Secondary 3 Additional Mathematics Algebra Functions Quiz

Secondary 3 Additional Mathematics Quiz - Algebra Functions

Answer Key and Marking Scheme

Total Marks: 60

Section A: Quadratic Functions and Equations (Questions 1–5)

1. ( 2x^2 - 12x + 7 = 2(x^2 - 6x) + 7 = 2[(x - 3)^2 - 9] + 7 = 2(x - 3)^2 - 18 + 7 = 2(x - 3)^2 - 11 ) Minimum value is ( -11 ), occurring at ( x = 3 ). [3 marks: 1 for completing square correctly, 1 for minimum value, 1 for x-value]

2. For two distinct real roots, discriminant ( > 0 ). ( \Delta = (k - 3)^2 - 4(1)(4) = k^2 - 6k + 9 - 16 = k^2 - 6k - 7 > 0 ) ( (k - 7)(k + 1) > 0 ) ( k < -1 ) or ( k > 7 ) [3 marks: 1 for discriminant expression, 1 for factorising, 1 for correct inequality solution]

3. (a) ( f(x) = -(x^2 - 6x) - 5 = -[(x - 3)^2 - 9] - 5 = -(x - 3)^2 + 9 - 5 = -(x - 3)^2 + 4 ) (b) Maximum value is 4, occurring at ( x = 3 ). (c) ( f(x) > 0 \implies -(x - 3)^2 + 4 > 0 \implies (x - 3)^2 < 4 \implies -2 < x - 3 < 2 \implies 1 < x < 5 ) [4 marks: 1 for (a), 1 for (b) max value, 1 for (b) x-value, 1 for (c)]

4. Substitute ( y = mx + 2 ) into ( y = x^2 + 3x + 1 ): ( mx + 2 = x^2 + 3x + 1 \implies x^2 + (3 - m)x - 1 = 0 ) For tangent, discriminant ( = 0 ): ( (3 - m)^2 - 4(1)(-1) = 0 \implies (3 - m)^2 + 4 = 0 ) Since ( (3 - m)^2 \ge 0 ), ( (3 - m)^2 + 4 > 0 ) always. No real value of ( m ) exists for which the line is a tangent. [3 marks: 1 for substitution, 1 for discriminant, 1 for conclusion]

5. From ( 2x^2 - 5x + 1 = 0 ): ( \alpha + \beta = \frac{5}{2} ), ( \alpha\beta = \frac{1}{2} ) (a) ( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{2}\right)^2 - 2\left(\frac{1}{2}\right) = \frac{25}{4} - 1 = \frac{21}{4} ) (b) ( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/2}{1/2} = 5 ) [2 marks: 1 for (a), 1 for (b)]

Section B: Polynomials and Partial Fractions (Questions 6–10)

6. ( P(2) = 0 \implies 8 + 4a + 2b - 6 = 0 \implies 4a + 2b = -2 \implies 2a + b = -1 ) ... (1) ( P(-1) = -4 \implies -1 + a - b - 6 = -4 \implies a - b = 3 ) ... (2) Solving (1) and (2): Adding: ( 3a = 2 \implies a = \frac{2}{3} ); ( b = a - 3 = \frac{2}{3} - 3 = -\frac{7}{3} ) [3 marks: 1 for each equation, 1 for solving]

7. Try ( x = 1 ): ( 2 - 3 - 3 + 2 = -2 \neq 0 ). Try ( x = -1 ): ( -2 - 3 + 3 + 2 = 0 ). So ( (x + 1) ) is a factor. Division: ( 2x^3 - 3x^2 - 3x + 2 = (x + 1)(2x^2 - 5x + 2) = (x + 1)(2x - 1)(x - 2) ) [3 marks: 1 for finding factor, 1 for division, 1 for complete factorisation]

8. Let ( \frac{4x + 1}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2} ) ( 4x + 1 = A(x + 2) + B(x - 1) ) ( x = 1 ): ( 5 = 3A \implies A = \frac{5}{3} ) ( x = -2 ): ( -7 = -3B \implies B = \frac{7}{3} ) ( \frac{4x + 1}{(x - 1)(x + 2)} = \frac{5}{3(x - 1)} + \frac{7}{3(x + 2)} ) [3 marks: 1 for form, 1 for finding A, 1 for finding B]

