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IB Maths AA HL · Paper 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

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Questions

Questions 1–2 · 55 marks

Question 1

28 marks

The family of curves $y^{2} = x^{3} + a x + b$ : shapes, inflexion and rational points

Functions
Calculus

(a)(i)

2 marks

The curve $y^{2} = x^{3}$ for $x \geq 0$ has a cusp at the origin and is symmetric in the $x$ -axis. It passes through $(0, 0)$ and $(1, \pm 1)$ , and meets $y = \pm 2$ at $x = 4^{1/3} \approx 1.59$ .

(a)(ii)

2 marks

The curve $y^{2} = x^{3} + 1$ for $x \geq - 1$ is symmetric in the $x$ -axis. It meets the $x$ -axis at $(- 1, 0)$ with a vertical tangent, and the $y$ -axis at $(0, \pm 1)$ . It is smooth everywhere.

(b)(i)

1 mark

The points of inflexion are where the curve changes concavity, on the $y$ -axis.

(0, 1) and (0, - 1)

(b)(ii)

1 mark

Two features: $y^{2} = x^{3}$ has a cusp at the origin whereas $y^{2} = x^{3} + 1$ is smooth, and $y^{2} = x^{3}$ meets the axes only at the origin whereas $y^{2} = x^{3} + 1$ has an $x$ -intercept at $(- 1, 0)$ and $y$ -intercepts at $(0, \pm 1)$ .

(c)

2 marks

Two features common to every curve $y^{2} = x^{3} + b$ with $b > 0$ : each is symmetric in the $x$ -axis, and each has an $x$ -intercept at $x = - 3 b$ with a vertical tangent there (no cusp).

(d)(i)

3 marks

Write $y = \pm (x^{3} + x)^{1/2}$ and use the chain rule.

\frac{d y}{d x} = \pm \frac{1}{2} (x^{3} + x)^{- 1/2} (3 x^{2} + 1) = \pm \frac{3 x ^{2} + 1}{2 x ^{3} + x}

(d)(ii)

1 mark

A turning point needs $\frac{d y}{d x} = 0$ . Since $3 x^{2} + 1 > 0$ for all $x$ , the derivative is never zero, so there are no local maxima or minima.

(e)

7 marks

Differentiate $2 y \frac{d y}{d x} = 3 x^{2} + 1$ implicitly. Using the product rule on the left:
At a point of inflexion $\frac{d ^{2} y}{d x ^{2}} = 0$ , so $(\frac{d y}{d x})^{2} = 3 x$ . Put in $\frac{d y}{d x} = \frac{3 x ^{2} + 1}{2 y}$ and $y^{2} = x^{3} + x$ .
Expand and simplify to a quadratic in $x^{2}$ .
Since $x^{2} > 0$ , take the positive root.

2 (\frac{d y}{d x})^{2} + 2 y \frac{d ^{2} y}{d x ^{2}} = 6 x \frac{( 3 x ^{2} + 1 ) ^{2}}{4 ( x ^{3} + x )} = 3 x \Rightarrow (3 x^{2} + 1)^{2} = 12 x (x^{3} + x) 12 x^{4} + 12 x^{2} = 9 x^{4} + 6 x^{2} + 1 \Rightarrow 3 x^{4} + 6 x^{2} - 1 = 0 x^{2} = \frac{- 6 \pm 36 + 12}{6} = \frac{- 3 \pm 2 3}{3} x = \frac{- 3 + 2 3}{3} (p = - 3, q = 2, r = 3) \approx 0.393

(f)(i)

2 marks

At $P (- 1, - 1)$ the gradient is $\frac{3 ( - 1 ) ^{2} + 0}{2 ( - 1 )} = - \frac{3}{2}$ .

y + 1 = - \frac{3}{2} (x + 1) \Rightarrow y = - \frac{3}{2} x - \frac{5}{2}

(f)(ii)

2 marks

Substitute the tangent into $y^{2} = x^{3} + 2$ .
The tangent gives a double root at $x = - 1$ , so factorise: $(x + 1)^{2} (x - \frac{17}{4}) = 0$ . The new intersection has $x = \frac{17}{4}$ .

