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IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2023 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

4 marks

Plant growth: linear regression and an estimate

Statistics and probability

(a)

2 marks

Do linear regression of $h$ on $d$ .

h = 1.93 d + 7.22

(b)

2 marks

Substitute $d = 20$ .

h = 1.9326 (20) + 7.2166 \approx 45.9 cm

Question 2

6 marks

Two buildings: the height from equal angles of elevation

Number and algebra

6 marks

Use the cosine rule in triangle APB.
The angles on the ground at P add to $18 0^{\circ}$ .
In the right triangle with hypotenuse $P A = 150$ :

cos A P B = \frac{15 0 ^{2} + 9 0 ^{2} - 15 4 ^{2}}{2 ( 150 ) ( 90 )} = \frac{6884}{27000} \Rightarrow A P B = 75. 2^{\circ} 2 θ = 180 - 75.2 = 104.8 \Rightarrow θ = 52. 4^{\circ} h = 150 sin 52. 4^{\circ} \approx 119 m

Question 3

8 marks

Bags of rice: normal probabilities and a binomial count

Statistics and probability

(a)

2 marks

Use $W \sim N (204, 5^{2})$ .

P (W > 210) = 0.115

(b)

2 marks

The three parts add to 1: below $w$ , between $w$ and 210 (80%), and above 210.

P (W < w) = 1 - 0.80 - 0.1151 = 0.0849

(c)

2 marks

Use the inverse normal with the lower tail probability 0.0849.

w = 204 + 5 (- 1.375) \approx 197

(d)

2 marks

The number of bags below $w$ is $X \sim B (10, 0.0849)$ .

P (X = 1) = 10 (0.0849) (0.9151)^{9} \approx 0.382

Question 4

7 marks

A geometric sequence in the binomial coefficients

Number and algebra

7 marks

The terms are $a = (1 8) h = 8 h$ , $b = (2 8) h^{2} = 28 h^{2}$ and $d = (4 8) h^{4} = 70 h^{4}$ .
In a geometric sequence the first three terms satisfy $\frac{b}{a} = \frac{d}{b}$ .

\frac{28 h ^{2}}{8 h} = \frac{70 h ^{4}}{28 h ^{2}} \Rightarrow 3.5 h = 2.5 h^{2} h = 1.4

Question 5

6 marks

A particle with decaying oscillating velocity

Calculus

(a)

2 marks

The acceleration is $v^{'} (t)$ .
Solve the first positive root.

v^{'} (t) = - \frac{4}{3} e^{- t /3} cos (\frac{t}{2} - \frac{π}{4}) - 2 e^{- t /3} sin (\frac{t}{2} - \frac{π}{4}) = 0 t_{1} = 0.395 s

(b)

2 marks

The particle is at rest when $v = 0$ , so $cos (\frac{t}{2} - \frac{π}{4}) = 0$ .
The second time is

t = \frac{3 π}{2} = 4.71 and t = \frac{7 π}{2} = 11.0 t_{2} = 11.0 s

(c)

2 marks

Integrate the speed from $t_{1}$ to $t_{2}$ , since $v$ changes sign at $t = \frac{3 π}{2}$ .

\int_{0.395}^{11.0} ∣ v (t) ∣ d t \approx 7.83 m

Question 6

8 marks

Two planes: the line of intersection and the nearest point to the origin

Geometry and trigonometry

(a)

3 marks

A general point on $L$ is $(λ, 2 - 2 λ, 4 - 2 λ)$ . Substitute it into each plane.
Both hold for every $λ$ , so every point of the line lies on both planes.

2 λ - (2 - 2 λ) + 2 (4 - 2 λ) = 6 and 4 λ + 3 (2 - 2 λ) - (4 - 2 λ) = 2 r = 024 + λ 1 - 2 - 2

(b)

5 marks

The point on $L$ nearest to $O$ has $O P \cdot d = 0$ .

