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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2021
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 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 1–9 · 55 marks

Question 1

7 marks

Regression line, a prediction and the mean point

Statistics and probability

(a)

2 marks

Enter the eight pairs in the calculator's linear regression (y on x).

a \approx 0.433, b \approx 4.50

(b)

2 marks

Substitute $x = 18$ into $y = a x + b$ .

y = 0.4331 \dots \times 18 + 4.5026 \dots \approx 12.3

Mentor tip

Using the stored unrounded values keeps the third significant figure safe.

(c)

1 mark

The regression line always passes through the point of means $(\overset{x}{ˉ}, \overset{y}{ˉ})$ .

\overset{x}{ˉ} = 15, \overset{y}{ˉ} = 11

(d)

2 marks

Draw a straight line through the mean point $(15, 11)$ that follows the trend of the points. It should meet the $y$ -axis at about 4.5 and rise by about 0.43 for each unit across.

Mentor tip

Use a ruler. Pivot the line about $(15, 11)$ so roughly equal numbers of points sit above and below it.

Question 2

6 marks

Rejected bags of sugar from a normal distribution

Statistics and probability

(a)

2 marks

Let $X \sim N (1000, 3. 5^{2})$ . The probability of a mass below 995 g comes from the normal cumulative distribution on the calculator.

P (X < 995) \approx 0.0766

(b)

1 mark

Multiply the probability by the sample size.

100 \times 0.0766 = 7.66 \approx 8 bags

(c)

3 marks

This is a conditional probability: the bag is not rejected, meaning $X \geq 995$ , and we want $X > 1005$ . The event $X > 1005$ is contained in $X \geq 995$ .

P (X > 1005 ∣ X \geq 995) = \frac{P ( X > 1005 )}{P ( X \geq 995 )} = \frac{0.07656}{1 - 0.07656} \approx 0.0829

Mentor tip

Rounding to 3 sf at the start (0.0766) changes the third figure. Keep the calculator's stored value.

Question 3

5 marks

A Ferris wheel modelled by a cosine

Number and algebra

5 marks

The amplitude is half the diameter, and starting at the lowest point means the cosine is reflected, so $a = - 55$ .
One revolution takes 20 minutes, so the period is 20 and $b = \frac{2 π}{20} = \frac{π}{10} \approx 0.314$ .
The middle height is the lowest height plus the radius, $10 + 55 = 65$ , so $c = 65$ .

h (t) = - 55 cos (\frac{π}{10} t) + 65

Mentor tip

Check: $h (0) = - 55 + 65 = 10$ , the lowest point. And $h (10) = 120$ , the top, 10 m plus the 110 m diameter.

Question 4

6 marks

Velocity, distance and acceleration from $v = t sin t - 3$

Calculus

(a)

2 marks

The particle is at rest when $v = 0$ , that is, $t sin t = 3$ . Solve on the calculator, or read the first axis crossing of the graph.

t \approx 6.74 s

(b)

2 marks

Total distance is the integral of the speed.

\int_{0}^{10} ∣ t sin t - 3∣ d t \approx 37.1 m

(c)

2 marks

Acceleration is the derivative of velocity. Differentiate with the product rule, then put in $t = 7$ .

a (t) = sin t + t cos t \Rightarrow a (7) = sin 7 + 7 cos 7 \approx 5.93 m s^{- 2}

Mentor tip

Radian mode. In degrees the answer is completely different.

Question 5

5 marks

A coefficient in a binomial expansion pins down $n$

Number and algebra

5 marks

The term in $x^{4}$ takes $x^{2}$ from two of the $n + 1$ brackets and 3 from the rest.
Set the coefficient equal to 20 412 and solve on the calculator, or by trying values.

(2 n + 1) 3^{n - 1} (x^{2})^{2} = \frac{n ( n + 1 )}{2} 3^{n - 1} x^{4} \frac{n ( n + 1 )}{2} 3^{n - 1} = 20412 \Rightarrow n = 7

Mentor tip

Check: $\frac{7 \cdot 8}{2} \cdot 3^{6} = 28 \times 729 = 20412$ .

Question 6

5 marks

A plane perpendicular to two planes, and the point where all three meet

Geometry and trigonometry

(a)

3 marks

The normals are $321$ and $1 - 2 1$ . The normal of the new plane is perpendicular to both, so take their cross product.
The plane passes through the origin, so there is no constant term.

