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

[Maximum number: 5]

Line $L_{1}$ is tangent to the graph of a function f(x) at the point P(3,-1). Line $L_{2}$ is given by the equation $y=-\frac{1}{2} x-\frac{5}{2}$ and is perpendicular to $L_{1}$ .

(a)

Write down the gradient of $L_{1}$ .

[ 1 ]

(b)

Find the equation of $L_{1}$ in the form y=m x+c.

[ 2 ]

attempt to substitute their part (a) and point (3,-1) into the slope-intercept form or point-slope form of an equation
$-1=2 \times 3+c \quad$ OR $\quad y+1=2(x-3)$
y=2 x-7
Note: Equation must be in the form y=m x+c for A1 to be awarded.

(c)

Show that $L_{2}$ is not the line that is normal to f(x) at point P .

[ 2 ]

METHOD 1
attempt to show that P does not lie on $L_{2}$
e.g. $-\frac{1}{2}(3)-\frac{5}{2} \quad \mathbf{OR}$ graph showing $L_{2}$ and P in approximate correct locations
$-1 \neq-\frac{1}{2}(3)-\frac{5}{2}(-1 \neq-4) \mathbf{OR}(3,-1)$ does not lie on the graph of $L_{2}$
hence $L_{2}$ is not the normal line to f(x) at point P

METHOD 2
attempt to find the equation of the normal line at (3,-1)
$\left(-1=-\frac{1}{2}(3)+c \quad\right.$ OR $\left.\quad y+1=-\frac{1}{2}(x-3)\right)$
the normal line is $y=-\frac{1}{2} x+\frac{1}{2}$
hence $L_{2}$ is not the normal line to f(x) at point P

METHOD 3
attempt to find the intersection of $L_{1}$ and $L_{2}$

Intersection of y=2 x-7 and $y=-\frac{1}{2} x-\frac{5}{2}$ is (1.8,-3.4)
$x=1.8 \neq 3 \quad$ OR $\quad y=-3.4 \neq-1$
hence $L_{2}$ is not the normal line to f(x) at point P

Note: Accept equivalent written arguments provided values are seen.
Methods 1 and 2 are independent of the answers in (a) and (b) but FT marks can be given for Method 3.

[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 $0 \leq x \leq 16$ .

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question (c)

Question 3