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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 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 · 56 marks

Question 1

6 marks

Waiting time against queue length: regression and interpretation

Statistics and probability

(a)

3 marks

Enter the five pairs into the calculator's regression (y on x). It also gives the correlation coefficient.

a \approx 0.805, b \approx 2.88, r \approx 0.978

(b)

1 mark

The gradient is the change in waiting time for each extra customer waiting ahead of Sarah.
So each extra customer ahead adds about 0.805 minutes to the waiting time, on average.

(c)

2 marks

Substitute $x = 7$ into the regression line.

y = 0.805 \times 7 + 2.88 \approx 8.52 minutes

Mentor tip

7 is inside the range of the data (3 to 11), so this is an interpolation and is reasonably trustworthy.

Question 2

5 marks

Where an arithmetic sequence reaches zero, and its largest sum

Number and algebra

(a)

2 marks

Write the $k^{th}$ term and set it to zero.

60 + (k - 1) (- 2.5) = 0 \Rightarrow k - 1 = 24 \Rightarrow k = 25

(b)

3 marks

The terms are positive up to $u_{24}$ and $u_{25} = 0$ , so the sum is largest at $n = 25$ (and equals $S_{24}$ ).

S_{25} = \frac{25}{2} (60 + 0) = 750

Mentor tip

Adding a zero term changes nothing, so either $S_{24}$ or $S_{25}$ gives 750.

Question 3

8 marks

Sport and theatre: a Venn diagram and independence

Statistics and probability

(a)

2 marks

The students doing at least one activity are $100% - 18% = 82%$ . Use inclusion–exclusion.

P (S \cup T) = 0.82 = 0.7 + 0.2 - P (S \cap T) \Rightarrow P (S \cap T) = 0.08

(b)

2 marks

Theatre only is all of theatre minus the overlap.

P (T \cap S^{'}) = 0.2 - 0.08 = 0.12

(c)

2 marks

Use $P (G \cap T) = P (G) P (T ∣ G)$ .

0.48 \times 0.25 = 0.12

(d)

2 marks

Compare $P (G) P (T)$ with $P (G \cap T)$ .
They are not equal, so $G$ and $T$ are not independent.

0.48 \times 0.2 = 0.096 \neq = 0.12

Mentor tip

Equivalent test: $P (T ∣ G) = 0.25$ is not equal to $P (T) = 0.2$ .

Question 4

6 marks

Range of a quadratic, then a constant that keeps a composite non-positive

Functions

(a)

2 marks

Complete the square to find the lowest point.

f (x) = 6 (x - 1)^{2} - 5 \Rightarrow range: f (x) \geq - 5

(b)

4 marks

Write the composite: $g (f (x)) = - f (x) + c$ . We need this at most 0 for all $x$ , that is, $c \leq f (x)$ for every $x$ .
The smallest value of $f$ is $- 5$ , so $c$ must be at most that.

c \leq - 5

Mentor tip

Check with $c = - 5$ : at $x = 1$ the composite is $5 - 5 = 0$ , which is allowed (the inequality is not strict).

Question 5

7 marks

Radiocarbon dating from an exponential decay model

Calculus

(a)

1 mark

At the moment of death $t = 0$ , so $A = A_{0} e^{0} = A_{0}$ . The question says a plant starts with 100 units.

A_{0} = 100

(b)

3 marks

After 5730 years half is left, so $A = 50$ .
Take natural logs, using $ln \frac{1}{2} = - ln 2$ .

50 = 100 e^{- 5730 k} \Rightarrow e^{- 5730 k} = \frac{1}{2} - 5730 k = - ln 2 \Rightarrow k = \frac{ln 2}{5730}

(c)

3 marks

If 25% has decayed, 75 units remain.
Correct to the nearest 10 years, that is about 2380 years.

75 = 100 e^{- k t} \Rightarrow t = \frac{ln ( 4/3 )}{k} = \frac{5730 ln ( 4/3 )}{ln 2} \approx 2378

Question 6

6 marks

Mean and variance of a power-law density

Calculus

(a)

2 marks

The density is $(n + 1) x^{n}$ on $[0, 1]$ . Integrate $x f (x)$ .

