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IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Session: November 2024
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2024 Global paper, which your school or the IB can provide.

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Section A · Short response

Questions 1–9 · 56 marks

Question 1

7 marks

A square root function: end values, roots and a sketch

Functions

(a)

2 marks

(i) Substitute $x = 0$ .

(ii) Substitute $x = 20$ .

f (0) = 0 - 0 - 11 = - 11 f (20) = 1120 - 40 - 11 \approx - 1.81

(b)

2 marks

Solve $11 x - 2 x - 11 = 0$ on the calculator.

x \approx 1.73 and x \approx 17.5

(c)

3 marks

The curve starts at $(0, - 11)$ , rises through the root $x \approx 1.73$ to a maximum at about $(7.6, 4.1)$ , then falls through the root $x \approx 17.5$ and ends at about $(20, - 1.8)$ .

Question 2

4 marks

The coefficient of x to the power 6 in (2x - 5) to the power 9

Number and algebra

4 marks

For $x^{6}$ we need $(2 x)^{6}$ , so the power of $- 5$ is $9 - 6 = 3$ .
The coefficient.

(6 9) (2 x)^{6} (- 5)^{3} = 84 \times 64 x^{6} \times (- 125) 84 \times 64 \times (- 125) = - 672000

Question 3

5 marks

A discrete distribution with an unknown constant and its mean

Statistics and probability

(a)

2 marks

Add the probabilities for $x = 3, 5, 8, 11$ .

\frac{k ( 3 + 5 + 8 + 11 )}{20} = \frac{27 k}{20} = 1 \Rightarrow k = \frac{20}{27} \approx 0.741

(b)

3 marks

The probabilities are $\frac{x}{27}$ , so use $E (X) = \sum x P (X = x)$ .

E (X) = \frac{3 ( 3 ) + 5 ( 5 ) + 8 ( 8 ) + 11 ( 11 )}{27} = \frac{9 + 25 + 64 + 121}{27} = \frac{219}{27} = \frac{73}{9} \approx 8.11

Question 4

8 marks

The roof of a house: angles and lengths of the beams

Geometry and trigonometry

(a)

3 marks

The beam $B A$ makes an angle $α$ with the horizontal, with $tan α = \frac{7}{12}$ , so $α \approx 30.2 6^{\circ}$ . The line $A E$ is vertical.

B A E = 9 0^{\circ} - 30.2 6^{\circ} \approx 59. 7^{\circ}

(b)

5 marks

(i) By symmetry, $D E$ goes across half of $D F$ , which is 350 cm horizontally. The slope $D E$ makes an angle of $30.2 6^{\circ}$ with the horizontal.

The points $C$ , $D$ , $E$ lie in a line, with $C D = 50$ .

(ii) The beam has thickness $B C = 30$ , with $B C ⊥ C E$ . Drop a perpendicular from $A$ to the line $C E$ , meeting it at $G$ . Then $C G = B A$ , and in the right triangle $A E G$ the angle at $E$ is $B A E = 59.7 4^{\circ}$ (the angle between the vertical and the beam).

D E = \frac{350}{cos 30.2 6 ^{\circ}} = \frac{350}{sin 59.7 4 ^{\circ}} \approx 405.2 C E = 405.2 + 50 \approx 455 cm E G = \frac{30}{tan 59.7 4 ^{\circ}} \approx 17.5 B A = C E + E G = 455.2 + 17.5 \approx 473 cm

Question 5

5 marks

When a logarithm is defined for every real number

Functions

5 marks

For $h (x)$ to be defined for all real $x$ , we need $4 x^{2} - r x + r - 1 > 0$ for every $x$ .
The parabola opens upwards, so it is always above the axis when its discriminant is negative.
Find the critical values.
The quadratic in $r$ is negative between the roots.