9. Let ( \frac{3x^2 + 5x + 2}{(x + 1)^2(x - 2)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x - 2} ) ( 3x^2 + 5x + 2 = A(x + 1)(x - 2) + B(x - 2) + C(x + 1)^2 ) ( x = -1 ): ( 3 - 5 + 2 = 0 = B(-3) \implies B = 0 ) ( x = 2 ): ( 12 + 10 + 2 = 24 = C(9) \implies C = \frac{8}{3} ) Coefficient of ( x^2 ): ( 3 = A + C \implies A = 3 - \frac{8}{3} = \frac{1}{3} ) ( \frac{3x^2 + 5x + 2}{(x + 1)^2(x - 2)} = \frac{1}{3(x + 1)} + \frac{8}{3(x - 2)} ) [3 marks: 1 for form, 1 for finding two constants, 1 for all correct]

10. ( Q(-2) = 0 \implies -16 + 4p - 2q - 12 = 0 \implies 4p - 2q = 28 \implies 2p - q = 14 ) ... (1) ( Q(3) = 0 \implies 54 + 9p + 3q - 12 = 0 \implies 9p + 3q = -42 \implies 3p + q = -14 ) ... (2) Adding (1) and (2): ( 5p = 0 \implies p = 0 ); ( q = -14 ) ( Q(x) = 2x^3 - 14x - 12 = 2(x^3 - 7x - 6) ) ( x = -1 ): ( -1 + 7 - 6 = 0 ), so ( (x + 1) ) is a factor. ( x^3 - 7x - 6 = (x + 1)(x^2 - x - 6) = (x + 1)(x - 3)(x + 2) ) ( Q(x) = 2(x + 1)(x - 3)(x + 2) ) [3 marks: 1 for equations, 1 for p and q, 1 for factorisation]

Section C: Surds, Indices, and Exponential/Logarithmic Functions (Questions 11–15)

11. ( \frac{3}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{3(\sqrt{5} + 2)}{5 - 4} = 3\sqrt{5} + 6 = 6 + 3\sqrt{5} ) [3 marks: 1 for conjugate, 1 for multiplication, 1 for simplified answer]

12. ( \sqrt{2x + 5} = 2 + \sqrt{x - 1} ) Square both sides: ( 2x + 5 = 4 + 4\sqrt{x - 1} + (x - 1) ) ( 2x + 5 = x + 3 + 4\sqrt{x - 1} ) ( x + 2 = 4\sqrt{x - 1} ) Square again: ( x^2 + 4x + 4 = 16(x - 1) = 16x - 16 ) ( x^2 - 12x + 20 = 0 \implies (x - 2)(x - 10) = 0 ) ( x = 2 ) or ( x = 10 ) Check: ( x = 2 ): LHS ( = \sqrt{9} - \sqrt{1} = 3 - 1 = 2 ) ✓ ( x = 10 ): LHS ( = \sqrt{25} - \sqrt{9} = 5 - 3 = 2 ) ✓ Both solutions are valid. [3 marks: 1 for isolating and squaring, 1 for solving quadratic, 1 for checking both]

13. ( 3^{2x+1} = 5^{x-2} ) Take lg: ( (2x + 1)\lg 3 = (x - 2)\lg 5 ) ( 2x\lg 3 + \lg 3 = x\lg 5 - 2\lg 5 ) ( 2x\lg 3 - x\lg 5 = -2\lg 5 - \lg 3 ) ( x(2\lg 3 - \lg 5) = -(2\lg 5 + \lg 3) ) ( x = \frac{-(2\lg 5 + \lg 3)}{2\lg 3 - \lg 5} = \frac{-(2(0.6990) + 0.4771)}{2(0.4771) - 0.6990} = \frac{-1.8751}{0.2552} = -7.35 ) (3 s.f.) [3 marks: 1 for taking logs, 1 for rearranging, 1 for correct answer]

