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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Session: November 2021
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2021 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 1–8 · 56 marks

Question 1

7 marks

Piano practice and diploma scores: correlation and regression

Statistics and probability

(a)

2 marks

Enter the eight pairs on the calculator.

r \approx 0.884

(b)

1 mark

The regression line of $D$ on $h$ :

a \approx 1.37, b \approx 64.5

(c)

2 marks

Five extra hours a week changes $D$ by $5 a$ .
So her score could have been expected to rise by about 7 marks.

5 \times 1.366 \approx 6.83

(d)

1 mark

A strong correlation does not mean practising causes the higher score. Natural ability or motivation could affect both, so Lucy's assertion isn't valid.

(e)

1 mark

Subtracting the same number of hours from every student shifts all the $h$ values but doesn't change how the points are arranged relative to each other, so $r$ is unchanged.

Question 2

5 marks

The ambiguous case: the smallest perimeter

Number and algebra

5 marks

Apply the cosine rule with $A B = c$ unknown, using the angle $A = 2 5^{\circ}$ opposite the side $C B = 7$ .
Both roots are positive and give valid triangles.
The smaller perimeter uses the shorter side.

7^{2} = 1 2^{2} + c^{2} - 2 (12) c cos 2 5^{\circ} \Rightarrow c^{2} - 21.75 c + 95 = 0 c \approx 6.05 or c \approx 15.70 12 + 7 + 6.05 \approx 25.1

Mentor tip

Check with the sine rule: $sin B = \frac{12 s i n 2 5 ^{\circ}}{7} \approx 0.7245$ gives $B \approx 46. 4^{\circ}$ or $133. 6^{\circ}$ , the two triangles.

Question 3

7 marks

Defective lamps: a binomial and a conditional probability

Statistics and probability

(a)

3 marks

The number of defective lamps is $X \sim B (30, 0.05)$ .

P (X \geq 1) = 1 - 0.9 5^{30} \approx 0.785

(b)

4 marks

This is a conditional probability. The events "at most two" and "at least one" overlap in $1 \leq X \leq 2$ .

P (X \leq 2 ∣ X \geq 1) = \frac{P ( 1 \leq X \leq 2 )}{P ( X \geq 1 )} = \frac{0.5975}{0.7854} \approx 0.761

Question 4

6 marks

A semicircle with two equal shaded areas

Geometry and trigonometry

(a)

5 marks

One shaded region is the segment cut off by the chord QR, which has area sector minus triangle. The other is the triangle POR, whose angle at O is $π - θ$ .
Set the two areas equal and cancel $\frac{1}{2} r^{2}$ .

segment = \frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} sin θ, triangle POR = \frac{1}{2} r^{2} sin (π - θ) = \frac{1}{2} r^{2} sin θ θ - sin θ = sin θ \Rightarrow θ = 2 sin θ

(b)

1 mark

Solve $θ = 2 sin θ$ on the calculator, in radians, for $0 < θ < π$ .

θ \approx 1.90

Question 5

9 marks

A geometric series given as a sum: first term, sum to infinity and when it's close

Number and algebra

(a)

2 marks

$S_{1}$ is the first term, so put $r = 1$ in the formula.

u_{1} = \frac{2}{3} \cdot \frac{7}{8} = \frac{7}{12}

(b)

3 marks

The common ratio is $\frac{7}{8}$ . Use $S_{\infty} = \frac{u _{1}}{1 - r}$ .

S_{\infty} = \frac{7/12}{1/8} = \frac{14}{3}

(c)

4 marks

The tail after $n$ terms is $S_{\infty} - S_{n} = \frac{14}{3} (\frac{7}{8})^{n}$ . Solve $\frac{14}{3} (\frac{7}{8})^{n} < 0.001$ .
At $n = 63$ the tail is 0.001036, which is too big. At $n = 64$ it is 0.000906.

n > \frac{ln ( 0.001 \times 3/14 )}{ln ( 7/8 )} \approx 63.3 n = 64

Question 6

8 marks

A cubic identity, and a new quadratic whose roots are reciprocal cubes

Number and algebra

(a)

2 marks

Expand $(p + q)^{3}$ and subtract.