λ 2 - 2 λ 4 - 2 λ \cdot 1 - 2 - 2 = λ - 4 + 4 λ - 8 + 4 λ = 9 λ - 12 = 0 λ = \frac{4}{3} P = (\frac{4}{3}, - \frac{2}{3}, \frac{4}{3})

Question 7

5 marks

Volume of revolution of an arctangent about the y-axis

Number and algebra

5 marks

From $y = arctan (x - 2)$ we get $x = tan y + 2$ .
Use $tan^{2} y = sec^{2} y - 1$ .

V = π \int_{0}^{π /3} (tan y + 2)^{2} d y = π \int_{0}^{π /3} (tan^{2} y + 4 tan y + 4) d y V = π [tan y + 3 y - 4 ln cos y]_{0}^{π /3} = π (3 + π + 4 ln 2) \approx 24.0

Mentor tip

Use radians. A calculator in degree mode gives a different, wrong number.

Question 8

7 marks

Range of a rational function and the sign of a product

Functions

(a)

4 marks

Write $y = \frac{2 x - 5}{x ^{2} - 3}$ as $y x^{2} - 2 x - 3 y + 5 = 0$ . The values of $y$ that occur are those for which this quadratic in $x$ has a real root.
Solve $3 y^{2} - 5 y + 1 \geq 0$ .

Δ = 4 - 4 y (5 - 3 y) = 12 y^{2} - 20 y + 4 \geq 0 y \leq \frac{5 - 13}{6} \approx 0.232 or y \geq \frac{5 + 13}{6} \approx 1.43

(b)

3 marks

Since $cos t < 0$ , $h (x) \leq 0$ means $g (∣ x ∣) \geq 0$ , that is $\frac{2∣ x ∣ - 5}{x ^{2} - 3} \geq 0$ .
The fraction is positive when both parts are negative ( $∣ x ∣ < 3$ ) or both positive ( $∣ x ∣ > \frac{5}{2}$ ), and it is zero at $∣ x ∣ = \frac{5}{2}$ .

x \leq - \frac{5}{2}, - 3 < x < 3, x \geq \frac{5}{2}

Question 9

5 marks

Selections of three numbers from 1 to 30 with a sum divisible by 3

Number and algebra

5 marks

There are 10 numbers with remainder 0, 10 with remainder 1 and 10 with remainder 2 when divided by 3.
The sum is divisible by 3 if all three come from the same group, or one comes from each group.

same group: 3 (3 10) = 3 (120) = 360 one from each: 10 \times 10 \times 10 = 1000 360 + 1000 = 1360

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

13 marks

A weight on a spring: a cosine model, times and probability

Functions

(a)

2 marks

The period of $cos (7.8 t)$ is $\frac{2 π}{7.8}$ .

0.806 s

(b)

3 marks

The minimum is 1 and the maximum is 1.8.

a = - \frac{1.8 - 1}{2} = - 0.4, b = \frac{1.8 + 1}{2} = 1.4

(c)

2 marks

The maximum heights occur at $t = \frac{( 2 k - 1 ) π}{7.8} = 0.403 (2 k - 1)$ . Find how many are at most 5.
The weight reaches its maximum 6 times.

0.403 (2 k - 1) \leq 5 \Rightarrow 2 k - 1 \leq 12.4 \Rightarrow k = 1, \dots, 6

(d)

2 marks

Solve $H (t) = 1.5$ .

- 0.4 cos (7.8 t) + 1.4 = 1.5 \Rightarrow cos (7.8 t) = - 0.25 t = \frac{arccos ( - 0.25 )}{7.8} = 0.234 s

(e)

4 marks

The height is above 1.5 when $cos (7.8 t) < - 0.25$ . Find the fraction of the first 5 seconds when this holds, using a graph or by adding the intervals.

P (H > 1.5) \approx 0.406

Question 11

20 marks

Drawing balls without replacement: forming an equation and solving it

Statistics and probability

(a)

4 marks

The probability of two yellow balls without replacement is
Clear the fractions and expand.