E (X) = \int_{0}^{1} (n + 1) x^{n + 1} d x = \frac{n + 1}{n + 2} [x^{n + 2}]_{0}^{1} = \frac{n + 1}{n + 2}

(b)

4 marks

First find $E (X^{2})$ in the same way.
Then subtract the square of the mean and put everything over $(n + 2)^{2} (n + 3)$ .
The bracket simplifies: $(n^{2} + 4 n + 4) - (n^{2} + 4 n + 3) = 1$ .

E (X^{2}) = \int_{0}^{1} (n + 1) x^{n + 2} d x = \frac{n + 1}{n + 3} Var (X) = \frac{n + 1}{n + 3} - \frac{( n + 1 ) ^{2}}{( n + 2 ) ^{2}} = \frac{( n + 1 ) [ ( n + 2 ) ^{2} - ( n + 1 ) ( n + 3 ) ]}{( n + 2 ) ^{2} ( n + 3 )} Var (X) = \frac{n + 1}{( n + 2 ) ^{2} ( n + 3 )}

Mentor tip

Factorise $(n + 1)$ out early. It stops the expansion from getting messy.

Question 7

5 marks

Arranging eight runners with two of them linked

Statistics and probability

(a)

2 marks

Treat Andrea followed directly by Jack as one block. Then there are 7 objects (the block and the other 6 runners) to arrange.

7! = 5040

(b)

3 marks

There are $8!$ finishing orders. In exactly half of them Andrea is ahead of Jack, since swapping the two runners turns each such order into one where Jack is ahead.

\frac{8 !}{2} = 20160

Question 8

5 marks

The real part of $\frac{1 + z}{1 - z}$ on the unit circle

Number and algebra

5 marks

Since $z = cos θ + i sin θ$ has modulus 1, $z \overset{z}{ˉ} = 1$ . Multiply top and bottom by $1 - \overset{z}{ˉ}$ .
The top becomes $1 - 1 + (z - \overset{z}{ˉ})$ , and $z - \overset{z}{ˉ} = 2 i sin θ$ is purely imaginary. The bottom $∣1 - z ∣^{2}$ is a positive real number (since $z \neq = 1$ ).

\frac{1 + z}{1 - z} = \frac{( 1 + z ) ( 1 - z ˉ )}{( 1 - z ) ( 1 - z ˉ )} = \frac{1 - z ˉ + z - z z ˉ}{∣1 - z ∣ ^{2}} \frac{1 + z}{1 - z} = \frac{2 i sin θ}{2 - 2 cos θ} \Rightarrow Re (\frac{1 + z}{1 - z}) = 0

Mentor tip

Writing out the real and imaginary parts, as in $\frac{1 + cos θ + i sin θ}{1 - cos θ - i sin θ}$ , gives the same result and is only longer.

Question 9

8 marks

Maclaurin series for sec, then a limit

Calculus

(a)

1 mark

The binomial series for a power of $- 1$ alternates in sign.

(1 + t)^{- 1} = 1 - t + t^{2} - \dots

(b)

4 marks

Write $sec x = \frac{1}{cos x}$ with $cos x = 1 + t$ , where $t = - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ . Use part (a) and keep terms up to $x^{4}$ .

sec x = 1 - (- \frac{x ^{2}}{2} + \frac{x ^{4}}{24}) + (\frac{x ^{2}}{2})^{2} + \dots sec x = 1 + \frac{x ^{2}}{2} - \frac{x ^{4}}{24} + \frac{x ^{4}}{4} = 1 + \frac{x ^{2}}{2} + \frac{5 x ^{4}}{24} + \dots

(c)

3 marks

Use $arctan 2 x = 2 x - \frac{( 2 x ) ^{3}}{3} + \dots = 2 x - \frac{8 x ^{3}}{3} + \dots$ . Then $x arctan 2 x = 2 x^{2} - \frac{8 x ^{4}}{3} + \dots$ and $sec x - 1 = \frac{x ^{2}}{2} + \frac{5 x ^{4}}{24} + \dots$ .
Divide top and bottom by $x^{2}$ , then let $x \to 0$ .

x \to 0 lim \frac{2 - \frac{8 x ^{2}}{3} + \dots}{\frac{1}{2} + \frac{5 x ^{2}}{24} + \dots} = \frac{2}{1/2} = 4

Mentor tip

The $x^{2}$ terms vanish in the limit, so only the leading coefficients matter.