(p + q)^{3} - 3 pq (p + q) = p^{3} + 3 p^{2} q + 3 p q^{2} + q^{3} - 3 p^{2} q - 3 p q^{2} = p^{3} + q^{3}

(b)

6 marks

For $2 x^{2} - 5 x + 1 = 0$ we have $α + β = \frac{5}{2}$ and $α β = \frac{1}{2}$ .
The product of the new roots is $n$ .
The sum of the new roots is $- m$ . Use part (a) to find $α^{3} + β^{3}$ .

n = \frac{1}{α ^{3}} \cdot \frac{1}{β ^{3}} = \frac{1}{( α β ) ^{3}} = 8 α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β) = \frac{125}{8} - 3 \cdot \frac{1}{2} \cdot \frac{5}{2} = \frac{95}{8} \frac{1}{α ^{3}} + \frac{1}{β ^{3}} = \frac{α ^{3} + β ^{3}}{( α β ) ^{3}} = \frac{95}{8} \times 8 = 95 \Rightarrow m = - 95 x^{2} - 95 x + 8 = 0

Question 7

6 marks

A density function built from arccosine: its median and a symmetric interval

Calculus

(a)

2 marks

The median $m$ satisfies $\int_{0}^{m} arccos x d x = 0.5$ . With the antiderivative $x arccos x - 1 - x^{2}$ :
Solve numerically.

[x arccos x - 1 - x^{2}]_{0}^{m} = m arccos m - 1 - m^{2} + 1 = 0.5 m \approx 0.360

(b)

4 marks

The probability that $X$ is within $a$ of the median is the integral from $m - a$ to $m + a$ .
Substitute the antiderivative, with $m = 0.360034$ , and solve for $a$ on the calculator.

\int_{m - a}^{m + a} arccos x d x = 0.3 a \approx 0.125

Question 8

8 marks

Implicit differentiation of $y = x - x y ln (x y)$ and a tangent

Calculus

(a)

3 marks

Differentiate $y = x - x y ln (x y)$ with respect to $x$ . The derivative of $x y ln (x y)$ is
So
Move the $\frac{d y}{d x}$ terms to the left.

(y + x \frac{d y}{d x}) ln (x y) + x y \cdot \frac{1}{x y} (y + x \frac{d y}{d x}) \frac{d y}{d x} = 1 - (y + x \frac{d y}{d x}) ln (x y) - (y + x \frac{d y}{d x}) \frac{d y}{d x} (1 + x (1 + ln (x y))) + y (1 + ln (x y)) = 1

(b)

5 marks

At $x = 1$ the curve has $y = 1 - y ln y$ , so $y (1 + ln y) = 1$ , which gives $y = 1$ .
Put $x = 1$ , $y = 1$ into the result of part (a), where $ln 1 = 0$ .
The tangent is horizontal through $(1, 1)$ .

\frac{d y}{d x} (1 + 1) + 1 = 1 \Rightarrow \frac{d y}{d x} = 0 y = 1

Section B · Extended response

Questions 9–11 · 54 marks

Question 9

15 marks

Tides in Dungeness harbour: finding and using a sine model

Functions

(a)

1 mark

The time between high tides is 12 hours, so that is the period.

b = \frac{2 π}{12} = \frac{π}{6}

(b)

2 marks

The amplitude is half the range.

a = \frac{6.8 - 2.2}{2} = 2.3

(c)

2 marks

The midline is halfway between the maximum and minimum.

d = \frac{6.8 + 2.2}{2} = 4.5

(d)

3 marks

At the first high tide, $t = 4.5$ and the sine equals 1, so its argument is $\frac{π}{2}$ .

\frac{π}{6} (4.5 - c) = \frac{π}{2} \Rightarrow 4.5 - c = 3 \Rightarrow c = 1.5

(e)

2 marks

Noon is $t = 12$ .

H (12) = 2.3 sin (\frac{π}{6} \times 10.5) + 4.5 \approx 2.87 m

(f)

3 marks

Solve $2.3 sin (\frac{π}{6} (t - 1.5)) + 4.5 = 5$ on the calculator. In the first 12 hours the tide is above 5 m from about 1.92 h to 7.08 h.
That happens in each of the two high tides in 24 hours.