(- \frac{3}{2} x - \frac{5}{2})^{2} = x^{3} + 2 \Rightarrow x^{3} - \frac{9}{4} x^{2} - \frac{15}{2} x - \frac{17}{4} = 0 y = - \frac{3}{2} \cdot \frac{17}{4} - \frac{5}{2} = - \frac{71}{8}, Q (\frac{17}{4}, - \frac{71}{8})

(g)

5 marks

The line QS has gradient $\frac{1 - ( - 71/8 )}{- 1 - 17/4} = \frac{79/8}{- 21/4} = - \frac{79}{42}$ . Its equation is $y = 1 - \frac{79}{42} (x + 1)$ .
Substituting $y = m x + c$ into $y^{2} = x^{3} + 2$ gives a cubic in $x$ whose roots add up to $m^{2}$ . Two roots are $\frac{17}{4}$ (Q) and $- 1$ (S).
Then $y_{3} = 1 - \frac{79}{42} (\frac{127}{441} + 1) = - \frac{13175}{9261}$ .

x_{3} = m^{2} - \frac{17}{4} + 1 = (\frac{79}{42})^{2} - \frac{13}{4} = \frac{6241}{1764} - \frac{5733}{1764} = \frac{127}{441} (\frac{127}{441}, - \frac{13175}{9261}) \approx (0.288, - 1.42)

Question 2

27 marks

When do polynomial equations have complex roots? Cubics, quartics and conditions on the coefficients

Number and algebra

(a)

3 marks

Expand $(x - α) (x - β) (x - γ)$ .
Compare with $x^{3} + p x^{2} + q x + r$ .

x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ p = - (α + β + γ), q = α β + β γ + γ α, r = - α β γ

(b)(i)

3 marks

Square $α + β + γ$ and use $q = α β + β γ + γ α$ .

p^{2} = (α + β + γ)^{2} = α^{2} + β^{2} + γ^{2} + 2 (α β + β γ + γ α) p^{2} - 2 q = α^{2} + β^{2} + γ^{2}

(b)(ii)

3 marks

Expand the three squares and collect.
Replace the two brackets using part (b)(i) and the value of $q$ .

(α - β)^{2} + (β - γ)^{2} + (γ - α)^{2} = 2 (α^{2} + β^{2} + γ^{2}) - 2 (α β + β γ + γ α) 2 (p^{2} - 2 q) - 2 q = 2 p^{2} - 6 q

(c)

2 marks

If $p^{2} < 3 q$ then $2 p^{2} - 6 q < 0$ , so by part (b)(ii) the sum $(α - β)^{2} + (β - γ)^{2} + (γ - α)^{2}$ is negative. If $α, β, γ$ were all real, this sum of squares would be at least 0. That is a contradiction, so the roots can't all be real.

(d)

2 marks

For $x^{3} - 7 x^{2} + 17 x + 1 = 0$ : $p = - 7$ and $q = 17$ .
By part (c) at least one root is complex (not real).

p^{2} = 49 < 3 q = 51

(e)(i)

2 marks

The condition is $p^{2} \geq 3 q$ , that is $49 \geq 3 q$ , so $q \leq 16$ . Try $q = 1, 2, 3, \dots$ on the calculator (graph or roots) and find the first for which the cubic has complex roots.
So Noah is wrong, and the smallest positive integer that shows it is

q = 1, \dots, 11 : three real roots, q = 12 : one real and two complex roots q = 12

(e)(ii)

1 mark

Complex roots of a real polynomial come in conjugate pairs, so a cubic with a complex pair has a third root that is real. Equivalently, a cubic curve always crosses the $x$ -axis.

