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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2021
Marks: 80
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2021 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 1–6 · 34 marks

Question 1

5 marks

Reading values off a graph, then transforming it

Functions

(a)

2 marks

Read $f (2)$ straight off the graph: the curve peaks at $x = 2$ , so $f (2) = 6$ .
For the composite, feed that output back in. $(f \circ f) (2) = f (6)$ , and the graph ends at $x = 6$ with $y = - 2$ .

f (2) = 6 (f \circ f) (2) = f (f (2)) = f (6) = - 2

Mentor tip

Work a composite from the inside out, and write the middle value down so the second reading is clearly $f (6)$ .

(b)

3 marks

The rule $g (x) = \frac{1}{2} f (x) + 1$ keeps every $x$ the same and sends each height $y$ to $\frac{1}{2} y + 1$ : halve it, then add 1.
Apply that to the key points of $f$ : the flat piece at height 4 on $- 4 \leq x \leq 0$ , the peak at $(2, 6)$ and the right-hand end at $(6, - 2)$ .
So the sketch is a flat segment at height 3 from $x = - 4$ to $x = 0$ , then a gentler downward-opening curve rising to its peak at $(2, 4)$ and ending at $(6, 0)$ .

(x, y) ⟼ (x, \frac{1}{2} y + 1) flat piece: y = 4 \to 3, (0, 4) \to (0, 3), (2, 6) \to (2, 4), (6, - 2) \to (6, 0)

Mentor tip

Label the end points and the peak on your sketch. They are the points that show you transformed correctly.

Question 2

4 marks

A planet's volume in standard form

Number and algebra

(a)

1 mark

The radius is half the diameter.

r = 3 \times 1 0^{4} km

(b)

3 marks

Use $V = \frac{4}{3} π r^{3}$ .
Write 36 as $3.6 \times 1 0^{1}$ to put it in the form $π (a \times 1 0^{k})$ with $1 \leq a < 10$ .

V = \frac{4}{3} π (3 \times 1 0^{4})^{3} = \frac{4}{3} π \cdot 27 \times 1 0^{12} = π \cdot 36 \times 1 0^{12} V = π (3.6 \times 1 0^{13}) \Rightarrow a = 3.6, k = 13

Question 3

5 marks

First term and common difference from a term and a sum

Number and algebra

5 marks

Write both facts using $u_{1}$ and $d$ . For the term, $u_{8} = u_{1} + 7 d$ . For the sum, $S_{8} = \frac{8}{2} (2 u_{1} + 7 d)$ .
Subtract the first equation from the second: that removes the $7 d$ and leaves $u_{1}$ alone.

u_{1} + 7 d = 8 4 (2 u_{1} + 7 d) = 8 \Rightarrow 2 u_{1} + 7 d = 2 (2 u_{1} + 7 d) - (u_{1} + 7 d) = 2 - 8 \Rightarrow u_{1} = - 6 - 6 + 7 d = 8 \Rightarrow d = 2

Mentor tip

Check in the other formula: $S_{8} = 4 (- 12 + 14) = 8$ . It takes ten seconds and catches a sign slip.

Question 4

5 marks

Smallest possible quartile from the outlier rule

Statistics and probability

(a)

3 marks

Since there are no outliers, the largest value, 75, cannot be above the upper fence. That fence is $U + 1.5 \times IQR$ , with an IQR of 20.
So the smallest possible value is $U = 45$ .

75 \leq U + 1.5 \times 20 = U + 30 U \geq 45

Mentor tip

"No outliers" is an inequality. Say so, then read off the extreme case that the question wants.

(b)

2 marks

The interquartile range is $U - L = 20$ , so use the value of $U$ from part (a).

L = U - 20 = 45 - 20 = 25

Mentor tip

Quick sanity check: the smallest value, 10, is above the lower fence $25 - 30 = - 5$ , so there is no outlier at that end either.

