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IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2021
Marks: 110
Duration: 2 hours
Calculator: Not 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

4 marks

Difference of consecutive squares equals their sum

Number and algebra

4 marks

Subtract the squares in the right order, expanding $(n + 1)^{2}$ .
The sum of the two integers is $n + (n + 1) = 2 n + 1$ , the same expression.

(n + 1)^{2} - n^{2} = n^{2} + 2 n + 1 - n^{2} = 2 n + 1

Mentor tip

Factorising works too: $(n + 1)^{2} - n^{2} = ((n + 1) - n) ((n + 1) + n) = 1 \cdot (2 n + 1)$ .

Question 2

7 marks

A trig equation turned into a quadratic in sine

Geometry and trigonometry

7 marks

Replace $cos^{2} x$ using $cos^{2} x = 1 - sin^{2} x$ , and collect everything on one side.
Factorise as a quadratic in $sin x$ .
$sin x = 2$ is impossible. That leaves $sin x = \frac{1}{2}$ in $0 \leq x \leq 2 π$ .

2 (1 - sin^{2} x) + 5 sin x = 4 \Rightarrow 2 sin^{2} x - 5 sin x + 2 = 0 (2 sin x - 1) (sin x - 2) = 0 x = \frac{π}{6}, \frac{5 π}{6}

Mentor tip

Say why the second factor is rejected: a sine can't be 2.

Question 3

5 marks

Finding a constant from one binomial coefficient

Number and algebra

5 marks

In $(x + k)^{7}$ the term in $x^{5}$ comes from choosing $x$ five times and $k$ twice.
Set the coefficient equal to 63.

(2 7) x^{5} k^{2} = 21 k^{2} x^{5} 21 k^{2} = 63 \Rightarrow k^{2} = 3 \Rightarrow k = \pm 3

Question 4

9 marks

A logarithmic curve: where it crosses and where it is steep

Calculus

(a)

3 marks

The graph meets the $x$ -axis where $f (x) = 0$ , that is, where the log's argument is 1.
Since $x > 4$ , only the positive root is allowed, so $a = 17$ .

ln (x^{2} - 16) = 0 \Rightarrow x^{2} - 16 = 1 \Rightarrow x = 17

(b)

6 marks

Differentiate with the chain rule, then set the gradient equal to $\frac{1}{3}$ .
Cross-multiply and solve the quadratic.

f^{'} (x) = \frac{2 x}{x ^{2} - 16} = \frac{1}{3} 6 x = x^{2} - 16 \Rightarrow x^{2} - 6 x - 16 = 0 \Rightarrow (x - 8) (x + 2) = 0 x = 8 (since x > 4)

Mentor tip

The domain check is the last line of the answer. Do it every time.

Question 5

4 marks

The magnitude of a cross product, squared

Number and algebra

4 marks

Use $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ$ and square it.
Distribute, and replace $∣ a ∣^{2} ∣ b ∣^{2} cos^{2} θ$ with $(a \cdot b)^{2}$ using $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ .

∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} sin^{2} θ = ∣ a ∣^{2} ∣ b ∣^{2} (1 - cos^{2} θ) ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}

Question 6

7 marks

Area of a triangle with a ratio of sides and a known cosine

Number and algebra

7 marks

Apply the cosine rule at $C$ , opposite the side $A B = 10$ , with $B C = x$ and $A C = 2 x$ .
From $cos C = \frac{3}{4}$ , $sin C = 1 - \frac{9}{16} = \frac{7}{4}$ .
The area is $\frac{1}{2} \cdot B C \cdot A C \cdot sin C = \frac{1}{2} \cdot x \cdot 2 x \cdot sin C = x^{2} sin C$ .

100 = x^{2} + 4 x^{2} - 2 (x) (2 x) \cdot \frac{3}{4} = 5 x^{2} - 3 x^{2} = 2 x^{2} x^{2} = 50 Area = 50 \cdot \frac{7}{4} = \frac{25}{2} 7

Mentor tip

Check the form asked for: a rational number times a square root, $p q$ with $p = \frac{25}{2}$ and $q = 7$ .

Question 7

5 marks

Roots of a cubic given in terms of each other

Number and algebra

5 marks

Let $s = α + β$ . The sum of the three roots is $α + β + (α + β) = 2 s$ , and it equals $k$ .
The $x$ coefficient is 0, so the sum of the products in pairs is 0: $α β + α s + β s = α β + s^{2} = 0$ . So $α β = - s^{2}$ .
The product of the roots is $- 3 k$ (minus the constant term), which is $α β s$ .
Divide by $- k$ (and $k > 0$ ).

