Secondary 3 Additional Mathematics Algebra Functions Quiz

Secondary 3 Additional Mathematics Quiz - Algebra Functions

ANSWER KEY AND MARKING SCHEME

Total Marks: 50

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Section A: Short Answer (10 marks)

1. Solve $2x^2 - 5x - 3 = 0$ by factorisation. [2 marks]

Answer: $2x^2 - 5x - 3 = 0$ $(2x + 1)(x - 3) = 0$ [M1 - correct factorisation] $x = -\frac{1}{2}$ or $x = 3$ [A1 - both correct]

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2. Express $x^2 - 6x + 10$ in the form $(x - p)^2 + q$. [2 marks]

Answer: $x^2 - 6x + 10$ $= (x^2 - 6x + 9) + 1$ [M1 - completing the square] $= (x - 3)^2 + 1$ [A1] $p = 3$, $q = 1$

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3. Find the range of values of $k$ for which $x^2 + kx + 9 = 0$ has no real roots. [2 marks]

Answer: For no real roots: discriminant $< 0$ $\Delta = k^2 - 4(1)(9) = k^2 - 36$ [M1] $k^2 - 36 < 0$ $(k - 6)(k + 6) < 0$ $-6 < k < 6$ [A1]

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4. Given $(x + 2)$ is a factor of $f(x) = 2x^3 + 3x^2 - 8x - 12$, find the remaining quadratic factor. [2 marks]

Answer: By polynomial division or synthetic division: $2x^3 + 3x^2 - 8x - 12 = (x + 2)(2x^2 - x - 6)$ [M1 - correct division] Remaining quadratic factor: $2x^2 - x - 6$ [A1]

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5. Simplify $\sqrt{75} - \sqrt{12} + \sqrt{27}$ in the form $a\sqrt{3}$. [2 marks]

Answer: $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$ $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$ [M1 - simplifying each surd] $5\sqrt{3} - 2\sqrt{3} + 3\sqrt{3} = 6\sqrt{3}$ [A1]

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Section B: Structured Questions (24 marks)

6. $x^2 - 4x + 1 = 0$ has roots $\alpha$ and $\beta$.

(a) Find $\alpha + \beta$ and $\alpha\beta$. [2 marks]

Answer: $\alpha + \beta = -\frac{-4}{1} = 4$ [A1] $\alpha\beta = \frac{1}{1} = 1$ [A1]

(b) Find the quadratic equation whose roots are $\alpha^2$ and $\beta^2$. [4 marks]

Answer: Sum of new roots: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ $= 4^2 - 2(1) = 16 - 2 = 14$ [M1, A1]

Product of new roots: $\alpha^2\beta^2 = (\alpha\beta)^2 = 1^2 = 1$ [M1]

New equation: $x^2 - (\text{sum})x + (\text{product}) = 0$ $x^2 - 14x + 1 = 0$ [A1]

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7. $P(x) = x^3 + ax^2 + bx - 6$

(a) Write down two equations connecting $a$ and $b$. [3 marks]

Answer: $(x - 1)$ is a factor $\implies P(1) = 0$ $1 + a + b - 6 = 0$ $a + b = 5$ ... (1) [M1, A1]

Remainder when divided by $(x + 2)$ is $-12 \implies P(-2) = -12$ $(-2)^3 + a(-2)^2 + b(-2) - 6 = -12$ $-8 + 4a - 2b - 6 = -12$ $4a - 2b - 14 = -12$ $4a - 2b = 2$ $2a - b = 1$ ... (2) [M1, A1]

(b) Hence find $a$ and $b$. [2 marks]

Answer: From (1): $b = 5 - a$ Substitute into (2): $2a - (5 - a) = 1$ $2a - 5 + a = 1$ $3a = 6$ $a = 2$ [M1, A1] $b = 5 - 2 = 3$ [A1]

(c) Factorise $P(x)$ completely. [3 marks]

Answer: $P(x) = x^3 + 2x^2 + 3x - 6$ Since $(x - 1)$ is a factor, divide: $x^3 + 2x^2 + 3x - 6 = (x - 1)(x^2 + 3x + 6)$ [M1, A1]

Check discriminant of $x^2 + 3x + 6$: $\Delta = 9 - 24 = -15 < 0$, so it cannot be factorised further over real numbers. [A1]

$P(x) = (x - 1)(x^2 + 3x + 6)$

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8. (a) Expand $(2 - 3x)^4$ in ascending powers of $x$. [4 marks]

Answer: Using binomial theorem: $(a + b)^n$ with $a = 2$, $b = -3x$, $n = 4$

$(2 - 3x)^4 = \binom{4}{0}(2)^4(-3x)^0 + \binom{4}{1}(2)^3(-3x)^1 + \binom{4}{2}(2)^2(-3x)^2 + \binom{4}{3}(2)^1(-3x)^3 + \binom{4}{4}(2)^0(-3x)^4$

$= 1 \cdot 16 \cdot 1 + 4 \cdot 8 \cdot (-3x) + 6 \cdot 4 \cdot 9x^2 + 4 \cdot 2 \cdot (-27x^3) + 1 \cdot 1 \cdot 81x^4$ [M1, M1]