321 \times 1 - 2 1 = 4 - 2 - 8 \propto 2 - 1 - 4 2 x - y - 4 z = 0

(b)

2 marks

Solve the three equations together. Adding the first two plane equations gives $4 x + 2 z = 10$ , so $z = 5 - 2 x$ .
Put these into $3 x + 2 y + z = 6$ .

2 x - y - 4 (5 - 2 x) = 0 \Rightarrow y = 10 x - 20 3 x + 2 (10 x - 20) + (5 - 2 x) = 6 \Rightarrow 21 x = 41 (\frac{41}{21}, - \frac{10}{21}, \frac{23}{21})

Mentor tip

Check in the second plane: $\frac{41}{21} + \frac{20}{21} + \frac{23}{21} = \frac{84}{21} = 4$ .

Question 7

7 marks

A probability density function with a parameter to find

Calculus

(a)

5 marks

The total probability is 1, so integrate $\frac{x}{( x ^{2} + k ) ^{3}} = x (x^{2} + k)^{- 3/2}$ over $0 \leq x \leq 4$ . The inside function is $x^{2} + k$ , whose derivative is $2 x$ , so the antiderivative is a multiple of $(x^{2} + k)^{- 1/2}$ .
Set this equal to 1 and put the left side over a common denominator.

\int_{0}^{4} x (x^{2} + k)^{- 3/2} d x = [- (x^{2} + k)^{- 1/2}]_{0}^{4} = \frac{1}{k} - \frac{1}{16 + k} \frac{16 + k - k}{k 16 + k} = 1 16 + k - k = k 16 + k

(b)

2 marks

Solve the equation from part (a) numerically on the calculator, or graph both sides.

k \approx 0.645

Mentor tip

Check in the equation: $\frac{1}{0.645} - \frac{1}{16.645} \approx 1.245 - 0.245 = 1$ .

Question 8

7 marks

Product of complex numbers in modulus-argument form

Number and algebra

(a)

1 mark

The modulus of a product is the product of the moduli.

∣ z w ∣ = ∣ z ∣∣ w ∣ = 2 \times 8 = 16

(b)

2 marks

Since $w = 8 (cos \frac{2 k π}{5} - i sin \frac{2 k π}{5})$ , its argument is $- \frac{2 k π}{5}$ . Add the arguments.

ar g (z w) = \frac{π}{5} - \frac{2 k π}{5} = \frac{( 1 - 2 k ) π}{5}

(c)

4 marks

For $z w$ to be a real integer, its argument must be a multiple of $π$ . So $\frac{1 - 2 k}{5}$ is an integer, meaning $1 - 2 k$ is a multiple of 5. Try $k = 1, 2, 3$ : the first that works is $k = 3$ , because $1 - 6 = - 5$ .
Then $ar g (z w) = - π$ , and the number is on the negative real axis with modulus 16.

k = 3 z w = 16 (cos (- π) + i sin (- π)) = - 16

Mentor tip

$k = 1$ gives argument $- \frac{π}{5}$ and $k = 2$ gives $- \frac{3 π}{5}$ . Neither is a multiple of $π$ .

Question 9

7 marks

Two boats and a changing bearing (related rates)

Geometry and trigonometry

(a)

1 mark

From the diagram, boat B is 50 m east of A, and $θ$ is measured from north. The extra distance travelled by B is $y - x$ , and $tan θ = \frac{50}{y - x}$ .

y - x = 50 cot θ \Rightarrow y = x + 50 cot θ

(b)

6 marks

Differentiate with respect to $t$ .
At time $T$ , B has gone 10 m further, so $50 cot θ = 10$ , giving $cot θ = 0.2$ and $cosec^{2} θ = 1 + 0. 2^{2} = 1.04$ . Also $\frac{d y}{d t} = 2 \frac{d x}{d t}$ and $\frac{d θ}{d t} = - 0.1$ .

\frac{d y}{d t} = \frac{d x}{d t} - 50 cosec^{2} θ \frac{d θ}{d t} 2 \frac{d x}{d t} = \frac{d x}{d t} - 50 (1.04) (- 0.1) = \frac{d x}{d t} + 5.2 \frac{d x}{d t} = 5.2 m s^{- 1}

Mentor tip

The negative rate of $θ$ is why the extra term is positive. Keep the sign through the substitution.

