Secondary 4 Additional Mathematics Algebra Functions Quiz

Secondary 4 Additional Mathematics Quiz - Algebra Functions

Answer Key

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Section A: Short Answer Questions (20 marks)

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1. (a) Express $f(x) = 2x^2 - 8x + 5$ in the form $a(x - h)^2 + k$. (2 marks)

Working: $f(x) = 2x^2 - 8x + 5$ $= 2(x^2 - 4x) + 5$ $= 2\left[(x - 2)^2 - 4\right] + 5$ $= 2(x - 2)^2 - 8 + 5$ $\boxed{f(x) = 2(x - 2)^2 - 3}$

Marking: 1 mark for correct factorisation of 2; 1 mark for correct completed square form.

(b) Coordinates of the minimum point. (2 marks)

From part (a), $h = 2$ and $k = -3$. Since $a = 2 > 0$, the parabola opens upwards and the minimum occurs at the vertex.

$\boxed{\text{Minimum point: } (2, -3)}$

Marking: 1 mark for identifying $x = 2$; 1 mark for $y = -3$.

Common mistake: Students may forget that the sign changes when reading $h$ from $(x - h)^2$. For $(x - 2)^2$, $h = 2$, not $-2$.

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2. Find the range of values of $k$ for which $x^2 + 4x + k = 0$ has no real roots. (4 marks)

Working:

For no real roots, the discriminant $D < 0$.

$D = b^2 - 4ac = 4^2 - 4(1)(k) = 16 - 4k$

For no real roots: $16 - 4k < 0$ $16 < 4k$ $k > 4$

$\boxed{k > 4}$

Marking: 1 mark for correct discriminant formula; 1 mark for correct substitution; 1 mark for correct inequality; 1 mark for final answer.

Common mistake: Students may write $k < 4$ instead of $k > 4$ due to sign error when dividing by $-4$.

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3. The quadratic function $f(x) = x^2 + px + 9$ is always positive for all real $x$. Find the range of values of $p$. (4 marks)

Working:

For $f(x)$ to be always positive for all real $x$:

• The coefficient of $x^2$ is $1 > 0$ ✓ (parabola opens upwards)
• The discriminant $D < 0$ (no real roots, so the graph never touches or crosses the $x$-axis)

$D = p^2 - 4(1)(9) = p^2 - 36$

For always positive: $p^2 - 36 < 0$ $p^2 < 36$ $-6 < p < 6$

$\boxed{-6 < p < 6}$

Marking: 1 mark for stating the condition $D < 0$; 1 mark for correct discriminant; 1 mark for solving the inequality; 1 mark for final answer.

Common mistake: Students may forget to check that $a > 0$ (though here $a = 1$ is given). They may also write $p < 6$ instead of the compound inequality.

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4. The line $y = 3x - 7$ is tangent to the curve $y = x^2 + ax + b$ at $x = 2$. Find $a$ and $b$. (4 marks)

Working:

Since the line is tangent at $x = 2$:

Condition 1: The point lies on both the line and the curve.

At $x = 2$, on the line: $y = 3(2) - 7 = -1$.

So the point $(2, -1)$ lies on the curve: $4 + 2a + b = -1 \quad \Rightarrow \quad 2a + b = -5 \quad \text{...(i)}$

Condition 2: The gradient of the curve equals the gradient of the line at $x = 2$.

Gradient of line $= 3$.

Gradient of curve: $\frac{dy}{dx} = 2x + a$.

At $x = 2$: $2(2) + a = 3$, so $4 + a = 3$, giving $a = -1$.

Substituting into (i): $2(-1) + b = -5$, so $-2 + b = -5$, giving $b = -3$.

$\boxed{a = -1, \quad b = -3}$

Marking: 1 mark for finding the point of contact; 1 mark for using the gradient condition; 1 mark for solving for $a$; 1 mark for solving for $b$.

Common mistake: Students may only use one condition and not realise both the point and the gradient must match for tangency.

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5. The equation $mx^2 - 6x + 2 = 0$ has two distinct real roots. Find the range of values of $m$. (4 marks)

Working:

For two distinct real roots, $D > 0$ and $m \neq 0$ (must be quadratic).

