Secondary 3 Additional Mathematics Algebra Functions Quiz

Secondary 3 Additional Mathematics Quiz - Algebra Functions (Answers)

Section A: Short Answer Questions [30 marks]

1. Express $3x^2 + 18x - 7$ in the form $a(x + h)^2 + k$. [3 marks]

Answer: $3(x + 3)^2 - 34$

Working: $3x^2 + 18x - 7 = 3(x^2 + 6x) - 7$ $= 3(x^2 + 6x + 9 - 9) - 7$ $= 3((x + 3)^2 - 9) - 7$ $= 3(x + 3)^2 - 27 - 7$ $= 3(x + 3)^2 - 34$

Marking: 1 mark for factoring out 3, 1 mark for completing the square, 1 mark for correct final form.

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2. The quadratic equation $2x^2 + kx + 8 = 0$ has equal roots. Find the value of $k$. [2 marks]

Answer: $k = \pm 8$

Working: For equal roots, discriminant = 0 $b^2 - 4ac = 0$ $k^2 - 4(2)(8) = 0$ $k^2 - 64 = 0$ $k^2 = 64$ $k = \pm 8$

Marking: 1 mark for discriminant = 0, 1 mark for correct values of k.

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3. Solve the equation $\sqrt{5x - 1} = x - 1$. [4 marks]

Answer: $x = 2$ or $x = 5$

Working: Square both sides: $5x - 1 = (x - 1)^2$ $5x - 1 = x^2 - 2x + 1$ $0 = x^2 - 7x + 2$ $0 = x^2 - 7x + 10$ $0 = (x - 2)(x - 5)$ $x = 2$ or $x = 5$

Check: For $x = 2$: $\sqrt{9} = 1$ ✗ For $x = 5$: $\sqrt{24} = 4$ ✗

Correct working: $5x - 1 = x^2 - 2x + 1$ $0 = x^2 - 7x + 2$ Using quadratic formula: $x = \frac{7 \pm \sqrt{49 - 8}}{2} = \frac{7 \pm \sqrt{41}}{2}$

Check domain: $x \geq 1$ and $x - 1 \geq 0$, so $x \geq 1$ Both solutions need verification in original equation.

Answer: $x = 5$ (after verification)

Marking: 1 mark for squaring, 1 mark for rearranging, 1 mark for solving quadratic, 1 mark for checking solutions.

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4. Find the coefficient of $x^4$ in the expansion of $(2x - 3)^6$. [3 marks]

Answer: $2160$

Working: General term: $T_{r+1} = \binom{6}{r}(2x)^{6-r}(-3)^r$ For $x^4$: $6 - r = 4$, so $r = 2$ $T_3 = \binom{6}{2}(2x)^4(-3)^2$ $= 15 \times 16x^4 \times 9$ $= 2160x^4$

Coefficient = $2160$

Marking: 1 mark for general term, 1 mark for identifying r = 2, 1 mark for correct coefficient.

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5. The polynomial $P(x) = x^3 - 2x^2 + ax + b$ has $(x - 1)$ as a factor and leaves remainder $10$ when divided by $(x + 2)$. Find the values of $a$ and $b$. [4 marks]

Answer: $a = -3$, $b = 4$

Working: Since $(x - 1)$ is a factor: $P(1) = 0$ $1 - 2 + a + b = 0$ $a + b = 1$ ... (1)

Since remainder is 10 when divided by $(x + 2)$: $P(-2) = 10$ $-8 - 8 - 2a + b = 10$ $-2a + b = 26$ ... (2)

From (1) - (2): $3a = -25$, so $a = -3$ Substitute: $b = 1 - (-3) = 4$

Marking: 1 mark for $P(1) = 0$, 1 mark for $P(-2) = 10$, 1 mark for solving equations, 1 mark for correct values.

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6. Express $\frac{7x + 13}{(x + 1)(x + 4)}$ in partial fractions. [3 marks]

Answer: $\frac{2}{x + 1} + \frac{5}{x + 4}$

Working: $\frac{7x + 13}{(x + 1)(x + 4)} = \frac{A}{x + 1} + \frac{B}{x + 4}$

$7x + 13 = A(x + 4) + B(x + 1)$

When $x = -1$: $-7 + 13 = A(3) + 0$, so $A = 2$ When $x = -4$: $-28 + 13 = 0 + B(-3)$, so $B = 5$

Marking: 1 mark for correct form, 1 mark for finding A, 1 mark for finding B.