7.081 - 1.919 = 5.163 hours 2 \times 5.163 \approx 10.3 hours

(g)

2 marks

The Folkestone high tide is 50 minutes earlier, at $4.5 - \frac{5}{6} = \frac{11}{3}$ . So $c$ becomes $\frac{11}{3} - 3 = \frac{2}{3}$ .

H (t) = 2.3 sin (\frac{π}{6} (t - \frac{2}{3})) + 4.5

Question 10

18 marks

A rational function with an oblique asymptote, and an integral of its reciprocal

Functions

(a)

3 marks

(i) The $x$ -intercepts are where the numerator is zero.

(ii) The $y$ -intercept is $f (0) = \frac{- 12}{- 15}$ .

x^{2} - x - 12 = (x - 4) (x + 3) = 0 \Rightarrow (- 3, 0), (4, 0) (0, \frac{4}{5})

(b)

1 mark

The denominator is zero at

x = \frac{15}{2}

(c)

4 marks

Divide $x^{2} - x - 12$ by $2 x - 15$ : the quotient is $\frac{1}{2} x + \frac{13}{4}$ with a remainder, which tends to 0 relative to the quotient as $x$ grows.

\frac{x ^{2} - x - 12}{2 x - 15} = \frac{1}{2} x + \frac{13}{4} + \frac{remainder}{2 x - 15} \Rightarrow a = \frac{1}{2}, b = \frac{13}{4}

(d)

3 marks

The graph has two branches, one each side of the vertical asymptote $x = 7.5$ , and the oblique asymptote $y = 0.5 x + 3.25$ . The left branch follows the oblique asymptote from below as $x \to - \infty$ , rises through $(- 3, 0)$ to a maximum, passes through $(0, 0.8)$ , comes down through $(4, 0)$ and falls to $- \infty$ as $x \to 7. 5^{-}$ . The right branch starts at $+ \infty$ just right of $x = 7.5$ , falls to a minimum, and then rises to follow the oblique asymptote from above.

(e)

7 marks

(i) The reciprocal is $\frac{2 x - 15}{( x - 4 ) ( x + 3 )}$ . Split into partial fractions.

(ii) Integrate from 0 to 3.

2 x - 15 = A (x - 4) + B (x + 3) \Rightarrow B = - 1 (x = 4), A = 3 (x = - 3) \frac{1}{f ( x )} = \frac{3}{x + 3} - \frac{1}{x - 4} \int_{0}^{3} (\frac{3}{x + 3} - \frac{1}{x - 4}) d x = [3 ln ∣ x + 3∣ - ln ∣ x - 4∣]_{0}^{3} (3 ln 6 - ln 1) - (3 ln 3 - ln 4) = 3 ln 2 + ln 4 = ln 8 + ln 4 = ln 32

Question 11

21 marks

Planes, their line of intersection, and a reflection in a plane

Geometry and trigonometry

(a)

7 marks

(i) Subtract coordinates.

(ii) The normal is the cross product.

So $2 x - 3 y - z = d$ , and the point $A (3, 0, 0)$ gives $d = 6$ .

A B = - 3 - 2 0, A C = - 2 1 - 7 A B \times A C = 14 - 21 - 7 \propto 2 - 3 - 1 2 x - 3 y - z = 6

(b)

2 marks

The point $(0, - 2, 0)$ is in both planes: $Π_{1}$ gives $0 + 6 - 0 = 6$ and $Π_{2}$ gives $0 + 2 + 0 = 2$ .
The direction $(1, 1, - 1)$ is perpendicular to both normals: $2 - 3 + 1 = 0$ and $3 - 1 - 2 = 0$ . So this vector equation gives L.

r = 0 - 2 0 + λ 11 - 1

(c)

3 marks

(i) On L the coordinates are $(λ, - 2 + λ, - λ)$ . Substitute into $2 x - 2 z = 3$ .

(ii) Put $λ = \frac{3}{4}$ back into L.

