SCSA Mathematics Specialist Functions and sketching graphs

1 sample question with marking guides and sample answers

2022

SCSA

Paper 1

6 marks

Consider functions $f(x) = \sqrt{4 - x}$ and $g(x) = \frac{1}{x^2}$ .

Q1a

2 marks

Determine the exact value of $g(f(-5))$ .

$g(f(-5)) = g(\sqrt{4-(-5)}) = g(3) = \frac{1}{9}$

Marking Criteria

Descriptor	Marks
determines $f(-5)$ correctly	1
obtains the correct value for $g(f(-5))$	1

Q1b

3 marks

Determine the domain for $f(g(x))$ .

$f(g(x)) = \sqrt{4-\frac{1}{x^2}}$ This will be defined when $4-\frac{1}{x^2} \ge 0$ , $x \ne 0$ since $g(x)$ must exist.

i.e. $\frac{1}{x^2} \le 4$ i.e. $x^2 \ge \frac{1}{4}$

$\therefore D_{fog} = \{x | x \ge \frac{1}{2} \cup x \le -\frac{1}{2}\}$

Marking Criteria

Descriptor	Marks
identifies that $f(g(x))$ is defined when $4-\frac{1}{x^2} \ge 0$	1
states $x \ge \frac{1}{2}$	1
states $x \le -\frac{1}{2}$	1

Q1c

1 mark

Explain why function $g$ is not a one-to-one function.

$g(-2) = g(2) = \frac{1}{4}$ This shows that $g$ maps two values of $x$ to a single value.

Hence $g$ is NOT a one-to-one function BUT is a MANY-to-one function.

Marking Criteria

Descriptor	Marks
justifies why $g$ is not a one-to-one function	1

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