IB Maths AA HL 2 Functions

Practise IB Maths AA HL functions through shared-core and HL graphs, transformations, inverses, equations and advanced models with exact reasoning.

Covered syllabus points

2.1 Functions - SL content 2.2 Functions - AHL content

[Maximum number: 15]

This question investigates a ratio of lengths found from the line passing through the points of inflexion of a quartic curve of the form $y=x^{4}-m x^{3}+n x$ .
The curve $y=x^{4}-3 x^{3}+3 x$ has points of inflexion at B and C . The line passing through B and C intersects the curve again at points A and D . This is shown in the following graph.

Figure for Question 2 Functions question 1 — IB Maths AA HL

(a)

Show that the equation of the line through B and C is y=-0.375 x.

[ 2 ]

$\quad y$ -intercept =0 (as line passes through (0,0) )

Note: Award R1 for correctly substituting point B or point C to show that y=0.

Award R1 for y=-0.375 x+0.

Note: Award AOFT if their answer to (b) doesn't lead to the given answer, but condone $\frac{-0.563}{1.5}$ .

so equation is y=-0.375 x
AG
Note: Award at most A1 R0 for working backwards to verify points B and C lie on the given line.

(b)

Find, correct to three decimal places, the x-coordinate of D .

Now consider the general curve $y=x^{4}-m x^{3}+n x$ , where $m, n \in \mathbb{R}$ and m>0.

[ 2 ]

$x^{4}-3 x^{3}+3 x=-0.375 x$
2.427

Note: Correct answer must be given to three decimal places for the A1 to be awarded.

Award (M1)A1 for (2.427,-0.910).
Award (M1)A0 for an unsupported answer of 2.43.

(c)

Show that the equation of the line through B and C is $y=\left(-\frac{m^{3}}{8}+n\right) x$ .

[ 2 ]

attempt to divide their $y_{C}$ by their $x_{C}$ OR substitute into $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

\begin{aligned} & \text { e.g. }\left(-\frac{m^{4}}{16}+\frac{n m}{2}\right) \div \frac{m}{2} \\ & \left(-\frac{m^{4}}{16}+\frac{n m}{2}\right) \times \frac{2}{m} \text { OR equivalent manipulation leading to given answer } \end{aligned}

Note: Award AO FT if their answer does not lead to the AG.

=-\frac{m^{3}}{8}+n \text { and } y \text {-intercept }=0

\text { so equation is } y=\left(-\frac{m^{3}}{8}+n\right) x

Note: Award at most M1A0 for working backwards to verify points B and C lie on the given line.

(d)

Show that $x_{A}=\frac{m}{4}-\frac{m}{4} \sqrt{5}$ .

[ 7 ]

(h) METHOD 1

x^{4}-m x^{3}+n x=-\frac{m^{3}}{8} x+n x

attempt to rearrange this equation to equal zero

x^{4}-m x^{3}+\frac{m^{3}}{8} x=0

recognizing their $\left(x-x_{C}\right)$ is a factor of this expression

x\left(x-\frac{m}{2}\right)\left(x^{2}-\frac{m}{2} x-\frac{m^{2}}{4}\right)(=0) \quad \text { OR equivalent }

A1M1A1

Note: Award A1 for $x\left(x-\frac{m}{2}\right)$ , M1A1 for $\left(x^{2}-\frac{m}{2} x-\frac{m^{2}}{4}\right)$ .
If a candidate divides by x without justification, do not award the first A1, but all subsequent marks can still be awarded.
use of quadratic formula to find roots, $X_{A}$ (and $X_{D}$ ), of the quadratic M1

x_{A}=\frac{\frac{m}{2}-\sqrt{\frac{m^{2}}{4}+m^{2}}}{2} \quad \text { OR equivalent }

Note: Condone a ± in place of the minus sign, provided given answer is restated.

x_{A}=\frac{m}{4}-\frac{m}{4} \sqrt{5}

Note: Award (M1)(M0)A0M0A0M0A0 for attempting to verify the given answer satisfies

x^{4}-m x^{3}+n x=-\frac{m^{3}}{8} x+n x

METHOD 2

x^{4}-m x^{3}+n x=-\frac{m^{3}}{8} x+n x

attempt to rearrange this equation to equal zero

x^{4}-m x^{3}+\frac{m^{3}}{8} x=0

x\left(x^{3}-m x^{2}+\frac{m^{3}}{8}\right)=0

Recognise that their $x_{C}$ is a root of this equation

Attempt to find sum and product of roots

x_{A}+x_{D}+\frac{m}{2}=m \quad \text { AND } \quad x_{A} \times x_{D} \times \frac{m}{2}=-\frac{m^{3}}{8}

x_{A} \times\left(\frac{m}{2}-x_{A}\right) \times \frac{m}{2}=-\frac{m^{3}}{8}

x_{A}^{2}-\frac{m}{2} x_{A}-\frac{m^{2}}{4}=0

x_{A}=\frac{\frac{m}{2}-\sqrt{\frac{m^{2}}{4}+m^{2}}}{2} \quad \text { OR equivalent }

Note: Condone a ± in place of the minus sign, provided given answer is restated.

x_{A}=\frac{m}{4}-\frac{m}{4} \sqrt{5}

(e)

Hence, find the exact value of $\frac{x_{B}-x_{A}}{x_{C}-x_{B}}$ .