Δ = r^{2} - 4 (4) (r - 1) = r^{2} - 16 r + 16 < 0 r = \frac{16 \pm 256 - 64}{2} = 8 \pm 43 8 - 43 < r < 8 + 43 (1.07 < r < 14.9)

Question 6

8 marks

A piecewise density: mean, a linear combination and the median

Statistics and probability

(a)

3 marks

Integrate over both pieces.

E (X) = \int_{0}^{2} \frac{x}{5} d x + \int_{2}^{8} x (- \frac{x}{30} + \frac{4}{15}) d x = \frac{2}{5} + \frac{12}{5} = \frac{14}{5} = 2.8

(b)

2 marks

Use the linear rule.

E (c - 2 X) = c - 2 E (X) = 0 \Rightarrow c = 2 (2.8) = 5.6

(c)

3 marks

The area from 0 to 2 is $\frac{2}{5} = 0.4$ , which is less than 0.5, so the median is in $[2, 8]$ .
Solve on the calculator.

\int_{2}^{m} (- \frac{x}{30} + \frac{4}{15}) d x = 0.1 m \approx 2.52

Question 7

6 marks

Seating three friends in a row of twelve seats

Number and algebra

(a)

3 marks

The number of ways to seat the three people in any of the 12 seats.
Sitting together: there are 10 positions for a block of three seats, and the three can swap in $3!$ ways.
Subtract.

^{12} P_{3} = 12 \times 11 \times 10 = 1320 10 \times 3! = 60 1320 - 60 = 1260

(b)

3 marks

The group of three can be placed in $10 \times 3! = 60$ ways.
The other three people are seated in the 9 remaining seats, in $9 \times 8 \times 7$ ways.

60 \times 504 = 30240

Question 8

6 marks

A complex number times its conjugate, and two equations in z and w

Number and algebra

(a)

2 marks

Let $w = x + i y$ .

w w^{*} = (x + i y) (x - i y) = x^{2} + y^{2} = ∣ w ∣^{2}

(b)

4 marks

Multiply the first equation by $w$ .
Use $w w^{*} = ∣ w ∣^{2} = (25)^{2} = 20$ , so $100 = (1 - 2 i) z^{2} w$ .
Use $z w = 10 + 10 i$ to replace $z w$ in $z^{2} w = z (z w)$ .
Multiply out the denominator.
Divide.

5 w w^{*} = (1 - 2 i) z^{2} w 100 = (1 - 2 i) z (10 + 10 i) \Rightarrow z = \frac{10}{( 1 - 2 i ) ( 1 + i )} (1 - 2 i) (1 + i) = 1 + i - 2 i + 2 = 3 - i z = \frac{10}{3 - i} = \frac{10 ( 3 + i )}{10} = 3 + i a = 3, b = 1

Question 9

7 marks

A geometric series that converges: range of c and a tolerance

Number and algebra

(a)

3 marks

The sum is $S_{n} = \sum_{r = 0}^{n - 1} 5 (lo g_{2} c)^{r}$ , which is geometric with first term 5 and common ratio $lo g_{2} c$ .
It converges when $∣ lo g_{2} c ∣ < 1$ .

- 1 < lo g_{2} c < 1 \Rightarrow \frac{1}{2} < c < 2

(b)

4 marks

With $c = 1.5$ the ratio is $r = lo g_{2} 1.5 \approx 0.58496$ and the sum to infinity is
The difference between the infinite sum and the sum of $n$ terms.
Check by calculator: $n = 8$ gives $0.165$ , which is too big, and $n = 9$ gives $0.0966$ , which is less than $0.1$ .

S_{\infty} = \frac{5}{1 - 0.58496} \approx 12.047 S_{\infty} - S_{n} = \frac{5 r ^{n}}{1 - r} < 0.1 n = 9

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

15 marks

Modelling the population of Canada: linear, exponential and logistic models

Functions

(a)

3 marks

(i) Use linear regression on the calculator.

(ii) The value of $a$ means that the population is increasing by about 0.358 million (358 000) people per year, on average.

a \approx 0.358, b \approx 30.5

(b)

1 mark

The prediction for 2030 is outside the range of the data (2000 to 2020), so it is an extrapolation, and so it may not be reliable.

