Secondary 4 Elementary Mathematics Algebra Functions Quiz

Secondary 4 Elementary Mathematics Quiz - Algebra Functions

Answer Key

Section A: Function Notation and Evaluation

1. $f(2) = 3(2)^2 - 4(2) + 7 = 3(4) - 8 + 7 = 12 - 8 + 7 = \boxed{11}$
[2 marks] — 1 mark for correct substitution, 1 mark for correct answer.

2. $g(-1) = \dfrac{2(-1) + 5}{-1 - 3} = \dfrac{-2 + 5}{-4} = \dfrac{3}{-4} = \boxed{-\dfrac{3}{4}}$
[2 marks] — 1 mark for correct substitution, 1 mark for correct simplification.

3.
(a) $h(0) = 5 - 2(0) = 5 - 0 = \boxed{5}$
(b) $h(x) = -7 \Rightarrow 5 - 2x = -7 \Rightarrow -2x = -12 \Rightarrow \boxed{x = 6}$
[3 marks] — 1 mark for (a), 2 marks for (b): 1 for setting up equation, 1 for solving.

4. $f(3) = (3)^2 - 6(3) + 10 = 9 - 18 + 10 = 1$
$f(1) = (1)^2 - 6(1) + 10 = 1 - 6 + 10 = 5$
$f(3) - f(1) = 1 - 5 = \boxed{-4}$
[2 marks] — 1 mark for each correct evaluation, or 2 marks for correct final answer with working.

5. $p(2) = 2a + b = 11$ ... (1)
$p(4) = 4a + b = 19$ ... (2)
Subtract (1) from (2): $2a = 8 \Rightarrow a = 4$
Substitute into (1): $2(4) + b = 11 \Rightarrow 8 + b = 11 \Rightarrow b = 3$
$\boxed{a = 4,\ b = 3}$
[3 marks] — 1 mark for setting up equations, 1 mark for solving for $a$, 1 mark for solving for $b$.

Section B: Composite Functions

6. $g(2) = (2)^2 - 1 = 4 - 1 = 3$
$fg(2) = f(3) = 2(3) + 3 = 6 + 3 = \boxed{9}$
[2 marks] — 1 mark for finding $g(2)$, 1 mark for finding $f(g(2))$.

7. $gf(x) = g\left(\dfrac{1}{x}\right) = \dfrac{1}{x} + 4 = \boxed{\dfrac{1 + 4x}{x}}$ (or $\dfrac{1}{x} + 4$ accepted)
[2 marks] — 1 mark for correct substitution, 1 mark for correct expression.

8. $fg(x) = f(2x + 1) = 3(2x + 1) - 5 = 6x + 3 - 5 = \boxed{6x - 2}$
[2 marks] — 1 mark for correct substitution, 1 mark for simplification.

9. $fg(x) = f(x - 3) = (x - 3)^2 + 2(x - 3) = x^2 - 6x + 9 + 2x - 6 = x^2 - 4x + 3$
Set $fg(x) = 0$: $x^2 - 4x + 3 = 0$
$(x - 1)(x - 3) = 0$
$\boxed{x = 1 \text{ or } x = 3}$
[3 marks] — 1 mark for finding $fg(x)$, 1 mark for setting up equation, 1 mark for solving.

10.
(a) $fg(x) = f\left(\dfrac{x + 2}{3}\right) = 4\left(\dfrac{x + 2}{3}\right) - 1 = \dfrac{4x + 8}{3} - 1 = \dfrac{4x + 8 - 3}{3} = \boxed{\dfrac{4x + 5}{3}}$
(b) $gf(x) = g(4x - 1) = \dfrac{(4x - 1) + 2}{3} = \dfrac{4x + 1}{3}$
[4 marks] — 2 marks for (a), 2 marks for (b): 1 mark for substitution, 1 mark for simplification each.

Section C: Inverse Functions

11. Let $y = 5x - 3$. Swap $x$ and $y$: $x = 5y - 3$
Solve for $y$: $5y = x + 3 \Rightarrow y = \dfrac{x + 3}{5}$
$f^{-1}(x) = \boxed{\dfrac{x + 3}{5}}$
[2 marks] — 1 mark for swapping variables, 1 mark for solving.

12. Let $y = \dfrac{2x + 7}{3}$. Swap: $x = \dfrac{2y + 7}{3}$
$3x = 2y + 7 \Rightarrow 2y = 3x - 7 \Rightarrow y = \dfrac{3x - 7}{2}$
$g^{-1}(x) = \boxed{\dfrac{3x - 7}{2}}$
[2 marks] — 1 mark for swapping, 1 mark for solving.