P (Y Y) = \frac{y}{r + y} \cdot \frac{y - 1}{r + y - 1} = \frac{1}{3} 3 y (y - 1) = (r + y) (r + y - 1) \Rightarrow 3 y^{2} - 3 y = r^{2} + 2 r y + y^{2} - r - y 2 y^{2} - 2 (r + 1) y + r - r^{2} = 0

(b)

4 marks

Solve the quadratic in $y$ .
For $r > 1$ we have $3 r^{2} + 1 > r + 1$ , so the root with the minus sign is negative and cannot be a number of balls. Keep the plus sign.

y = \frac{2 ( r + 1 ) \pm 4 ( r + 1 ) ^{2} - 8 ( r - r ^{2} )}{4} = \frac{( r + 1 ) \pm 3 r ^{2} + 1}{2} y = \frac{( r + 1 ) + 3 r ^{2} + 1}{2}

(c)

4 marks

Test small positive integers $r$ so that $3 r^{2} + 1$ is a perfect square and $y$ is an integer.
Check: $\frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$ and $\frac{6}{10} \cdot \frac{5}{9} = \frac{1}{3}$ .

r = 1 : y = \frac{2 + 2}{2} = 2, r = 4 : y = \frac{5 + 7}{2} = 6 (r, y) = (1, 2) and (4, 6)

(d)

3 marks

With 10 red and $y$ yellow balls the total is $y + 10$ .

P (Y Y Y) = \frac{y}{y + 10} \cdot \frac{y - 1}{y + 9} \cdot \frac{y - 2}{y + 8}

(e)

5 marks

With one more yellow ball the probability is $\frac{( y + 1 ) y ( y - 1 )}{( y + 11 ) ( y + 10 ) ( y + 9 )}$ , and this is twice the expression in (d).
Cancel the common factors.

\frac{( y + 1 ) y ( y - 1 )}{( y + 11 ) ( y + 10 ) ( y + 9 )} = \frac{2 y ( y - 1 ) ( y - 2 )}{( y + 10 ) ( y + 9 ) ( y + 8 )} \frac{y + 1}{y + 11} = \frac{2 ( y - 2 )}{y + 8} \Rightarrow (y + 1) (y + 8) = 2 (y - 2) (y + 11) y^{2} + 9 y - 52 = 0 \Rightarrow (y - 4) (y + 13) = 0 \Rightarrow y = 4

Question 12

21 marks

A homogeneous differential equation: Euler, exact solution and an inflexion

Calculus

(a)

2 marks

Euler's method with $h = 0.1$ from $(1, 2)$ : $y_{1} = y_{0} + h \frac{x _{0}^{2} + 3 y _{0}^{2}}{x _{0} y _{0}}$ .

y_{1} = 2 + 0.1 \cdot \frac{1 + 12}{2} = 2.65

(b)

8 marks

Put $y = v x$ , so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
Separate the variables and integrate.
At $x = 1$ , $y = 2$ , so $v = 2$ .

v + x \frac{d v}{d x} = \frac{x ^{2} + 3 v ^{2} x ^{2}}{v x ^{2}} = \frac{1 + 3 v ^{2}}{v} \Rightarrow x \frac{d v}{d x} = \frac{1 + 2 v ^{2}}{v} \int \frac{v}{1 + 2 v ^{2}} d v = \int \frac{d x}{x} \Rightarrow \frac{1}{4} ln (1 + 2 v^{2}) = ln x + c \frac{1}{4} ln 9 = c ln (1 + 2 v^{2}) = ln (9 x^{4}) \Rightarrow v^{2} = \frac{9 x ^{4} - 1}{2} y = x \frac{9 x ^{4} - 1}{2}

(c)

1 mark

Substitute $x = 1.1$ into the exact solution.

y (1.1) = 1.1 \frac{9 ( 1.1 ) ^{4} - 1}{2} \approx 2.714

(d)

2 marks

The second derivative formula shows $\frac{d ^{2} y}{d x ^{2}} > 0$ on $1 \leq x \leq 1.1$ (the graph is concave up).
A tangent to a concave-up curve lies below the curve, so the Euler value 2.65 is smaller than the true value 2.714.