$D = (-6)^2 - 4(m)(2) = 36 - 8m$

For two distinct real roots: $36 - 8m > 0$ $36 > 8m$ $m < \frac{9}{2}$

Also, for the equation to be quadratic, $m \neq 0$.

$\boxed{m < \frac{9}{2} \text{ and } m \neq 0}$

Marking: 1 mark for correct discriminant; 1 mark for solving $D > 0$; 1 mark for $m < \frac{9}{2}$; 1 mark for stating $m \neq 0$.

Common mistake: Students may forget that $m \neq 0$ is required for the equation to remain quadratic.

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Section B: Structured Questions (24 marks)

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6. $h(t) = -5t^2 + 30t + 10$, $t \geq 0$.

(a) Express $h(t)$ in the form $a(t - p)^2 + q$. (2 marks)

$h(t) = -5t^2 + 30t + 10$ $= -5(t^2 - 6t) + 10$ $= -5\left[(t - 3)^2 - 9\right] + 10$ $= -5(t - 3)^2 + 45 + 10$ $\boxed{h(t) = -5(t - 3)^2 + 55}$

Marking: 1 mark for correct factorisation; 1 mark for correct completed square form.

(b) Maximum height. (2 marks)

Since $a = -5 < 0$, the parabola opens downwards. The maximum occurs at the vertex $(3, 55)$.

$\boxed{\text{Maximum height} = 55 \text{ metres}}$

Marking: 1 mark for identifying the vertex; 1 mark for the correct value.

(c) Time at maximum height. (2 marks)

$\boxed{t = 3 \text{ seconds}}$

Marking: 2 marks for correct answer (or follow-through from part (a)).

(d) Height of the platform. (2 marks)

The platform height is $h(0)$: $h(0) = -5(0)^2 + 30(0) + 10 = 10$

$\boxed{\text{Platform height} = 10 \text{ metres}}$

Marking: 2 marks for correct answer.

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7. $f(x) = x^2 - 6x + 13$.

(a) Express $f(x)$ in the form $(x - a)^2 + b$. (2 marks)

$f(x) = x^2 - 6x + 13$ $= (x - 3)^2 - 9 + 13$ $\boxed{f(x) = (x - 3)^2 + 4}$

Marking: 1 mark for $(x - 3)^2$; 1 mark for $+4$.

(b) Least value and where it occurs. (2 marks)

Since $(x - 3)^2 \geq 0$, the least value is $4$ when $x = 3$.

$\boxed{\text{Least value} = 4 \text{ at } x = 3}$

Marking: 1 mark for the value; 1 mark for the $x$-value.

(c) Range of $c$ for which $f(x) = c$ has two distinct real roots. (2 marks)

$f(x) = c \implies (x - 3)^2 + 4 = c \implies (x - 3)^2 = c - 4$

For two distinct real roots, we need $c - 4 > 0$, so $c > 4$.

$\boxed{c > 4}$

Marking: 1 mark for setting up the equation; 1 mark for $c > 4$.

(d) Range of $m$ for which $y = mx + 1$ intersects $y = f(x)$ at two distinct points. (2 marks)

Setting $f(x) = mx + 1$: $x^2 - 6x + 13 = mx + 1$ $x^2 - (6 + m)x + 12 = 0$

For two distinct intersections, $D > 0$: $[-(6 + m)]^2 - 4(1)(12) > 0$ $(6 + m)^2 - 48 > 0$ $(6 + m)^2 > 48$ $6 + m > \sqrt{48} \quad \text{or} \quad 6 + m < -\sqrt{48}$ $m > 4\sqrt{3} - 6 \quad \text{or} \quad m < -4\sqrt{3} - 6$

$\boxed{m > 4\sqrt{3} - 6 \text{ or } m < -4\sqrt{3} - 6}$

Marking: 1 mark for correct quadratic in $x$; 1 mark for correct range of $m$.

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8. $f(x) = ax^2 + bx + 8$ passes through $(1, 5)$ and $(-2, 20)$.