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

15 marks

An exponential curve, a tangent of gradient $- 1$ and enclosed areas

Calculus

(a)

2 marks

Solve $90 e^{- 0.5 x} = x$ on the calculator, or find the crossing point of the two graphs.

x \approx 5.57

(b)

4 marks

The gradient of $f$ is $f^{'} (x) = - 45 e^{- 0.5 x}$ . Set it equal to $- 1$ and solve exactly.
Then $y = 90 \cdot \frac{1}{45} = 2$ .

- 45 e^{- 0.5 x} = - 1 \Rightarrow e^{- 0.5 x} = \frac{1}{45} \Rightarrow x = 2 ln 45 Q (2 ln 45, 2)

(c)

2 marks

Use the point-gradient form with gradient $- 1$ through $Q$ .

y - 2 = - (x - 2 ln 45) \Rightarrow y = - x + 2 ln 45 + 2

(d)

5 marks

(i) The line $L$ meets $y = x$ where $x = - x + 2 ln 45 + 2$ , so $x = ln 45 + 1 \approx 4.81$ .

(ii) The region has two parts. From $x \approx 4.81$ to $x \approx 5.57$ it lies between $y = x$ (above) and $L$ (below). From $x \approx 5.57$ to $x = 2 ln 45 \approx 7.61$ it lies between $f$ (above) and $L$ (below).

A = \int_{4.81}^{5.57} (x - L (x)) d x + \int_{5.57}^{7.61} (90 e^{- 0.5 x} - L (x)) d x \approx 1.52

Mentor tip

Keep the limits unrounded in the calculator. Rounding them to 3 sf shifts the area in the third figure.

(e)

2 marks

By symmetry in $y = x$ the shaded area between $f$ , $f^{- 1}$ and $L$ is twice the area in part (d).

2 \times 1.52 \approx 3.04

Question 11

20 marks

A rational function, its graph, and a comparison with its reciprocal

Functions
Calculus

(a)

2 marks

$f$ is undefined where the denominator is 0.

4 x^{2} - 1 = 0 \Rightarrow p = \frac{1}{2}, q = - \frac{1}{2}

(b)

3 marks

Use the quotient rule.

f^{'} (x) = \frac{3 ( 4 x ^{2} - 1 ) - ( 3 x + 2 ) ( 8 x )}{( 4 x ^{2} - 1 ) ^{2}} = - \frac{12 x ^{2} + 16 x + 3}{( 4 x ^{2} - 1 ) ^{2}}

(c)

2 marks

A point of inflexion is where $f^{''}$ changes sign. Solve $f^{''} (x) = 0$ on the calculator.

x \approx - 1.60

(d)

5 marks

Set the numerator of $f^{'}$ to zero to find the turning points: $12 x^{2} + 16 x + 3 = 0$ gives $x \approx - 1.11$ and $x \approx - 0.226$ .
Intercepts: $f = 0$ when $x = - \frac{2}{3}$ , and $f (0) = - 2$ . Asymptotes: vertical lines $x = \frac{1}{2}$ and $x = - \frac{1}{2}$ , and horizontal $y = 0$ .
The sketch has three branches: on the left it comes up from just below the axis to the minimum at $x \approx - 1.11$ , then rises through $(- \frac{2}{3}, 0)$ to $+ \infty$ at $x = - \frac{1}{2}$ . In the middle it comes up from $- \infty$ to the maximum at $x \approx - 0.226$ , passes through $(0, - 2)$ , and goes back to $- \infty$ at $x = \frac{1}{2}$ . On the right it falls from $+ \infty$ toward the axis.

local minimum (- 1.11, - 0.339), local maximum (- 0.226, - 1.66)

Mentor tip

Put the coordinates of every turning point and intercept on the sketch. The details are what make a sketch complete.

(e)

4 marks

The vertical asymptote of $g$ is where its denominator is 0: $x = - \frac{2}{3}$ .
Since the numerator is one degree higher than the denominator, there is an oblique asymptote. Divide.

\frac{4 x ^{2} - 1}{3 x + 2} = \frac{4}{3} x - \frac{8}{9} + \frac{7/9}{3 x + 2} x = - \frac{2}{3}, y = \frac{4}{3} x - \frac{8}{9}

(f)

4 marks

The graphs cross where $(3 x + 2)^{2} = (4 x^{2} - 1)^{2}$ . Taking square roots gives $4 x^{2} - 3 x - 3 = 0$ (the other sign has no real solutions), so $x = \frac{3 \pm 57}{8}$ , which is $x \approx - 0.569$ or $1.32$ .
Test each interval between these crossing points and the asymptotes $x = - \frac{2}{3}$ , $- \frac{1}{2}$ , $\frac{1}{2}$ . The inequality $f < g$ holds on three of them.

- \frac{2}{3} < x < - 0.569, - \frac{1}{2} < x < \frac{1}{2}, x > 1.32

Mentor tip

Check one test point in each interval: $x = 0$ gives $f = - 2$ and $g = - \frac{1}{2}$ , so $f < g$ there.

Question 12

20 marks

Partial fractions, then a logistic model of an ant colony

Calculus

(a)

3 marks

Put the fractions over a common denominator and compare numerators.
Putting $x = 0$ gives $a = \frac{1}{k}$ and putting $x = k$ gives $b = \frac{1}{k}$ .

\frac{1}{x ( k - x )} = \frac{a}{x} + \frac{b}{k - x} \Rightarrow 1 = a (k - x) + b x

(b)

3 marks

Integrate the two terms from part (a).

f (x) = \frac{1}{k} \int (\frac{1}{x} + \frac{1}{k - x}) d x = \frac{1}{k} (ln ∣ x ∣ - ln ∣ k - x ∣) + c

(c)

8 marks

Separate the variables, use part (b) with $\frac{1}{P ( k - P )} = \frac{1}{k} (\frac{1}{P} + \frac{1}{k - P})$ , and integrate.
Use $P = 1200$ at $t = 0$ to fix the constant, then divide by 5 and take exponentials.
Solve for $P$ : cross-multiply, collect the terms in $P$ , and divide by $e^{t /5}$ to get a negative exponent.

\int \frac{5 k}{P ( k - P )} d P = \int 1 d t \Rightarrow 5 (ln P - ln (k - P)) = t + c ln \frac{P}{k - P} = \frac{t}{5} + ln \frac{1200}{k - 1200} \Rightarrow \frac{P}{k - P} = \frac{1200}{k - 1200} e^{t /5} P [(k - 1200) + 1200 e^{t /5}] = 1200 k e^{t /5} \Rightarrow P = \frac{1200 k}{( k - 1200 ) e ^{- t /5} + 1200}

(d)

3 marks

At $t = 10$ the population is 2400. Substitute and solve for $k$ .

2400 = \frac{1200 k}{( k - 1200 ) e ^{- 2} + 1200} \Rightarrow k = \frac{2400 ( 1 - e ^{- 2} )}{1 - 2 e ^{- 2}} \approx 2845

(e)

3 marks

The rate $\frac{d P}{d t} = \frac{P ( k - P )}{5 k}$ is largest when $P = \frac{k}{2}$ . Put $P = \frac{k}{2}$ into the formula for $P$ and solve for $t$ .

(k - 1200) e^{- t /5} + 1200 = 2400 \Rightarrow t = 5 ln \frac{k - 1200}{1200} \approx 1.58 days

Mentor tip

The rate is a quadratic in $P$ with its peak halfway between 0 and $k$ . No need to differentiate again.

Our analysis

Where the marks are

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

Calculus53 marks · 48%
Q4, Q7, Q10, Q11, Q12
Number and algebra17 marks · 15%
Q3, Q5, Q8
Functions15 marks · 14%
Q11
Statistics and probability13 marks · 12%
Q1, Q2
Geometry and trigonometry12 marks · 11%
Q6, Q9

Mentor insights

Traps to watch for

01
Predict from the equation, not the picture · Q1
Part (b) wants a calculated value from $y = a x + b$ with unrounded or 3 sf values, not a reading off the scatter diagram.
02
Total probability is 1 · Q7
A density function must integrate to 1 over its range. That single condition produces the equation for $k$ .
03
Use the symmetry in part (e) · Q10
The line $L$ is tangent to both $f$ and $f^{- 1}$ and the graphs are reflections in $y = x$ . The area in (e) is exactly twice the area in (d).
04
Reciprocals share crossing points · Q11
$f$ and $g$ are reciprocals, so they are equal when $f = \pm 1$ . Compare the two by finding where $(3 x + 2)^{2} = (4 x^{2} - 1)^{2}$ and then testing the intervals.
05
Solve first, then use the data · Q12
Solve the differential equation with the general $k$ and the starting value 1200. Only then use $P (10) = 2400$ to find $k$ .

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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2021
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 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 1–9 · 55 marks

Question 1

7 marks

Regression line, a prediction and the mean point

Statistics and probability

(a)

2 marks

Enter the eight pairs in the calculator's linear regression (y on x).

a \approx 0.433, b \approx 4.50

(b)

2 marks

Substitute $x = 18$ into $y = a x + b$ .

y = 0.4331 \dots \times 18 + 4.5026 \dots \approx 12.3

Mentor tip

Using the stored unrounded values keeps the third significant figure safe.

(c)

1 mark

The regression line always passes through the point of means $(\overset{x}{ˉ}, \overset{y}{ˉ})$ .

\overset{x}{ˉ} = 15, \overset{y}{ˉ} = 11

(d)

2 marks

Draw a straight line through the mean point $(15, 11)$ that follows the trend of the points. It should meet the $y$ -axis at about 4.5 and rise by about 0.43 for each unit across.

Mentor tip

Use a ruler. Pivot the line about $(15, 11)$ so roughly equal numbers of points sit above and below it.

Question 2

6 marks

Rejected bags of sugar from a normal distribution

Statistics and probability

(a)

2 marks

Let $X \sim N (1000, 3. 5^{2})$ . The probability of a mass below 995 g comes from the normal cumulative distribution on the calculator.

P (X < 995) \approx 0.0766

(b)

1 mark

Multiply the probability by the sample size.

100 \times 0.0766 = 7.66 \approx 8 bags

(c)

3 marks

This is a conditional probability: the bag is not rejected, meaning $X \geq 995$ , and we want $X > 1005$ . The event $X > 1005$ is contained in $X \geq 995$ .

P (X > 1005 ∣ X \geq 995) = \frac{P ( X > 1005 )}{P ( X \geq 995 )} = \frac{0.07656}{1 - 0.07656} \approx 0.0829

Mentor tip

Rounding to 3 sf at the start (0.0766) changes the third figure. Keep the calculator's stored value.

Question 3

5 marks

A Ferris wheel modelled by a cosine

Number and algebra

5 marks

The amplitude is half the diameter, and starting at the lowest point means the cosine is reflected, so $a = - 55$ .
One revolution takes 20 minutes, so the period is 20 and $b = \frac{2 π}{20} = \frac{π}{10} \approx 0.314$ .
The middle height is the lowest height plus the radius, $10 + 55 = 65$ , so $c = 65$ .

h (t) = - 55 cos (\frac{π}{10} t) + 65

Mentor tip

Check: $h (0) = - 55 + 65 = 10$ , the lowest point. And $h (10) = 120$ , the top, 10 m plus the 110 m diameter.

Question 4

6 marks

Velocity, distance and acceleration from $v = t sin t - 3$

Calculus

(a)

2 marks

The particle is at rest when $v = 0$ , that is, $t sin t = 3$ . Solve on the calculator, or read the first axis crossing of the graph.

t \approx 6.74 s

(b)

2 marks

Total distance is the integral of the speed.

\int_{0}^{10} ∣ t sin t - 3∣ d t \approx 37.1 m

(c)

2 marks

Acceleration is the derivative of velocity. Differentiate with the product rule, then put in $t = 7$ .

a (t) = sin t + t cos t \Rightarrow a (7) = sin 7 + 7 cos 7 \approx 5.93 m s^{- 2}

Mentor tip

Radian mode. In degrees the answer is completely different.

Question 5

5 marks

A coefficient in a binomial expansion pins down $n$

Number and algebra

5 marks

The term in $x^{4}$ takes $x^{2}$ from two of the $n + 1$ brackets and 3 from the rest.
Set the coefficient equal to 20 412 and solve on the calculator, or by trying values.

(2 n + 1) 3^{n - 1} (x^{2})^{2} = \frac{n ( n + 1 )}{2} 3^{n - 1} x^{4} \frac{n ( n + 1 )}{2} 3^{n - 1} = 20412 \Rightarrow n = 7

Mentor tip

Check: $\frac{7 \cdot 8}{2} \cdot 3^{6} = 28 \times 729 = 20412$ .