Question 5

7 marks

A common tangent forces two equations

Calculus

(a)

1 mark

Differentiate term by term: the bracket is squared, so the chain rule brings down a factor of $- 2$ times the inner derivative, which is 1.

f^{'} (x) = - 2 (x - h)

(b)

3 marks

Equal tangents means equal gradients at $x = 3$ . First find $g^{'}$ .
Set $f^{'} (3) = g^{'} (3)$ and solve for $h$ .

g^{'} (x) = e^{x - 2} \Rightarrow g^{'} (3) = e - 2 (3 - h) = e \Rightarrow 3 - h = - \frac{e}{2} \Rightarrow h = 3 + \frac{e}{2} = \frac{e + 6}{2}

Mentor tip

In a "show that" question you finish on the stated result. Write the last line exactly as it is printed.

(c)

3 marks

The graphs also meet at $x = 3$ , so $f (3) = g (3)$ . From part (b), $3 - h = - \frac{e}{2}$ , so $(3 - h)^{2} = \frac{e ^{2}}{4}$ .
Collect the $k$ terms on one side.

- \frac{e ^{2}}{4} + 2 k = e + k k = e + \frac{e ^{2}}{4} = \frac{e ^{2} + 4 e}{4}

Mentor tip

Reusing $(3 - h)$ from part (b) saves you expanding $h$ again, which is where the arithmetic goes wrong.

Question 6

8 marks

A double-angle identity, then a trig equation

Geometry and trigonometry

(a)

2 marks

Rewrite $sin 2 x$ and $cos 2 x$ with the double-angle formulas, choosing the version $cos 2 x = 1 - 2 sin^{2} x$ so the constant cancels.

sin 2 x + cos 2 x - 1 = 2 sin x cos x + (1 - 2 sin^{2} x) - 1 = 2 sin x cos x - 2 sin^{2} x = 2 sin x (cos x - sin x)

Mentor tip

Starting from one side and arriving at the other is cleaner than working both sides together.

(b)

6 marks

The first three terms are exactly the left side of part (a). Replace them, then notice the common factor $(cos x - sin x)$ .
Each factor gives its own set of solutions in $0 < x < 2 π$ . From $cos x = sin x$ we get $tan x = 1$ . From $2 sin x + 1 = 0$ we get $sin x = - \frac{1}{2}$ .

2 sin x (cos x - sin x) + (cos x - sin x) = (cos x - sin x) (2 sin x + 1) = 0 tan x = 1 : x = \frac{π}{4}, \frac{5 π}{4} sin x = - \frac{1}{2} : x = \frac{7 π}{6}, \frac{11 π}{6}

Mentor tip

Four solutions is a good sign here: one equation in $tan$ and one in $sin$ each give two angles in a full turn.

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

14 marks

A quadratic tangent to a line, then its three forms

Number and algebra

(a)

6 marks

Set the two expressions equal and rearrange into a quadratic in $x$ .
Exactly one solution means the discriminant is zero.
If $m = 0$ , then $f (x) = 0$ and the line is $y = - 9$ , which never meet. So $m \neq = 0$ .

m x^{2} - 2 m x = m x - 9 \Rightarrow m x^{2} - 3 m x + 9 = 0 Δ = 9 m^{2} - 36 m = 9 m (m - 4) = 0 \Rightarrow m = 0 or m = 4 m = 4

(b)

2 marks

With $m = 4$ , factorise.

f (x) = 4 x^{2} - 8 x = 4 x (x - 2) \Rightarrow p = 0, q = 2

(c)

3 marks

Complete the square, or use the vertex halfway between the roots at $x = 1$ .

f (x) = 4 (x - 1)^{2} - 4 \Rightarrow h = 1, k = - 4

(d)

3 marks

$f$ is negative between its roots, $0 < x < 2$ . It is increasing to the right of the vertex, $x > 1$ . Both hold for

1 < x < 2

Question 8

16 marks

$y = \frac{ln x}{x ^{4}}$ : derivative, maximum and sketch

Calculus
Functions

(a)

3 marks

Use the quotient rule with $u = ln x$ and $v = x^{4}$ .

\frac{d y}{d x} = \frac{\frac{1}{x} \cdot x ^{4} - ln x \cdot 4 x ^{3}}{x ^{8}} = \frac{x ^{3} - 4 x ^{3} ln x}{x ^{8}} = \frac{1 - 4 ln x}{x ^{5}}

(b)

5 marks

A horizontal tangent needs $f^{'} (x) = 0$ .
Substitute to find the $y$ -coordinate.