(e)

2 marks

A point of inflexion is where $\frac{d ^{2} y}{d x ^{2}}$ changes sign, which is a local extreme value of $\frac{d y}{d x}$ . Sketch the graph of $\frac{d ^{2} y}{d x ^{2}}$ against $x$ and find where it crosses the axis between $\frac{3}{3}$ and 1.

x \approx 0.656

(f)

6 marks

At the inflexion $\frac{d ^{2} y}{d x ^{2}} = 0$ , so $- x^{4} + x^{2} y^{2} + 6 y^{4} = 0$ . Put $y = m x$ .
Factorise.
The other factor has no real solution, and $m > 0$ because $x, y > 0$ . So P lies on the line $y = \frac{3}{3} x$ .

- x^{4} + m^{2} x^{4} + 6 m^{4} x^{4} = 0 \Rightarrow 6 m^{4} + m^{2} - 1 = 0 (3 m^{2} - 1) (2 m^{2} + 1) = 0 \Rightarrow m^{2} = \frac{1}{3} m = \frac{1}{3} = \frac{3}{3}

Our analysis

Where the marks are

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

Statistics and probability32 marks · 29%
Q1, Q3, Q11
Calculus27 marks · 25%
Q5, Q12
Number and algebra23 marks · 21%
Q2, Q4, Q7, Q9
Functions20 marks · 18%
Q8, Q10
Geometry and trigonometry8 marks · 7%
Q6

Mentor insights

Traps to watch for

01
Three regions of the normal curve · Q3
The probability below $w$ is what is left after removing the 80% and the upper tail.
02
Check the point and the direction · Q6
To verify a line of intersection, show that its point lies on both planes and that its direction is perpendicular to both normals.
03
Released at the minimum · Q10
At $t = 0$ the weight is at its lowest point, so the cosine model needs a negative $a$ .
04
Without replacement · Q11
After the first yellow ball there are $y - 1$ yellow balls and $r + y - 1$ balls in total.
05
Substitution y = vx · Q12
Then $\frac{d y}{d x} = v + x \frac{d v}{d x}$ , and the equation separates.

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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2023
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2023 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

4 marks

Plant growth: linear regression and an estimate

Statistics and probability

(a)

2 marks

Do linear regression of $h$ on $d$ .

h = 1.93 d + 7.22

(b)

2 marks

Substitute $d = 20$ .

h = 1.9326 (20) + 7.2166 \approx 45.9 cm

Question 2

6 marks

Two buildings: the height from equal angles of elevation

Number and algebra

6 marks

Use the cosine rule in triangle APB.
The angles on the ground at P add to $18 0^{\circ}$ .
In the right triangle with hypotenuse $P A = 150$ :

cos A P B = \frac{15 0 ^{2} + 9 0 ^{2} - 15 4 ^{2}}{2 ( 150 ) ( 90 )} = \frac{6884}{27000} \Rightarrow A P B = 75. 2^{\circ} 2 θ = 180 - 75.2 = 104.8 \Rightarrow θ = 52. 4^{\circ} h = 150 sin 52. 4^{\circ} \approx 119 m

Question 3

8 marks

Bags of rice: normal probabilities and a binomial count

Statistics and probability

(a)

2 marks

Use $W \sim N (204, 5^{2})$ .

P (W > 210) = 0.115

(b)

2 marks

The three parts add to 1: below $w$ , between $w$ and 210 (80%), and above 210.

P (W < w) = 1 - 0.80 - 0.1151 = 0.0849

(c)

2 marks

Use the inverse normal with the lower tail probability 0.0849.

w = 204 + 5 (- 1.375) \approx 197

(d)

2 marks

The number of bags below $w$ is $X \sim B (10, 0.0849)$ .