(c)

3 marks

(i) The year 2100 is $t = 100$ .

(ii) The number 1.005 means the population grows by 0.5% each year.

B (100) = 33.5 (1.005)^{100} \approx 55.2 million

(d)

1 mark

Put $t = 100$ .

C (100) = \frac{62}{1 + e ^{- 2}} \approx 54.6 million

(e)

3 marks

The difference in the predictions is $C (t) - B (t)$ . Find where it is greatest on the calculator, for $25 \leq t \leq 100$ .
That is the year 2058 (or 2059 if rounding up).

t \approx 58.6

(f)

2 marks

(i) Differentiate Benoit's model, or use the calculator.

(ii) Do the same for Cecilia's model.

B^{'} (75) \approx 0.243 C^{'} (75) \approx 0.185

(g)

2 marks

Since $B^{'} (75) > C^{'} (75)$ , in 2075 the population is growing faster in Benoit's model (about 0.243 million per year) than in Cecilia's model (about 0.185 million per year).

Question 11

18 marks

A line and a plane: distance from the origin, containment and a parallel line

Number and algebra

(a)

3 marks

Set $- \frac{x}{2} + 1 = y + 4 = \frac{z}{3} = λ$ .

x = 2 - 2 λ, y = - 4 + λ, z = 3 λ r = 2 - 4 0 + λ - 2 13

(b)

5 marks

Take $a = (2, - 4, 0)$ and $b = (- 2, 1, 3)$ .
The distance.

a \times b = - 12 - 6 - 6, ∣ a \times b ∣ = 144 + 36 + 36 = 216 ∣ b ∣ = 14 d = \frac{216}{14} = \frac{108}{7} = \frac{6 21}{7} \approx 3.93

(c)

3 marks

Substitute the line into the plane equation.
This equals 24 for every $λ$ , so every point of $L$ is on the plane, and the plane contains $L$ .

6 (2 - 2 λ) - 3 (- 4 + λ) + 5 (3 λ) = 12 - 12 λ + 12 - 3 λ + 15 λ = 24

(d)

7 marks

The line $M$ meets the $z$ -axis at $(0, 0, z)$ , so its direction is $(4, 1, 2) - (0, 0, z) = (4, 1, 2 - z)$ .
$M$ is parallel to the plane, so its direction is perpendicular to the normal $(6, - 3, 5)$ .
The direction is $(4, 1, - \frac{21}{5})$ , which is a multiple of $(20, 5, - 21)$ .

6 (4) - 3 (1) + 5 (2 - z) = 0 \Rightarrow 31 - 5 z = 0 \Rightarrow z = \frac{31}{5} s = 412 + μ 205 - 21

Question 12

21 marks

A rational curve with a parameter: stationary points, asymptotes and a sketch

Calculus

(a)

4 marks

Use the quotient rule with $u = 2 x^{2} + 6 x - 3$ and $v = x + k$ .
Expand the numerator.

\frac{d y}{d x} = \frac{( 4 x + 6 ) ( x + k ) - ( 2 x ^{2} + 6 x - 3 )}{( x + k ) ^{2}} 4 x^{2} + 4 k x + 6 x + 6 k - 2 x^{2} - 6 x + 3 = 2 x^{2} + 4 k x + 6 k + 3 \frac{d y}{d x} = \frac{2 x ^{2} + 4 k x + 6 k + 3}{( x + k ) ^{2}}

(b)

4 marks

A local maximum or minimum needs $\frac{d y}{d x} = 0$ , which means $2 x^{2} + 4 k x + 6 k + 3 = 0$ has two distinct real roots.
The critical values are $k = \frac{3 \pm 15}{2}$ , and $k$ is positive, so

(4 k)^{2} - 4 (2) (6 k + 3) > 0 \Rightarrow 16 k^{2} - 48 k - 24 > 0 \Rightarrow 2 k^{2} - 6 k - 3 > 0 k > \frac{3 + 15}{2} \approx 3.44

(c)

1 mark

With $k = 2$ the denominator is $x + 2$ .

x = - 2

(d)

4 marks

Divide $2 x^{2} + 6 x - 3$ by $x + 2$ .
As $x \to \pm \infty$ the last term goes to 0.