(f)(i)

3 marks

Square $α + β + γ + δ$ .

p^{2} = (α + β + γ + δ)^{2} = α^{2} + β^{2} + γ^{2} + δ^{2} + 2 q α^{2} + β^{2} + γ^{2} + δ^{2} = p^{2} - 2 q

(f)(ii)

1 mark

If $p^{2} < 2 q$ , then the sum of the squares of the four roots is negative, which is impossible if all four are real.

p^{2} < 2 q

(g)

1 mark

For $x^{4} - 2 x^{3} + 3 x^{2} - 4 x + 5 = 0$ : $p = - 2$ and $q = 3$ , so $p^{2} = 4 < 6 = 2 q$ . By part (f)(ii) there is at least one complex root.

(h)(i)

1 mark

Here $p = - 9$ and $q = 24$ , so $p^{2} = 81$ and $2 q = 48$ . Since $81 > 48$ , the condition $p^{2} < 2 q$ is not satisfied, so the result tells us nothing about whether there is a complex root.

(h)(ii)

1 mark

Try $x = - 1$ .

1 + 9 + 24 - 22 - 12 = 0 \Rightarrow x = - 1

(h)(iii)

4 marks

Divide by $(x + 1)$ .
For the cubic factor, $p = - 10$ and $q = 34$ , so $p^{2} = 100 < 102 = 3 q$ . By part (c) the cubic has at least one complex root, and so does the original equation.

x^{4} - 9 x^{3} + 24 x^{2} + 22 x - 12 = (x + 1) (x^{3} - 10 x^{2} + 34 x - 12)

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 algebra27 marks · 49%
Q2
Calculus21 marks · 38%
Q1
Functions7 marks · 13%
Q1

Mentor insights

Traps to watch for

01
Symmetry in the $x$ -axis · Q1
Every curve $y^{2} = f (x)$ is symmetric in the $x$ -axis, since $y$ and $- y$ give the same equation.
02
Rational points from a line · Q1
The line through two points on the curve meets the cubic in a third point. Use the sum of the roots of the cubic in $x$ to find its coordinate.
03
Implicit differentiation · Q1
For $y^{2} = x^{3} + a x + b$ the gradient is $\frac{3 x ^{2} + a}{2 y}$ . Use it for tangents at rational points.
04
Squares of real numbers are non-negative · Q2
The argument in (c) and (f) is always that a sum of squares of real numbers can't be negative. If the expression you've shown is negative, some root can't be real.
05
Be careful about the converse · Q2
In (e) the condition $p^{2} \geq 3 q$ does not guarantee three real roots. A single counterexample, a value of $q$ that satisfies the inequality but gives complex roots, shows Noah is wrong.

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 3

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ2 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

28 marks

The family of curves $y^{2} = x^{3} + a x + b$ : shapes, inflexion and rational points

Functions
Calculus

(a)(i)

2 marks

(a)(ii)

2 marks

(b)(i)

1 mark

The points of inflexion are where the curve changes concavity, on the $y$ -axis.

(0, 1) and (0, - 1)

(b)(ii)

1 mark

(c)

2 marks

Two features common to every curve $y^{2} = x^{3} + b$ with $b > 0$ : each is symmetric in the $x$ -axis, and each has an $x$ -intercept at $x = - 3 b$ with a vertical tangent there (no cusp).

(d)(i)

3 marks

Write $y = \pm (x^{3} + x)^{1/2}$ and use the chain rule.

\frac{d y}{d x} = \pm \frac{1}{2} (x^{3} + x)^{- 1/2} (3 x^{2} + 1) = \pm \frac{3 x ^{2} + 1}{2 x ^{3} + x}

(d)(ii)

1 mark

A turning point needs $\frac{d y}{d x} = 0$ . Since $3 x^{2} + 1 > 0$ for all $x$ , the derivative is never zero, so there are no local maxima or minima.

