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(a) If the roots are \(\tfrac{1}{2}\) and \(-7\), the equation is

\[ \left(x - \tfrac{1}{2}\right)(x + 7) = 0. \]

Multiply the first factor by 2 to clear the fraction (using \((2x-1)\)):

\[ (2x - 1)(x + 7) = 0 \implies 2x^2 + 14x - x - 7 = 0. \] \[ 2x^2 + 13x - 7 = 0. \]

This is the smallest equation with integer coefficients.

(b) Let the daughter's present age be \(d\) and the father's present age be \(f\).

"Three years ago the father was four times as old as the daughter is now":

\[ f - 3 = 4d \implies f = 4d + 3. \]

"The product of their present ages is 430":

\[ f \cdot d = 430 \implies (4d + 3)d = 430 \implies 4d^2 + 3d - 430 = 0. \]

Using the quadratic formula, \(\;d = \dfrac{-3 \pm \sqrt{3^2 + 4(4)(430)}}{2(4)} = \dfrac{-3 \pm \sqrt{6889}}{8} = \dfrac{-3 \pm 83}{8}.\)

Taking the positive root, \(d = \dfrac{80}{8} = 10\), so \(f = 4(10) + 3 = 43\).

The daughter is 10 years old and the father is 43 years old (product \(= 430\)).

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From the diagram, \(O\) is the centre, radius \(r = 3.2\ \text{cm}\), and \(\angle PRQ = 42^\circ\) is an angle at the circumference standing on chord \(PQ\).

Angle at the centre. The angle subtended at the centre is twice the angle at the circumference on the same arc:

\[ \angle POQ = 2 \times 42^\circ = 84^\circ \]

(a)(i) Area of minor sector POQ.

\[ A_{\text{sector}} = \frac{\theta}{360^\circ}\,\pi r^2 = \frac{84}{360}\times \pi \times (3.2)^2 \]

\[ = \frac{84}{360}\times \pi \times 10.24 = 0.23333 \times 32.1699 = 7.51\ \text{cm}^2 \]

(a)(ii) Area of the shaded part. The shaded region is the minor segment cut off by chord \(PQ\); it equals the sector minus triangle \(POQ\).

\[ A_{\triangle POQ} = \tfrac{1}{2} r^2 \sin\theta = \tfrac{1}{2}\times 10.24 \times \sin 84^\circ = 5.12 \times 0.99452 = 5.0920\ \text{cm}^2 \]

\[ A_{\text{shaded}} = 7.5063 - 5.0920 = 2.41\ \text{cm}^2 \]

(b) Base radius of the cone. When the sector is rolled into a cone, its arc length becomes the circumference of the base, while the sector radius becomes the slant height.

Arc length of sector:

\[ \ell = \frac{\theta}{360^\circ}\times 2\pi r = \frac{84}{360}\times 2\pi \times 3.2 = 4.6915\ \text{cm} \]

Set equal to base circumference \(2\pi R\):

\[ 2\pi R = 4.6915 \;\Rightarrow\; R = \frac{4.6915}{2\pi} = 0.7467 \approx 0.7\ \text{cm} \]

Answers: (i) \(7.51\ \text{cm}^2\); (ii) \(2.41\ \text{cm}^2\); (b) base radius \(\approx 0.7\ \text{cm}\).

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Table of values for \(d = 60t - 10t^2\):

\(t\) (s)	1	2	3	4	5
\(d\) (m)	50	80	90	80	50

Plot these points (2 cm to 1 unit on the \(t\)-axis, 2 cm to 20 units on the \(d\)-axis) and join with a smooth curve (a downward parabola).

(a)(i) Time to reach 70 m: Draw the line \(d = 70\); it cuts the curve twice. Solving \(60t - 10t^2 = 70\) gives \(t^2 - 6t + 7 = 0\), so \(t = 3 \pm \sqrt{2}\), i.e. \(t \approx \mathbf{1.6\text{ s}}\) (rising) and \(t \approx \mathbf{4.4\text{ s}}\) (falling).

(ii) Height after 5 s: from the curve, \(d = \mathbf{50\text{ m}}\).

(iii) Time at maximum height: the curve peaks at the vertex, \(t = \dfrac{60}{2(10)} = \mathbf{3\text{ s}}\) (height 90 m).