\frac{2 x ^{2} + 6 x - 3}{x + 2} = 2 x + 2 - \frac{7}{x + 2} y = 2 x + 2

(e)

4 marks

With $k = 2$ , differentiate $y = 2 x + 2 - 7 (x + 2)^{- 1}$ .
Since $\frac{7}{( x + 2 ) ^{2}} > 0$ for all $x \neq = - 2$ , we have $\frac{d y}{d x} > 2$ .

\frac{d y}{d x} = 2 + \frac{7}{( x + 2 ) ^{2}}

(f)

4 marks

Draw the vertical asymptote $x = - 2$ and the oblique asymptote $y = 2 x + 2$ . The curve is always increasing.
For $x > - 2$ the curve is below the oblique line (since $- \frac{7}{x + 2} < 0$ ). It comes up from $- \infty$ next to the vertical asymptote and approaches the oblique line from below as $x \to \infty$ .
For $x < - 2$ the curve is above the oblique line. It comes in from the left, approaching the oblique line from above as $x \to - \infty$ , and goes up to $+ \infty$ as $x \to - 2$ from the left.

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 algebra41 marks · 37%
Q2, Q7, Q8, Q9, Q11
Functions27 marks · 25%
Q1, Q5, Q10
Calculus21 marks · 19%
Q12
Statistics and probability13 marks · 12%
Q3, Q6
Geometry and trigonometry8 marks · 7%
Q4

Mentor insights

Traps to watch for

01
Gradient and angle · Q4
A gradient of $\frac{7}{12}$ means $tan (slope angle) = \frac{7}{12}$ .
02
Mean · Q6
$E (X) = \int x f (x) d x$ .
03
Extrapolation · Q10
Predicting outside the range of the data is unreliable.
04
Parametric form · Q11
Set each part of the symmetric form equal to $λ$ .
05
Quotient rule · Q12
$(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ .

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

Session: November 2024
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 November 2024 Global 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

7 marks

A square root function: end values, roots and a sketch

Functions

(a)

2 marks

(i) Substitute $x = 0$ .

(ii) Substitute $x = 20$ .

f (0) = 0 - 0 - 11 = - 11 f (20) = 1120 - 40 - 11 \approx - 1.81

(b)

2 marks

Solve $11 x - 2 x - 11 = 0$ on the calculator.

x \approx 1.73 and x \approx 17.5

(c)

3 marks

The curve starts at $(0, - 11)$ , rises through the root $x \approx 1.73$ to a maximum at about $(7.6, 4.1)$ , then falls through the root $x \approx 17.5$ and ends at about $(20, - 1.8)$ .

Question 2

4 marks

The coefficient of x to the power 6 in (2x - 5) to the power 9

Number and algebra

4 marks

For $x^{6}$ we need $(2 x)^{6}$ , so the power of $- 5$ is $9 - 6 = 3$ .
The coefficient.

(6 9) (2 x)^{6} (- 5)^{3} = 84 \times 64 x^{6} \times (- 125) 84 \times 64 \times (- 125) = - 672000

Question 3

5 marks

A discrete distribution with an unknown constant and its mean

Statistics and probability

(a)

2 marks

Add the probabilities for $x = 3, 5, 8, 11$ .

\frac{k ( 3 + 5 + 8 + 11 )}{20} = \frac{27 k}{20} = 1 \Rightarrow k = \frac{20}{27} \approx 0.741

(b)

3 marks

The probabilities are $\frac{x}{27}$ , so use $E (X) = \sum x P (X = x)$ .