2 s = k \Rightarrow s = \frac{k}{2} α β s = - s^{3} = - \frac{k ^{3}}{8} = - 3 k k^{2} = 24 \Rightarrow k = 26

Question 8

8 marks

Parallel lines in space: no intersection, then the gap

Geometry and trigonometry

(a)

3 marks

Write the lines as $r = (3, 2, - 1) + λ (2, - 2, 2)$ and $r = (2, 0, 4) + μ (1, - 1, 1)$ , and equate the components.
Add the first two equations. The $λ$ and $μ$ cancel and leave $5 = 2$ , which is false. So there is no solution, and the lines do not intersect.

3 + 2 λ = 2 + μ, 2 - 2 λ = - μ, - 1 + 2 λ = 4 + μ

Mentor tip

The direction vectors are multiples of each other, so the lines are parallel. The contradiction shows they are distinct.

(b)

5 marks

The lines are parallel, so the distance is the same everywhere: take $A (3, 2, - 1)$ on the first line and $B (2, 0, 4)$ on the second, and use the direction $v = (1, - 1, 1)$ .
The shortest distance is $\frac{∣ A B \times v ∣}{∣ v ∣}$ .

A B = - 1 - 2 5 A B \times v = 363, ∣ A B \times v ∣ = 54 d = \frac{54}{3} = 18 = 32

Mentor tip

Check with calculus: the squared distance from $A$ to the general point $B (2 + μ, - μ, 4 + μ)$ is $3 μ^{2} + 12 μ + 30$ , minimised at $μ = - 2$ with value 18.

Question 9

7 marks

Integrating with the substitution $u = sin x$

Number and algebra

7 marks

Substituting $u = sin x$ and $d u = cos x d x$ turns the integral into a rational one.
Split into partial fractions. Putting $u = 2$ gives $B = \frac{2}{3}$ and putting $u = - 1$ gives $A = \frac{1}{3}$ .
Integrate, then put $u = sin x$ back.

\int \frac{u}{u ^{2} - u - 2} d u = \int \frac{u}{( u + 1 ) ( u - 2 )} d u \frac{u}{( u + 1 ) ( u - 2 )} = \frac{1/3}{u + 1} + \frac{2/3}{u - 2} \frac{1}{3} ln ∣ sin x + 1∣ + \frac{2}{3} ln ∣ sin x - 2∣ + C

Mentor tip

Don't leave the answer in $u$ . The question was about $x$ .

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

15 marks

Area under a cosine curve, then a cone with the same surface area

Calculus
Geometry and trigonometry

(a)

3 marks

The graph touches the $x$ -axis where $f (x) = 0$ , that is, where $cos x = - 1$ in $0 \leq x \leq 4 π$ .

6 + 6 cos x = 0 \Rightarrow cos x = - 1 \Rightarrow x = π, 3 π

(b)

5 marks

The area is the integral between those two points.

\int_{π}^{3 π} (6 + 6 cos x) d x = [6 x + 6 sin x]_{π}^{3 π} (18 π + 0) - (6 π + 0) = 12 π

Mentor tip

Both sine terms vanish at multiples of $π$ , so write that rather than skipping it.

(c)

3 marks

The total surface area is the base plus the curved surface, $π r^{2} + π r l$ , with $r = 2$ . Set it equal to $12 π$ .

4 π + 2 π l = 12 π \Rightarrow l = 4

(d)

4 marks

Find the height from $r^{2} + h^{2} = l^{2}$ , then use $V = \frac{1}{3} π r^{2} h$ .

h = 16 - 4 = 23 V = \frac{1}{3} π (2^{2}) (23) = \frac{8 3 π}{3}

Question 11

20 marks

A particle whose acceleration depends on its velocity

Calculus

(a)

6 marks

Write the acceleration as $\frac{d v}{d t} = - (1 + v)$ and separate the variables.
At $t = 0$ , $v = v_{0}$ , so $C = ln (1 + v_{0})$ . Rearrange for $v$ . (No modulus is needed, since $v > - 1$ .)