(b) Slope at \(t = 4\) s: the gradient of the tangent. Since \(\dfrac{dd}{dt} = 60 - 20t\), at \(t = 4\): \(60 - 20(4) = \mathbf{-20}\) (m/s). A tangent drawn at \(t = 4\) on the graph gives approximately this value.

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(a) Let the first term be \(a\) and the common difference be \(d\). The \(n\)th term is \(T_n = a + (n-1)d\).

The 8th and 4th terms are \(T_8 = a + 7d\) and \(T_4 = a + 3d\).

(i) Their difference is 20:

\[ (a + 7d) - (a + 3d) = 20 \implies 4d = 20 \implies d = 5. \]

The common difference is 5.

(ii) The 8th term is \(1\tfrac{1}{2}\) times the 4th term:

\[ a + 7d = \tfrac{3}{2}(a + 3d). \]

Substitute \(d = 5\):

\[ a + 35 = \tfrac{3}{2}(a + 15) \implies 2(a + 35) = 3(a + 15) \implies 2a + 70 = 3a + 45. \] \[ a = 25. \]

The first term is 25.

(b) A 5% depreciation each year multiplies the value by \(0.95\) annually. After 9 years:

\[ V = 10250 \times (0.95)^{9}. \]

Now \((0.95)^{9} = 0.6302\) (to 4 d.p.), so

\[ V = 10250 \times 0.6302 = 6460.06. \]

Correct to three significant figures, the value in January 1989 is N6 460.00.

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(a)(i) \(\cos\alpha = 0.6717\). From the cosine tables, \(\alpha = \cos^{-1}(0.6717) \approx \mathbf{47.8°}\).

(ii) Using \(\sin^2\alpha + \cos^2\alpha = 1\):

\(\sin\alpha = \sqrt{1 - (0.6717)^2} = \sqrt{1 - 0.4512} = \sqrt{0.5488} = \mathbf{0.7408}\)

(Equivalently, \(\sin 47.8° = 0.7408\).)

(b) The angle of depression of \(P\) from the top \(T\) equals the angle of elevation of \(T\) from \(P\) (alternate angles), \(= 23.6°\). The horizontal distance from the foot of the building to \(P\) is \(50\text{ m}\).

\(\tan 23.6° = \dfrac{\text{height}}{50}\)

\(\text{height} = 50 \times \tan 23.6° = 50 \times 0.4369 = 21.85\text{ m}\)

Height of the building \(\approx \mathbf{22\text{ m}}\) (to the nearest metre).

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Surface area of the cone-on-hemisphere structure.

From the diagram, the common radius (from centre \(O\) to the base edge) is \(r=7\text{ cm}\), and the vertical height of the cone is \(h=24\text{ cm}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere; the flat circular join is internal and is not counted.

Slant height of the cone:

\[l=\sqrt{r^{2}+h^{2}}=\sqrt{7^{2}+24^{2}}=\sqrt{49+576}=\sqrt{625}=25\text{ cm}\]

Curved surface area of the cone:

\[\pi r l=\frac{22}{7}\times7\times25=22\times25=550\text{ cm}^{2}\]

Curved surface area of the hemisphere:

\[2\pi r^{2}=2\times\frac{22}{7}\times7^{2}=2\times\frac{22}{7}\times49=2\times22\times7=308\text{ cm}^{2}\]

Total surface area:

\[550+308=858\text{ cm}^{2}\]

Correct to three significant figures, the surface area is \(\mathbf{858\text{ cm}^{2}}\).

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```html

Construction of Triangle ABC and the Required Loci

Construct triangle ABC using a ruler and pair of compasses only, given that:

• |AB| = 8.4 cm
• |BC| = 6.5 cm
• ∠ABC = 30°

1. Construct triangle ABC

1. Draw a line segment BC = 6.5 cm.
2. At point B, construct a 60° angle using the equilateral-triangle construction: with centre B and any suitable radius, draw an arc; using the same radius from a point on the arc, mark the intersection to form a 60° ray.
3. Bisect the 60° angle to obtain a ray making 30° with BC.
4. With centre B and radius 8.4 cm, draw an arc cutting the 30° ray at A.
5. Join A to C. This completes triangle ABC.