2 λ + 2 λ = 3 \Rightarrow λ = \frac{3}{4} P (\frac{3}{4}, - \frac{5}{4}, - \frac{3}{4})

(d)

9 marks

(i) The normal to $Π_{3}$ is $20 - 2$ . The line through B along it is $(2 μ, - 2, - 2 μ)$ . It meets $Π_{3}$ where $4 μ + 4 μ = 3$ , so $μ = \frac{3}{8}$ , at the point $(\frac{3}{4}, - 2, - \frac{3}{4})$ . The reflection goes the same distance again, so use $2 μ = \frac{3}{4}$ .

(ii) The reflected line passes through P (on both L and the plane) and B'.

B^{'} = (\frac{3}{2}, - 2, - \frac{3}{2}) P B^{'} = \frac{3}{4} - \frac{3}{4} - \frac{3}{4} r = \frac{3}{4} - \frac{5}{4} - \frac{3}{4} + λ 1 - 1 - 1

Mentor tip

Check: reflecting the direction $(1, 1, - 1)$ in the plane with normal $(1, 0, - 1)$ gives $(- 1, 1, 1)$ , which is parallel to $(1, - 1, - 1)$ .

Our analysis

Where the marks are

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

Functions33 marks · 30%
Q9, Q10
Geometry and trigonometry27 marks · 25%
Q4, Q11
Number and algebra22 marks · 20%
Q2, Q5, Q6
Statistics and probability14 marks · 13%
Q1, Q3
Calculus14 marks · 13%
Q7, Q8

Mentor insights

Traps to watch for

01
The inequality is on the remainder · Q5
$S_{\infty} - S_{n}$ is the tail of the series, which is $\frac{u _{1} r ^{n}}{1 - r}$ . Solve that against 0.001 with logs or a table, not $u_{n} < 0.001$ .
02
Use symmetric functions of roots · Q6
Don't solve $2 x^{2} - 5 x + 1 = 0$ . The new roots are $\frac{1}{α ^{3}}$ and $\frac{1}{β ^{3}}$ , whose sum and product come from $α + β$ and $α β$ .
03
Read the data carefully · Q9
"The first high tide at 04:30" means the maximum occurs at $t = 4.5$ . The sine reaches its maximum when its argument is $\frac{π}{2}$ , which gives $c$ .
04
Both asymptotes · Q10
A rational function with a numerator one degree higher than the denominator has an oblique asymptote, found by division. The vertical asymptote is where the denominator is zero.
05
A normal from a cross product · Q11
The normal to a plane through three points is the cross product of two vectors in it. The direction of the intersection line is perpendicular to both normals.

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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Session: November 2021
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2021 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 1–8 · 56 marks

Question 1

7 marks

Piano practice and diploma scores: correlation and regression

Statistics and probability

(a)

2 marks

Enter the eight pairs on the calculator.

r \approx 0.884

(b)

1 mark

The regression line of $D$ on $h$ :

a \approx 1.37, b \approx 64.5

(c)

2 marks

Five extra hours a week changes $D$ by $5 a$ .
So her score could have been expected to rise by about 7 marks.

5 \times 1.366 \approx 6.83

(d)

1 mark

A strong correlation does not mean practising causes the higher score. Natural ability or motivation could affect both, so Lucy's assertion isn't valid.

(e)

1 mark

Subtracting the same number of hours from every student shifts all the $h$ values but doesn't change how the points are arranged relative to each other, so $r$ is unchanged.

Question 2

5 marks

The ambiguous case: the smallest perimeter

Number and algebra

5 marks

Apply the cosine rule with $A B = c$ unknown, using the angle $A = 2 5^{\circ}$ opposite the side $C B = 7$ .
Both roots are positive and give valid triangles.
The smaller perimeter uses the shorter side.

7^{2} = 1 2^{2} + c^{2} - 2 (12) c cos 2 5^{\circ} \Rightarrow c^{2} - 21.75 c + 95 = 0 c \approx 6.05 or c \approx 15.70 12 + 7 + 6.05 \approx 25.1

Mentor tip

Check with the sine rule: $sin B = \frac{12 s i n 2 5 ^{\circ}}{7} \approx 0.7245$ gives $B \approx 46. 4^{\circ}$ or $133. 6^{\circ}$ , the two triangles.

