 !} x^{3} + \dots$
At $x = 0.15$ :

y = 3 - 3 x + \frac{3}{2} x^{2} + \frac{3}{2} x^{3} y \approx 2.58881

(d)

7 marks

Separate the variables, using the given identity $\frac{x ^{2} - 1}{x ^{2} + 1} = 1 - \frac{2}{x ^{2} + 1}$ .
Use $y = 3$ when $x = 0$ .
At $x = 0.15$ :

\int \frac{d y}{y} = \int (1 - \frac{2}{x ^{2} + 1}) d x \Rightarrow ln y = x - 2 arctan x + c ln 3 = c y = 3 e^{x - 2 a r c t a n x} y = 3 e^{0.15 - 2 a r c t a n 0.15} \approx 2.58786

(e)

2 marks

Differentiate $\frac{y ^{'}}{y} = 1 - \frac{2}{x ^{2} + 1}$ .
For $y > 0$ and $x \geq 0$ the bracket is positive, so $y^{''} > 0$ : the graph is concave up. Each Euler step follows a tangent, and a tangent to a concave-up curve lies below it, so every Euler estimate is too small.

\frac{y ^{''}}{y} - (\frac{y ^{'}}{y})^{2} = \frac{4 x}{( x ^{2} + 1 ) ^{2}} \Rightarrow y^{''} = y [(\frac{y ^{'}}{y})^{2} + \frac{4 x}{( x ^{2} + 1 ) ^{2}}]

Our analysis

Where the marks are

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

Calculus35 marks · 32%
Q3, Q10, Q11, Q12
Number and algebra32 marks · 29%
Q4, Q5, Q6, Q8, Q9
Statistics and probability23 marks · 21%
Q3, Q7, Q11
Geometry and trigonometry15 marks · 14%
Q1, Q10
Functions5 marks · 5%
Q2

Mentor insights

Traps to watch for

01
Regress x on y · Q3
The vendor wants to predict ice creams ( $x$ ) from the number of children ( $y$ ), so use the regression line of $x$ on $y$ .
02
Pick the right term · Q5
In $(a x^{3} + b)^{n}$ the $x^{6}$ term needs $(a x^{3})^{2}$ , which is the term with $r = 2$ .
03
The sheet width fixes the arc · Q10
The 45 cm sheet is two vertical sides of 10 cm plus the curved arc, so the arc is 25 cm.
04
Check the density first · Q11
$\frac{6}{π 16 - x ^{2}}$ integrates to 1 on $[0, 2]$ because $arcsin \frac{1}{2} = \frac{π}{6}$ .
05
Euler uses the old values · Q12
Each step is $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ with $h = 0.03$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers