IB Maths AI SL 2.1 Functions Sl Content Questions

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

Exam points

Represent functions with correct notation, domain, range and inverse relationships, evaluating and composing them in context.
Construct, sketch and transform function graphs, identifying intercepts, extrema, symmetry, asymptotes and intersections.
Select and use function models, interpret parameters and use technology to fit, solve or predict within a valid domain.
Use graph and model features to answer thresholds, optimisation and applied questions with appropriate units.
Check a function or model using substitution, graph behaviour, data fit, assumptions 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

Learn this topic

Study 2.1 Functions - SL content conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

[Maximum number: 10]

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

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.

[ 1 ]

(c)

State appropriate units for the gradient.

[ 1 ]

$\quad{ }^{\circ} \mathrm{C} /$ 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.

[ 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)

Thandizo uses his regression line to predict the year when the mean annual temperature will first exceed $15^{\circ} \mathrm{C}$ .

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
$(-\infty,-3.67] \cup[-2.6, \infty)$
OR
$y \leq-3.67$ or $y \geq-2.6$
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 $x \mathrm{~cm}$ .

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

Learn this topic

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question 2

Question (a)

Question (b)

Question (c)

Question 3