\int \frac{1}{1 + v} d v = - \int 1 d t \Rightarrow ln (1 + v) = - t + C 1 + v = (1 + v_{0}) e^{- t} \Rightarrow v (t) = (1 + v_{0}) e^{- t} - 1

(b)

7 marks

At the farthest point the particle is momentarily at rest, so $v (T) = 0$ .
Integrate $v$ to get the displacement, using $s = 0$ at $t = 0$ to find the constant.
Put in $t = T = ln (1 + v_{0})$ and use $e^{- T} = \frac{1}{1 + v _{0}}$ .

(1 + v_{0}) e^{- T} = 1 \Rightarrow e^{T} = 1 + v_{0} s (t) = - (1 + v_{0}) e^{- t} - t + D, D = 1 + v_{0} s (t) = (1 + v_{0}) (1 - e^{- t}) - t s_{m a x} = (1 + v_{0}) (1 - \frac{1}{1 + v _{0}}) - ln (1 + v_{0}) = v_{0} - ln (1 + v_{0})

(c)

2 marks

Split the exponent: $e^{- (T - k)} = e^{- T} e^{k}$ , and use $(1 + v_{0}) e^{- T} = 1$ .

v (T - k) = (1 + v_{0}) e^{- T} e^{k} - 1 = e^{k} - 1

(d)

2 marks

Do the same with $T + k$ .

v (T + k) = (1 + v_{0}) e^{- T} e^{- k} - 1 = e^{- k} - 1

(e)

3 marks

Add the two results. The $- 1$ terms combine to $- 2$ , and the rest is a perfect square.

v (T - k) + v (T + k) = e^{k} + e^{- k} - 2 = (e^{k /2} - e^{- k /2})^{2} \geq 0

Mentor tip

A square is never negative, so the sum is at least 0, with equality only at $k = 0$ .

Question 12

19 marks

Transforming arctan, an addition formula, and an induction on a sum

Functions
Geometry and trigonometry
Number and algebra

(a)

3 marks

The graph is $y = arctan (2 x + 1) + \frac{π}{4}$ and starts from $y = arctan x$ . Do the horizontal changes first: a translation 1 unit left, then a horizontal stretch with scale factor $\frac{1}{2}$ (which replaces $x$ by $2 x$ ).
Finish with a vertical translation up by $\frac{π}{4}$ . This moves the asymptotes from $\pm \frac{π}{2}$ to $- \frac{π}{4}$ and $\frac{3 π}{4}$ .

arctan x \to arctan (x + 1) \to arctan (2 x + 1)

Mentor tip

A stretch of factor $\frac{1}{2}$ and a shift of $\frac{1}{2}$ left, in that order, also gives $arctan (2 x + 1)$ .

(b)

4 marks

Let $α = arctan p$ and $β = arctan q$ , so $tan α = p$ and $tan β = q$ . Use the addition formula for tangent.
Since $p, q > 0$ and $pq < 1$ , the angle $α + β$ lies between 0 and $\frac{π}{2}$ , where arctan undoes tan.

tan (α + β) = \frac{p + q}{1 - pq} arctan p + arctan q = arctan (\frac{p + q}{1 - pq})

(c)

3 marks

Use part (b) with $p = 1$ and $q = \frac{x}{x + 1}$ . Here $pq = \frac{x}{x + 1} < 1$ for $x > 0$ , so the condition holds.
Since $arctan 1 = \frac{π}{4}$ , this says exactly the identity asked for.

\frac{1 + \frac{x}{x + 1}}{1 - \frac{x}{x + 1}} = \frac{( x + 1 ) + x}{( x + 1 ) - x} = 2 x + 1 arctan (2 x + 1) = \frac{π}{4} + arctan \frac{x}{x + 1}

(d)

9 marks

For $n = 1$ , the left side is $arctan \frac{1}{2}$ and the right side is $arctan \frac{1}{1 + 1} = arctan \frac{1}{2}$ . So it holds for $n = 1$ .
Assume it holds for $n = k$ :
For $n = k + 1$ , add the next term and use part (b) with $p = \frac{k}{k + 1}$ and $q = \frac{1}{2 ( k + 1 ) ^{2}}$ .
Multiply the top and the bottom by $2 (k + 1)^{3}$ and simplify. The top becomes $(k + 1) (2 k^{2} + 2 k + 1)$ and the bottom becomes $2 (k + 1)^{3} - k = 2 k^{3} + 6 k^{2} + 5 k + 2 = (k + 2) (2 k^{2} + 2 k + 1)$ .
This is the formula for $n = k + 1$ . It is true for $n = 1$ and true for $k + 1$ whenever it is true for $k$ , so by induction it is true for all positive integers $n$ .

r = 1 \sum k arctan \frac{1}{2 r ^{2}} = arctan \frac{k}{k + 1} arctan \frac{k}{k + 1} + arctan \frac{1}{2 ( k + 1 ) ^{2}} = arctan \frac{\frac{k}{k + 1} + \frac{1}{2 ( k + 1 ) ^{2}}}{1 - \frac{k}{k + 1} \cdot \frac{1}{2 ( k + 1 ) ^{2}}} arctan \frac{( k + 1 ) ( 2 k ^{2} + 2 k + 1 )}{( k + 2 ) ( 2 k ^{2} + 2 k + 1 )} = arctan \frac{k + 1}{k + 2}

Mentor tip

Cancelling the common factor $2 k^{2} + 2 k + 1$ is what makes the denominator come out as $k + 2$ . Look for it before expanding.

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%
Q1, Q3, Q5, Q6, Q7, Q9, Q12
Calculus37 marks · 34%
Q4, Q10, Q11
Geometry and trigonometry29 marks · 26%
Q2, Q8, Q10, Q12
Functions3 marks · 3%
Q12

Mentor insights

Traps to watch for

01
Reject the root outside the domain · Q4
In part (b) the quadratic has $x = 8$ and $x = - 2$ . The function is only defined for $x > 4$ , so only 8 is valid.
02
Parallel is not the same as equal · Q8
Show the lines don't meet by a contradiction in their equations, not by saying the directions match. Parallel lines can be the same line.
03
Total surface area includes the base · Q10
The cone's surface area is the curved part $π r l$ plus the circular base $π r^{2}$ . Using only $π r l$ gives the wrong slant height.
04
Use the result given · Q11
Parts (c) to (e) rely on the formula for $v$ from part (a) and on $e^{T} = 1 + v_{0}$ from part (b). Each is a short step if you substitute the earlier result instead of starting again.
05
The induction must name the assumption · Q12
In part (d), state the assumption for $n = k$ and say where you use it. The final line must also say it is true for $n = 1$ and that truth passes from $k$ to $k + 1$ .

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Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2021
Marks: 110
Duration: 2 hours
Calculator: Not 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

4 marks

Difference of consecutive squares equals their sum

Number and algebra

4 marks

Subtract the squares in the right order, expanding $(n + 1)^{2}$ .
The sum of the two integers is $n + (n + 1) = 2 n + 1$ , the same expression.

(n + 1)^{2} - n^{2} = n^{2} + 2 n + 1 - n^{2} = 2 n + 1

Mentor tip

Factorising works too: $(n + 1)^{2} - n^{2} = ((n + 1) - n) ((n + 1) + n) = 1 \cdot (2 n + 1)$ .

Question 2

7 marks

A trig equation turned into a quadratic in sine

Geometry and trigonometry

7 marks

Replace $cos^{2} x$ using $cos^{2} x = 1 - sin^{2} x$ , and collect everything on one side.
Factorise as a quadratic in $sin x$ .
$sin x = 2$ is impossible. That leaves $sin x = \frac{1}{2}$ in $0 \leq x \leq 2 π$ .

2 (1 - sin^{2} x) + 5 sin x = 4 \Rightarrow 2 sin^{2} x - 5 sin x + 2 = 0 (2 sin x - 1) (sin x - 2) = 0 x = \frac{π}{6}, \frac{5 π}{6}

Mentor tip

Say why the second factor is rejected: a sine can't be 2.

Question 3

5 marks

Finding a constant from one binomial coefficient

Number and algebra

5 marks

In $(x + k)^{7}$ the term in $x^{5}$ comes from choosing $x$ five times and $k$ twice.
Set the coefficient equal to 63.

(2 7) x^{5} k^{2} = 21 k^{2} x^{5} 21 k^{2} = 63 \Rightarrow k^{2} = 3 \Rightarrow k = \pm 3

Question 4

9 marks

A logarithmic curve: where it crosses and where it is steep

Calculus

(a)

3 marks

The graph meets the $x$ -axis where $f (x) = 0$ , that is, where the log's argument is 1.
Since $x > 4$ , only the positive root is allowed, so $a = 17$ .

ln (x^{2} - 16) = 0 \Rightarrow x^{2} - 16 = 1 \Rightarrow x = 17

(b)

6 marks

Differentiate with the chain rule, then set the gradient equal to $\frac{1}{3}$ .
Cross-multiply and solve the quadratic.

f^{'} (x) = \frac{2 x}{x ^{2} - 16} = \frac{1}{3} 6 x = x^{2} - 16 \Rightarrow x^{2} - 6 x - 16 = 0 \Rightarrow (x - 8) (x + 2) = 0 x = 8 (since x > 4)

Mentor tip

The domain check is the last line of the answer. Do it every time.

Question 5

4 marks

The magnitude of a cross product, squared

Number and algebra

4 marks

Use $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ$ and square it.
Distribute, and replace $∣ a ∣^{2} ∣ b ∣^{2} cos^{2} θ$ with $(a \cdot b)^{2}$ using $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ .

∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} sin^{2} θ = ∣ a ∣^{2} ∣ b ∣^{2} (1 - cos^{2} θ) ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}

Question 6

7 marks

Area of a triangle with a ratio of sides and a known cosine

Number and algebra

7 marks

Apply the cosine rule at $C$ , opposite the side $A B = 10$ , with $B C = x$ and $A C = 2 x$ .
From $cos C = \frac{3}{4}$ , $sin C = 1 - \frac{9}{16} = \frac{7}{4}$ .
The area is $\frac{1}{2} \cdot B C \cdot A C \cdot sin C = \frac{1}{2} \cdot x \cdot 2 x \cdot sin C = x^{2} sin C$ .

100 = x^{2} + 4 x^{2} - 2 (x) (2 x) \cdot \frac{3}{4} = 5 x^{2} - 3 x^{2} = 2 x^{2} x^{2} = 50 Area = 50 \cdot \frac{7}{4} = \frac{25}{2} 7

Mentor tip

Check the form asked for: a rational number times a square root, $p q$ with $p = \frac{25}{2}$ and $q = 7$ .

Question 7

5 marks

Roots of a cubic given in terms of each other

Number and algebra

5 marks

Let $s = α + β$ . The sum of the three roots is $α + β + (α + β) = 2 s$ , and it equals $k$ .
The $x$ coefficient is 0, so the sum of the products in pairs is 0: $α β + α s + β s = α β + s^{2} = 0$ . So $α β = - s^{2}$ .
The product of the roots is $- 3 k$ (minus the constant term), which is $α β s$ .
Divide by $- k$ (and $k > 0$ ).

2 s = k \Rightarrow s = \frac{k}{2} α β s = - s^{3} = - \frac{k ^{3}}{8} = - 3 k k^{2} = 24 \Rightarrow k = 26

Question 8

8 marks

Parallel lines in space: no intersection, then the gap

Geometry and trigonometry

(a)

3 marks

Write the lines as $r = (3, 2, - 1) + λ (2, - 2, 2)$ and $r = (2, 0, 4) + μ (1, - 1, 1)$ , and equate the components.
Add the first two equations. The $λ$ and $μ$ cancel and leave $5 = 2$ , which is false. So there is no solution, and the lines do not intersect.

3 + 2 λ = 2 + μ, 2 - 2 λ = - μ, - 1 + 2 λ = 4 + μ

Mentor tip

The direction vectors are multiples of each other, so the lines are parallel. The contradiction shows they are distinct.

(b)

5 marks

The lines are parallel, so the distance is the same everywhere: take $A (3, 2, - 1)$ on the first line and $B (2, 0, 4)$ on the second, and use the direction $v = (1, - 1, 1)$ .
The shortest distance is $\frac{∣ A B \times v ∣}{∣ v ∣}$ .

A B = - 1 - 2 5 A B \times v = 363, ∣ A B \times v ∣ = 54 d = \frac{54}{3} = 18 = 32

Mentor tip

Check with calculus: the squared distance from $A$ to the general point $B (2 + μ, - μ, 4 + μ)$ is $3 μ^{2} + 12 μ + 30$ , minimised at $μ = - 2$ with value 18.

Question 9

7 marks

Integrating with the substitution $u = sin x$

Number and algebra

7 marks

Substituting $u = sin x$ and $d u = cos x d x$ turns the integral into a rational one.
Split into partial fractions. Putting $u = 2$ gives $B = \frac{2}{3}$ and putting $u = - 1$ gives $A = \frac{1}{3}$ .
Integrate, then put $u = sin x$ back.

\int \frac{u}{u ^{2} - u - 2} d u = \int \frac{u}{( u + 1 ) ( u - 2 )} d u \frac{u}{( u + 1 ) ( u - 2 )} = \frac{1/3}{u + 1} + \frac{2/3}{u - 2} \frac{1}{3} ln ∣ sin x + 1∣ + \frac{2}{3} ln ∣ sin x - 2∣ + C

Mentor tip

Don't leave the answer in $u$ . The question was about $x$ .

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

15 marks

Area under a cosine curve, then a cone with the same surface area

Calculus
Geometry and trigonometry

(a)

3 marks

The graph touches the $x$ -axis where $f (x) = 0$ , that is, where $cos x = - 1$ in $0 \leq x \leq 4 π$ .

6 + 6 cos x = 0 \Rightarrow cos x = - 1 \Rightarrow x = π, 3 π

(b)

5 marks

The area is the integral between those two points.

\int_{π}^{3 π} (6 + 6 cos x) d x = [6 x + 6 sin x]_{π}^{3 π} (18 π + 0) - (6 π + 0) = 12 π

Mentor tip

Both sine terms vanish at multiples of $π$ , so write that rather than skipping it.

(c)

3 marks

The total surface area is the base plus the curved surface, $π r^{2} + π r l$ , with $r = 2$ . Set it equal to $12 π$ .

4 π + 2 π l = 12 π \Rightarrow l = 4

(d)

4 marks

Find the height from $r^{2} + h^{2} = l^{2}$ , then use $V = \frac{1}{3} π r^{2} h$ .

h = 16 - 4 = 23 V = \frac{1}{3} π (2^{2}) (23) = \frac{8 3 π}{3}

Question 11

20 marks

A particle whose acceleration depends on its velocity

Calculus

(a)

6 marks

Write the acceleration as $\frac{d v}{d t} = - (1 + v)$ and separate the variables.
At $t = 0$ , $v = v_{0}$ , so $C = ln (1 + v_{0})$ . Rearrange for $v$ . (No modulus is needed, since $v > - 1$ .)

\int \frac{1}{1 + v} d v = - \int 1 d t \Rightarrow ln (1 + v) = - t + C 1 + v = (1 + v_{0}) e^{- t} \Rightarrow v (t) = (1 + v_{0}) e^{- t} - 1

(b)

7 marks

At the farthest point the particle is momentarily at rest, so $v (T) = 0$ .
Integrate $v$ to get the displacement, using $s = 0$ at $t = 0$ to find the constant.
Put in $t = T = ln (1 + v_{0})$ and use $e^{- T} = \frac{1}{1 + v _{0}}$ .

(1 + v_{0}) e^{- T} = 1 \Rightarrow e^{T} = 1 + v_{0} s (t) = - (1 + v_{0}) e^{- t} - t + D, D = 1 + v_{0} s (t) = (1 + v_{0}) (1 - e^{- t}) - t s_{m a x} = (1 + v_{0}) (1 - \frac{1}{1 + v _{0}}) - ln (1 + v_{0}) = v_{0} - ln (1 + v_{0})

(c)

2 marks

Split the exponent: $e^{- (T - k)} = e^{- T} e^{k}$ , and use $(1 + v_{0}) e^{- T} = 1$ .

v (T - k) = (1 + v_{0}) e^{- T} e^{k} - 1 = e^{k} - 1

(d)

2 marks

Do the same with $T + k$ .

v (T + k) = (1 + v_{0}) e^{- T} e^{- k} - 1 = e^{- k} - 1

(e)

3 marks

Add the two results. The $- 1$ terms combine to $- 2$ , and the rest is a perfect square.

v (T - k) + v (T + k) = e^{k} + e^{- k} - 2 = (e^{k /2} - e^{- k /2})^{2} \geq 0

Mentor tip

A square is never negative, so the sum is at least 0, with equality only at $k = 0$ .

Question 12

19 marks

Transforming arctan, an addition formula, and an induction on a sum

Functions
Geometry and trigonometry
Number and algebra

(a)

3 marks

The graph is $y = arctan (2 x + 1) + \frac{π}{4}$ and starts from $y = arctan x$ . Do the horizontal changes first: a translation 1 unit left, then a horizontal stretch with scale factor $\frac{1}{2}$ (which replaces $x$ by $2 x$ ).
Finish with a vertical translation up by $\frac{π}{4}$ . This moves the asymptotes from $\pm \frac{π}{2}$ to $- \frac{π}{4}$ and $\frac{3 π}{4}$ .

arctan x \to arctan (x + 1) \to arctan (2 x + 1)

Mentor tip

A stretch of factor $\frac{1}{2}$ and a shift of $\frac{1}{2}$ left, in that order, also gives $arctan (2 x + 1)$ .