(e)

7 marks

Differentiate $2 y \frac{d y}{d x} = 3 x^{2} + 1$ implicitly. Using the product rule on the left:
At a point of inflexion $\frac{d ^{2} y}{d x ^{2}} = 0$ , so $(\frac{d y}{d x})^{2} = 3 x$ . Put in $\frac{d y}{d x} = \frac{3 x ^{2} + 1}{2 y}$ and $y^{2} = x^{3} + x$ .
Expand and simplify to a quadratic in $x^{2}$ .
Since $x^{2} > 0$ , take the positive root.

2 (\frac{d y}{d x})^{2} + 2 y \frac{d ^{2} y}{d x ^{2}} = 6 x \frac{( 3 x ^{2} + 1 ) ^{2}}{4 ( x ^{3} + x )} = 3 x \Rightarrow (3 x^{2} + 1)^{2} = 12 x (x^{3} + x) 12 x^{4} + 12 x^{2} = 9 x^{4} + 6 x^{2} + 1 \Rightarrow 3 x^{4} + 6 x^{2} - 1 = 0 x^{2} = \frac{- 6 \pm 36 + 12}{6} = \frac{- 3 \pm 2 3}{3} x = \frac{- 3 + 2 3}{3} (p = - 3, q = 2, r = 3) \approx 0.393

(f)(i)

2 marks

At $P (- 1, - 1)$ the gradient is $\frac{3 ( - 1 ) ^{2} + 0}{2 ( - 1 )} = - \frac{3}{2}$ .

y + 1 = - \frac{3}{2} (x + 1) \Rightarrow y = - \frac{3}{2} x - \frac{5}{2}

(f)(ii)

2 marks

Substitute the tangent into $y^{2} = x^{3} + 2$ .
The tangent gives a double root at $x = - 1$ , so factorise: $(x + 1)^{2} (x - \frac{17}{4}) = 0$ . The new intersection has $x = \frac{17}{4}$ .

(- \frac{3}{2} x - \frac{5}{2})^{2} = x^{3} + 2 \Rightarrow x^{3} - \frac{9}{4} x^{2} - \frac{15}{2} x - \frac{17}{4} = 0 y = - \frac{3}{2} \cdot \frac{17}{4} - \frac{5}{2} = - \frac{71}{8}, Q (\frac{17}{4}, - \frac{71}{8})

(g)

5 marks

The line QS has gradient $\frac{1 - ( - 71/8 )}{- 1 - 17/4} = \frac{79/8}{- 21/4} = - \frac{79}{42}$ . Its equation is $y = 1 - \frac{79}{42} (x + 1)$ .
Substituting $y = m x + c$ into $y^{2} = x^{3} + 2$ gives a cubic in $x$ whose roots add up to $m^{2}$ . Two roots are $\frac{17}{4}$ (Q) and $- 1$ (S).
Then $y_{3} = 1 - \frac{79}{42} (\frac{127}{441} + 1) = - \frac{13175}{9261}$ .

x_{3} = m^{2} - \frac{17}{4} + 1 = (\frac{79}{42})^{2} - \frac{13}{4} = \frac{6241}{1764} - \frac{5733}{1764} = \frac{127}{441} (\frac{127}{441}, - \frac{13175}{9261}) \approx (0.288, - 1.42)

Question 2

27 marks

When do polynomial equations have complex roots? Cubics, quartics and conditions on the coefficients

Number and algebra

(a)

3 marks

Expand $(x - α) (x - β) (x - γ)$ .
Compare with $x^{3} + p x^{2} + q x + r$ .

x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ p = - (α + β + γ), q = α β + β γ + γ α, r = - α β γ

(b)(i)

3 marks

Square $α + β + γ$ and use $q = α β + β γ + γ α$ .

p^{2} = (α + β + γ)^{2} = α^{2} + β^{2} + γ^{2} + 2 (α β + β γ + γ α) p^{2} - 2 q = α^{2} + β^{2} + γ^{2}

(b)(ii)

3 marks

Expand the three squares and collect.
Replace the two brackets using part (b)(i) and the value of $q$ .