14. ( \log_2[(x + 3)(x - 1)] = 3 ) ( (x + 3)(x - 1) = 2^3 = 8 ) ( x^2 + 2x - 3 = 8 \implies x^2 + 2x - 11 = 0 ) ( x = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2} = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3} ) Check domain: ( x + 3 > 0 ) and ( x - 1 > 0 \implies x > 1 ) ( x = -1 + 2\sqrt{3} \approx 2.46 > 1 ) ✓ ( x = -1 - 2\sqrt{3} \approx -4.46 < 1 ) ✗ Solution: ( x = -1 + 2\sqrt{3} ) [3 marks: 1 for combining logs, 1 for solving quadratic, 1 for checking domain]

15. ( \lg y = \lg a + x \lg b ) Gradient ( = \lg b = 0.255 \implies b = 10^{0.255} \approx 1.8 ) (2 s.f.) Intercept ( = \lg a = 0.523 \implies a = 10^{0.523} \approx 3.3 ) (2 s.f.) [3 marks: 1 for identifying relationships, 1 for b, 1 for a]

Section D: Binomial Expansions (Questions 16–20)

16. General term: ( T_{r+1} = \binom{5}{r} (2)^{5-r} (3x)^r = \binom{5}{r} 2^{5-r} 3^r x^r ) For ( x^3 ), ( r = 3 ): ( \binom{5}{3} 2^2 3^3 = 10 \times 4 \times 27 = 1080 ) [3 marks: 1 for general term, 1 for identifying r, 1 for coefficient]

17. General term: ( T_{r+1} = \binom{n}{r} (1)^{n-r} (-2x)^r = \binom{n}{r} (-2)^r x^r ) Coefficient of ( x^2 ): ( \binom{n}{2} (-2)^2 = \binom{n}{2} \times 4 = 60 ) ( \binom{n}{2} = 15 \implies \frac{n(n-1)}{2} = 15 \implies n(n-1) = 30 \implies n^2 - n - 30 = 0 ) ( (n - 6)(n + 5) = 0 \implies n = 6 ) (since ( n > 0 )) [3 marks: 1 for general term, 1 for equation, 1 for n]

18. General term: ( T_{r+1} = \binom{6}{r} (2x^2)^{6-r} \left(-\frac{1}{x}\right)^r = \binom{6}{r} 2^{6-r} (-1)^r x^{12-2r-r} = \binom{6}{r} 2^{6-r} (-1)^r x^{12-3r} ) For term independent of ( x ): ( 12 - 3r = 0 \implies r = 4 ) Term: ( \binom{6}{4} 2^2 (-1)^4 = 15 \times 4 \times 1 = 60 ) [3 marks: 1 for general term, 1 for finding r, 1 for term value]

19. ( (1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4 ) ( (1 - 2x)^3 = 1 + 3(-2x) + 3(-2x)^2 + (-2x)^3 = 1 - 6x + 12x^2 - 8x^3 ) Product up to ( x^2 ): Constant: ( 1 \times 1 = 1 ) ( x ): ( 1 \times (-6x) + 4x \times 1 = -6x + 4x = -2x ) ( x^2 ): ( 1 \times 12x^2 + 4x \times (-6x) + 6x^2 \times 1 = 12x^2 - 24x^2 + 6x^2 = -6x^2 ) ( (1 + x)^4(1 - 2x)^3 = 1 - 2x - 6x^2 + \cdots ) [3 marks: 1 for each expansion, 1 for correct multiplication up to x^2]

20. General term: ( T_{r+1} = \binom{8}{r} (1)^{8-r} (kx)^r = \binom{8}{r} k^r x^r ) Coefficient of ( x^3 ): ( \binom{8}{3} k^3 = 56k^3 ) Coefficient of ( x^4 ): ( \binom{8}{4} k^4 = 70k^4 ) Equal: ( 56k^3 = 70k^4 \implies 56k^3 - 70k^4 = 0 \implies 14k^3(4 - 5k) = 0 ) ( k = 0 ) or ( k = \frac{4}{5} ) Since ( k \neq 0 ) (otherwise trivial), ( k = \frac{4}{5} ) [3 marks: 1 for coefficients, 1 for equation, 1 for k]

End of Answer Key