P (X = 1) = 10 (0.0849) (0.9151)^{9} \approx 0.382

Question 4

7 marks

A geometric sequence in the binomial coefficients

Number and algebra

7 marks

The terms are $a = (1 8) h = 8 h$ , $b = (2 8) h^{2} = 28 h^{2}$ and $d = (4 8) h^{4} = 70 h^{4}$ .
In a geometric sequence the first three terms satisfy $\frac{b}{a} = \frac{d}{b}$ .

\frac{28 h ^{2}}{8 h} = \frac{70 h ^{4}}{28 h ^{2}} \Rightarrow 3.5 h = 2.5 h^{2} h = 1.4

Question 5

6 marks

A particle with decaying oscillating velocity

Calculus

(a)

2 marks

The acceleration is $v^{'} (t)$ .
Solve the first positive root.

v^{'} (t) = - \frac{4}{3} e^{- t /3} cos (\frac{t}{2} - \frac{π}{4}) - 2 e^{- t /3} sin (\frac{t}{2} - \frac{π}{4}) = 0 t_{1} = 0.395 s

(b)

2 marks

The particle is at rest when $v = 0$ , so $cos (\frac{t}{2} - \frac{π}{4}) = 0$ .
The second time is

t = \frac{3 π}{2} = 4.71 and t = \frac{7 π}{2} = 11.0 t_{2} = 11.0 s

(c)

2 marks

Integrate the speed from $t_{1}$ to $t_{2}$ , since $v$ changes sign at $t = \frac{3 π}{2}$ .

\int_{0.395}^{11.0} ∣ v (t) ∣ d t \approx 7.83 m

Question 6

8 marks

Two planes: the line of intersection and the nearest point to the origin

Geometry and trigonometry

(a)

3 marks

A general point on $L$ is $(λ, 2 - 2 λ, 4 - 2 λ)$ . Substitute it into each plane.
Both hold for every $λ$ , so every point of the line lies on both planes.

2 λ - (2 - 2 λ) + 2 (4 - 2 λ) = 6 and 4 λ + 3 (2 - 2 λ) - (4 - 2 λ) = 2 r = 024 + λ 1 - 2 - 2

(b)

5 marks

The point on $L$ nearest to $O$ has $O P \cdot d = 0$ .

λ 2 - 2 λ 4 - 2 λ \cdot 1 - 2 - 2 = λ - 4 + 4 λ - 8 + 4 λ = 9 λ - 12 = 0 λ = \frac{4}{3} P = (\frac{4}{3}, - \frac{2}{3}, \frac{4}{3})

Question 7

5 marks

Volume of revolution of an arctangent about the y-axis

Number and algebra

5 marks

From $y = arctan (x - 2)$ we get $x = tan y + 2$ .
Use $tan^{2} y = sec^{2} y - 1$ .

V = π \int_{0}^{π /3} (tan y + 2)^{2} d y = π \int_{0}^{π /3} (tan^{2} y + 4 tan y + 4) d y V = π [tan y + 3 y - 4 ln cos y]_{0}^{π /3} = π (3 + π + 4 ln 2) \approx 24.0

Mentor tip

Use radians. A calculator in degree mode gives a different, wrong number.

Question 8

7 marks

Range of a rational function and the sign of a product

Functions

(a)

4 marks

Write $y = \frac{2 x - 5}{x ^{2} - 3}$ as $y x^{2} - 2 x - 3 y + 5 = 0$ . The values of $y$ that occur are those for which this quadratic in $x$ has a real root.
Solve $3 y^{2} - 5 y + 1 \geq 0$ .

Δ = 4 - 4 y (5 - 3 y) = 12 y^{2} - 20 y + 4 \geq 0 y \leq \frac{5 - 13}{6} \approx 0.232 or y \geq \frac{5 + 13}{6} \approx 1.43

(b)

3 marks

Since $cos t < 0$ , $h (x) \leq 0$ means $g (∣ x ∣) \geq 0$ , that is $\frac{2∣ x ∣ - 5}{x ^{2} - 3} \geq 0$ .
The fraction is positive when both parts are negative ( $∣ x ∣ < 3$ ) or both positive ( $∣ x ∣ > \frac{5}{2}$ ), and it is zero at $∣ x ∣ = \frac{5}{2}$ .