Question 6

5 marks

A plane perpendicular to two planes, and the point where all three meet

Geometry and trigonometry

(a)

3 marks

The normals are $321$ and $1 - 2 1$ . The normal of the new plane is perpendicular to both, so take their cross product.
The plane passes through the origin, so there is no constant term.

321 \times 1 - 2 1 = 4 - 2 - 8 \propto 2 - 1 - 4 2 x - y - 4 z = 0

(b)

2 marks

Solve the three equations together. Adding the first two plane equations gives $4 x + 2 z = 10$ , so $z = 5 - 2 x$ .
Put these into $3 x + 2 y + z = 6$ .

2 x - y - 4 (5 - 2 x) = 0 \Rightarrow y = 10 x - 20 3 x + 2 (10 x - 20) + (5 - 2 x) = 6 \Rightarrow 21 x = 41 (\frac{41}{21}, - \frac{10}{21}, \frac{23}{21})

Mentor tip

Check in the second plane: $\frac{41}{21} + \frac{20}{21} + \frac{23}{21} = \frac{84}{21} = 4$ .

Question 7

7 marks

A probability density function with a parameter to find

Calculus

(a)

5 marks

The total probability is 1, so integrate $\frac{x}{( x ^{2} + k ) ^{3}} = x (x^{2} + k)^{- 3/2}$ over $0 \leq x \leq 4$ . The inside function is $x^{2} + k$ , whose derivative is $2 x$ , so the antiderivative is a multiple of $(x^{2} + k)^{- 1/2}$ .
Set this equal to 1 and put the left side over a common denominator.

\int_{0}^{4} x (x^{2} + k)^{- 3/2} d x = [- (x^{2} + k)^{- 1/2}]_{0}^{4} = \frac{1}{k} - \frac{1}{16 + k} \frac{16 + k - k}{k 16 + k} = 1 16 + k - k = k 16 + k

(b)

2 marks

Solve the equation from part (a) numerically on the calculator, or graph both sides.

k \approx 0.645

Mentor tip

Check in the equation: $\frac{1}{0.645} - \frac{1}{16.645} \approx 1.245 - 0.245 = 1$ .

Question 8

7 marks

Product of complex numbers in modulus-argument form

Number and algebra

(a)

1 mark

The modulus of a product is the product of the moduli.

∣ z w ∣ = ∣ z ∣∣ w ∣ = 2 \times 8 = 16

(b)

2 marks

Since $w = 8 (cos \frac{2 k π}{5} - i sin \frac{2 k π}{5})$ , its argument is $- \frac{2 k π}{5}$ . Add the arguments.

ar g (z w) = \frac{π}{5} - \frac{2 k π}{5} = \frac{( 1 - 2 k ) π}{5}

(c)

4 marks

For $z w$ to be a real integer, its argument must be a multiple of $π$ . So $\frac{1 - 2 k}{5}$ is an integer, meaning $1 - 2 k$ is a multiple of 5. Try $k = 1, 2, 3$ : the first that works is $k = 3$ , because $1 - 6 = - 5$ .
Then $ar g (z w) = - π$ , and the number is on the negative real axis with modulus 16.

k = 3 z w = 16 (cos (- π) + i sin (- π)) = - 16

Mentor tip

$k = 1$ gives argument $- \frac{π}{5}$ and $k = 2$ gives $- \frac{3 π}{5}$ . Neither is a multiple of $π$ .

Question 9

7 marks

Two boats and a changing bearing (related rates)

Geometry and trigonometry

(a)

1 mark

From the diagram, boat B is 50 m east of A, and $θ$ is measured from north. The extra distance travelled by B is $y - x$ , and $tan θ = \frac{50}{y - x}$ .

y - x = 50 cot θ \Rightarrow y = x + 50 cot θ

(b)

6 marks

Differentiate with respect to $t$ .
At time $T$ , B has gone 10 m further, so $50 cot θ = 10$ , giving $cot θ = 0.2$ and $cosec^{2} θ = 1 + 0. 2^{2} = 1.04$ . Also $\frac{d y}{d t} = 2 \frac{d x}{d t}$ and $\frac{d θ}{d t} = - 0.1$ .

\frac{d y}{d t} = \frac{d x}{d t} - 50 cosec^{2} θ \frac{d θ}{d t} 2 \frac{d x}{d t} = \frac{d x}{d t} - 50 (1.04) (- 0.1) = \frac{d x}{d t} + 5.2 \frac{d x}{d t} = 5.2 m s^{- 1}

Mentor tip

The negative rate of $θ$ is why the extra term is positive. Keep the sign through the substitution.