1 - 4 ln x = 0 \Rightarrow ln x = \frac{1}{4} \Rightarrow x = e^{1/4} y = \frac{\frac{1}{4}}{e} = \frac{1}{4 e} P (e^{1/4}, \frac{1}{4 e}) \approx (1.28, 0.092)

(c)

3 marks

Substitute into the given second derivative at $x = e^{1/4}$ .
A negative second derivative means P is a local maximum.

f^{''} (e^{1/4}) = \frac{20 \cdot \frac{1}{4} - 9}{e ^{3/2}} = - \frac{4}{e ^{3/2}} < 0

(d)

2 marks

Since $x^{4} > 0$ , the sign of $f$ is the sign of $ln x$ .

ln x > 0 \Rightarrow x > 1

(e)

3 marks

The graph crosses the $x$ -axis only at $(1, 0)$ . For $0 < x < 1$ it is negative and plunges to $- \infty$ as $x \to 0^{+}$ . It rises through $(1, 0)$ to the maximum at P, about $(1.28, 0.092)$ , and then decreases, approaching the $x$ -axis from above as $x \to \infty$ .

Question 9

16 marks

A biased four-sided die and a game between two dice

Statistics and probability

(a)

2 marks

The probabilities must sum to 1.

p + p + p + \frac{1}{2} p = 1 \Rightarrow \frac{7}{2} p = 1 \Rightarrow p = \frac{2}{7}

(b)

2 marks

Multiply each score by its probability and add.

E (X) = p + 2 p + 3 p + 4 \cdot \frac{1}{2} p = 8 p = \frac{16}{7}

(c)

3 marks

A probability lies between 0 and 1, so $0 \leq r \leq 1$ .
The probabilities of die B sum to 1, so $3 q + r = 1$ and $q = \frac{1 - r}{3}$ . Putting $r$ at each end gives the range of $q$ .

0 \leq q \leq \frac{1}{3}

(d)

3 marks

Write $E (Y)$ using $r = 1 - 3 q$ , so that only one letter is left.
This decreases as $q$ increases, so the ends of the range come from $q = \frac{1}{3}$ and $q = 0$ .

E (Y) = q + 2 q + 3 q + 4 r = 6 q + 4 (1 - 3 q) = 4 - 6 q 2 \leq E (Y) \leq 4

(e)

6 marks

Agnes's score is less than Barbara's in three ways: Agnes rolls 1 and Barbara 2 or more, Agnes 2 and Barbara 3 or more, or Agnes 3 and Barbara 4. Write each as a product.
Put in $p = \frac{2}{7}$ and set this equal to $\frac{1}{2}$ to find $r$ , then $q$ , then $E (Y)$ .

P (A < B) = p (1 - q) + p (q + r) + p r = p (1 + 2 r) \frac{2}{7} (1 + 2 r) = \frac{1}{2} \Rightarrow r = \frac{3}{8}, q = \frac{1 - 3/8}{3} = \frac{5}{24} E (Y) = 4 - 6 q = 4 - \frac{5}{4} = \frac{11}{4}

Mentor tip

Check that $q = \frac{5}{24}$ lies in the range from part (c), and that $3 q + r = \frac{5}{8} + \frac{3}{8} = 1$ .

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 algebra23 marks · 29%
Q2, Q3, Q7
Statistics and probability21 marks · 26%
Q4, Q9
Calculus18 marks · 23%
Q5, Q8
Functions10 marks · 13%
Q1, Q8
Geometry and trigonometry8 marks · 10%
Q6

Mentor insights

Traps to watch for

01
Same point and same slope · Q5
A common tangent at $x = 3$ means the two graphs meet there and also have equal gradients there. Use both facts, in that order, to pin down $h$ and then $k$ .
02
Factor, don't divide · Q6
In part (b) it is tempting to divide by $cos x - sin x$ . That throws away the solutions where $cos x = sin x$ . Factor it out instead.
03
Exactly one point means discriminant zero · Q7
A line meeting a curve at exactly one point gives a quadratic with a repeated root. Reject $m = 0$ , because then the line and the curve don't meet at all.
04
Show the working for the maximum · Q8
For a local maximum, show $f^{'} = 0$ , then give the sign of $f^{''}$ there. A negative second derivative at the stationary point is the evidence.
05
Range of values, not one value · Q9
Part (d) asks for a range of $E (Y)$ . Write $E (Y)$ in terms of $q$ alone, then use the allowed range of $q$ to find both ends.