$= 16 - 96x + 216x^2 - 216x^3 + 81x^4$ [A2 - 1 mark per two correct terms]

(b) Find the coefficient of $x^2$ in $(1 + 2x)(2 - 3x)^4$. [2 marks]

Answer: $(1 + 2x)(16 - 96x + 216x^2 - 216x^3 + 81x^4)$

$x^2$ terms come from: $1 \cdot 216x^2$ and $2x \cdot (-96x) = -192x^2$ [M1]

Coefficient of $x^2 = 216 - 192 = 24$ [A1]

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9. Solve $\sqrt{2x + 5} - x = 1$. [4 marks]

Answer: $\sqrt{2x + 5} = x + 1$ [M1 - isolating surd]

Square both sides: $2x + 5 = (x + 1)^2$ $2x + 5 = x^2 + 2x + 1$ [M1] $0 = x^2 - 4$ $x^2 = 4$ $x = 2$ or $x = -2$ [M1]

Check in original equation: For $x = 2$: $\sqrt{2(2) + 5} - 2 = \sqrt{9} - 2 = 3 - 2 = 1$ ✓ For $x = -2$: $\sqrt{2(-2) + 5} - (-2) = \sqrt{1} + 2 = 1 + 2 = 3 \neq 1$ ✗ [A1 - both checks with correct conclusion]

$\therefore x = 2$ only.

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10. Find the set of values of $x$ for which $f(x) \leq 0$, where $f(x) = x^2 - 2x - 8$. [4 marks]

Answer: $x^2 - 2x - 8 \leq 0$ $(x - 4)(x + 2) \leq 0$ [M1 - factorisation]

Critical values: $x = -2$ and $x = 4$ [M1]

Sketch or sign analysis: For $x < -2$: $(-)(-) = (+) > 0$ For $-2 < x < 4$: $(-)(+) = (-) < 0$ [M1] For $x > 4$: $(+)(+) = (+) > 0$

$\therefore -2 \leq x \leq 4$ [A1]

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Section C: Application & Proof (16 marks)

11. $Q(x) = 2x^3 - 7x^2 + 7x - 2$

(a) Verify $(x - 2)$ is a factor. [1 mark]

Answer: $Q(2) = 2(8) - 7(4) + 7(2) - 2$ $= 16 - 28 + 14 - 2 = 0$ [A1] Since $Q(2) = 0$, $(x - 2)$ is a factor by the Factor Theorem.

(b) Factorise $Q(x)$ completely. [4 marks]

Answer: Divide $Q(x)$ by $(x - 2)$: $2x^3 - 7x^2 + 7x - 2 = (x - 2)(2x^2 - 3x + 1)$ [M1, A1]

Factorise the quadratic: $2x^2 - 3x + 1 = (2x - 1)(x - 1)$ [M1, A1]

$\therefore Q(x) = (x - 2)(2x - 1)(x - 1)$

(c) Hence solve $2x^3 - 7x^2 + 7x - 2 = 0$. [2 marks]

Answer: $(x - 2)(2x - 1)(x - 1) = 0$ $x - 2 = 0 \implies x = 2$ $2x - 1 = 0 \implies x = \frac{1}{2}$ [M1] $x - 1 = 0 \implies x = 1$

$\therefore x = \frac{1}{2}, 1, 2$ [A1 - all three]

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12. (a) Rationalise $\frac{5}{2\sqrt{3} - 1}$. [3 marks]

Answer: $\frac{5}{2\sqrt{3} - 1} \times \frac{2\sqrt{3} + 1}{2\sqrt{3} + 1}$ [M1 - multiplying by conjugate]

$= \frac{5(2\sqrt{3} + 1)}{(2\sqrt{3})^2 - 1^2}$ $= \frac{10\sqrt{3} + 5}{12 - 1}$ [M1] $= \frac{10\sqrt{3} + 5}{11}$ $= \frac{10}{11}\sqrt{3} + \frac{5}{11}$ [A1]

$a = \frac{10}{11}$, $b = \frac{5}{11}$ (or $a = 10$, $b = 5$ if denominator kept as 11)

(b) Simplify $\frac{5}{2\sqrt{3} - 1} - \frac{5}{2\sqrt{3} + 1}$. [2 marks]

Answer: Using result from (a): $\frac{5}{2\sqrt{3} - 1} = \frac{10\sqrt{3} + 5}{11}$

Similarly: $\frac{5}{2\sqrt{3} + 1} = \frac{5(2\sqrt{3} - 1)}{(2\sqrt{3})^2 - 1} = \frac{10\sqrt{3} - 5}{11}$ [M1]

Difference: $\frac{10\sqrt{3} + 5}{11} - \frac{10\sqrt{3} - 5}{11} = \frac{10}{11}$ [A1]

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13. The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}(2a + (n-1)d)$. The sum of the first 10 terms is 145, and the sum of the first 20 terms is 590. Find the first term $a$ and the common difference $d$. [4 marks]

Answer: $S_{10} = \frac{10}{2}(2a + 9d) = 5(2a + 9d) = 145$ $2a + 9d = 29$ ... (1) [M1, A1]