E (X) = \frac{3 ( 3 ) + 5 ( 5 ) + 8 ( 8 ) + 11 ( 11 )}{27} = \frac{9 + 25 + 64 + 121}{27} = \frac{219}{27} = \frac{73}{9} \approx 8.11

Question 4

8 marks

The roof of a house: angles and lengths of the beams

Geometry and trigonometry

(a)

3 marks

The beam $B A$ makes an angle $α$ with the horizontal, with $tan α = \frac{7}{12}$ , so $α \approx 30.2 6^{\circ}$ . The line $A E$ is vertical.

B A E = 9 0^{\circ} - 30.2 6^{\circ} \approx 59. 7^{\circ}

(b)

5 marks

(i) By symmetry, $D E$ goes across half of $D F$ , which is 350 cm horizontally. The slope $D E$ makes an angle of $30.2 6^{\circ}$ with the horizontal.

The points $C$ , $D$ , $E$ lie in a line, with $C D = 50$ .

D E = \frac{350}{cos 30.2 6 ^{\circ}} = \frac{350}{sin 59.7 4 ^{\circ}} \approx 405.2 C E = 405.2 + 50 \approx 455 cm E G = \frac{30}{tan 59.7 4 ^{\circ}} \approx 17.5 B A = C E + E G = 455.2 + 17.5 \approx 473 cm

Question 5

5 marks

When a logarithm is defined for every real number

Functions

5 marks

For $h (x)$ to be defined for all real $x$ , we need $4 x^{2} - r x + r - 1 > 0$ for every $x$ .
The parabola opens upwards, so it is always above the axis when its discriminant is negative.
Find the critical values.
The quadratic in $r$ is negative between the roots.

Δ = r^{2} - 4 (4) (r - 1) = r^{2} - 16 r + 16 < 0 r = \frac{16 \pm 256 - 64}{2} = 8 \pm 43 8 - 43 < r < 8 + 43 (1.07 < r < 14.9)

Question 6

8 marks

A piecewise density: mean, a linear combination and the median

Statistics and probability

(a)

3 marks

Integrate over both pieces.

E (X) = \int_{0}^{2} \frac{x}{5} d x + \int_{2}^{8} x (- \frac{x}{30} + \frac{4}{15}) d x = \frac{2}{5} + \frac{12}{5} = \frac{14}{5} = 2.8

(b)

2 marks

Use the linear rule.

E (c - 2 X) = c - 2 E (X) = 0 \Rightarrow c = 2 (2.8) = 5.6

(c)

3 marks

The area from 0 to 2 is $\frac{2}{5} = 0.4$ , which is less than 0.5, so the median is in $[2, 8]$ .
Solve on the calculator.

\int_{2}^{m} (- \frac{x}{30} + \frac{4}{15}) d x = 0.1 m \approx 2.52

Question 7

6 marks

Seating three friends in a row of twelve seats

Number and algebra

(a)

3 marks

The number of ways to seat the three people in any of the 12 seats.
Sitting together: there are 10 positions for a block of three seats, and the three can swap in $3!$ ways.
Subtract.

^{12} P_{3} = 12 \times 11 \times 10 = 1320 10 \times 3! = 60 1320 - 60 = 1260

(b)

3 marks

The group of three can be placed in $10 \times 3! = 60$ ways.
The other three people are seated in the 9 remaining seats, in $9 \times 8 \times 7$ ways.

60 \times 504 = 30240

Question 8

6 marks

A complex number times its conjugate, and two equations in z and w

Number and algebra

(a)

2 marks

Let $w = x + i y$ .

w w^{*} = (x + i y) (x - i y) = x^{2} + y^{2} = ∣ w ∣^{2}

(b)

4 marks

Multiply the first equation by $w$ .
Use $w w^{*} = ∣ w ∣^{2} = (25)^{2} = 20$ , so $100 = (1 - 2 i) z^{2} w$ .
Use $z w = 10 + 10 i$ to replace $z w$ in $z^{2} w = z (z w)$ .
Multiply out the denominator.
Divide.

