O Level Additional Mathematics Algebra Functions Quiz

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

Total Marks: 40
Topic: Algebra & Functions

Section A: Quadratic Functions and Equations

1. [2 marks]
$f(x) = x^2 - 6x + 4$
Complete the square:
$x^2 - 6x = (x - 3)^2 - 9$
So $f(x) = (x - 3)^2 - 9 + 4 = (x - 3)^2 - 5$
Answer: $(x - 3)^2 - 5$
Teaching note: Completing the square reveals the vertex at (3, –5). Award 1 mark for correct bracket, 1 mark for correct constant.

2. [2 marks]
For $g(x) = 2x^2 + px + 5$ always positive, we need $a > 0$ (true, $a=2$) and discriminant $< 0$.
Discriminant $\Delta = p^2 - 4(2)(5) = p^2 - 40$.
Condition: $p^2 - 40 < 0$ i.e. $p^2 < 40$.
Answer: $p^2 - 40 < 0$ or $p^2 < 40$
Teaching note: Always-positive quadratic means graph lies above x-axis; no real roots → Δ < 0.

3. [2 marks]
$3x^2 - 5x + 1 = 0$: $a=3, b=-5, c=1$
$\Delta = (-5)^2 - 4(3)(1) = 25 - 12 = 13$
Since $\Delta > 0$, two distinct real roots.
Answer: Discriminant = 13; two real roots
Common mistake: Sign error on $b^2$ when $b$ negative.

4. [2 marks]
Substitute $y = x+3$ into $y = x^2 - 2x - 1$:
$x + 3 = x^2 - 2x - 1 \Rightarrow x^2 - 3x - 4 = 0$
$(x - 4)(x + 1) = 0 \Rightarrow x = 4$ or $x = -1$
When $x=4, y=7$; when $x=-1, y=2$.
Answer: $(4, 7)$ and $(-1, 2)$

5. [2 marks]
$x^2 - 4x - 5 < 0 \Rightarrow (x - 5)(x + 1) < 0$
Critical values: $x = -1, 5$.
Inequality holds between roots: $-1 < x < 5$.
Number line: open circles at –1 and 5, shaded between.
Answer: $-1 < x < 5$

Section B: Surds, Polynomials and Partial Fractions

6. [2 marks]
$\sqrt{12} = 2\sqrt{3}, \sqrt{27} = 3\sqrt{3}$
Sum = $5\sqrt{3}$
Answer: $5\sqrt{3}$

7. [2 marks]
$\dfrac{2}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} = \dfrac{2(\sqrt{3} + 1)}{3 - 1} = \dfrac{2(\sqrt{3} + 1)}{2} = \sqrt{3} + 1$
Answer: $\sqrt{3} + 1$

8. [2 marks]
Remainder = $P(2) = 2^3 - 4(2)^2 + 2 + 6 = 8 - 16 + 2 + 6 = 0$
Answer: 0
Note: Remainder Theorem: remainder when divided by $(x-a)$ is $P(a)$.

9. [2 marks]
$P(-1) = (-1)^3 + 2(-1)^2 - (-1) - 2 = -1 + 2 + 1 - 2 = 0$
Since $P(-1)=0$, $(x+1)$ is a factor.
Answer: Shown via $P(-1)=0$

10. [2 marks]
$\dfrac{5}{(x+1)(x-2)} = \dfrac{A}{x+1} + \dfrac{B}{x-2}$
$5 = A(x-2) + B(x+1)$
Let $x = -1$: $5 = A(-3) \Rightarrow A = -\frac{5}{3}$
Let $x = 2$: $5 = B(3) \Rightarrow B = \frac{5}{3}$
Answer: $-\dfrac{5/3}{x+1} + \dfrac{5/3}{x-2}$ or $\dfrac{-5}{3(x+1)} + \dfrac{5}{3(x-2)}$

Section C: Binomial Expansions

11. [2 marks]
$(1+2x)^4 = 1 + \binom{4}{1}(2x) + \binom{4}{2}(2x)^2 + \cdots$
$= 1 + 4(2x) + 6(4x^2) = 1 + 8x + 24x^2$
Answer: $1 + 8x + 24x^2$

12. [2 marks]
General term: $\binom{6}{r} x^{6-r} (x^{-1})^r = \binom{6}{r} x^{6-2r}$
Independent of $x$ → $6 - 2r = 0 \Rightarrow r = 3$
Term = $\binom{6}{3} = 20$
Answer: 20

13. [2 marks]
$(2 - y)^3 = 2^3 - 3(2^2)y + 3(2)y^2 - y^3 = 8 - 12y + 6y^2 - y^3$
Answer: $8 - 12y + 6y^2 - y^3$

14. [2 marks]
Term with $x^3$: $\binom{5}{3}(1)^{2}(-3x)^3 = 10 \times (-27x^3) = -270x^3$
Coefficient = –270
Answer: –270

15. [2 marks]
4th term, $r=3$: $\binom{5}{3}(2x)^{2}(1)^3 = 10 \times 4x^2 = 40x^2$
Answer: $40x^2$

Section D: Exponential and Logarithmic Functions

16. [2 marks]
$2^x = 16 = 2^4 \Rightarrow x = 4$
Answer: $x = 4$

17. [2 marks]
$\log_2(x+1) = 3 \Rightarrow x+1 = 2^3 = 8 \Rightarrow x = 7$
Answer: $x = 7$

18. [2 marks]
$\log_3 81 - \log_3 9 = \log_3(81/9) = \log_3 9 = 2$
Answer: $\log_3 9 = 2$

19. [2 marks]
$\log_4 8 = \dfrac{\log_2 8}{\log_2 4} = \dfrac{3}{2} = 1.5$
Answer: $\frac{3}{2}$ or $1.5$

20. [2 marks]
$e^{2x} = 5 \Rightarrow 2x = \ln 5 \Rightarrow x = \frac{\ln 5}{2} \approx 0.8047$
To 3 s.f.: $0.805$
Answer: $x = 0.805$