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.

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Worked solutions

IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2

Session: May 2021
Marks: 80
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2021 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 1–6 · 34 marks

Question 1

5 marks

Reading values off a graph, then transforming it

Functions

(a)

2 marks

Read $f (2)$ straight off the graph: the curve peaks at $x = 2$ , so $f (2) = 6$ .
For the composite, feed that output back in. $(f \circ f) (2) = f (6)$ , and the graph ends at $x = 6$ with $y = - 2$ .

f (2) = 6 (f \circ f) (2) = f (f (2)) = f (6) = - 2

Mentor tip

Work a composite from the inside out, and write the middle value down so the second reading is clearly $f (6)$ .

(b)

3 marks

The rule $g (x) = \frac{1}{2} f (x) + 1$ keeps every $x$ the same and sends each height $y$ to $\frac{1}{2} y + 1$ : halve it, then add 1.
Apply that to the key points of $f$ : the flat piece at height 4 on $- 4 \leq x \leq 0$ , the peak at $(2, 6)$ and the right-hand end at $(6, - 2)$ .
So the sketch is a flat segment at height 3 from $x = - 4$ to $x = 0$ , then a gentler downward-opening curve rising to its peak at $(2, 4)$ and ending at $(6, 0)$ .

(x, y) ⟼ (x, \frac{1}{2} y + 1) flat piece: y = 4 \to 3, (0, 4) \to (0, 3), (2, 6) \to (2, 4), (6, - 2) \to (6, 0)

Mentor tip

Label the end points and the peak on your sketch. They are the points that show you transformed correctly.

Question 2

4 marks

A planet's volume in standard form

Number and algebra

(a)

1 mark

The radius is half the diameter.

r = 3 \times 1 0^{4} km

(b)

3 marks

Use $V = \frac{4}{3} π r^{3}$ .
Write 36 as $3.6 \times 1 0^{1}$ to put it in the form $π (a \times 1 0^{k})$ with $1 \leq a < 10$ .

V = \frac{4}{3} π (3 \times 1 0^{4})^{3} = \frac{4}{3} π \cdot 27 \times 1 0^{12} = π \cdot 36 \times 1 0^{12} V = π (3.6 \times 1 0^{13}) \Rightarrow a = 3.6, k = 13

Question 3

5 marks

First term and common difference from a term and a sum

Number and algebra

5 marks

Write both facts using $u_{1}$ and $d$ . For the term, $u_{8} = u_{1} + 7 d$ . For the sum, $S_{8} = \frac{8}{2} (2 u_{1} + 7 d)$ .
Subtract the first equation from the second: that removes the $7 d$ and leaves $u_{1}$ alone.

u_{1} + 7 d = 8 4 (2 u_{1} + 7 d) = 8 \Rightarrow 2 u_{1} + 7 d = 2 (2 u_{1} + 7 d) - (u_{1} + 7 d) = 2 - 8 \Rightarrow u_{1} = - 6 - 6 + 7 d = 8 \Rightarrow d = 2

Mentor tip

Check in the other formula: $S_{8} = 4 (- 12 + 14) = 8$ . It takes ten seconds and catches a sign slip.

Question 4

5 marks

Smallest possible quartile from the outlier rule

Statistics and probability

(a)

3 marks

Since there are no outliers, the largest value, 75, cannot be above the upper fence. That fence is $U + 1.5 \times IQR$ , with an IQR of 20.
So the smallest possible value is $U = 45$ .

75 \leq U + 1.5 \times 20 = U + 30 U \geq 45

Mentor tip

"No outliers" is an inequality. Say so, then read off the extreme case that the question wants.

(b)

2 marks

The interquartile range is $U - L = 20$ , so use the value of $U$ from part (a).

L = U - 20 = 45 - 20 = 25

Mentor tip

Quick sanity check: the smallest value, 10, is above the lower fence $25 - 30 = - 5$ , so there is no outlier at that end either.