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

15 marks

Flight times from a normal distribution

Statistics and probability

(a)

3 marks

The upper 2% of flights is above 82 minutes, so 82 is $z =$ inverse normal of 0.98, about 2.054, standard deviations above the mean of 75.

\frac{82 - 75}{σ} = 2.0537 \Rightarrow σ \approx 3.41

(b)

2 marks

With $T \sim N (75, 3.40 8^{2})$ , find the area above 80 on the calculator.

P (T > 80) \approx 0.0712

(c)

4 marks

Use the conditional probability formula. The numerator is $P (80 < T < 82) = 0.0712 - 0.02$ .

P (T < 82 ∣ T > 80) = \frac{P ( 80 < T < 82 )}{P ( T > 80 )} = \frac{0.0512}{0.0712} \approx 0.719

(d)

3 marks

The number of long flights follows $X \sim B (64, 0.0712)$ , so the expected value is $n p$ .

64 \times 0.0712 \approx 4.56

(e)

3 marks

"More than 6" is the complement of "at most 6".

P (X > 6) = 1 - P (X \leq 6) \approx 1 - 0.8309 = 0.169

Mentor tip

Store the unrounded probability from part (b). It feeds parts (c) to (e).

Question 11

18 marks

The volume of a bowl from a solid of revolution

Calculus

(a)

6 marks

The volume is $π \int_{0}^{l n 16} y^{2} d x$ , and $y^{2} = \frac{k ^{2} e ^{x}}{( 1 + e ^{x} ) ^{2}}$ .
Let $u = 1 + e^{x}$ , so $d u = e^{x} d x$ . The limits $x = 0$ and $x = ln 16$ become $u = 2$ and $u = 17$ .

V = π k^{2} \int_{2}^{17} \frac{1}{u ^{2}} d u = π k^{2} [- \frac{1}{u}]_{2}^{17} = π k^{2} (\frac{1}{2} - \frac{1}{17}) V = \frac{15 π k ^{2}}{34}

(b)

2 marks

Set the volume equal to 300 and solve for $k$ (positive).

\frac{15 π k ^{2}}{34} = 300 \Rightarrow k^{2} = \frac{680}{π} \Rightarrow k \approx 14.7

(c)

4 marks

The top radius OA is $f (0)$ and the base radius BC is $f (ln 16)$ . Note $e^{(l n 16) /2} = 4$ .

OA = f (0) = \frac{k}{2} \approx 7.36 cm BC = f (ln 16) = \frac{4 k}{17} \approx 3.46 cm

(d)

6 marks

The radius $f (x)$ decreases fastest where its slope $f^{'} (x)$ is most negative. That is the minimum of the graph of $f^{'}$ , which occurs where $f^{''} (x) = 0$ . Sketch $y = f^{'} (x)$ and read off the lowest point, or solve $f^{''} (x) = 0$ on the calculator.
The radius there is $f (1.76)$ .

x = ln (3 + 22) \approx 1.76 f (1.76) \approx 5.20 cm

Mentor tip

The bowl's radius falls from 7.36 to 3.46, and the steepest part of that fall is in the middle of the range, which agrees with 5.20 being between the two.

Question 12

21 marks

An even function built from an arcsin, and its inverse

Functions
Calculus

(a)

1 mark

Replacing $x$ by $- x$ changes nothing, because $x$ appears only as $x^{2}$ .

f (- x) = arcsin \frac{( - x ) ^{2} - 1}{( - x ) ^{2} + 1} = f (x), so f is even

(b)

2 marks

As $x \to \pm \infty$ , divide the top and bottom of the fraction by $x^{2}$ : it tends to 1.
So the graph has a horizontal asymptote $y = \frac{π}{2}$ .

\frac{x ^{2} - 1}{x ^{2} + 1} \to 1, f (x) \to arcsin 1 = \frac{π}{2}

(c)

9 marks

Let $u = \frac{x ^{2} - 1}{x ^{2} + 1}$ . The quotient rule gives $u^{'} = \frac{4 x}{( x ^{2} + 1 ) ^{2}}$ . Also $1 - u^{2} = \frac{4 x ^{2}}{( x ^{2} + 1 ) ^{2}}$ , so $1 - u^{2} = \frac{2∣ x ∣}{x ^{2} + 1}$ .
For $x < 0$ we have $∣ x ∣ = - x$ , so the derivative simplifies.
This is negative for every $x < 0$ , so $f$ is decreasing there.

f^{'} (x) = \frac{u ^{'}}{1 - u ^{2}} = \frac{4 x}{( x ^{2} + 1 ) ^{2}} \cdot \frac{x ^{2} + 1}{2∣ x ∣} = \frac{2 x}{∣ x ∣ ( x ^{2} + 1 )} f^{'} (x) = - \frac{2}{x ^{2} + 1} < 0

(d)

5 marks

Write $y = arcsin \frac{x ^{2} - 1}{x ^{2} + 1}$ , so $sin y = \frac{x ^{2} - 1}{x ^{2} + 1}$ , and solve for $x^{2}$ .
The domain of $g$ is $x \geq 0$ , so the inverse's outputs are non-negative: take the positive square root.

x^{2} sin y + sin y = x^{2} - 1 \Rightarrow x^{2} = \frac{1 + sin y}{1 - sin y} g^{- 1} (x) = \frac{1 + sin x}{1 - sin x}

(e)

1 mark

The domain of $g^{- 1}$ is the range of $g$ . For $x \geq 0$ the fraction runs from $- 1$ (at $x = 0$ ) up towards 1 but never reaches it, so $g$ takes values from $- \frac{π}{2}$ up to but not including $\frac{π}{2}$ .

- \frac{π}{2} \leq x < \frac{π}{2}

(f)

3 marks

The graph starts at $(- \frac{π}{2}, 0)$ , because $sin (- \frac{π}{2}) = - 1$ makes the numerator 0. It passes through $(0, 1)$ and increases without bound, with the vertical asymptote $x = \frac{π}{2}$ .

Mentor tip

The inverse reflects $g$ in $y = x$ , so $g$ 's horizontal asymptote $y = \frac{π}{2}$ becomes the vertical asymptote $x = \frac{π}{2}$ .

Our analysis

Where the marks are

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

Calculus48 marks · 44%
Q5, Q6, Q9, Q11, Q12
Statistics and probability34 marks · 31%
Q1, Q3, Q7, Q10
Functions18 marks · 16%
Q4, Q12
Number and algebra10 marks · 9%
Q2, Q8

Mentor insights

Traps to watch for

01
Independence is a test · Q3
Two events are independent only if $P (G \cap T) = P (G) P (T)$ . Work out both sides and compare them in the answer.
02
Square the whole bracket · Q9
Part (b) needs $t^{2}$ with $t = - \frac{x ^{2}}{2} + \frac{x ^{4}}{24}$ , so keep the cross terms up to $x^{4}$ and drop everything above.
03
Conditional means a smaller sample space · Q10
In (c) the flight is already known to take more than 80 minutes, so divide the probability of $80 < T < 82$ by $P (T > 80)$ .
04
Substitute before you integrate · Q11
In (a) the integrand $\frac{k ^{2} e ^{x}}{( 1 + e ^{x} ) ^{2}}$ is exactly $\frac{k ^{2}}{u ^{2}}$ once you set $u = 1 + e^{x}$ , since $d u = e^{x} d x$ . Change the limits too.
05
Use the sign of $x$ · Q12
The derivative contains $∣ x ∣$ , which hides the sign. Split into $x < 0$ and $x > 0$ so you can say what $\frac{x}{∣ x ∣}$ is.

