O Level Additional Mathematics Algebra Functions Quiz

Answer Key - Algebra Functions Quiz

1. Discriminant $\Delta < 0$: $(k+1)^2 - 4(2)(3) < 0 \implies k^2 + 2k + 1 - 24 < 0 \implies k^2 + 2k - 23 < 0$. Roots of $k^2+2k-23=0$ are $k = \frac{-2 \pm \sqrt{4 - 4(1)(-23)}}{2} = \frac{-2 \pm \sqrt{96}}{2} = -1 \pm 2\sqrt{6}$. Range: $-1 - 2\sqrt{6} < k < -1 + 2\sqrt{6}$ (or $-5.90 < k < 3.90$). [3 marks]

2. $3(x^2 - 4x) + 15 = 3[(x-2)^2 - 4] + 15 = 3(x-2)^2 - 12 + 15 = 3(x-2)^2 + 3$. Form: $3(x-2)^2 + 3$. Min point: $(2, 3)$. [3 marks]

3. $(2x-8)(x+1.5)$ approx $\implies (2x+3)(x-4) < 0$. Critical values: $x = -1.5, x = 4$. Range: $-1.5 < x < 4$. [3 marks]

4. $\Delta = 0 \implies (m-2)^2 - 4(1)(4) = 0 \implies (m-2)^2 = 16$. $m-2 = \pm 4 \implies m = 6$ or $m = -2$. [3 marks]

5. $x^2 + 6x + 11 = (x+3)^2 + 2$. Since $(x+3)^2 \ge 0$ for all real $x$, $(x+3)^2 + 2 \ge 2$. Therefore, the expression is always positive. [3 marks]

6. Substitute $y$: $x^2 + (2x-3)^2 = 10 \implies x^2 + 4x^2 - 12x + 9 = 10 \implies 5x^2 - 12x - 1 = 0$. $x = \frac{12 \pm \sqrt{144 - 4(5)(-1)}}{10} = \frac{12 \pm \sqrt{164}}{10} = 1.2 \pm 1.28$. $x_1 \approx 2.48, y_1 \approx 1.96$; $x_2 \approx -0.08, y_2 \approx -3.16$. [4 marks]

7. $x^2 + 3x + 5 = px - 1 \implies x^2 + (3-p)x + 6 = 0$. No intersection $\implies \Delta < 0 \implies (3-p)^2 - 4(1)(6) < 0$. $p^2 - 6p + 9 - 24 < 0 \implies p^2 - 6p - 15 < 0$. Roots: $p = \frac{6 \pm \sqrt{36 - 4(-15)}}{2} = \frac{6 \pm \sqrt{96}}{2} = 3 \pm 2\sqrt{6}$. Range: $3 - 2\sqrt{6} < p < 3 + 2\sqrt{6}$. [4 marks]

8. $P(2) = 0 \implies 2(2)^3 + a(2)^2 - 5(2) + 6 = 0 \implies 16 + 4a - 10 + 6 = 0 \implies 4a + 12 = 0 \implies a = -3$. [3 marks]

9. $f(-1) = 3(-1)^3 - 2(-1)^2 + (-1) - 5 = -3 - 2 - 1 - 5 = -11$. Remainder: $-11$. [3 marks]

10. $(x-3)(x^2 + 3x + 9)$. [2 marks]

11. $x^3 - 4x^2 - 7x + 10 = (x-1)(x^2 - 3x - 10) = (x-1)(x-5)(x+2)$. Solutions: $x = 1, x = 5, x = -2$. [4 marks]

12. $\frac{5x-1}{(x-2)(x+3)} = \frac{A}{x-2} + \frac{B}{x+3} \implies 5x-1 = A(x+3) + B(x-2)$. Let $x=2: 9 = 5A \implies A = 1.8$. Let $x=-3: -16 = -5B \implies B = 3.2$. Answer: $\frac{1.8}{x-2} + \frac{3.2}{x+3}$. [4 marks]

13. $\frac{x+7}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2} \implies x+7 = A(x+2) + B(x-3)$. Let $x=3: 10 = 5A \implies A = 2$. Let $x=-2: 5 = -5B \implies B = -1$. Answer: $\frac{2}{x-3} - \frac{1}{x+2}$. [4 marks]

14. $\frac{3x^2+x+1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} \implies 3x^2+x+1 = A(x^2+1) + (Bx+C)(x-1)$. Let $x=1: 5 = 2A \implies A = 2.5$. Compare $x^2$ coeff: $3 = A + B \implies 3 = 2.5 + B \implies B = 0.5$. Compare const: $1 = A - C \implies 1 = 2.5 - C \implies C = 1.5$. Answer: $\frac{2.5}{x-1} + \frac{0.5x + 1.5}{x^2+1}$. [5 marks]

15. $\binom{5}{0}(3)^5 + \binom{5}{1}(3)^4(2x) + \binom{5}{2}(3)^3(2x)^2 = 243 + 5(81)(2x) + 10(27)(4x^2) = 243 + 810x + 1080x^2$. [4 marks]

16. Term: $\binom{6}{3}(3x)^3(-1)^3 = 20 \cdot 27x^3 \cdot (-1) = -540x^3$. Coefficient: $-540$. [3 marks]

17. $\log_2((x+3)(x-1)) = 5 \implies x^2 + 2x - 3 = 2^5 \implies x^2 + 2x - 35 = 0$. $(x+7)(x-5) = 0 \implies x = -7$ or $x = 5$. Since $x > 1$ (from $\log_2(x-1)$), $x = 5$. [4 marks]

18. Let $3^x = u$. $3u^2 - 10u + 3 = 0$. $(3u-1)(u-3) = 0 \implies u = 1/3$ or $u = 3$. $3^x = 3^{-1} \implies x = -1$; $3^x = 3^1 \implies x = 1$. Solutions: $x = -1, x = 1$. [5 marks]

19. $\log_a 12 = \log_a(2^2 \cdot 3) = 2\log_a 2 + \log_a 3 = 2(0.301) + 0.477 = 0.602 + 0.477 = 1.079$. [3 marks]

20. $\ln(x^2) - \ln(x-1) = \ln 4 \implies \ln(\frac{x^2}{x-1}) = \ln 4 \implies \frac{x^2}{x-1} = 4$. $x^2 = 4x - 4 \implies x^2 - 4x + 4 = 0 \implies (x-2)^2 = 0 \implies x = 2$. [5 marks]