(a) Form two simultaneous equations. (2 marks)

At $(1, 5)$: $a(1)^2 + b(1) + 8 = 5 \implies a + b = -3$ ...(i)

At $(-2, 20)$: $a(-2)^2 + b(-2) + 8 = 20 \implies 4a - 2b = 12$ ...(ii)

$\boxed{a + b = -3 \quad \text{and} \quad 4a - 2b = 12}$

Marking: 1 mark for each correct equation.

(b) Solve for $a$ and $b$. (2 marks)

From (i): $b = -3 - a$.

Substituting into (ii): $4a - 2(-3 - a) = 12$ $4a + 6 + 2a = 12$ $6a = 6 \implies a = 1$

Then $b = -3 - 1 = -4$.

$\boxed{a = 1, \quad b = -4}$

Marking: 1 mark for correct substitution; 1 mark for correct values.

(c) Express $f(x)$ in the form $p(x - q)^2 + r$. (2 marks)

$f(x) = x^2 - 4x + 8$ $= (x - 2)^2 - 4 + 8$ $\boxed{f(x) = (x - 2)^2 + 4}$

Marking: 1 mark for correct process; 1 mark for correct answer.

(d) Coordinates of the vertex. (2 marks)

$\boxed{\text{Vertex: } (2, 4)}$

Marking: 2 marks for correct answer (or follow-through from part (c)).

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Section C: Application and Problem Solving (16 marks)

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9. Rectangular garden fenced on three sides, 40 m of fencing.

(a) Show that $A = 40x - 2x^2$. (2 marks)

Let $x$ = length perpendicular to wall, and $y$ = length parallel to wall.

Total fencing: $2x + y = 40$, so $y = 40 - 2x$.

Area: $A = xy = x(40 - 2x) = 40x - 2x^2$.

$\boxed{A = 40x - 2x^2}$

Marking: 1 mark for expressing $y$ in terms of $x$; 1 mark for the area expression.

(b) Express $A$ in the form $a(x - h)^2 + k$. (2 marks)

$A = 40x - 2x^2 = -2x^2 + 40x$ $= -2(x^2 - 20x)$ $= -2\left[(x - 10)^2 - 100\right]$ $= -2(x - 10)^2 + 200$

$\boxed{A = -2(x - 10)^2 + 200}$

Marking: 1 mark for correct factorisation; 1 mark for correct completed square form.

(c) Maximum possible area. (2 marks)

Since $a = -2 < 0$, maximum occurs at $x = 10$: $A_{\max} = -2(0)^2 + 200 = 200$

$\boxed{\text{Maximum area} = 200 \text{ m}^2}$

Marking: 1 mark for identifying $x = 10$; 1 mark for the maximum area.

(d) Dimensions when area is maximum. (2 marks)

When $x = 10$: $y = 40 - 2(10) = 20$.

$\boxed{\text{Dimensions: } 10 \text{ m} \times 20 \text{ m}}$

Marking: 1 mark for finding $y$; 1 mark for stating both dimensions.

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10. $f(x) = x^2 - 2kx + k^2 - 4$.

(a) Express in the form $(x - a)^2 + b$. (2 marks)

$f(x) = x^2 - 2kx + k^2 - 4$ $= (x - k)^2 - 4$

$\boxed{f(x) = (x - k)^2 - 4}$

Marking: 2 marks for correct answer.

(b) Coordinates of the minimum point. (2 marks)

Since $(x - k)^2 \geq 0$, the minimum value is $-4$ when $x = k$.

$\boxed{\text{Minimum point: } (k, -4)}$

Marking: 1 mark for $x = k$; 1 mark for $y = -4$.

(c) Roots of $f(x) = 0$. (2 marks)

$(x - k)^2 - 4 = 0$ $(x - k)^2 = 4$ $x - k = \pm 2$ $x = k + 2 \quad \text{or} \quad x = k - 2$

$\boxed{x = k + 2 \quad \text{or} \quad x = k - 2}$

Marking: 1 mark for $(x - k)^2 = 4$; 1 mark for both roots.

(d) Length $PQ$. (2 marks)

Points $P$ and $Q$ are at $x = k - 2$ and $x = k + 2$ on the $x$-axis.