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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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 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 · 56 marks

Question 1

6 marks

Waiting time against queue length: regression and interpretation

Statistics and probability

(a)

3 marks

Enter the five pairs into the calculator's regression (y on x). It also gives the correlation coefficient.

a \approx 0.805, b \approx 2.88, r \approx 0.978

(b)

1 mark

The gradient is the change in waiting time for each extra customer waiting ahead of Sarah.
So each extra customer ahead adds about 0.805 minutes to the waiting time, on average.

(c)

2 marks

Substitute $x = 7$ into the regression line.

y = 0.805 \times 7 + 2.88 \approx 8.52 minutes

Mentor tip

7 is inside the range of the data (3 to 11), so this is an interpolation and is reasonably trustworthy.

Question 2

5 marks

Where an arithmetic sequence reaches zero, and its largest sum

Number and algebra

(a)

2 marks

Write the $k^{th}$ term and set it to zero.

60 + (k - 1) (- 2.5) = 0 \Rightarrow k - 1 = 24 \Rightarrow k = 25

(b)

3 marks

The terms are positive up to $u_{24}$ and $u_{25} = 0$ , so the sum is largest at $n = 25$ (and equals $S_{24}$ ).

S_{25} = \frac{25}{2} (60 + 0) = 750

Mentor tip

Adding a zero term changes nothing, so either $S_{24}$ or $S_{25}$ gives 750.

Question 3

8 marks

Sport and theatre: a Venn diagram and independence

Statistics and probability

(a)

2 marks

The students doing at least one activity are $100% - 18% = 82%$ . Use inclusion–exclusion.

P (S \cup T) = 0.82 = 0.7 + 0.2 - P (S \cap T) \Rightarrow P (S \cap T) = 0.08

(b)

2 marks

Theatre only is all of theatre minus the overlap.

P (T \cap S^{'}) = 0.2 - 0.08 = 0.12

(c)

2 marks

Use $P (G \cap T) = P (G) P (T ∣ G)$ .

0.48 \times 0.25 = 0.12

(d)

2 marks

Compare $P (G) P (T)$ with $P (G \cap T)$ .
They are not equal, so $G$ and $T$ are not independent.

0.48 \times 0.2 = 0.096 \neq = 0.12

Mentor tip

Equivalent test: $P (T ∣ G) = 0.25$ is not equal to $P (T) = 0.2$ .

Question 4

6 marks

Range of a quadratic, then a constant that keeps a composite non-positive

Functions

(a)

2 marks

Complete the square to find the lowest point.

f (x) = 6 (x - 1)^{2} - 5 \Rightarrow range: f (x) \geq - 5

(b)

4 marks

Write the composite: $g (f (x)) = - f (x) + c$ . We need this at most 0 for all $x$ , that is, $c \leq f (x)$ for every $x$ .
The smallest value of $f$ is $- 5$ , so $c$ must be at most that.

c \leq - 5

Mentor tip

Check with $c = - 5$ : at $x = 1$ the composite is $5 - 5 = 0$ , which is allowed (the inequality is not strict).

Question 5

7 marks

Radiocarbon dating from an exponential decay model

Calculus

(a)

1 mark

At the moment of death $t = 0$ , so $A = A_{0} e^{0} = A_{0}$ . The question says a plant starts with 100 units.

A_{0} = 100

(b)

3 marks

After 5730 years half is left, so $A = 50$ .
Take natural logs, using $ln \frac{1}{2} = - ln 2$ .

50 = 100 e^{- 5730 k} \Rightarrow e^{- 5730 k} = \frac{1}{2} - 5730 k = - ln 2 \Rightarrow k = \frac{ln 2}{5730}

(c)

3 marks

If 25% has decayed, 75 units remain.
Correct to the nearest 10 years, that is about 2380 years.

75 = 100 e^{- k t} \Rightarrow t = \frac{ln ( 4/3 )}{k} = \frac{5730 ln ( 4/3 )}{ln 2} \approx 2378

Question 6

6 marks

Mean and variance of a power-law density

Calculus

(a)

2 marks

The density is $(n + 1) x^{n}$ on $[0, 1]$ . Integrate $x f (x)$ .

E (X) = \int_{0}^{1} (n + 1) x^{n + 1} d x = \frac{n + 1}{n + 2} [x^{n + 2}]_{0}^{1} = \frac{n + 1}{n + 2}

(b)

4 marks

First find $E (X^{2})$ in the same way.
Then subtract the square of the mean and put everything over $(n + 2)^{2} (n + 3)$ .
The bracket simplifies: $(n^{2} + 4 n + 4) - (n^{2} + 4 n + 3) = 1$ .

E (X^{2}) = \int_{0}^{1} (n + 1) x^{n + 2} d x = \frac{n + 1}{n + 3} Var (X) = \frac{n + 1}{n + 3} - \frac{( n + 1 ) ^{2}}{( n + 2 ) ^{2}} = \frac{( n + 1 ) [ ( n + 2 ) ^{2} - ( n + 1 ) ( n + 3 ) ]}{( n + 2 ) ^{2} ( n + 3 )} Var (X) = \frac{n + 1}{( n + 2 ) ^{2} ( n + 3 )}

Mentor tip

Factorise $(n + 1)$ out early. It stops the expansion from getting messy.

Question 7

5 marks

Arranging eight runners with two of them linked

Statistics and probability

(a)

2 marks

Treat Andrea followed directly by Jack as one block. Then there are 7 objects (the block and the other 6 runners) to arrange.

7! = 5040

(b)

3 marks

There are $8!$ finishing orders. In exactly half of them Andrea is ahead of Jack, since swapping the two runners turns each such order into one where Jack is ahead.

\frac{8 !}{2} = 20160

Question 8

5 marks

The real part of $\frac{1 + z}{1 - z}$ on the unit circle

Number and algebra

5 marks

Since $z = cos θ + i sin θ$ has modulus 1, $z \overset{z}{ˉ} = 1$ . Multiply top and bottom by $1 - \overset{z}{ˉ}$ .
The top becomes $1 - 1 + (z - \overset{z}{ˉ})$ , and $z - \overset{z}{ˉ} = 2 i sin θ$ is purely imaginary. The bottom $∣1 - z ∣^{2}$ is a positive real number (since $z \neq = 1$ ).

\frac{1 + z}{1 - z} = \frac{( 1 + z ) ( 1 - z ˉ )}{( 1 - z ) ( 1 - z ˉ )} = \frac{1 - z ˉ + z - z z ˉ}{∣1 - z ∣ ^{2}} \frac{1 + z}{1 - z} = \frac{2 i sin θ}{2 - 2 cos θ} \Rightarrow Re (\frac{1 + z}{1 - z}) = 0

Mentor tip

Writing out the real and imaginary parts, as in $\frac{1 + cos θ + i sin θ}{1 - cos θ - i sin θ}$ , gives the same result and is only longer.

Question 9

8 marks

Maclaurin series for sec, then a limit

Calculus

(a)

1 mark

The binomial series for a power of $- 1$ alternates in sign.

(1 + t)^{- 1} = 1 - t + t^{2} - \dots

(b)

4 marks

Write $sec x = \frac{1}{cos x}$ with $cos x = 1 + t$ , where $t = - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ . Use part (a) and keep terms up to $x^{4}$ .

sec x = 1 - (- \frac{x ^{2}}{2} + \frac{x ^{4}}{24}) + (\frac{x ^{2}}{2})^{2} + \dots sec x = 1 + \frac{x ^{2}}{2} - \frac{x ^{4}}{24} + \frac{x ^{4}}{4} = 1 + \frac{x ^{2}}{2} + \frac{5 x ^{4}}{24} + \dots