5 w w^{*} = (1 - 2 i) z^{2} w 100 = (1 - 2 i) z (10 + 10 i) \Rightarrow z = \frac{10}{( 1 - 2 i ) ( 1 + i )} (1 - 2 i) (1 + i) = 1 + i - 2 i + 2 = 3 - i z = \frac{10}{3 - i} = \frac{10 ( 3 + i )}{10} = 3 + i a = 3, b = 1

Question 9

7 marks

A geometric series that converges: range of c and a tolerance

Number and algebra

(a)

3 marks

The sum is $S_{n} = \sum_{r = 0}^{n - 1} 5 (lo g_{2} c)^{r}$ , which is geometric with first term 5 and common ratio $lo g_{2} c$ .
It converges when $∣ lo g_{2} c ∣ < 1$ .

- 1 < lo g_{2} c < 1 \Rightarrow \frac{1}{2} < c < 2

(b)

4 marks

With $c = 1.5$ the ratio is $r = lo g_{2} 1.5 \approx 0.58496$ and the sum to infinity is
The difference between the infinite sum and the sum of $n$ terms.
Check by calculator: $n = 8$ gives $0.165$ , which is too big, and $n = 9$ gives $0.0966$ , which is less than $0.1$ .

S_{\infty} = \frac{5}{1 - 0.58496} \approx 12.047 S_{\infty} - S_{n} = \frac{5 r ^{n}}{1 - r} < 0.1 n = 9

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

15 marks

Modelling the population of Canada: linear, exponential and logistic models

Functions

(a)

3 marks

(i) Use linear regression on the calculator.

(ii) The value of $a$ means that the population is increasing by about 0.358 million (358 000) people per year, on average.

a \approx 0.358, b \approx 30.5

(b)

1 mark

The prediction for 2030 is outside the range of the data (2000 to 2020), so it is an extrapolation, and so it may not be reliable.

(c)

3 marks

(i) The year 2100 is $t = 100$ .

(ii) The number 1.005 means the population grows by 0.5% each year.

B (100) = 33.5 (1.005)^{100} \approx 55.2 million

(d)

1 mark

Put $t = 100$ .

C (100) = \frac{62}{1 + e ^{- 2}} \approx 54.6 million

(e)

3 marks

The difference in the predictions is $C (t) - B (t)$ . Find where it is greatest on the calculator, for $25 \leq t \leq 100$ .
That is the year 2058 (or 2059 if rounding up).

t \approx 58.6

(f)

2 marks

(i) Differentiate Benoit's model, or use the calculator.

(ii) Do the same for Cecilia's model.

B^{'} (75) \approx 0.243 C^{'} (75) \approx 0.185

(g)

2 marks

Since $B^{'} (75) > C^{'} (75)$ , in 2075 the population is growing faster in Benoit's model (about 0.243 million per year) than in Cecilia's model (about 0.185 million per year).

Question 11

18 marks

A line and a plane: distance from the origin, containment and a parallel line

Number and algebra

(a)

3 marks

Set $- \frac{x}{2} + 1 = y + 4 = \frac{z}{3} = λ$ .

x = 2 - 2 λ, y = - 4 + λ, z = 3 λ r = 2 - 4 0 + λ - 2 13

(b)

5 marks

Take $a = (2, - 4, 0)$ and $b = (- 2, 1, 3)$ .
The distance.

a \times b = - 12 - 6 - 6, ∣ a \times b ∣ = 144 + 36 + 36 = 216 ∣ b ∣ = 14 d = \frac{216}{14} = \frac{108}{7} = \frac{6 21}{7} \approx 3.93

(c)

3 marks

Substitute the line into the plane equation.
This equals 24 for every $λ$ , so every point of $L$ is on the plane, and the plane contains $L$ .