Question 5

7 marks

A common tangent forces two equations

Calculus

(a)

1 mark

Differentiate term by term: the bracket is squared, so the chain rule brings down a factor of $- 2$ times the inner derivative, which is 1.

f^{'} (x) = - 2 (x - h)

(b)

3 marks

Equal tangents means equal gradients at $x = 3$ . First find $g^{'}$ .
Set $f^{'} (3) = g^{'} (3)$ and solve for $h$ .

g^{'} (x) = e^{x - 2} \Rightarrow g^{'} (3) = e - 2 (3 - h) = e \Rightarrow 3 - h = - \frac{e}{2} \Rightarrow h = 3 + \frac{e}{2} = \frac{e + 6}{2}

Mentor tip

In a "show that" question you finish on the stated result. Write the last line exactly as it is printed.

(c)

3 marks

The graphs also meet at $x = 3$ , so $f (3) = g (3)$ . From part (b), $3 - h = - \frac{e}{2}$ , so $(3 - h)^{2} = \frac{e ^{2}}{4}$ .
Collect the $k$ terms on one side.

- \frac{e ^{2}}{4} + 2 k = e + k k = e + \frac{e ^{2}}{4} = \frac{e ^{2} + 4 e}{4}

Mentor tip

Reusing $(3 - h)$ from part (b) saves you expanding $h$ again, which is where the arithmetic goes wrong.

Question 6

8 marks

A double-angle identity, then a trig equation

Geometry and trigonometry

(a)

2 marks

Rewrite $sin 2 x$ and $cos 2 x$ with the double-angle formulas, choosing the version $cos 2 x = 1 - 2 sin^{2} x$ so the constant cancels.

sin 2 x + cos 2 x - 1 = 2 sin x cos x + (1 - 2 sin^{2} x) - 1 = 2 sin x cos x - 2 sin^{2} x = 2 sin x (cos x - sin x)

Mentor tip

Starting from one side and arriving at the other is cleaner than working both sides together.

(b)

6 marks

The first three terms are exactly the left side of part (a). Replace them, then notice the common factor $(cos x - sin x)$ .
Each factor gives its own set of solutions in $0 < x < 2 π$ . From $cos x = sin x$ we get $tan x = 1$ . From $2 sin x + 1 = 0$ we get $sin x = - \frac{1}{2}$ .

2 sin x (cos x - sin x) + (cos x - sin x) = (cos x - sin x) (2 sin x + 1) = 0 tan x = 1 : x = \frac{π}{4}, \frac{5 π}{4} sin x = - \frac{1}{2} : x = \frac{7 π}{6}, \frac{11 π}{6}

Mentor tip

Four solutions is a good sign here: one equation in $tan$ and one in $sin$ each give two angles in a full turn.

Section B · Extended response

Questions 7–9 · 46 marks

Question 7

14 marks

A quadratic tangent to a line, then its three forms

Number and algebra

(a)

6 marks

Set the two expressions equal and rearrange into a quadratic in $x$ .
Exactly one solution means the discriminant is zero.
If $m = 0$ , then $f (x) = 0$ and the line is $y = - 9$ , which never meet. So $m \neq = 0$ .

m x^{2} - 2 m x = m x - 9 \Rightarrow m x^{2} - 3 m x + 9 = 0 Δ = 9 m^{2} - 36 m = 9 m (m - 4) = 0 \Rightarrow m = 0 or m = 4 m = 4

(b)

2 marks

With $m = 4$ , factorise.

f (x) = 4 x^{2} - 8 x = 4 x (x - 2) \Rightarrow p = 0, q = 2

(c)

3 marks

Complete the square, or use the vertex halfway between the roots at $x = 1$ .

f (x) = 4 (x - 1)^{2} - 4 \Rightarrow h = 1, k = - 4

(d)

3 marks

$f$ is negative between its roots, $0 < x < 2$ . It is increasing to the right of the vertex, $x > 1$ . Both hold for

1 < x < 2

Question 8

16 marks

$y = \frac{ln x}{x ^{4}}$ : derivative, maximum and sketch

Calculus
Functions

(a)

3 marks

Use the quotient rule with $u = ln x$ and $v = x^{4}$ .

\frac{d y}{d x} = \frac{\frac{1}{x} \cdot x ^{4} - ln x \cdot 4 x ^{3}}{x ^{8}} = \frac{x ^{3} - 4 x ^{3} ln x}{x ^{8}} = \frac{1 - 4 ln x}{x ^{5}}

(b)

5 marks

A horizontal tangent needs $f^{'} (x) = 0$ .
Substitute to find the $y$ -coordinate.