(c)

3 marks

Use $arctan 2 x = 2 x - \frac{( 2 x ) ^{3}}{3} + \dots = 2 x - \frac{8 x ^{3}}{3} + \dots$ . Then $x arctan 2 x = 2 x^{2} - \frac{8 x ^{4}}{3} + \dots$ and $sec x - 1 = \frac{x ^{2}}{2} + \frac{5 x ^{4}}{24} + \dots$ .
Divide top and bottom by $x^{2}$ , then let $x \to 0$ .

x \to 0 lim \frac{2 - \frac{8 x ^{2}}{3} + \dots}{\frac{1}{2} + \frac{5 x ^{2}}{24} + \dots} = \frac{2}{1/2} = 4

Mentor tip

The $x^{2}$ terms vanish in the limit, so only the leading coefficients matter.

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

15 marks

Flight times from a normal distribution

Statistics and probability

(a)

3 marks

The upper 2% of flights is above 82 minutes, so 82 is $z =$ inverse normal of 0.98, about 2.054, standard deviations above the mean of 75.

\frac{82 - 75}{σ} = 2.0537 \Rightarrow σ \approx 3.41

(b)

2 marks

With $T \sim N (75, 3.40 8^{2})$ , find the area above 80 on the calculator.

P (T > 80) \approx 0.0712

(c)

4 marks

Use the conditional probability formula. The numerator is $P (80 < T < 82) = 0.0712 - 0.02$ .

P (T < 82 ∣ T > 80) = \frac{P ( 80 < T < 82 )}{P ( T > 80 )} = \frac{0.0512}{0.0712} \approx 0.719

(d)

3 marks

The number of long flights follows $X \sim B (64, 0.0712)$ , so the expected value is $n p$ .

64 \times 0.0712 \approx 4.56

(e)

3 marks

"More than 6" is the complement of "at most 6".

P (X > 6) = 1 - P (X \leq 6) \approx 1 - 0.8309 = 0.169

Mentor tip

Store the unrounded probability from part (b). It feeds parts (c) to (e).

Question 11

18 marks

The volume of a bowl from a solid of revolution

Calculus

(a)

6 marks

The volume is $π \int_{0}^{l n 16} y^{2} d x$ , and $y^{2} = \frac{k ^{2} e ^{x}}{( 1 + e ^{x} ) ^{2}}$ .
Let $u = 1 + e^{x}$ , so $d u = e^{x} d x$ . The limits $x = 0$ and $x = ln 16$ become $u = 2$ and $u = 17$ .

V = π k^{2} \int_{2}^{17} \frac{1}{u ^{2}} d u = π k^{2} [- \frac{1}{u}]_{2}^{17} = π k^{2} (\frac{1}{2} - \frac{1}{17}) V = \frac{15 π k ^{2}}{34}

(b)

2 marks

Set the volume equal to 300 and solve for $k$ (positive).

\frac{15 π k ^{2}}{34} = 300 \Rightarrow k^{2} = \frac{680}{π} \Rightarrow k \approx 14.7

(c)

4 marks

The top radius OA is $f (0)$ and the base radius BC is $f (ln 16)$ . Note $e^{(l n 16) /2} = 4$ .

OA = f (0) = \frac{k}{2} \approx 7.36 cm BC = f (ln 16) = \frac{4 k}{17} \approx 3.46 cm

(d)

6 marks

The radius $f (x)$ decreases fastest where its slope $f^{'} (x)$ is most negative. That is the minimum of the graph of $f^{'}$ , which occurs where $f^{''} (x) = 0$ . Sketch $y = f^{'} (x)$ and read off the lowest point, or solve $f^{''} (x) = 0$ on the calculator.
The radius there is $f (1.76)$ .

x = ln (3 + 22) \approx 1.76 f (1.76) \approx 5.20 cm

Mentor tip

The bowl's radius falls from 7.36 to 3.46, and the steepest part of that fall is in the middle of the range, which agrees with 5.20 being between the two.