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

15 marks

An exponential curve, a tangent of gradient $- 1$ and enclosed areas

Calculus

(a)

2 marks

Solve $90 e^{- 0.5 x} = x$ on the calculator, or find the crossing point of the two graphs.

x \approx 5.57

(b)

4 marks

The gradient of $f$ is $f^{'} (x) = - 45 e^{- 0.5 x}$ . Set it equal to $- 1$ and solve exactly.
Then $y = 90 \cdot \frac{1}{45} = 2$ .

- 45 e^{- 0.5 x} = - 1 \Rightarrow e^{- 0.5 x} = \frac{1}{45} \Rightarrow x = 2 ln 45 Q (2 ln 45, 2)

(c)

2 marks

Use the point-gradient form with gradient $- 1$ through $Q$ .

y - 2 = - (x - 2 ln 45) \Rightarrow y = - x + 2 ln 45 + 2

(d)

5 marks

(i) The line $L$ meets $y = x$ where $x = - x + 2 ln 45 + 2$ , so $x = ln 45 + 1 \approx 4.81$ .

A = \int_{4.81}^{5.57} (x - L (x)) d x + \int_{5.57}^{7.61} (90 e^{- 0.5 x} - L (x)) d x \approx 1.52

Mentor tip

Keep the limits unrounded in the calculator. Rounding them to 3 sf shifts the area in the third figure.

(e)

2 marks

By symmetry in $y = x$ the shaded area between $f$ , $f^{- 1}$ and $L$ is twice the area in part (d).

2 \times 1.52 \approx 3.04

Question 11

20 marks

A rational function, its graph, and a comparison with its reciprocal

Functions
Calculus

(a)

2 marks

$f$ is undefined where the denominator is 0.

4 x^{2} - 1 = 0 \Rightarrow p = \frac{1}{2}, q = - \frac{1}{2}

(b)

3 marks

Use the quotient rule.

f^{'} (x) = \frac{3 ( 4 x ^{2} - 1 ) - ( 3 x + 2 ) ( 8 x )}{( 4 x ^{2} - 1 ) ^{2}} = - \frac{12 x ^{2} + 16 x + 3}{( 4 x ^{2} - 1 ) ^{2}}

(c)

2 marks

A point of inflexion is where $f^{''}$ changes sign. Solve $f^{''} (x) = 0$ on the calculator.

x \approx - 1.60

(d)

5 marks

Set the numerator of $f^{'}$ to zero to find the turning points: $12 x^{2} + 16 x + 3 = 0$ gives $x \approx - 1.11$ and $x \approx - 0.226$ .
Intercepts: $f = 0$ when $x = - \frac{2}{3}$ , and $f (0) = - 2$ . Asymptotes: vertical lines $x = \frac{1}{2}$ and $x = - \frac{1}{2}$ , and horizontal $y = 0$ .
The sketch has three branches: on the left it comes up from just below the axis to the minimum at $x \approx - 1.11$ , then rises through $(- \frac{2}{3}, 0)$ to $+ \infty$ at $x = - \frac{1}{2}$ . In the middle it comes up from $- \infty$ to the maximum at $x \approx - 0.226$ , passes through $(0, - 2)$ , and goes back to $- \infty$ at $x = \frac{1}{2}$ . On the right it falls from $+ \infty$ toward the axis.

local minimum (- 1.11, - 0.339), local maximum (- 0.226, - 1.66)

Mentor tip

Put the coordinates of every turning point and intercept on the sketch. The details are what make a sketch complete.

(e)

4 marks

The vertical asymptote of $g$ is where its denominator is 0: $x = - \frac{2}{3}$ .
Since the numerator is one degree higher than the denominator, there is an oblique asymptote. Divide.

\frac{4 x ^{2} - 1}{3 x + 2} = \frac{4}{3} x - \frac{8}{9} + \frac{7/9}{3 x + 2} x = - \frac{2}{3}, y = \frac{4}{3} x - \frac{8}{9}

(f)

4 marks

The graphs cross where $(3 x + 2)^{2} = (4 x^{2} - 1)^{2}$ . Taking square roots gives $4 x^{2} - 3 x - 3 = 0$ (the other sign has no real solutions), so $x = \frac{3 \pm 57}{8}$ , which is $x \approx - 0.569$ or $1.32$ .
Test each interval between these crossing points and the asymptotes $x = - \frac{2}{3}$ , $- \frac{1}{2}$ , $\frac{1}{2}$ . The inequality $f < g$ holds on three of them.

