IB Mathematics AI SL 2.1 Functions Question Bank

Practise IB Mathematics AI SL 2.1 by modelling functions, graphs, transformations, inverse relationships and technology-supported real-world problems.

Exam points

Interpret function notation, domains, ranges and key graph features from tables, graphs and formulas.
Apply transformations, inverse relationships and technology regression to build and test models.
Use graphing technology while checking restrictions, intersections, units and contextual reasonableness.

Covered syllabus points

SL 2.1—Straight lines SL 2.2—Functions and inverses SL 2.3—Graphing functions SL 2.4—Key graph features SL 2.5—Modelling with functions SL 2.6—Modelling process

[Maximum number: 12]

The mean annual temperatures for Earth, recorded at fifty-year intervals, are shown in the table.

Table for Question 2.1 Functions - SL content question 1 — IB Maths AI SL

Tami creates a linear model for this data by finding the equation of the straight line passing through the points with coordinates (1708,8.73) and (1958,9.45).

(a)

Calculate the gradient of the straight line that passes through these two points.

[ 2 ]

$\frac{9.45-8.73}{1958-1708}$

=0.00288\left(\frac{9}{3125}\right)

(b)

Interpret the meaning of the gradient in the context of the question.

[ 2 ]

(c)

State appropriate units for the gradient.

[ 2 ]

$\quad{ }^{\circ} \mathrm{C} /$ year OR degrees C per year A1

Note: Do not follow through from part (b)(i) into (b)(ii).

(d)

Find the equation of this line giving your answer in the form y=m x+c.

[ 2 ]

attempt to substitute point and gradient into appropriate formula $8.73=0.00288 \times 1708+c \Rightarrow c=3.81096 \ldots$
or $9.45=0.00288 \times 1958+c \Rightarrow c=3.81096$ .
equation is y=0.00288 x+3.81

(e)

Use Tami's model to estimate the mean annual temperature in the year 2000.

Thandizo uses linear regression to obtain a model for the data.

[ 2 ]

attempt to substitute 2000 into their part (c)
$0.00288 \times 2000+3.81096 \ldots$ $=9.57\left({ }^{\circ} \mathrm{C}\right)(9.57096 \ldots)$

(f)

State two reasons why Thandizo's prediction may not be valid.

[ 2 ]

[Maximum number: 6]

A function is defined by $f(x)=\frac{4}{x}-3$ , for $-6 \leq x \leq 10, x \neq 0$ .

(a)

Find f(10).

[ 1 ]

$-2.6\left(\frac{-13}{5}\right)$

(b)

Find the range of f.

[ 3 ]

EITHER
$(-\infty,-3.67] \cup[-2.6, \infty)$
A1A1A1
OR
$y \leq-3.67$ or $y \geq-2.6$
A1A1A1

Note: Award A1 for $-3.67\left(-3.66666 \ldots,-\frac{11}{3}\right)$ seen
A1 for their $-2.6\left(-\frac{13}{5}\right)$ seen and both sets of inequalities/brackets correct;
A1 if completely correct, including union symbol or "or".
It is possible to award A0A1A0 for correctly presenting their -3.67 (and their -2.6 ).
The -3.67 maybe seen as part of a sketch.

Award A1A0A0 for $-3.67 \leq y \leq-2.6$ or [-3.67,-2.6].

(c)

Find $f^{-1}(-2)$ .

[ 2 ]

setting f(x) equal to -2 f(x)=-2 OR $-2=\frac{4}{x}-3$ OR sketch of graph with line labelled -2

[Maximum number: 2]

A hollow chocolate box is manufactured in the form of a right prism with a regular hexagonal base. The height of the prism is $h \mathrm{~cm}$ , and the top and base of the prism have sides of length $x \mathrm{~cm}$ .

Figure for Question 2.1 Functions - SL content question 3 — IB Maths AI SL

Sketch the graph of $V=300 \sqrt{3} x-\frac{9}{4} x^{3}$ , for $0 \leq x \leq 16$ .

All question bank results loaded

Covered syllabus points

2.1 Functions - SL content question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

2.1 Functions - SL content question 2

Question (a)

Question (b)

Question (c)

2.1 Functions - SL content question 3