Question 3

7 marks

Defective lamps: a binomial and a conditional probability

Statistics and probability

(a)

3 marks

The number of defective lamps is $X \sim B (30, 0.05)$ .

P (X \geq 1) = 1 - 0.9 5^{30} \approx 0.785

(b)

4 marks

This is a conditional probability. The events "at most two" and "at least one" overlap in $1 \leq X \leq 2$ .

P (X \leq 2 ∣ X \geq 1) = \frac{P ( 1 \leq X \leq 2 )}{P ( X \geq 1 )} = \frac{0.5975}{0.7854} \approx 0.761

Question 4

6 marks

A semicircle with two equal shaded areas

Geometry and trigonometry

(a)

5 marks

One shaded region is the segment cut off by the chord QR, which has area sector minus triangle. The other is the triangle POR, whose angle at O is $π - θ$ .
Set the two areas equal and cancel $\frac{1}{2} r^{2}$ .

segment = \frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} sin θ, triangle POR = \frac{1}{2} r^{2} sin (π - θ) = \frac{1}{2} r^{2} sin θ θ - sin θ = sin θ \Rightarrow θ = 2 sin θ

(b)

1 mark

Solve $θ = 2 sin θ$ on the calculator, in radians, for $0 < θ < π$ .

θ \approx 1.90

Question 5

9 marks

A geometric series given as a sum: first term, sum to infinity and when it's close

Number and algebra

(a)

2 marks

$S_{1}$ is the first term, so put $r = 1$ in the formula.

u_{1} = \frac{2}{3} \cdot \frac{7}{8} = \frac{7}{12}

(b)

3 marks

The common ratio is $\frac{7}{8}$ . Use $S_{\infty} = \frac{u _{1}}{1 - r}$ .

S_{\infty} = \frac{7/12}{1/8} = \frac{14}{3}

(c)

4 marks

The tail after $n$ terms is $S_{\infty} - S_{n} = \frac{14}{3} (\frac{7}{8})^{n}$ . Solve $\frac{14}{3} (\frac{7}{8})^{n} < 0.001$ .
At $n = 63$ the tail is 0.001036, which is too big. At $n = 64$ it is 0.000906.

n > \frac{ln ( 0.001 \times 3/14 )}{ln ( 7/8 )} \approx 63.3 n = 64

Question 6

8 marks

A cubic identity, and a new quadratic whose roots are reciprocal cubes

Number and algebra

(a)

2 marks

Expand $(p + q)^{3}$ and subtract.

(p + q)^{3} - 3 pq (p + q) = p^{3} + 3 p^{2} q + 3 p q^{2} + q^{3} - 3 p^{2} q - 3 p q^{2} = p^{3} + q^{3}

(b)

6 marks

For $2 x^{2} - 5 x + 1 = 0$ we have $α + β = \frac{5}{2}$ and $α β = \frac{1}{2}$ .
The product of the new roots is $n$ .
The sum of the new roots is $- m$ . Use part (a) to find $α^{3} + β^{3}$ .

n = \frac{1}{α ^{3}} \cdot \frac{1}{β ^{3}} = \frac{1}{( α β ) ^{3}} = 8 α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β) = \frac{125}{8} - 3 \cdot \frac{1}{2} \cdot \frac{5}{2} = \frac{95}{8} \frac{1}{α ^{3}} + \frac{1}{β ^{3}} = \frac{α ^{3} + β ^{3}}{( α β ) ^{3}} = \frac{95}{8} \times 8 = 95 \Rightarrow m = - 95 x^{2} - 95 x + 8 = 0

Question 7

6 marks

A density function built from arccosine: its median and a symmetric interval

Calculus

(a)

2 marks

The median $m$ satisfies $\int_{0}^{m} arccos x d x = 0.5$ . With the antiderivative $x arccos x - 1 - x^{2}$ :
Solve numerically.

[x arccos x - 1 - x^{2}]_{0}^{m} = m arccos m - 1 - m^{2} + 1 = 0.5 m \approx 0.360

(b)

4 marks

The probability that $X$ is within $a$ of the median is the integral from $m - a$ to $m + a$ .
Substitute the antiderivative, with $m = 0.360034$ , and solve for $a$ on the calculator.