$S_{20} = \frac{20}{2}(2a + 19d) = 10(2a + 19d) = 590$ $2a + 19d = 59$ ... (2) [M1, A1]

(2) - (1): $10d = 30 \implies d = 3$ [M1] Substitute into (1): $2a + 9(3) = 29 \implies 2a + 27 = 29 \implies 2a = 2 \implies a = 1$ [A1]

$\therefore a = 1$, $d = 3$

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14. Solve the simultaneous equations: $y = x^2 - 3x + 4$ $y = 2x + 1$ [4 marks]

Answer: Equate: $x^2 - 3x + 4 = 2x + 1$ $x^2 - 5x + 3 = 0$ [M1] Using quadratic formula: $x = \frac{5 \pm \sqrt{25 - 12}}{2} = \frac{5 \pm \sqrt{13}}{2}$ [M1, A1]

Substitute into $y = 2x + 1$: $y = 2\left(\frac{5 \pm \sqrt{13}}{2}\right) + 1 = 5 \pm \sqrt{13} + 1 = 6 \pm \sqrt{13}$ [M1, A1]

Solutions: $\left( \frac{5 + \sqrt{13}}{2}, 6 + \sqrt{13} \right)$ and $\left( \frac{5 - \sqrt{13}}{2}, 6 - \sqrt{13} \right)$

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15. Given that $\log_2 x = a$ and $\log_2 y = b$, express $\log_2 \left( \frac{8x^3}{\sqrt{y}} \right)$ in terms of $a$ and $b$. [3 marks]

Answer: $\log_2 \left( \frac{8x^3}{\sqrt{y}} \right) = \log_2 8 + \log_2 x^3 - \log_2 \sqrt{y}$ [M1] $= \log_2 2^3 + 3\log_2 x - \log_2 y^{1/2}$ [M1] $= 3 + 3a - \frac{1}{2}b$ [A1]

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Section D: Problem Solving (10 marks)

16. A curve has equation $y = x^3 - 6x^2 + 9x + 1$. Find the coordinates of the stationary points and determine their nature. [5 marks]

Answer: $\frac{dy}{dx} = 3x^2 - 12x + 9$ [M1] Set $\frac{dy}{dx} = 0$: $3x^2 - 12x + 9 = 0 \implies x^2 - 4x + 3 = 0$ [M1] $(x - 1)(x - 3) = 0 \implies x = 1, 3$ [A1]

When $x = 1$: $y = 1 - 6 + 9 + 1 = 5$. Point: $(1, 5)$ When $x = 3$: $y = 27 - 54 + 27 + 1 = 1$. Point: $(3, 1)$ [A1]

$\frac{d^2y}{dx^2} = 6x - 12$ At $x = 1$: $\frac{d^2y}{dx^2} = 6 - 12 = -6 < 0 \implies$ maximum point $(1, 5)$ At $x = 3$: $\frac{d^2y}{dx^2} = 18 - 12 = 6 > 0 \implies$ minimum point $(3, 1)$ [A1]

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17. The roots of the quadratic equation $2x^2 - 3x + 5 = 0$ are $\alpha$ and $\beta$. Find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$. [2 marks]

Answer: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$ [M1] $\alpha + \beta = -\frac{-3}{2} = \frac{3}{2}$, $\alpha\beta = \frac{5}{2}$ $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3/2}{5/2} = \frac{3}{5}$ [A1]

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18. Solve the inequality $\frac{x-1}{x+2} > 0$. [3 marks]

Answer: Critical values: $x = 1$ and $x = -2$ [M1] Sign analysis: $x < -2$: $\frac{-}{-} = + > 0$ $-2 < x < 1$: $\frac{-}{+} = - < 0$ $x > 1$: $\frac{+}{+} = + > 0$ [M1] Solution: $x < -2$ or $x > 1$ [A1]

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19. Express $\frac{3x+5}{(x-1)(x+2)}$ in partial fractions. [3 marks]

Answer: Let $\frac{3x+5}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$ [M1] $3x + 5 = A(x+2) + B(x-1)$ Set $x = 1$: $3(1) + 5 = A(3) \implies 8 = 3A \implies A = \frac{8}{3}$ [M1] Set $x = -2$: $3(-2) + 5 = B(-3) \implies -1 = -3B \implies B = \frac{1}{3}$ [M1] $\frac{3x+5}{(x-1)(x+2)} = \frac{8/3}{x-1} + \frac{1/3}{x+2}$ [A1]

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20. Given that $f(x) = 3x^2 - 12x + 7$, express $f(x)$ in the form $a(x - h)^2 + k$ and hence state the minimum value of $f(x)$ and the value of $x$ at which it occurs. [3 marks]

Answer: $f(x) = 3(x^2 - 4x) + 7$ $= 3[(x - 2)^2 - 4] + 7$ [M1] $= 3(x - 2)^2 - 12 + 7$ $= 3(x - 2)^2 - 5$ [A1] Minimum value is $-5$, occurring at $x = 2$. [A1]

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END OF ANSWER KEY