2. Construct locus l<sub>1</sub>

The locus l<sub>1</sub> of points equidistant from the lines AB and BC, and lying within ∠ABC, is the internal angle bisector of ∠ABC.

1. With centre B, draw an arc cutting BA and BC at points D and E.
2. With the same compass radius, draw arcs from D and E to meet at F inside ∠ABC.
3. Draw the line BF. This line is locus l<sub>1</sub>.

3. Construct locus l<sub>2</sub>

The locus l<sub>2</sub> of points equidistant from points B and C is the perpendicular bisector of BC.

1. Set the compass radius to more than half of BC.
2. With centres B and C, draw arcs above and below BC so that they intersect.
3. Join the two arc-intersection points. The resulting line is locus l<sub>2</sub>.

Therefore:

• l<sub>1</sub> is the internal bisector of ∠ABC.
• l<sub>2</sub> is the perpendicular bisector of BC.
• I = l<sub>1</sub> ∩ l<sub>2</sub>.

```

Antwoorddetails

(a) \(6\log(x+4) = \log 64\)

\(\log(x+4)^6 = \log 64\)

\((x+4)^6 = 64 = 2^6\)

Taking the positive sixth root, \(x + 4 = 2\), so \(\mathbf{x = -2}\). (Check: \(x+4 = 2 > 0\), so the logarithm is defined.)

(b) \(U = \{1,2,3,4,5,6,7,8,9,10\}\), \(X = \{1,2,4,6,7,8,9\}\), \(Y = \{1,2,3,4,7,9\}\), \(Z = \{2,3,4,7,9\}\).

First, \(Z' = U \setminus Z = \{1,5,6,8,10\}\).

Next, \(X \cap Y = \{1,2,4,7,9\}\).

Then \((X \cap Y) \cap Z' = \{1,2,4,7,9\} \cap \{1,5,6,8,10\} = \mathbf{\{1\}}\).

Antwoorddetails

(a) Bar chart

(b)(i) Mode

The highest frequency is 4, corresponding to a weight of 61 kg.

\[\boxed{\text{Mode}=61\text{ kg}}\]

(b)(ii) Median

There are \(12\) students. Therefore, the median is the mean of the 6th and 7th observations.

The ordered weights are:

\[55,\ 55,\ 57,\ 59,\ 59,\ \underbrace{61,\ 61}_{\text{6th and 7th}},\ 61,\ 61,\ 63,\ 63,\ 63\]

\[\text{Median}=\frac{61+61}{2}=\boxed{61\text{ kg}}\]

(c) Mean weight

\[\begin{aligned}\sum f&=2+1+2+4+3=12,\\\sum fx&=55(2)+57(1)+59(2)+61(4)+63(3)\\&=110+57+118+244+189\\&=718.\end{aligned}\]

\[\text{Mean}=\frac{\sum fx}{\sum f}=\frac{718}{12}=59.833\ldots\text{ kg}\]

\[\boxed{\text{Mean weight}=60\text{ kg, correct to the nearest kilogram}}\]

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(a)(i) Theorem: the angle an arc subtends at the centre is twice the angle it subtends at the circumference.

Let \(O\) be the centre and let arc \(PQ\) subtend \(\angle POQ\) at the centre and \(\angle PRQ\) at a point \(R\) on the major arc. Join \(RO\) and produce it to a point \(X\).

In triangle \(OPR\), \(OP = OR\) (radii), so it is isosceles and \(\angle OPR = \angle ORP\). The exterior angle of a triangle equals the sum of the two interior opposite angles, so

\[\angle POX = \angle OPR + \angle ORP = 2\,\angle ORP.\]

Similarly, in triangle \(OQR\), \(OQ = OR\), giving \(\angle QOX = 2\,\angle ORQ\). Adding,

\[\angle POQ = \angle POX + \angle QOX = 2(\angle ORP + \angle ORQ) = 2\,\angle PRQ.\]

Hence the central angle is twice the inscribed angle standing on the same arc. (Q.E.D.)

(a)(ii) Calculate \(\angle QRS\). From the diagram, \(O\) is the centre, \(PT\) is a diameter, \(\angle PTQ = 22^\circ\), and the arc \(TR\) (through \(S\)) subtends \(\angle TOR = 98^\circ\) at the centre with \(\angle TOS = 22^\circ\).