6 (2 - 2 λ) - 3 (- 4 + λ) + 5 (3 λ) = 12 - 12 λ + 12 - 3 λ + 15 λ = 24

(d)

7 marks

The line $M$ meets the $z$ -axis at $(0, 0, z)$ , so its direction is $(4, 1, 2) - (0, 0, z) = (4, 1, 2 - z)$ .
$M$ is parallel to the plane, so its direction is perpendicular to the normal $(6, - 3, 5)$ .
The direction is $(4, 1, - \frac{21}{5})$ , which is a multiple of $(20, 5, - 21)$ .

6 (4) - 3 (1) + 5 (2 - z) = 0 \Rightarrow 31 - 5 z = 0 \Rightarrow z = \frac{31}{5} s = 412 + μ 205 - 21

Question 12

21 marks

A rational curve with a parameter: stationary points, asymptotes and a sketch

Calculus

(a)

4 marks

Use the quotient rule with $u = 2 x^{2} + 6 x - 3$ and $v = x + k$ .
Expand the numerator.

\frac{d y}{d x} = \frac{( 4 x + 6 ) ( x + k ) - ( 2 x ^{2} + 6 x - 3 )}{( x + k ) ^{2}} 4 x^{2} + 4 k x + 6 x + 6 k - 2 x^{2} - 6 x + 3 = 2 x^{2} + 4 k x + 6 k + 3 \frac{d y}{d x} = \frac{2 x ^{2} + 4 k x + 6 k + 3}{( x + k ) ^{2}}

(b)

4 marks

A local maximum or minimum needs $\frac{d y}{d x} = 0$ , which means $2 x^{2} + 4 k x + 6 k + 3 = 0$ has two distinct real roots.
The critical values are $k = \frac{3 \pm 15}{2}$ , and $k$ is positive, so

(4 k)^{2} - 4 (2) (6 k + 3) > 0 \Rightarrow 16 k^{2} - 48 k - 24 > 0 \Rightarrow 2 k^{2} - 6 k - 3 > 0 k > \frac{3 + 15}{2} \approx 3.44

(c)

1 mark

With $k = 2$ the denominator is $x + 2$ .

x = - 2

(d)

4 marks

Divide $2 x^{2} + 6 x - 3$ by $x + 2$ .
As $x \to \pm \infty$ the last term goes to 0.

\frac{2 x ^{2} + 6 x - 3}{x + 2} = 2 x + 2 - \frac{7}{x + 2} y = 2 x + 2

(e)

4 marks

With $k = 2$ , differentiate $y = 2 x + 2 - 7 (x + 2)^{- 1}$ .
Since $\frac{7}{( x + 2 ) ^{2}} > 0$ for all $x \neq = - 2$ , we have $\frac{d y}{d x} > 2$ .

\frac{d y}{d x} = 2 + \frac{7}{( x + 2 ) ^{2}}

(f)

4 marks

Draw the vertical asymptote $x = - 2$ and the oblique asymptote $y = 2 x + 2$ . The curve is always increasing.
For $x > - 2$ the curve is below the oblique line (since $- \frac{7}{x + 2} < 0$ ). It comes up from $- \infty$ next to the vertical asymptote and approaches the oblique line from below as $x \to \infty$ .
For $x < - 2$ the curve is above the oblique line. It comes in from the left, approaching the oblique line from above as $x \to - \infty$ , and goes up to $+ \infty$ as $x \to - 2$ from the left.

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 algebra41 marks · 37%
Q2, Q7, Q8, Q9, Q11
Functions27 marks · 25%
Q1, Q5, Q10
Calculus21 marks · 19%
Q12
Statistics and probability13 marks · 12%
Q3, Q6
Geometry and trigonometry8 marks · 7%
Q4

Mentor insights

Traps to watch for

01
Gradient and angle · Q4
A gradient of $\frac{7}{12}$ means $tan (slope angle) = \frac{7}{12}$ .
02
Mean · Q6
$E (X) = \int x f (x) d x$ .
03
Extrapolation · Q10
Predicting outside the range of the data is unreliable.
04
Parametric form · Q11
Set each part of the symmetric form equal to $λ$ .
05
Quotient rule · Q12
$(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ .

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