$PQ = (k + 2) - (k - 2) = 4$

$\boxed{PQ = 4 \text{ units}}$

Marking: 1 mark for identifying the $x$-coordinates; 1 mark for the length.

Common mistake: Students may try to use the distance formula with $y$-coordinates, but since both points are on the $x$-axis, the distance is simply the difference in $x$-coordinates.

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Section D: Further Practice (20 marks)

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11. Express $x^2 + 6x - 4$ in the form $(x + a)^2 + b$. (2 marks)

$x^2 + 6x - 4 = (x + 3)^2 - 9 - 4 = (x + 3)^2 - 13$

$\boxed{a = 3, \quad b = -13}$

Marking: 1 mark for $a = 3$; 1 mark for $b = -13$.

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12. Find the discriminant of $3x^2 - 5x + 1 = 0$. (2 marks)

$D = (-5)^2 - 4(3)(1) = 25 - 12 = 13$

$\boxed{D = 13}$

Marking: 1 mark for correct formula; 1 mark for correct value.

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13. Condition for two equal real roots. (2 marks)

$\boxed{D = 0 \quad \text{or} \quad b^2 - 4ac = 0}$

Marking: 2 marks for correct condition.

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14. Least value of $f(x) = x^2 - 4x + 7$. (2 marks)

$f(x) = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3$

Least value $= 3$ when $x = 2$.

$\boxed{3}$

Marking: 1 mark for completing the square; 1 mark for the least value.

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15. Find the range of $x$ for which $x^2 - 3x - 10 < 0$. (2 marks)

$x^2 - 3x - 10 = (x - 5)(x + 2) = 0 \implies x = 5 \text{ or } x = -2$

Since the parabola opens upwards, $x^2 - 3x - 10 < 0$ between the roots.

$\boxed{-2 < x < 5}$

Marking: 1 mark for finding the roots; 1 mark for the correct inequality.

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16. $x^2 + kx + 16 = 0$ has two equal roots. Find $k$. (2 marks)

For equal roots, $D = 0$: $k^2 - 4(1)(16) = 0$ $k^2 = 64$ $k = \pm 8$

$\boxed{k = 8 \text{ or } k = -8}$

Marking: 1 mark for $D = 0$; 1 mark for both values.

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17. Maximum value of $f(x) = -2x^2 + 12x - 7$. (2 marks)

$f(x) = -2(x^2 - 6x) - 7 = -2[(x - 3)^2 - 9] - 7 = -2(x - 3)^2 + 18 - 7 = -2(x - 3)^2 + 11$

Maximum value $= 11$ when $x = 3$.

$\boxed{11}$

Marking: 1 mark for completing the square; 1 mark for the maximum value.

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18. Vertex of $y = x^2 + 2x - 8$. (2 marks)

$y = (x + 1)^2 - 1 - 8 = (x + 1)^2 - 9$

Vertex: $(-1, -9)$.

$\boxed{(-1, -9)}$

Marking: 1 mark for $x = -1$; 1 mark for $y = -9$.

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19. $x$-coordinates of intersection of $y = 2x + 1$ and $y = x^2 - 3x + 5$. (2 marks)

$x^2 - 3x + 5 = 2x + 1$ $x^2 - 5x + 4 = 0$ $(x - 1)(x - 4) = 0$ $x = 1 \text{ or } x = 4$

$\boxed{x = 1 \text{ or } x = 4}$

Marking: 1 mark for correct equation; 1 mark for both values.

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20. Quadratic with minimum $-3$ at $x = 2$, passing through $(0, 5)$. (2 marks)

Using vertex form: $f(x) = a(x - 2)^2 - 3$.

At $(0, 5)$: $a(0 - 2)^2 - 3 = 5$, so $4a = 8$, giving $a = 2$.

$f(x) = 2(x - 2)^2 - 3 = 2(x^2 - 4x + 4) - 3 = 2x^2 - 8x + 8 - 3 = 2x^2 - 8x + 5$

$\boxed{f(x) = 2x^2 - 8x + 5}$

Marking: 1 mark for using vertex form; 1 mark for correct equation.

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End of Answer Key