Arc PQ. \(\angle PTQ = 22^\circ\) is an inscribed angle on arc \(PQ\), so the central angle \(\angle POQ = 2(22^\circ) = 44^\circ\); thus arc \(PQ = 44^\circ\).

Arc QT (lower semicircle). Since \(PT\) is a diameter, \(P,O,T\) are collinear, so

\[\angle QOT = 180^\circ - \angle POQ = 180^\circ - 44^\circ = 136^\circ\;\Rightarrow\;\text{arc } QT = 136^\circ.\]

Splitting the lower arc. arc \(TR = 98^\circ\) and arc \(TS = 22^\circ\), so

\[\text{arc } QR = \text{arc } QT - \text{arc } TR = 136^\circ - 98^\circ = 38^\circ,\qquad \text{arc } SR = 98^\circ - 22^\circ = 76^\circ.\]

Angle QRS. \(\angle QRS\) is an inscribed angle at \(R\) standing on chord \(QS\); it equals half the arc \(QS\) that does not contain \(R\). That arc runs \(Q\to P\to T\to S\) over the top:

\[\text{arc } QPTS = \text{arc }QP + \text{arc }PT_{(\text{diameter, top})} + \text{arc }TS = 44^\circ + 180^\circ + 22^\circ = 246^\circ.\]\[\angle QRS = \tfrac{1}{2}\times 246^\circ = \boxed{123^\circ}.\]

(b) Cyclic quadrilateral \(ABCD\), diagonals meet at \(H\); \(\angle DAC = 41^\circ\), \(\angle AHB = 70^\circ\). Find \(\angle ABC\).

\(\angle AHB\) is an exterior angle of triangle \(AHD\), so it equals the sum of the two remote interior angles:

\[\angle AHB = \angle DAC + \angle ADB \;\Rightarrow\; 70^\circ = 41^\circ + \angle ADB \;\Rightarrow\; \angle ADB = 29^\circ.\]

Using the arcs cut by the intersecting diagonals, \(\angle AHB = \tfrac12(\text{arc }AB + \text{arc }DC)\). Since \(\angle DAC = 41^\circ\) stands on arc \(DC\), arc \(DC = 82^\circ\), giving

\[70^\circ = \tfrac12(\text{arc }AB + 82^\circ)\;\Rightarrow\;\text{arc }AB = 58^\circ,\qquad \angle ACB = \tfrac12(58^\circ)=29^\circ.\]

Also \(\angle DBC = \angle DAC = 41^\circ\) (same arc \(DC\)). Taking arc \(BC = 80^\circ\) as fixed by the figure, \(\angle BAC = 40^\circ\) and \(\angle ABD = 110^\circ-40^\circ = 70^\circ\), so

\[\angle ABC = \angle ABD + \angle DBC = 70^\circ + 41^\circ = \boxed{111^\circ}.\]

Equivalently, the opposite angle \(\angle ADC = 180^\circ - 111^\circ = 69^\circ\), consistent with a cyclic quadrilateral.

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(a) Two dice give \(6 \times 6 = 36\) equally likely outcomes. A sum of "at least 9" means a sum of 9, 10, 11 or 12.

• Sum 9: (3,6),(4,5),(5,4),(6,3) = 4 ways
• Sum 10: (4,6),(5,5),(6,4) = 3 ways
• Sum 11: (5,6),(6,5) = 2 ways
• Sum 12: (6,6) = 1 way

Favourable outcomes \(= 4+3+2+1 = 10\).

\[ P(\text{sum} \ge 9) = \frac{10}{36} = \frac{5}{18}. \]

(b) Let \(P(\text{owns a car}) = \frac{1}{6}\), so \(P(\text{does not}) = \frac{5}{6}\). Selections are independent.

(i) Both A and B own a car:

\[ \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}. \]

(ii) Of C and D, exactly one owns a car (owns-then-not, or not-then-owns):

\[ 2 \times \frac{1}{6} \times \frac{5}{6} = \frac{10}{36} = \frac{5}{18}. \]

(iii) Of X, Y and Z, exactly one owns a car. Choose which one (\(\binom{3}{1}=3\) ways):

\[ 3 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^{2} = 3 \times \frac{25}{216} = \frac{75}{216} = \frac{25}{72}. \]