Secondary 3 Elementary Mathematics Algebra Functions Quiz

Secondary 3 Elementary Mathematics Quiz - Algebra Functions (Answer Key)

1. $f(4) = 3(4) - 5 = 12 - 5 = 7$
Answer: 7 [1]

2. $g(-3) = (-3)^2 + 2 = 9 + 2 = 11$
Answer: 11 [1]

3. Division by zero is undefined. $x = 0$.
Answer: 0 [1]

4. Since $x \ge 0$, the minimum value of $2x$ is 0.
Minimum $k(x) = 2(0) + 1 = 1$.
As $x$ increases, $k(x)$ increases without bound.
Answer: $k(x) \ge 1$ or $[1, \infty)$ [2]

5. $5 - 2x = 11$
$-2x = 11 - 5$
$-2x = 6$
$x = -3$
Answer: $x = -3$ [2]

6. Substitute each domain element into $x^2 - 4$:
$x = -2 \Rightarrow (-2)^2 - 4 = 0$
$x = -1 \Rightarrow (-1)^2 - 4 = -3$
$x = 0 \Rightarrow 0^2 - 4 = -4$
$x = 1 \Rightarrow 1^2 - 4 = -3$
$x = 2 \Rightarrow 2^2 - 4 = 0$
Unique values: $\{-4, -3, 0\}$
Answer: $\{-4, -3, 0\}$ [3]

7. $fg(x) = f(g(x)) = f(x - 5)$
$= 2(x - 5) + 3$
$= 2x - 10 + 3$
$= 2x - 7$
Answer: $2x - 7$ [2]

8. First find $f(4)$: $f(4) = 2(4) + 3 = 11$.
Then find $g(11)$: $g(11) = 11 - 5 = 6$.
Answer: 6 [2]

9. $qp(x) = q(p(x)) = q(\frac{x}{3})$
$= 4(\frac{x}{3}) - 1$
$= \frac{4x}{3} - 1$
Answer: $\frac{4x}{3} - 1$ [2]

10. Let $y = 3x - 7$.
Swap $x$ and $y$: $x = 3y - 7$.
Make $y$ the subject:
$x + 7 = 3y$
$y = \frac{x + 7}{3}$
Answer: $f^{-1}(x) = \frac{x + 7}{3}$ [2]

11. Let $y = \frac{2x + 1}{5}$.
Swap $x$ and $y$: $x = \frac{2y + 1}{5}$.
Make $y$ the subject:
$5x = 2y + 1$
$5x - 1 = 2y$
$y = \frac{5x - 1}{2}$
Answer: $g^{-1}(x) = \frac{5x - 1}{2}$ [3]

12. Let $y = x^2 + 2$.
Swap $x$ and $y$: $x = y^2 + 2$.
Make $y$ the subject:
$y^2 = x - 2$
$y = \pm\sqrt{x - 2}$
Since the original domain was $x \ge 0$, the range of $f$ is $y \ge 2$. The domain of $f^{-1}$ is $x \ge 2$. The range of $f^{-1}$ must match the domain of $f$ ($y \ge 0$). Thus, we take the positive root.
Answer: $h^{-1}(x) = \sqrt{x - 2}$ [3]

13. $f^{-1}(x) = \frac{x - 1}{2}$ (derived similarly to Q10/11).
$f^{-1}(f(x)) = f^{-1}(2x + 1)$
$= \frac{(2x + 1) - 1}{2}$
$= \frac{2x}{2}$
$= x$
Answer: Verified [3]

14. (a) Let $y = \frac{1}{x-2}$. Swap $x, y$: $x = \frac{1}{y-2}$.
$x(y - 2) = 1 \Rightarrow y - 2 = \frac{1}{x} \Rightarrow y = \frac{1}{x} + 2$.
$f^{-1}(x) = \frac{1}{x} + 2$.
(b) The domain of $f^{-1}$ is the range of $f$. Since $x > 2$, $x - 2 > 0$, so $\frac{1}{x-2} > 0$. Thus $f(x) > 0$.
Answer: (a) $\frac{1}{x} + 2$, (b) $x > 0$ [3]

15. Vertex: $2x - 4 = 0 \Rightarrow x = 2, y = 0$. Vertex $(2, 0)$.
Y-intercept: $x = 0 \Rightarrow y = |-4| = 4$. Point $(0, 4)$.
Endpoint $x = -1 \Rightarrow y = |-2 - 4| = 6$. Point $(-1, 6)$.
Endpoint $x = 4 \Rightarrow y = |8 - 4| = 4$. Point $(4, 4)$.
V-shape graph with vertex at $(2,0)$, passing through $(0,4)$ and $(4,4)$.
Answer: Correct sketch with labels [3]

16. $f(x) = x^2 - 4x + 3 = (x - 2)^2 - 1$.
Vertex at $(2, -1)$. Since $0 \le 2 \le 5$, the minimum is at the vertex.
(a) Min value = $-1$.
(b) Check endpoints:
$f(0) = 3$.
$f(5) = 25 - 20 + 3 = 8$.
Max value is 8.
Answer: (a) -1, (b) 8 [4]

17. (a) Perimeter $2(l + w) = 20 \Rightarrow l + w = 10$.
$w = 10 - x$.
(b) Area $A = l \times w = x(10 - x) = 10x - x^2$.
(c) $A(x) = -(x^2 - 10x) = -(x - 5)^2 + 25$.
Max occurs at vertex $x = 5$.
Answer: (a) $10 - x$, (b) Shown, (c) $x = 5$ [5]

18. (a) $2x^2 - 8 = x + 2$
$2x^2 - x - 10 = 0$
$(2x - 5)(x + 2) = 0$
$x = 2.5$ or $x = -2$.
(b) If $x = 2.5, y = 2.5 + 2 = 4.5$. Point $(2.5, 4.5)$.
If $x = -2, y = -2 + 2 = 0$. Point $(-2, 0)$.
Answer: (a) $x = 2.5, -2$, (b) $(2.5, 4.5)$ and $(-2, 0)$ [4]

19. (a) $50 + 2n = 4n$
$50 = 2n \Rightarrow n = 25$.
(b) $S(100) = 400$. $C(100) = 50 + 200 = 250$.
Profit $= 400 - 250 = 150$.
Answer: (a) 25 items, (b) $150 [4]

20. (a) Vertical asymptote where denominator is zero: $x = -1$.
(b) As $x \to \infty$, $y \to 0$. Horizontal asymptote: $y = 0$.
(c) Y-intercept: $x = 0 \Rightarrow y = 3/1 = 3$. Point $(0, 3)$.
Hyperbola in 1st and 3rd quadrants relative to asymptotes.
Answer: (a) $x = -1$, (b) $y = 0$, (c) Correct sketch [4]