1 - 4 ln x = 0 \Rightarrow ln x = \frac{1}{4} \Rightarrow x = e^{1/4} y = \frac{\frac{1}{4}}{e} = \frac{1}{4 e} P (e^{1/4}, \frac{1}{4 e}) \approx (1.28, 0.092)

(c)

3 marks

Substitute into the given second derivative at $x = e^{1/4}$ .
A negative second derivative means P is a local maximum.

f^{''} (e^{1/4}) = \frac{20 \cdot \frac{1}{4} - 9}{e ^{3/2}} = - \frac{4}{e ^{3/2}} < 0

(d)

2 marks

Since $x^{4} > 0$ , the sign of $f$ is the sign of $ln x$ .

ln x > 0 \Rightarrow x > 1

(e)

3 marks

Question 9

16 marks

A biased four-sided die and a game between two dice

Statistics and probability

(a)

2 marks

The probabilities must sum to 1.

p + p + p + \frac{1}{2} p = 1 \Rightarrow \frac{7}{2} p = 1 \Rightarrow p = \frac{2}{7}

(b)

2 marks

Multiply each score by its probability and add.

E (X) = p + 2 p + 3 p + 4 \cdot \frac{1}{2} p = 8 p = \frac{16}{7}

(c)

3 marks

A probability lies between 0 and 1, so $0 \leq r \leq 1$ .
The probabilities of die B sum to 1, so $3 q + r = 1$ and $q = \frac{1 - r}{3}$ . Putting $r$ at each end gives the range of $q$ .

0 \leq q \leq \frac{1}{3}

(d)

3 marks

Write $E (Y)$ using $r = 1 - 3 q$ , so that only one letter is left.
This decreases as $q$ increases, so the ends of the range come from $q = \frac{1}{3}$ and $q = 0$ .

E (Y) = q + 2 q + 3 q + 4 r = 6 q + 4 (1 - 3 q) = 4 - 6 q 2 \leq E (Y) \leq 4

(e)

6 marks

Agnes's score is less than Barbara's in three ways: Agnes rolls 1 and Barbara 2 or more, Agnes 2 and Barbara 3 or more, or Agnes 3 and Barbara 4. Write each as a product.
Put in $p = \frac{2}{7}$ and set this equal to $\frac{1}{2}$ to find $r$ , then $q$ , then $E (Y)$ .

P (A < B) = p (1 - q) + p (q + r) + p r = p (1 + 2 r) \frac{2}{7} (1 + 2 r) = \frac{1}{2} \Rightarrow r = \frac{3}{8}, q = \frac{1 - 3/8}{3} = \frac{5}{24} E (Y) = 4 - 6 q = 4 - \frac{5}{4} = \frac{11}{4}

Mentor tip

Check that $q = \frac{5}{24}$ lies in the range from part (c), and that $3 q + r = \frac{5}{8} + \frac{3}{8} = 1$ .

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 algebra23 marks · 29%
Q2, Q3, Q7
Statistics and probability21 marks · 26%
Q4, Q9
Calculus18 marks · 23%
Q5, Q8
Functions10 marks · 13%
Q1, Q8
Geometry and trigonometry8 marks · 10%
Q6

Mentor insights

Traps to watch for

01
Same point and same slope · Q5
A common tangent at $x = 3$ means the two graphs meet there and also have equal gradients there. Use both facts, in that order, to pin down $h$ and then $k$ .
02
Factor, don't divide · Q6
In part (b) it is tempting to divide by $cos x - sin x$ . That throws away the solutions where $cos x = sin x$ . Factor it out instead.
03
Exactly one point means discriminant zero · Q7
A line meeting a curve at exactly one point gives a quadratic with a repeated root. Reject $m = 0$ , because then the line and the curve don't meet at all.
04
Show the working for the maximum · Q8
For a local maximum, show $f^{'} = 0$ , then give the sign of $f^{''}$ there. A negative second derivative at the stationary point is the evidence.
05
Range of values, not one value · Q9
Part (d) asks for a range of $E (Y)$ . Write $E (Y)$ in terms of $q$ alone, then use the allowed range of $q$ to find both ends.

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