x \leq - \frac{5}{2}, - 3 < x < 3, x \geq \frac{5}{2}

Question 9

5 marks

Selections of three numbers from 1 to 30 with a sum divisible by 3

Number and algebra

5 marks

There are 10 numbers with remainder 0, 10 with remainder 1 and 10 with remainder 2 when divided by 3.
The sum is divisible by 3 if all three come from the same group, or one comes from each group.

same group: 3 (3 10) = 3 (120) = 360 one from each: 10 \times 10 \times 10 = 1000 360 + 1000 = 1360

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

13 marks

A weight on a spring: a cosine model, times and probability

Functions

(a)

2 marks

The period of $cos (7.8 t)$ is $\frac{2 π}{7.8}$ .

0.806 s

(b)

3 marks

The minimum is 1 and the maximum is 1.8.

a = - \frac{1.8 - 1}{2} = - 0.4, b = \frac{1.8 + 1}{2} = 1.4

(c)

2 marks

The maximum heights occur at $t = \frac{( 2 k - 1 ) π}{7.8} = 0.403 (2 k - 1)$ . Find how many are at most 5.
The weight reaches its maximum 6 times.

0.403 (2 k - 1) \leq 5 \Rightarrow 2 k - 1 \leq 12.4 \Rightarrow k = 1, \dots, 6

(d)

2 marks

Solve $H (t) = 1.5$ .

- 0.4 cos (7.8 t) + 1.4 = 1.5 \Rightarrow cos (7.8 t) = - 0.25 t = \frac{arccos ( - 0.25 )}{7.8} = 0.234 s

(e)

4 marks

The height is above 1.5 when $cos (7.8 t) < - 0.25$ . Find the fraction of the first 5 seconds when this holds, using a graph or by adding the intervals.

P (H > 1.5) \approx 0.406

Question 11

20 marks

Drawing balls without replacement: forming an equation and solving it

Statistics and probability

(a)

4 marks

The probability of two yellow balls without replacement is
Clear the fractions and expand.

P (Y Y) = \frac{y}{r + y} \cdot \frac{y - 1}{r + y - 1} = \frac{1}{3} 3 y (y - 1) = (r + y) (r + y - 1) \Rightarrow 3 y^{2} - 3 y = r^{2} + 2 r y + y^{2} - r - y 2 y^{2} - 2 (r + 1) y + r - r^{2} = 0

(b)

4 marks

Solve the quadratic in $y$ .
For $r > 1$ we have $3 r^{2} + 1 > r + 1$ , so the root with the minus sign is negative and cannot be a number of balls. Keep the plus sign.

y = \frac{2 ( r + 1 ) \pm 4 ( r + 1 ) ^{2} - 8 ( r - r ^{2} )}{4} = \frac{( r + 1 ) \pm 3 r ^{2} + 1}{2} y = \frac{( r + 1 ) + 3 r ^{2} + 1}{2}

(c)

4 marks

Test small positive integers $r$ so that $3 r^{2} + 1$ is a perfect square and $y$ is an integer.
Check: $\frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$ and $\frac{6}{10} \cdot \frac{5}{9} = \frac{1}{3}$ .

r = 1 : y = \frac{2 + 2}{2} = 2, r = 4 : y = \frac{5 + 7}{2} = 6 (r, y) = (1, 2) and (4, 6)

(d)

3 marks

With 10 red and $y$ yellow balls the total is $y + 10$ .

P (Y Y Y) = \frac{y}{y + 10} \cdot \frac{y - 1}{y + 9} \cdot \frac{y - 2}{y + 8}

(e)

5 marks

With one more yellow ball the probability is $\frac{( y + 1 ) y ( y - 1 )}{( y + 11 ) ( y + 10 ) ( y + 9 )}$ , and this is twice the expression in (d).
Cancel the common factors.

\frac{( y + 1 ) y ( y - 1 )}{( y + 11 ) ( y + 10 ) ( y + 9 )} = \frac{2 y ( y - 1 ) ( y - 2 )}{( y + 10 ) ( y + 9 ) ( y + 8 )} \frac{y + 1}{y + 11} = \frac{2 ( y - 2 )}{y + 8} \Rightarrow (y + 1) (y + 8) = 2 (y - 2) (y + 11) y^{2} + 9 y - 52 = 0 \Rightarrow (y - 4) (y + 13) = 0 \Rightarrow y = 4

Question 12

21 marks

A homogeneous differential equation: Euler, exact solution and an inflexion

Calculus

(a)

2 marks

Euler's method with $h = 0.1$ from $(1, 2)$ : $y_{1} = y_{0} + h \frac{x _{0}^{2} + 3 y _{0}^{2}}{x _{0} y _{0}}$ .

y_{1} = 2 + 0.1 \cdot \frac{1 + 12}{2} = 2.65

(b)

8 marks

Put $y = v x$ , so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
Separate the variables and integrate.
At $x = 1$ , $y = 2$ , so $v = 2$ .

v + x \frac{d v}{d x} = \frac{x ^{2} + 3 v ^{2} x ^{2}}{v x ^{2}} = \frac{1 + 3 v ^{2}}{v} \Rightarrow x \frac{d v}{d x} = \frac{1 + 2 v ^{2}}{v} \int \frac{v}{1 + 2 v ^{2}} d v = \int \frac{d x}{x} \Rightarrow \frac{1}{4} ln (1 + 2 v^{2}) = ln x + c \frac{1}{4} ln 9 = c ln (1 + 2 v^{2}) = ln (9 x^{4}) \Rightarrow v^{2} = \frac{9 x ^{4} - 1}{2} y = x \frac{9 x ^{4} - 1}{2}

(c)

1 mark

Substitute $x = 1.1$ into the exact solution.

y (1.1) = 1.1 \frac{9 ( 1.1 ) ^{4} - 1}{2} \approx 2.714

(d)

2 marks

The second derivative formula shows $\frac{d ^{2} y}{d x ^{2}} > 0$ on $1 \leq x \leq 1.1$ (the graph is concave up).
A tangent to a concave-up curve lies below the curve, so the Euler value 2.65 is smaller than the true value 2.714.

(e)

2 marks

x \approx 0.656

(f)

6 marks

At the inflexion $\frac{d ^{2} y}{d x ^{2}} = 0$ , so $- x^{4} + x^{2} y^{2} + 6 y^{4} = 0$ . Put $y = m x$ .
Factorise.
The other factor has no real solution, and $m > 0$ because $x, y > 0$ . So P lies on the line $y = \frac{3}{3} x$ .

- x^{4} + m^{2} x^{4} + 6 m^{4} x^{4} = 0 \Rightarrow 6 m^{4} + m^{2} - 1 = 0 (3 m^{2} - 1) (2 m^{2} + 1) = 0 \Rightarrow m^{2} = \frac{1}{3} m = \frac{1}{3} = \frac{3}{3}

Our analysis

Where the marks are

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

Statistics and probability32 marks · 29%
Q1, Q3, Q11
Calculus27 marks · 25%
Q5, Q12
Number and algebra23 marks · 21%
Q2, Q4, Q7, Q9
Functions20 marks · 18%
Q8, Q10
Geometry and trigonometry8 marks · 7%
Q6

Mentor insights

Traps to watch for

01
Three regions of the normal curve · Q3
The probability below $w$ is what is left after removing the 80% and the upper tail.
02
Check the point and the direction · Q6
To verify a line of intersection, show that its point lies on both planes and that its direction is perpendicular to both normals.
03
Released at the minimum · Q10
At $t = 0$ the weight is at its lowest point, so the cosine model needs a negative $a$ .
04
Without replacement · Q11
After the first yellow ball there are $y - 1$ yellow balls and $r + y - 1$ balls in total.
05
Substitution y = vx · Q12
Then $\frac{d y}{d x} = v + x \frac{d v}{d x}$ , and the equation separates.

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