Question 12

21 marks

An even function built from an arcsin, and its inverse

Functions
Calculus

(a)

1 mark

Replacing $x$ by $- x$ changes nothing, because $x$ appears only as $x^{2}$ .

f (- x) = arcsin \frac{( - x ) ^{2} - 1}{( - x ) ^{2} + 1} = f (x), so f is even

(b)

2 marks

As $x \to \pm \infty$ , divide the top and bottom of the fraction by $x^{2}$ : it tends to 1.
So the graph has a horizontal asymptote $y = \frac{π}{2}$ .

\frac{x ^{2} - 1}{x ^{2} + 1} \to 1, f (x) \to arcsin 1 = \frac{π}{2}

(c)

9 marks

Let $u = \frac{x ^{2} - 1}{x ^{2} + 1}$ . The quotient rule gives $u^{'} = \frac{4 x}{( x ^{2} + 1 ) ^{2}}$ . Also $1 - u^{2} = \frac{4 x ^{2}}{( x ^{2} + 1 ) ^{2}}$ , so $1 - u^{2} = \frac{2∣ x ∣}{x ^{2} + 1}$ .
For $x < 0$ we have $∣ x ∣ = - x$ , so the derivative simplifies.
This is negative for every $x < 0$ , so $f$ is decreasing there.

f^{'} (x) = \frac{u ^{'}}{1 - u ^{2}} = \frac{4 x}{( x ^{2} + 1 ) ^{2}} \cdot \frac{x ^{2} + 1}{2∣ x ∣} = \frac{2 x}{∣ x ∣ ( x ^{2} + 1 )} f^{'} (x) = - \frac{2}{x ^{2} + 1} < 0

(d)

5 marks

Write $y = arcsin \frac{x ^{2} - 1}{x ^{2} + 1}$ , so $sin y = \frac{x ^{2} - 1}{x ^{2} + 1}$ , and solve for $x^{2}$ .
The domain of $g$ is $x \geq 0$ , so the inverse's outputs are non-negative: take the positive square root.

x^{2} sin y + sin y = x^{2} - 1 \Rightarrow x^{2} = \frac{1 + sin y}{1 - sin y} g^{- 1} (x) = \frac{1 + sin x}{1 - sin x}

(e)

1 mark

- \frac{π}{2} \leq x < \frac{π}{2}

(f)

3 marks

Mentor tip

The inverse reflects $g$ in $y = x$ , so $g$ 's horizontal asymptote $y = \frac{π}{2}$ becomes the vertical asymptote $x = \frac{π}{2}$ .

Our analysis

Where the marks are

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

Calculus48 marks · 44%
Q5, Q6, Q9, Q11, Q12
Statistics and probability34 marks · 31%
Q1, Q3, Q7, Q10
Functions18 marks · 16%
Q4, Q12
Number and algebra10 marks · 9%
Q2, Q8

Mentor insights

Traps to watch for

01
Independence is a test · Q3
Two events are independent only if $P (G \cap T) = P (G) P (T)$ . Work out both sides and compare them in the answer.
02
Square the whole bracket · Q9
Part (b) needs $t^{2}$ with $t = - \frac{x ^{2}}{2} + \frac{x ^{4}}{24}$ , so keep the cross terms up to $x^{4}$ and drop everything above.
03
Conditional means a smaller sample space · Q10
In (c) the flight is already known to take more than 80 minutes, so divide the probability of $80 < T < 82$ by $P (T > 80)$ .
04
Substitute before you integrate · Q11
In (a) the integrand $\frac{k ^{2} e ^{x}}{( 1 + e ^{x} ) ^{2}}$ is exactly $\frac{k ^{2}}{u ^{2}}$ once you set $u = 1 + e^{x}$ , since $d u = e^{x} d x$ . Change the limits too.
05
Use the sign of $x$ · Q12
The derivative contains $∣ x ∣$ , which hides the sign. Split into $x < 0$ and $x > 0$ so you can say what $\frac{x}{∣ x ∣}$ is.

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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