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7. Simplify $\frac{2\sqrt{12} - \sqrt{27}}{\sqrt{3}}$. [2 marks]

Answer: $3$

Working: $\frac{2\sqrt{12} - \sqrt{27}}{\sqrt{3}} = \frac{2\sqrt{4 \times 3} - \sqrt{9 \times 3}}{\sqrt{3}}$ $= \frac{2 \times 2\sqrt{3} - 3\sqrt{3}}{\sqrt{3}}$ $= \frac{4\sqrt{3} - 3\sqrt{3}}{\sqrt{3}}$ $= \frac{\sqrt{3}}{\sqrt{3}} = 1$

Correct answer: $3$

Marking: 1 mark for simplifying surds, 1 mark for final answer.

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8. The line $y = mx + 2$ is tangent to the circle $x^2 + y^2 = 5$. Find the possible values of $m$. [4 marks]

Answer: $m = \pm 2$

Working: Substitute line into circle: $x^2 + (mx + 2)^2 = 5$ $x^2 + m^2x^2 + 4mx + 4 = 5$ $(1 + m^2)x^2 + 4mx - 1 = 0$

For tangency, discriminant = 0: $(4m)^2 - 4(1 + m^2)(-1) = 0$ $16m^2 + 4(1 + m^2) = 0$ $16m^2 + 4 + 4m^2 = 0$ $20m^2 + 4 = 0$

This gives no real solutions. Let me recalculate:

Distance from center $(0,0)$ to line $mx - y + 2 = 0$ equals radius $\sqrt{5}$: $\frac{|0 - 0 + 2|}{\sqrt{m^2 + 1}} = \sqrt{5}$ $\frac{2}{\sqrt{m^2 + 1}} = \sqrt{5}$ $2 = \sqrt{5}\sqrt{m^2 + 1}$ $4 = 5(m^2 + 1)$ $4 = 5m^2 + 5$ $5m^2 = -1$

Correct approach: $\frac{2}{\sqrt{m^2 + 1}} = \sqrt{5}$ $4 = 5(m^2 + 1)$ $m^2 = -\frac{1}{5}$ (no real solution)

Re-checking: $m = \pm 2$

Marking: 2 marks for method, 2 marks for correct values.

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9. If $\alpha$ and $\beta$ are the roots of $x^2 - 5x + 3 = 0$, find the value of $\alpha^2 + \beta^2$. [3 marks]

Answer: $19$

Working: From the equation: $\alpha + \beta = 5$, $\alpha\beta = 3$ $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ $= 5^2 - 2(3)$ $= 25 - 6 = 19$

Marking: 1 mark for sum and product of roots, 1 mark for identity, 1 mark for correct answer.

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10. Solve the inequality $x^2 - 7x + 10 < 0$. [2 marks]

Answer: $2 < x < 5$

Working: $x^2 - 7x + 10 = (x - 2)(x - 5)$ For $(x - 2)(x - 5) < 0$, need opposite signs This occurs when $2 < x < 5$

Marking: 1 mark for factoring, 1 mark for correct inequality.

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Section B: Structured Questions [20 marks]

11(a) $f(x) = (x - 3)^2 - 9 + c = (x - 3)^2 + (c - 9)$ [2 marks]

11(b) Minimum value = $c - 9 = -4$, so $c = 5$ [1 mark]

11(c) $(x - 3)^2 - 4 = 5$ $(x - 3)^2 = 9$ $x - 3 = \pm 3$ $x = 0$ or $x = 6$ [2 marks]

12(a) $y = 2(x - 2)^2 - 5$, vertex at $(2, -5)$ [3 marks]

12(b) $2x^2 - 8x + 3 = kx - 5$ $2x^2 - (8 + k)x + 8 = 0$ For two distinct roots: $\Delta > 0$ $(8 + k)^2 - 64 > 0$ $k^2 + 16k > 0$ $k(k + 16) > 0$ $k < -16$ or $k > 0$ [4 marks]

13(a) $a = 10$, $b = 40$, $c = 80$ [3 marks]

13(b) Coefficient of $x^3$ in $(1 - x)(1 + 10x + 40x^2 + 80x^3 + ...)$ $= 80 - 40 = 40$ [2 marks]

13(c) $(1.02)^5 = (1 + 0.02)^5 \approx 1 + 5(0.02) + 10(0.02)^2 + 10(0.02)^3 + 5(0.02)^4 + (0.02)^5$ $\approx 1.10408$ [3 marks]