[ 2 ]

$\frac{0-\left(\frac{m}{4}-\frac{m}{4} \sqrt{5}\right)}{\frac{m}{2}-0}$

=\frac{\sqrt{5}-1}{2}

Note: Answer must be exact.

[Maximum number: 4]

Consider a three-digit code a b c, where each of a, b and c is assigned one of the values 1,2,3,4 or 5 .

(a)

Express P(x) as a product of linear factors.

[ 1 ]

$x^{3}+4 x^{2}+5 x+2=\left(x^{2}+3 x+2\right)(x+1)$

=(x+2)(x+1)(x+1)

(b)

Hence or otherwise, sketch the graph of y=P(x), clearly showing the coordinates of any intercepts with the axes.

[ 3 ]

[Maximum number: 10]

A function f is defined by $f(x)=\arcsin \left(\frac{x^{2}-1}{x^{2}+1}\right), x \in \mathbb{R}$ .

(a)

Show that f is an even function.

[ 1 ]

EITHER
$f(-x)=\arcsin \left(\frac{(-x)^{2}-1}{(-x)^{2}+1}\right)=\arcsin \left(\frac{x^{2}-1}{x^{2}+1}\right)=f(x)$

OR
a sketch graph of y=f(x) with line symmetry in the y-axis indicated

THEN
so f(x) is an even function
AG

Marking guidance:

[1 mark]

(b)

Find an expression for $g^{-1}(x)$ , justifying your answer.

[ 5 ]

$\quad x=\arcsin \left(\frac{y^{2}-1}{y^{2}+1}\right)$

\sin x=\frac{y^{2}-1}{y^{2}+1} \Rightarrow y^{2} \sin x+\sin x=y^{2}-1

y^{2}=\frac{1+\sin x}{1-\sin x}

domain of g is $x \in \mathbb{R}, x \geq 0$ and so the range of $g^{-1}$ must be $y \in \mathbb{R}, y \geq 0$
hence the positive root is taken (or the negative root is rejected)

Note: The R1 is dependent on the above A1.

\text { so }\left(g^{-1}(x)=\right) \sqrt{\frac{1+\sin x}{1-\sin x}}

Note: The final A1 is not dependent on R1 mark.

(c)

State the domain of $g^{-1}$ .

[ 1 ]

domain is $-\frac{\pi}{2} \leq x<\frac{\pi}{2}$
Note: Accept correct alternative notations, for example, $\left\lfloor-\frac{\pi}{2}, \frac{\pi}{2}\right\rfloor$ or $\left\lfloor-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .

Marking guidance:

Accept [-1.57,1.57[ if correct to 3 s.f.

(d)

Sketch the graph of $y=g^{-1}(x)$ , clearly indicating any asymptotes with their equations and stating the values of any axes intercepts.

[ 3 ]

A1 for asymptotic behaviour $x \rightarrow \frac{\pi}{2}$
A1 for $x=\frac{\pi}{2}$
Coordinates are not required.

Figure for Answer to Question (d) — IB Maths AA HL

Note: A1 for correct domain and correct range and y-intercept at y=1
Do not accept x=1.57 or other inexact values.

[Maximum number: 5]

The following question explores features of a family of curves. The family is then linked to a homogeneous differential equation.
Consider the curve given by $y=\frac{x\left(x^{2}-16\right)}{x^{2}+16}$ .

(a)

Sketch the curve of y for $-10 \leq x \leq 10$ .

[ 1 ]

(b)

State the coordinates of the points where the curve crosses the x-axis.

[ 1 ]

$\quad(-4,0),(0,0),(4,0)$

Note: Award A0 if additional points are given.

(c)

State whether the function $f(x)=\frac{x\left(x^{2}-16\right)}{x^{2}+16}$ is odd, even or neither. Justify your answer. [2]

Now consider the general curve given by $y=\frac{x\left(x^{2}-A\right)}{x^{2}+A}$ , where A is a positive constant
and $x \in \mathbb{R}$ . and $x \in \mathbb{R}$ .

[ 2 ]

EITHER
odd A1
since $f(-x)=\frac{(-x)\left((-x)^{2}-16\right)}{(-x)^{2}+16}=\frac{-x\left(x^{2}-16\right)}{x^{2}+16}=-f(x)$ (or equivalent) $\quad \boldsymbol{R} 1$
OR
odd A1
since graph of f has rotational symmetry order 2 (of 180 degrees) about $(0,0) \quad \boldsymbol{R}$ 1

OR
odd A1 since graph has reflection symmetry over the x-axis followed by reflection symmetry over the y-axis (or vice versa)

R1

Note: Do not award A1R0.

(d)

Hence, determine the equation of the oblique asymptote to the curve.

[ 1 ]

hence oblique asymptote is y=x (since $\frac{2 A x}{x^{2}+A} \rightarrow 0$ as $x \rightarrow \pm \infty$ )

Note: Award AO if not given as an equation.

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Covered syllabus points

2 Functions question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

2 Functions question 2

Question (a)

Question (b)

2 Functions question 3

Question (a)

Question (b)

Question (c)

Question (d)

2 Functions question 4

Question (a)

Question (b)

Question (c)

Question (d)