(α - β)^{2} + (β - γ)^{2} + (γ - α)^{2} = 2 (α^{2} + β^{2} + γ^{2}) - 2 (α β + β γ + γ α) 2 (p^{2} - 2 q) - 2 q = 2 p^{2} - 6 q

(c)

2 marks

(d)

2 marks

For $x^{3} - 7 x^{2} + 17 x + 1 = 0$ : $p = - 7$ and $q = 17$ .
By part (c) at least one root is complex (not real).

p^{2} = 49 < 3 q = 51

(e)(i)

2 marks

The condition is $p^{2} \geq 3 q$ , that is $49 \geq 3 q$ , so $q \leq 16$ . Try $q = 1, 2, 3, \dots$ on the calculator (graph or roots) and find the first for which the cubic has complex roots.
So Noah is wrong, and the smallest positive integer that shows it is

q = 1, \dots, 11 : three real roots, q = 12 : one real and two complex roots q = 12

(e)(ii)

1 mark

Complex roots of a real polynomial come in conjugate pairs, so a cubic with a complex pair has a third root that is real. Equivalently, a cubic curve always crosses the $x$ -axis.

(f)(i)

3 marks

Square $α + β + γ + δ$ .

p^{2} = (α + β + γ + δ)^{2} = α^{2} + β^{2} + γ^{2} + δ^{2} + 2 q α^{2} + β^{2} + γ^{2} + δ^{2} = p^{2} - 2 q

(f)(ii)

1 mark

If $p^{2} < 2 q$ , then the sum of the squares of the four roots is negative, which is impossible if all four are real.

p^{2} < 2 q

(g)

1 mark

For $x^{4} - 2 x^{3} + 3 x^{2} - 4 x + 5 = 0$ : $p = - 2$ and $q = 3$ , so $p^{2} = 4 < 6 = 2 q$ . By part (f)(ii) there is at least one complex root.

(h)(i)

1 mark

Here $p = - 9$ and $q = 24$ , so $p^{2} = 81$ and $2 q = 48$ . Since $81 > 48$ , the condition $p^{2} < 2 q$ is not satisfied, so the result tells us nothing about whether there is a complex root.

(h)(ii)

1 mark

Try $x = - 1$ .

1 + 9 + 24 - 22 - 12 = 0 \Rightarrow x = - 1

(h)(iii)

4 marks

Divide by $(x + 1)$ .
For the cubic factor, $p = - 10$ and $q = 34$ , so $p^{2} = 100 < 102 = 3 q$ . By part (c) the cubic has at least one complex root, and so does the original equation.

x^{4} - 9 x^{3} + 24 x^{2} + 22 x - 12 = (x + 1) (x^{3} - 10 x^{2} + 34 x - 12)

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 algebra27 marks · 49%
Q2
Calculus21 marks · 38%
Q1
Functions7 marks · 13%
Q1

Mentor insights

Traps to watch for

01
Symmetry in the $x$ -axis · Q1
Every curve $y^{2} = f (x)$ is symmetric in the $x$ -axis, since $y$ and $- y$ give the same equation.
02
Rational points from a line · Q1
The line through two points on the curve meets the cubic in a third point. Use the sum of the roots of the cubic in $x$ to find its coordinate.
03
Implicit differentiation · Q1
For $y^{2} = x^{3} + a x + b$ the gradient is $\frac{3 x ^{2} + a}{2 y}$ . Use it for tangents at rational points.
04
Squares of real numbers are non-negative · Q2
The argument in (c) and (f) is always that a sum of squares of real numbers can't be negative. If the expression you've shown is negative, some root can't be real.
05
Be careful about the converse · Q2
In (e) the condition $p^{2} \geq 3 q$ does not guarantee three real roots. A single counterexample, a value of $q$ that satisfies the inequality but gives complex roots, shows Noah is wrong.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers