Secondary 4 Additional Mathematics Algebra Functions Quiz

Secondary 4 Additional Mathematics Quiz - Algebra Functions

ANSWER KEY AND MARKING SCHEME

Total Marks: 50

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Section A: Quadratic Functions and Equations (15 marks)

1. Express $y = 2x^2 + 8x - 3$ in the form $y = a(x + h)^2 + k$, and state the minimum value and where it occurs.

Answer: $y = 2(x^2 + 4x) - 3$ $y = 2[(x + 2)^2 - 4] - 3$ [M1 - completing square inside bracket] $y = 2(x + 2)^2 - 8 - 3$ $y = 2(x + 2)^2 - 11$ [A1 - correct form]

Minimum value of $y = -11$, occurring at $x = -2$. [A1 - both correct]

Marking: M1 for method of completing square, A1 for correct form, A1 for minimum value and x-value. [3 marks]

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2. Find the range of values of $k$ for which $x^2 + (k - 2)x + (k + 3) = 0$ has two distinct real roots.

Answer: For two distinct real roots, discriminant $b^2 - 4ac > 0$. Here $a = 1$, $b = k - 2$, $c = k + 3$.

$(k - 2)^2 - 4(1)(k + 3) > 0$ [M1 - correct discriminant set up] $k^2 - 4k + 4 - 4k - 12 > 0$ $k^2 - 8k - 8 > 0$ [M1 - simplification]

Solve $k^2 - 8k - 8 = 0$: $k = \frac{8 \pm \sqrt{64 + 32}}{2} = \frac{8 \pm \sqrt{96}}{2} = \frac{8 \pm 4\sqrt{6}}{2} = 4 \pm 2\sqrt{6}$

Since coefficient of $k^2$ is positive, $k^2 - 8k - 8 > 0$ when: $k < 4 - 2\sqrt{6}$ or $k > 4 + 2\sqrt{6}$ [A1 - correct range]

Marking: M1 for discriminant expression, M1 for simplifying inequality, A1 for correct range. [3 marks]

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3. Find $p$, $q$, $r$ given minimum at $(1, -2)$ and passes through $(3, 6)$.

Answer: $y = px^2 + qx + r$ At minimum $(1, -2)$: $\frac{dy}{dx} = 2px + q = 0$ at $x = 1$, so $2p + q = 0$ ... (1) [M1] Also $p(1)^2 + q(1) + r = -2$, so $p + q + r = -2$ ... (2) [M1]

At $(3, 6)$: $p(9) + q(3) + r = 6$, so $9p + 3q + r = 6$ ... (3) [M1]

From (1): $q = -2p$ Substitute into (2): $p - 2p + r = -2$, so $-p + r = -2$, $r = p - 2$ ... (4) Substitute into (3): $9p + 3(-2p) + r = 6$, so $9p - 6p + r = 6$, $3p + r = 6$ ... (5)

From (4) and (5): $3p + (p - 2) = 6$, $4p = 8$, $p = 2$ Then $q = -2(2) = -4$, $r = 2 - 2 = 0$

$\therefore p = 2$, $q = -4$, $r = 0$ [A1 - all correct]

Marking: M1 for derivative condition, M1 for point condition at minimum, M1 for point condition at (3,6), A1 for all three values. [4 marks]

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4. Solve $2x^2 - 5x - 3 \leq 0$ and represent on a number line.

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

Since coefficient of $x^2$ is positive, the parabola opens upward. $\therefore 2x^2 - 5x - 3 \leq 0$ when $-\frac{1}{2} \leq x \leq 3$ [A1 - correct inequality]

Number line: [A1 - correct representation]

    ←———|===========|———→
       -½           3

(Shaded region between $-\frac{1}{2}$ and 3 inclusive, with closed circles at endpoints)

Marking: M1 for finding critical values, A1 for correct inequality, A1 for correct number line. [3 marks]

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5. Find condition on $m$ such that $y = mx + 2$ is tangent to $y = x^2 + 3x + 1$.

Answer: At intersection: $mx + 2 = x^2 + 3x + 1$ $x^2 + (3 - m)x - 1 = 0$ [M1 - forming quadratic]

For tangency, discriminant = 0: $(3 - m)^2 - 4(1)(-1) = 0$ $(3 - m)^2 + 4 = 0$ [M1 - discriminant set to zero]

$(3 - m)^2 = -4$, which has no real solutions. $\therefore$ No real value of $m$ exists for which the line is tangent. [A1]

Alternative interpretation: If the question expects a condition, state that no such $m$ exists.

Marking: M1 for equating and forming quadratic, M1 for discriminant = 0, A1 for conclusion. [2 marks]

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Section B: Polynomials and Partial Fractions (15 marks)

6. Find $a$ and $b$ given $P(x) = 2x^3 + ax^2 + bx - 6$ has factor $(x + 2)$ and remainder 20 when divided by $(x - 1)$.

Answer: Factor $(x + 2)$ means $P(-2) = 0$: $2(-8) + a(4) + b(-2) - 6 = 0$ $-16 + 4a - 2b - 6 = 0$ $4a - 2b = 22$ $2a - b = 11$ ... (1) [M1]

Remainder when divided by $(x - 1)$ is $P(1) = 20$: $2(1) + a(1) + b(1) - 6 = 20$ $2 + a + b - 6 = 20$ $a + b = 24$ ... (2) [M1]

From (1): $b = 2a - 11$ Substitute into (2): $a + (2a - 11) = 24$ $3a = 35$ $a = \frac{35}{3}$ [A1]

$b = 2(\frac{35}{3}) - 11 = \frac{70}{3} - \frac{33}{3} = \frac{37}{3}$ [A1]

$\therefore a = \frac{35}{3}$, $b = \frac{37}{3}$

Marking: M1 for factor theorem, M1 for remainder theorem, A1 for $a$, A1 for $b$. [4 marks]

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7. Factorise completely $x^3 - 3x^2 - 4x + 12$.

Answer: Try $x = 2$: $8 - 12 - 8 + 12 = 0$, so $(x - 2)$ is a factor. [M1 - finding one factor]

Divide by $(x - 2)$: $x^3 - 3x^2 - 4x + 12 = (x - 2)(x^2 - x - 6)$ [M1 - polynomial division] $= (x - 2)(x - 3)(x + 2)$ [A1 - complete factorisation]

Marking: M1 for identifying a factor, M1 for division, A1 for complete factorisation. [3 marks]

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8. Express $\frac{3x + 5}{(x + 1)(x - 2)}$ in partial fractions.

Answer: Let $\frac{3x + 5}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}$ [M1 - correct form]

$3x + 5 = A(x - 2) + B(x + 1)$

When $x = -1$: $3(-1) + 5 = A(-3)$, $2 = -3A$, $A = -\frac{2}{3}$ [M1] When $x = 2$: $3(2) + 5 = B(3)$, $11 = 3B$, $B = \frac{11}{3}$ [M1]

$\therefore \frac{3x + 5}{(x + 1)(x - 2)} = -\frac{2}{3(x + 1)} + \frac{11}{3(x - 2)}$ [A1]

Marking: M1 for correct form, M1 for multiplying through, M1 for finding one constant, A1 for both correct. [3 marks]

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9. Express $\frac{2x^2 + 3x + 4}{(x + 1)(x^2 + 1)}$ in partial fractions.

Answer: Let $\frac{2x^2 + 3x + 4}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1}$ [M1 - correct form]

$2x^2 + 3x + 4 = A(x^2 + 1) + (Bx + C)(x + 1)$ $= Ax^2 + A + Bx^2 + Bx + Cx + C$ $= (A + B)x^2 + (B + C)x + (A + C)$ [M1 - expansion]

Comparing coefficients: $x^2$: $A + B = 2$ ... (1) $x$: $B + C = 3$ ... (2) Constant: $A + C = 4$ ... (3) [M1]

From (1): $B = 2 - A$ From (3): $C = 4 - A$ Substitute into (2): $(2 - A) + (4 - A) = 3$ $6 - 2A = 3$ $2A = 3$ $A = \frac{3}{2}$ [A1]

$B = 2 - \frac{3}{2} = \frac{1}{2}$ $C = 4 - \frac{3}{2} = \frac{5}{2}$ [A1]

$\therefore \frac{2x^2 + 3x + 4}{(x + 1)(x^2 + 1)} = \frac{3}{2(x + 1)} + \frac{x + 5}{2(x^2 + 1)}$

Marking: M1 for correct form, M1 for expansion, M1 for comparing coefficients, A1 for $A$, A1 for $B$ and $C$. [5 marks]

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10. Given that $f(x) = x^3 - 4x^2 + x + 6$, find the remainder when $f(x)$ is divided by $(x - 3)$.

Answer: By the Remainder Theorem, remainder = $f(3)$. [M1] $f(3) = (3)^3 - 4(3)^2 + 3 + 6$ $= 27 - 36 + 3 + 6$ $= 0$ [A1]

Marking: M1 for applying remainder theorem, A1 for correct remainder. [2 marks]

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Section C: Binomial Expansions (10 marks)

11. Find the coefficient of $x^3$ in $(2 - 3x)^5$.

Answer: General term: $\binom{5}{r}(2)^{5-r}(-3x)^r = \binom{5}{r}2^{5-r}(-3)^r x^r$ [M1]

For $x^3$, $r = 3$: Term = $\binom{5}{3}2^{2}(-3)^3 x^3$ [M1] $= 10 \times 4 \times (-27) x^3$ $= -1080x^3$

Coefficient of $x^3 = -1080$ [A1]

Marking: M1 for general term, M1 for substituting $r = 3$, A1 for correct coefficient. [3 marks]

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12. Given $(1 + px)^n = 1 + 12x + 60x^2 + ...$, find $n$ and $p$.

Answer: $(1 + px)^n = 1 + \binom{n}{1}px + \binom{n}{2}p^2x^2 + ...$ $= 1 + npx + \frac{n(n-1)}{2}p^2x^2 + ...$ [M1]

Comparing coefficients: $np = 12$ ... (1) [M1] $\frac{n(n-1)}{2}p^2 = 60$ ... (2) [M1]

From (1): $p = \frac{12}{n}$ Substitute into (2): $\frac{n(n-1)}{2} \cdot \frac{144}{n^2} = 60$ $\frac{144(n-1)}{2n} = 60$ $\frac{72(n-1)}{n} = 60$ $72n - 72 = 60n$ $12n = 72$ $n = 6$ [A1]

$p = \frac{12}{6} = 2$ [A1]

$\therefore n = 6$, $p = 2$

Marking: M1 for expansion form, M1 for equating first coefficient, M1 for equating second coefficient, A1 for $n$, A1 for $p$. [4 marks]

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13. Find the term independent of $x$ in $\left(2x^2 - \frac{1}{x}\right)^9$.

Answer: General term: $\binom{9}{r}(2x^2)^{9-r}\left(-\frac{1}{x}\right)^r$ $= \binom{9}{r}2^{9-r}(-1)^r x^{2(9-r)} \cdot x^{-r}$ $= \binom{9}{r}2^{9-r}(-1)^r x^{18-2r-r}$ $= \binom{9}{r}2^{9-r}(-1)^r x^{18-3r}$ [M1]

For term independent of $x$: $18 - 3r = 0$ $r = 6$ [M1]

Term = $\binom{9}{6}2^{3}(-1)^6 = 84 \times 8 \times 1 = 672$ [A1]

Marking: M1 for general term with correct power of $x$, M1 for solving $r$, A1 for correct term. [3 marks]

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Section D: Exponential and Logarithmic Functions (10 marks)

14. Solve $2^{2x+1} - 5(2^x) + 2 = 0$.

Answer: $2^{2x+1} = 2 \cdot 2^{2x} = 2(2^x)^2$ Let $y = 2^x$: [M1] $2y^2 - 5y + 2 = 0$ $(2y - 1)(y - 2) = 0$ [M1] $y = \frac{1}{2}$ or $y = 2$

When $y = \frac{1}{2}$: $2^x = 2^{-1}$, so $x = -1$ [A1] When $y = 2$: $2^x = 2^1$, so $x = 1$ [A1]

$\therefore x = -1$ or $x = 1$

Marking: M1 for substitution, M1 for solving quadratic, A1 for each correct solution. [4 marks]

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15. Solve $\log_2(x + 1) + \log_2(x - 1) = 3$.

Answer: $\log_2[(x + 1)(x - 1)] = 3$ [M1] $\log_2(x^2 - 1) = 3$ $x^2 - 1 = 2^3 = 8$ [M1] $x^2 = 9$ $x = 3$ or $x = -3$

Check domain: $x + 1 > 0$ and $x - 1 > 0 \implies x > 1$. $\therefore x = 3$ only. [A1]

Marking: M1 for combining logs, M1 for converting to exponential form, A1 for correct solution with domain check. [3 marks]

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16. Given $\log_a b = 3$ and $\log_a c = 2$, find $\log_a\left(\frac{b^2}{c^3}\right)$.

Answer: $\log_a\left(\frac{b^2}{c^3}\right) = \log_a(b^2) - \log_a(c^3)$ [M1] $= 2\log_a b - 3\log_a c$ [M1] $= 2(3) - 3(2) = 6 - 6 = 0$ [A1]

Marking: M1 for quotient rule, M1 for power rule, A1 for correct value. [3 marks]

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Section E: Surds (5 marks)

17. Simplify $\frac{3}{\sqrt{5} - 2}$, giving your answer in the form $a + b\sqrt{5}$.

Answer: $\frac{3}{\sqrt{5} - 2} = \frac{3(\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)}$ [M1] $= \frac{3\sqrt{5} + 6}{5 - 4}$ $= 3\sqrt{5} + 6$ [A1] $= 6 + 3\sqrt{5}$

Marking: M1 for rationalising denominator, A1 for correct simplified form. [2 marks]

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18. Solve $\sqrt{2x + 5} - \sqrt{x - 1} = 2$.

Answer: $\sqrt{2x + 5} = 2 + \sqrt{x - 1}$ Square both sides: $2x + 5 = 4 + 4\sqrt{x - 1} + (x - 1)$ [M1] $2x + 5 = x + 3 + 4\sqrt{x - 1}$ $x + 2 = 4\sqrt{x - 1}$ Square again: $(x + 2)^2 = 16(x - 1)$ [M1] $x^2 + 4x + 4 = 16x - 16$ $x^2 - 12x + 20 = 0$ $(x - 2)(x - 10) = 0$ $x = 2$ or $x = 10$

Check: For $x = 2$: $\sqrt{9} - \sqrt{1} = 3 - 1 = 2$ (valid) For $x = 10$: $\sqrt{25} - \sqrt{9} = 5 - 3 = 2$ (valid) [A1]

$\therefore x = 2$ or $x = 10$

Marking: M1 for isolating and squaring once, M1 for squaring again and solving quadratic, A1 for both correct solutions with check. [3 marks]

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19. Rationalise the denominator of $\frac{4}{\sqrt{7} + \sqrt{3}}$.

Answer: $\frac{4}{\sqrt{7} + \sqrt{3}} = \frac{4(\sqrt{7} - \sqrt{3})}{(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3})}$ [M1] $= \frac{4(\sqrt{7} - \sqrt{3})}{7 - 3}$ $= \frac{4(\sqrt{7} - \sqrt{3})}{4}$ $= \sqrt{7} - \sqrt{3}$ [A1]

Marking: M1 for multiplying by conjugate, A1 for correct simplified form. [2 marks]

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20. Solve $\sqrt{3x + 1} + \sqrt{x - 4} = 5$.

Answer: $\sqrt{3x + 1} = 5 - \sqrt{x - 4}$ Square both sides: $3x + 1 = 25 - 10\sqrt{x - 4} + (x - 4)$ [M1] $3x + 1 = x + 21 - 10\sqrt{x - 4}$ $2x - 20 = -10\sqrt{x - 4}$ $x - 10 = -5\sqrt{x - 4}$ Square again: $(x - 10)^2 = 25(x - 4)$ [M1] $x^2 - 20x + 100 = 25x - 100$ $x^2 - 45x + 200 = 0$ $(x - 5)(x - 40) = 0$ $x = 5$ or $x = 40$

Check: For $x = 5$: $\sqrt{16} + \sqrt{1} = 4 + 1 = 5$ (valid) For $x = 40$: $\sqrt{121} + \sqrt{36} = 11 + 6 = 17 \neq 5$ (extraneous) [A1]

$\therefore x = 5$

Marking: M1 for isolating and squaring once, M1 for squaring again and solving quadratic, A1 for correct solution with check. [3 marks]

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END OF ANSWER KEY