O Level Elementary Mathematics Algebra Functions Quiz

O-Level Elementary Mathematics Quiz - Algebra Functions (Answer Key)

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

2. $x^2 + 5 = 30 \Rightarrow x^2 = 25 \Rightarrow x = \pm 5$
Answer: $5, -5$ [2]
(1 mark for $x^2=25$, 1 mark for both roots)

3. Denominator cannot be zero. $x + 2 = 0 \Rightarrow x = -2$.
Answer: $-2$ [1]

4. $fg(x) = f(g(x)) = f(x-3) = 2(x-3) + 1 = 2x - 6 + 1 = 2x - 5$.
Answer: $2x - 5$ [2]

5. Gradient $a = \frac{9-5}{4-2} = \frac{4}{2} = 2$.
Using $(2,5)$: $5 = 2(2) + b \Rightarrow 5 = 4 + b \Rightarrow b = 1$.
Answer: $2x + 1$ [2]

6. Expression inside square root must be $\ge 0$.
$x - 3 \ge 0 \Rightarrow x \ge 3$.
Smallest integer is 3.
Answer: 3 [2]

7. Let $y = 5x + 2$. Swap $x$ and $y$: $x = 5y + 2$.
$5y = x - 2 \Rightarrow y = \frac{x-2}{5}$.
Answer: $f^{-1}(x) = \frac{x-2}{5}$ [2]

8. (a) Let $y = \frac{x}{3} - 4$. Swap $x$ and $y$: $x = \frac{y}{3} - 4$.
$x + 4 = \frac{y}{3} \Rightarrow y = 3(x+4) = 3x + 12$.
Answer: $g^{-1}(x) = 3x + 12$ [2]

(b) $3x + 12 = 10 \Rightarrow 3x = -2 \Rightarrow x = -\frac{2}{3}$.
Answer: $-\frac{2}{3}$ [2]

9. (a) $gf(x) = g(f(x)) = g(2x-1) = (2x-1)^2$.
Answer: $(2x-1)^2$ or $4x^2 - 4x + 1$ [2]

(b) $fg(x) = f(g(x)) = f(x^2) = 2(x^2) - 1 = 2x^2 - 1$.
Answer: $2x^2 - 1$ [2]

(c) $(2x-1)^2 = 2x^2 - 1$
$4x^2 - 4x + 1 = 2x^2 - 1$
$2x^2 - 4x + 2 = 0$
$x^2 - 2x + 1 = 0$
$(x-1)^2 = 0 \Rightarrow x = 1$.
Answer: $x = 1$ [3]

10. (a) Let $y = \frac{2x+1}{x-3}$.
$y(x-3) = 2x + 1$
$xy - 3y = 2x + 1$
$xy - 2x = 3y + 1$
$x(y-2) = 3y + 1$
$x = \frac{3y+1}{y-2}$
Answer: $h^{-1}(x) = \frac{3x+1}{x-2}$ [3]

(b) Denominator $x-2 \neq 0 \Rightarrow x \neq 2$.
Answer: 2 [1]

11. Find inverse of $f(x) = 3x + k$:
$y = 3x + k \Rightarrow x = \frac{y-k}{3} \Rightarrow f^{-1}(x) = \frac{x-k}{3}$.
Given $f^{-1}(x) = \frac{x-5}{3}$.
Comparing numerators: $-k = -5 \Rightarrow k = 5$.
Answer: 5 [2]

12. (a) Without restriction, $f(x) = x^2 - 4$ is not one-to-one (fails horizontal line test), so inverse is not a function. Restricting to $x \ge 0$ makes it one-to-one. [1]

(b) $y = x^2 - 4 \Rightarrow x^2 = y + 4 \Rightarrow x = \sqrt{y+4}$ (positive root since $x \ge 0$).
Answer: $f^{-1}(x) = \sqrt{x+4}$ [2]

13. Vertex at $2x-4=0 \Rightarrow x=2, y=0$. Coordinates $(2,0)$.
y-intercept at $x=0 \Rightarrow y=|-4|=4$. Coordinates $(0,4)$.
V-shape graph with vertex at $(2,0)$, passing through $(0,4)$ and $(4,4)$.
Answer: Sketch showing V-shape, vertex $(2,0)$, y-int $(0,4)$. [3]

14. Transformation is translation by vector $\begin{pmatrix} 0 \\ 3 \end{pmatrix}$.
New points: $(0, 5)$, $(2, 3)$, $(4, 9)$.
Answer: Sketch with points shifted up by 3 units. [3]

15. (a) $x^2 - 6x + 11 = (x-3)^2 - 9 + 11 = (x-3)^2 + 2$.
Answer: $(x-3)^2 + 2$ [2]

(b) Minimum value is 2 at $x = 3$.
Answer: Min value 2, $x=3$ [2]

16. $f(1) = a(1)^2 + b(1) = a + b = 5$ (Eq 1)
$f(2) = a(2)^2 + b(2) = 4a + 2b = 14 \Rightarrow 2a + b = 7$ (Eq 2)
Subtract Eq 1 from Eq 2: $(2a+b) - (a+b) = 7 - 5 \Rightarrow a = 2$.
Sub $a=2$ into Eq 1: $2 + b = 5 \Rightarrow b = 3$.
Answer: $a=2, b=3$ [3]

17. (a) $P(n) = (4n - 0.1n^2) - (50 + 2n) = -0.1n^2 + 2n - 50$.
Answer: $-0.1n^2 + 2n - 50$ [2]

(b) $-0.1n^2 + 2n - 50 = 0$. Multiply by -10: $n^2 - 20n + 500 = 0$.
Discriminant $\Delta = (-20)^2 - 4(1)(500) = 400 - 2000 = -1600$.
Since $\Delta < 0$, there are no real solutions.
Correction in question logic for student benefit: If the question implies a break-even is possible, check signs. Here, max revenue vertex is at $n = -4/(2(-0.1)) = 20$. $R(20) = 80 - 40 = 40$. Cost $C(20) = 50 + 40 = 90$. Loss is always incurred.
Alternative Interpretation: If the question meant $R(n) = 5n - 0.1n^2$? Let's stick to the math derived.
Wait, let's re-read standard O-Level patterns. Usually, they factorise.
Let's assume the question intended solvable numbers.
If $P(n) = -0.1n^2 + 2n - 50 = 0$, no solution.
Let's adjust the answer key to reflect the mathematical truth: "No break-even point exists as the discriminant is negative."
However, for a standard quiz, let's provide the working for the quadratic formula.
$n = \frac{-2 \pm \sqrt{-1600}}{-0.2}$ -> No real root.
Answer: No real solution (Company never breaks even with these parameters). [3]
(Note: If this were an exam, full marks for showing discriminant < 0)

18. Translation by vector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ (2 units right, 1 unit up).
Answer: Translation 2 units right and 1 unit up. [2]

19. (a) $f(3) = 2^3 = 8$. $f(0) = 2^0 = 1$. $8 - 1 = 7$.
Answer: 7 [1]

(b) $2^x = 32 \Rightarrow 2^x = 2^5 \Rightarrow x = 5$.
Answer: 5 [1]

20. $3x - 2 = \frac{x+1}{2}$
$2(3x - 2) = x + 1$
$6x - 4 = x + 1$
$5x = 5 \Rightarrow x = 1$.
Answer: 1 [2]