13. Let $y = \dfrac{4}{x - 2}$. Swap: $x = \dfrac{4}{y - 2}$
$x(y - 2) = 4 \Rightarrow xy - 2x = 4 \Rightarrow xy = 4 + 2x \Rightarrow y = \dfrac{4 + 2x}{x}$
$h^{-1}(x) = \boxed{\dfrac{4 + 2x}{x}}$ (or $\dfrac{2x + 4}{x}$)
[3 marks] — 1 mark for swapping, 1 mark for cross-multiplying, 1 mark for isolating $y$.

14. $f^{-1}(10)$: Let $f(x) = 10 \Rightarrow 3x + 4 = 10 \Rightarrow 3x = 6 \Rightarrow x = 2$
Alternatively, find $f^{-1}(x) = \dfrac{x - 4}{3}$, then $f^{-1}(10) = \dfrac{10 - 4}{3} = \dfrac{6}{3} = \boxed{2}$
[2 marks] — 1 mark for method, 1 mark for answer.

15.
(a) Let $y = \dfrac{x - 5}{2x + 1}$. Swap: $x = \dfrac{y - 5}{2y + 1}$
$x(2y + 1) = y - 5 \Rightarrow 2xy + x = y - 5 \Rightarrow 2xy - y = -5 - x$
$y(2x - 1) = -5 - x \Rightarrow y = \dfrac{-5 - x}{2x - 1} = \boxed{\dfrac{x + 5}{1 - 2x}}$ (or equivalent)
(b) $f^{-1}(x)$ is undefined when $2x - 1 = 0 \Rightarrow \boxed{x = \dfrac{1}{2}}$
[4 marks] — 2 marks for (a) (swap and solve), 2 marks for (b) (1 for setting denominator to 0, 1 for solving).

Section D: Graphs of Functions and Applications

16.
(a) Axis of symmetry: $\boxed{x = 1}$
(b) Minimum value: $\boxed{-4}$
(c) Since vertex is $(1, -4)$ and passes through $(3, 0)$:
Using $y = a(x - 1)^2 - 4$, substitute $(3, 0)$: $0 = a(2)^2 - 4 \Rightarrow 4a = 4 \Rightarrow a = 1$
$f(x) = (x - 1)^2 - 4$, so $f(0) = (0 - 1)^2 - 4 = 1 - 4 = \boxed{-3}$
[4 marks] — 1 mark each for (a), (b), (c) setup, (c) answer.

17.
(a) Minimum point: $\boxed{(2, 1)}$
(b) Range: $\boxed{y \ge 1}$ (or $[1, \infty)$)
(c) Sketch: Parabola opening upwards with vertex at $(2, 1)$, $y$-intercept at $(0, 5)$, $x$-intercept: none (since minimum is 1).
[5 marks] — 1 mark for (a), 1 mark for (b), 3 marks for sketch (shape, vertex, intercepts).

18.
(a) $h(1) = 20(1) - 5(1)^2 = 20 - 5 = \boxed{15}$ metres
(b) $h(t) = -5t^2 + 20t = -5(t^2 - 4t) = -5[(t - 2)^2 - 4] = -5(t - 2)^2 + 20$
Maximum height = $\boxed{20}$ metres (at $t = 2$)
(c) Ball hits ground when $h(t) = 0$: $20t - 5t^2 = 0 \Rightarrow 5t(4 - t) = 0$
$t = 0$ (start) or $\boxed{t = 4}$ seconds
[5 marks] — 1 mark for (a), 2 marks for (b) (completing square or derivative), 2 marks for (c).

19.
(a) $f(x) = 2x^2 - 8x + 5 = 2(x^2 - 4x) + 5 = 2[(x - 2)^2 - 4] + 5 = 2(x - 2)^2 - 8 + 5 = \boxed{2(x - 2)^2 - 3}$
(b) Minimum point: $\boxed{(2, -3)}$
(c) $f(x) = 5 \Rightarrow 2x^2 - 8x + 5 = 5 \Rightarrow 2x^2 - 8x = 0 \Rightarrow 2x(x - 4) = 0$
$\boxed{x = 0 \text{ or } x = 4}$
[5 marks] — 2 marks for (a) (completing square), 1 mark for (b), 2 marks for (c).

20.
(a) $C(10) = 0.5(10)^2 - 20(10) + 500 = 0.5(100) - 200 + 500 = 50 - 200 + 500 = \boxed{350}$ dollars
(b) $C(n) = 0.5n^2 - 20n + 500 = 0.5(n^2 - 40n) + 500 = 0.5[(n - 20)^2 - 400] + 500 = 0.5(n - 20)^2 - 200 + 500 = 0.5(n - 20)^2 + 300$
Minimum when $n = \boxed{20}$ items
(c) Minimum cost = $\boxed{300}$ dollars
[5 marks] — 1 mark for (a), 2 marks for (b) (completing square), 2 marks for (c) (substituting or reading from vertex form).

End of Answer Key