\int_{m - a}^{m + a} arccos x d x = 0.3 a \approx 0.125

Question 8

8 marks

Implicit differentiation of $y = x - x y ln (x y)$ and a tangent

Calculus

(a)

3 marks

Differentiate $y = x - x y ln (x y)$ with respect to $x$ . The derivative of $x y ln (x y)$ is
So
Move the $\frac{d y}{d x}$ terms to the left.

(y + x \frac{d y}{d x}) ln (x y) + x y \cdot \frac{1}{x y} (y + x \frac{d y}{d x}) \frac{d y}{d x} = 1 - (y + x \frac{d y}{d x}) ln (x y) - (y + x \frac{d y}{d x}) \frac{d y}{d x} (1 + x (1 + ln (x y))) + y (1 + ln (x y)) = 1

(b)

5 marks

At $x = 1$ the curve has $y = 1 - y ln y$ , so $y (1 + ln y) = 1$ , which gives $y = 1$ .
Put $x = 1$ , $y = 1$ into the result of part (a), where $ln 1 = 0$ .
The tangent is horizontal through $(1, 1)$ .

\frac{d y}{d x} (1 + 1) + 1 = 1 \Rightarrow \frac{d y}{d x} = 0 y = 1

Section B · Extended response

Questions 9–11 · 54 marks

Question 9

15 marks

Tides in Dungeness harbour: finding and using a sine model

Functions

(a)

1 mark

The time between high tides is 12 hours, so that is the period.

b = \frac{2 π}{12} = \frac{π}{6}

(b)

2 marks

The amplitude is half the range.

a = \frac{6.8 - 2.2}{2} = 2.3

(c)

2 marks

The midline is halfway between the maximum and minimum.

d = \frac{6.8 + 2.2}{2} = 4.5

(d)

3 marks

At the first high tide, $t = 4.5$ and the sine equals 1, so its argument is $\frac{π}{2}$ .

\frac{π}{6} (4.5 - c) = \frac{π}{2} \Rightarrow 4.5 - c = 3 \Rightarrow c = 1.5

(e)

2 marks

Noon is $t = 12$ .

H (12) = 2.3 sin (\frac{π}{6} \times 10.5) + 4.5 \approx 2.87 m

(f)

3 marks

Solve $2.3 sin (\frac{π}{6} (t - 1.5)) + 4.5 = 5$ on the calculator. In the first 12 hours the tide is above 5 m from about 1.92 h to 7.08 h.
That happens in each of the two high tides in 24 hours.

7.081 - 1.919 = 5.163 hours 2 \times 5.163 \approx 10.3 hours

(g)

2 marks

The Folkestone high tide is 50 minutes earlier, at $4.5 - \frac{5}{6} = \frac{11}{3}$ . So $c$ becomes $\frac{11}{3} - 3 = \frac{2}{3}$ .

H (t) = 2.3 sin (\frac{π}{6} (t - \frac{2}{3})) + 4.5

Question 10

18 marks

A rational function with an oblique asymptote, and an integral of its reciprocal

Functions

(a)

3 marks

(i) The $x$ -intercepts are where the numerator is zero.

(ii) The $y$ -intercept is $f (0) = \frac{- 12}{- 15}$ .

x^{2} - x - 12 = (x - 4) (x + 3) = 0 \Rightarrow (- 3, 0), (4, 0) (0, \frac{4}{5})

(b)

1 mark

The denominator is zero at

x = \frac{15}{2}

(c)

4 marks

Divide $x^{2} - x - 12$ by $2 x - 15$ : the quotient is $\frac{1}{2} x + \frac{13}{4}$ with a remainder, which tends to 0 relative to the quotient as $x$ grows.

\frac{x ^{2} - x - 12}{2 x - 15} = \frac{1}{2} x + \frac{13}{4} + \frac{remainder}{2 x - 15} \Rightarrow a = \frac{1}{2}, b = \frac{13}{4}

(d)

3 marks

(e)

7 marks

(i) The reciprocal is $\frac{2 x - 15}{( x - 4 ) ( x + 3 )}$ . Split into partial fractions.