(b)

4 marks

Let $α = arctan p$ and $β = arctan q$ , so $tan α = p$ and $tan β = q$ . Use the addition formula for tangent.
Since $p, q > 0$ and $pq < 1$ , the angle $α + β$ lies between 0 and $\frac{π}{2}$ , where arctan undoes tan.

tan (α + β) = \frac{p + q}{1 - pq} arctan p + arctan q = arctan (\frac{p + q}{1 - pq})

(c)

3 marks

Use part (b) with $p = 1$ and $q = \frac{x}{x + 1}$ . Here $pq = \frac{x}{x + 1} < 1$ for $x > 0$ , so the condition holds.
Since $arctan 1 = \frac{π}{4}$ , this says exactly the identity asked for.

\frac{1 + \frac{x}{x + 1}}{1 - \frac{x}{x + 1}} = \frac{( x + 1 ) + x}{( x + 1 ) - x} = 2 x + 1 arctan (2 x + 1) = \frac{π}{4} + arctan \frac{x}{x + 1}

(d)

9 marks

For $n = 1$ , the left side is $arctan \frac{1}{2}$ and the right side is $arctan \frac{1}{1 + 1} = arctan \frac{1}{2}$ . So it holds for $n = 1$ .
Assume it holds for $n = k$ :
For $n = k + 1$ , add the next term and use part (b) with $p = \frac{k}{k + 1}$ and $q = \frac{1}{2 ( k + 1 ) ^{2}}$ .
Multiply the top and the bottom by $2 (k + 1)^{3}$ and simplify. The top becomes $(k + 1) (2 k^{2} + 2 k + 1)$ and the bottom becomes $2 (k + 1)^{3} - k = 2 k^{3} + 6 k^{2} + 5 k + 2 = (k + 2) (2 k^{2} + 2 k + 1)$ .
This is the formula for $n = k + 1$ . It is true for $n = 1$ and true for $k + 1$ whenever it is true for $k$ , so by induction it is true for all positive integers $n$ .

r = 1 \sum k arctan \frac{1}{2 r ^{2}} = arctan \frac{k}{k + 1} arctan \frac{k}{k + 1} + arctan \frac{1}{2 ( k + 1 ) ^{2}} = arctan \frac{\frac{k}{k + 1} + \frac{1}{2 ( k + 1 ) ^{2}}}{1 - \frac{k}{k + 1} \cdot \frac{1}{2 ( k + 1 ) ^{2}}} arctan \frac{( k + 1 ) ( 2 k ^{2} + 2 k + 1 )}{( k + 2 ) ( 2 k ^{2} + 2 k + 1 )} = arctan \frac{k + 1}{k + 2}

Mentor tip

Cancelling the common factor $2 k^{2} + 2 k + 1$ is what makes the denominator come out as $k + 2$ . Look for it before expanding.

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%
Q1, Q3, Q5, Q6, Q7, Q9, Q12
Calculus37 marks · 34%
Q4, Q10, Q11
Geometry and trigonometry29 marks · 26%
Q2, Q8, Q10, Q12
Functions3 marks · 3%
Q12

Mentor insights

Traps to watch for

01
Reject the root outside the domain · Q4
In part (b) the quadratic has $x = 8$ and $x = - 2$ . The function is only defined for $x > 4$ , so only 8 is valid.
02
Parallel is not the same as equal · Q8
Show the lines don't meet by a contradiction in their equations, not by saying the directions match. Parallel lines can be the same line.
03
Total surface area includes the base · Q10
The cone's surface area is the curved part $π r l$ plus the circular base $π r^{2}$ . Using only $π r l$ gives the wrong slant height.
04
Use the result given · Q11
Parts (c) to (e) rely on the formula for $v$ from part (a) and on $e^{T} = 1 + v_{0}$ from part (b). Each is a short step if you substitute the earlier result instead of starting again.
05
The induction must name the assumption · Q12
In part (d), state the assumption for $n = k$ and say where you use it. The final line must also say it is true for $n = 1$ and that truth passes from $k$ to $k + 1$ .

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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