- \frac{2}{3} < x < - 0.569, - \frac{1}{2} < x < \frac{1}{2}, x > 1.32

Mentor tip

Check one test point in each interval: $x = 0$ gives $f = - 2$ and $g = - \frac{1}{2}$ , so $f < g$ there.

Question 12

20 marks

Partial fractions, then a logistic model of an ant colony

Calculus

(a)

3 marks

Put the fractions over a common denominator and compare numerators.
Putting $x = 0$ gives $a = \frac{1}{k}$ and putting $x = k$ gives $b = \frac{1}{k}$ .

\frac{1}{x ( k - x )} = \frac{a}{x} + \frac{b}{k - x} \Rightarrow 1 = a (k - x) + b x

(b)

3 marks

Integrate the two terms from part (a).

f (x) = \frac{1}{k} \int (\frac{1}{x} + \frac{1}{k - x}) d x = \frac{1}{k} (ln ∣ x ∣ - ln ∣ k - x ∣) + c

(c)

8 marks

Separate the variables, use part (b) with $\frac{1}{P ( k - P )} = \frac{1}{k} (\frac{1}{P} + \frac{1}{k - P})$ , and integrate.
Use $P = 1200$ at $t = 0$ to fix the constant, then divide by 5 and take exponentials.
Solve for $P$ : cross-multiply, collect the terms in $P$ , and divide by $e^{t /5}$ to get a negative exponent.

\int \frac{5 k}{P ( k - P )} d P = \int 1 d t \Rightarrow 5 (ln P - ln (k - P)) = t + c ln \frac{P}{k - P} = \frac{t}{5} + ln \frac{1200}{k - 1200} \Rightarrow \frac{P}{k - P} = \frac{1200}{k - 1200} e^{t /5} P [(k - 1200) + 1200 e^{t /5}] = 1200 k e^{t /5} \Rightarrow P = \frac{1200 k}{( k - 1200 ) e ^{- t /5} + 1200}

(d)

3 marks

At $t = 10$ the population is 2400. Substitute and solve for $k$ .

2400 = \frac{1200 k}{( k - 1200 ) e ^{- 2} + 1200} \Rightarrow k = \frac{2400 ( 1 - e ^{- 2} )}{1 - 2 e ^{- 2}} \approx 2845

(e)

3 marks

The rate $\frac{d P}{d t} = \frac{P ( k - P )}{5 k}$ is largest when $P = \frac{k}{2}$ . Put $P = \frac{k}{2}$ into the formula for $P$ and solve for $t$ .

(k - 1200) e^{- t /5} + 1200 = 2400 \Rightarrow t = 5 ln \frac{k - 1200}{1200} \approx 1.58 days

Mentor tip

The rate is a quadratic in $P$ with its peak halfway between 0 and $k$ . No need to differentiate again.

Our analysis

Where the marks are

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

Calculus53 marks · 48%
Q4, Q7, Q10, Q11, Q12
Number and algebra17 marks · 15%
Q3, Q5, Q8
Functions15 marks · 14%
Q11
Statistics and probability13 marks · 12%
Q1, Q2
Geometry and trigonometry12 marks · 11%
Q6, Q9

Mentor insights

Traps to watch for

01
Predict from the equation, not the picture · Q1
Part (b) wants a calculated value from $y = a x + b$ with unrounded or 3 sf values, not a reading off the scatter diagram.
02
Total probability is 1 · Q7
A density function must integrate to 1 over its range. That single condition produces the equation for $k$ .
03
Use the symmetry in part (e) · Q10
The line $L$ is tangent to both $f$ and $f^{- 1}$ and the graphs are reflections in $y = x$ . The area in (e) is exactly twice the area in (d).
04
Reciprocals share crossing points · Q11
$f$ and $g$ are reciprocals, so they are equal when $f = \pm 1$ . Compare the two by finding where $(3 x + 2)^{2} = (4 x^{2} - 1)^{2}$ and then testing the intervals.
05
Solve first, then use the data · Q12
Solve the differential equation with the general $k$ and the starting value 1200. Only then use $P (10) = 2400$ to find $k$ .

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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