(ii) Integrate from 0 to 3.

2 x - 15 = A (x - 4) + B (x + 3) \Rightarrow B = - 1 (x = 4), A = 3 (x = - 3) \frac{1}{f ( x )} = \frac{3}{x + 3} - \frac{1}{x - 4} \int_{0}^{3} (\frac{3}{x + 3} - \frac{1}{x - 4}) d x = [3 ln ∣ x + 3∣ - ln ∣ x - 4∣]_{0}^{3} (3 ln 6 - ln 1) - (3 ln 3 - ln 4) = 3 ln 2 + ln 4 = ln 8 + ln 4 = ln 32

Question 11

21 marks

Planes, their line of intersection, and a reflection in a plane

Geometry and trigonometry

(a)

7 marks

(i) Subtract coordinates.

(ii) The normal is the cross product.

So $2 x - 3 y - z = d$ , and the point $A (3, 0, 0)$ gives $d = 6$ .

A B = - 3 - 2 0, A C = - 2 1 - 7 A B \times A C = 14 - 21 - 7 \propto 2 - 3 - 1 2 x - 3 y - z = 6

(b)

2 marks

The point $(0, - 2, 0)$ is in both planes: $Π_{1}$ gives $0 + 6 - 0 = 6$ and $Π_{2}$ gives $0 + 2 + 0 = 2$ .
The direction $(1, 1, - 1)$ is perpendicular to both normals: $2 - 3 + 1 = 0$ and $3 - 1 - 2 = 0$ . So this vector equation gives L.

r = 0 - 2 0 + λ 11 - 1

(c)

3 marks

(i) On L the coordinates are $(λ, - 2 + λ, - λ)$ . Substitute into $2 x - 2 z = 3$ .

(ii) Put $λ = \frac{3}{4}$ back into L.

2 λ + 2 λ = 3 \Rightarrow λ = \frac{3}{4} P (\frac{3}{4}, - \frac{5}{4}, - \frac{3}{4})

(d)

9 marks

(ii) The reflected line passes through P (on both L and the plane) and B'.

B^{'} = (\frac{3}{2}, - 2, - \frac{3}{2}) P B^{'} = \frac{3}{4} - \frac{3}{4} - \frac{3}{4} r = \frac{3}{4} - \frac{5}{4} - \frac{3}{4} + λ 1 - 1 - 1

Mentor tip

Check: reflecting the direction $(1, 1, - 1)$ in the plane with normal $(1, 0, - 1)$ gives $(- 1, 1, 1)$ , which is parallel to $(1, - 1, - 1)$ .

Our analysis

Where the marks are

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

Functions33 marks · 30%
Q9, Q10
Geometry and trigonometry27 marks · 25%
Q4, Q11
Number and algebra22 marks · 20%
Q2, Q5, Q6
Statistics and probability14 marks · 13%
Q1, Q3
Calculus14 marks · 13%
Q7, Q8

Mentor insights

Traps to watch for

01
The inequality is on the remainder · Q5
$S_{\infty} - S_{n}$ is the tail of the series, which is $\frac{u _{1} r ^{n}}{1 - r}$ . Solve that against 0.001 with logs or a table, not $u_{n} < 0.001$ .
02
Use symmetric functions of roots · Q6
Don't solve $2 x^{2} - 5 x + 1 = 0$ . The new roots are $\frac{1}{α ^{3}}$ and $\frac{1}{β ^{3}}$ , whose sum and product come from $α + β$ and $α β$ .
03
Read the data carefully · Q9
"The first high tide at 04:30" means the maximum occurs at $t = 4.5$ . The sine reaches its maximum when its argument is $\frac{π}{2}$ , which gives $c$ .
04
Both asymptotes · Q10
A rational function with a numerator one degree higher than the denominator has an oblique asymptote, found by division. The vertical asymptote is where the denominator is zero.
05
A normal from a cross product · Q11
The normal to a plane through three points is the cross product of two vectors in it. The direction of the intersection line is perpendicular to both normals.

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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