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(a) Write both sides in base 3: \(81=3^{4},\ 27=3^{3}\).

\[\frac{1}{81^{x-2}}=3^{-4(x-2)},\qquad 27^{1-x}=3^{3(1-x)}.\]

Equating indices:

\[-4(x-2)=3(1-x)\Rightarrow -4x+8=3-3x\Rightarrow -x=-5\Rightarrow x=5.\]

(b) Rationalise each term:

\[\frac{5}{\sqrt7-\sqrt3}=\frac{5(\sqrt7+\sqrt3)}{7-3}=\frac{5(\sqrt7+\sqrt3)}{4},\qquad \frac{1}{\sqrt7+\sqrt3}=\frac{\sqrt7-\sqrt3}{4}.\]

\[\text{Sum}=\frac{5\sqrt7+5\sqrt3+\sqrt7-\sqrt3}{4}=\frac{6\sqrt7+4\sqrt3}{4}=\frac{3\sqrt7+2\sqrt3}{2}.\]

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(a) Factor the first bracket:

\[\frac{x^{2}}{2}-x+\frac12=\frac12\left(x^{2}-2x+1\right)=\frac12(x-1)^{2}.\]

\[\therefore\ \frac12(x-1)^{2}\cdot\frac{1}{x-1}=\frac{x-1}{2}.\]

(b) \(P\) is 40 km from \(Q\) on bearing \(061^{\circ}\) (measured from north).

(i) North of Q: \(40\cos61^{\circ}=40(0.4848)=19.4\text{ km}\ (1\text{ d.p.}).\)

(ii) East of Q: \(40\sin61^{\circ}=40(0.8746)=35.0\text{ km}\ (1\text{ d.p.}).\)

(c) Let the son's share be \(s\). Each daughter gets \(\tfrac34 s\); three daughters get \(3\times\tfrac34 s=\tfrac94 s\).

\[s+\frac94 s=5720\Rightarrow\frac{13}{4}s=5720\Rightarrow s=5720\times\frac{4}{13}=\mathbf{N1760}.\]

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(a) Substitute \(x=1,\ y=-2\):

\[p(1)+q(-2)=5\Rightarrow p-2q=5\quad(1)\]

\[q(1)+p(-2)=-10\Rightarrow -2p+q=-10\quad(2)\]

From (1), \(p=5+2q\). Substitute into (2): \(-2(5+2q)+q=-10\Rightarrow -10-4q+q=-10\Rightarrow -3q=0\Rightarrow q=0\).

Then \(p=5\). So \(\mathbf{p=5,\ q=0}\).

(b) Cross-multiply \(\dfrac{4r-3}{6r+1}=\dfrac{2r-1}{3r+4}\):

\[(4r-3)(3r+4)=(2r-1)(6r+1)\]

\[12r^{2}+7r-12=12r^{2}-4r-1\Rightarrow 7r-12=-4r-1\Rightarrow 11r=11\Rightarrow r=1.\]

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(a) \(O\) is the centre, so \(|OA|=|OB|=r\). Perimeter of \(\triangle AOB\):

\[2r+|AB|=52.2\Rightarrow 2r+24.2=52.2\Rightarrow r=14\text{ cm}.\]

By the cosine rule in \(\triangle AOB\):

\[\cos(\angle AOB)=\frac{r^{2}+r^{2}-|AB|^{2}}{2r^{2}}=\frac{196+196-585.64}{392}=\frac{-193.64}{392}=-0.4939.\]

\[\angle AOB=\cos^{-1}(-0.4939)=119.6^{\circ}\approx\mathbf{120^{\circ}}\ (\text{nearest degree}).\]

(b) Volume of tank \(=60\times80\times100=480{,}000\text{ cm}^{3}\). Half filled:

\[\tfrac12\times480{,}000=240{,}000\text{ cm}^{3}=\frac{240{,}000}{1000}=\mathbf{240\text{ litres}}\quad(1000\text{ cm}^{3}=1\text{ litre}).\]

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(a) Sides \(7:5:7\). The angle included between the two equal sides (each \(7k\)) is opposite the side \(5k\). By the cosine rule:

\[\cos\theta=\frac{7^{2}+7^{2}-5^{2}}{2(7)(7)}=\frac{49+49-25}{98}=\frac{73}{98}=0.7449.\]

\[\theta=\cos^{-1}(0.7449)=41.9^{\circ}\approx\mathbf{42^{\circ}}\ (\text{nearest degree}).\]

(b) Sum of interior angles \(=(n-2)\times180^{\circ}=1440^{\circ}\):

\[n-2=\frac{1440}{180}=8\Rightarrow \mathbf{n=10\text{ sides}}.\]

(ii) Each exterior angle \(=\dfrac{360^{\circ}}{n}=\dfrac{360^{\circ}}{10}=\mathbf{36^{\circ}}.\)

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(a) Area of the minor segment

The chord \(XY\) subtends \(\theta = 96^\circ\) at the centre and the radius is \(r = 5\text{ cm}\). The minor segment is the minor sector minus triangle \(OXY\).

Area of minor sector (with \(\pi = \tfrac{22}{7}\)):

\[\frac{\theta}{360}\pi r^2 = \frac{96}{360}\times\frac{22}{7}\times 5^2 = 20.95\text{ cm}^2\]

Area of triangle OXY:

\[\frac{1}{2}r^2\sin\theta = \frac{1}{2}\times 5^2\times\sin 96^\circ = 12.43\text{ cm}^2\]

Area of minor segment:

\[20.95 - 12.43 = 8.52\text{ cm}^2\]

Correct to three significant figures, the area of the minor segment is \(\mathbf{8.52\text{ cm}^2}\).

(b) Total area of the shaded portions

The square has side 14 cm, so the inscribed circle has radius \(r = 7\text{ cm}\).

Shaded sector of the circle (angle \(80^\circ\)):

\[\frac{80}{360}\times\frac{22}{7}\times 7^2 = 34.22\text{ cm}^2\]

Shaded portion of the square (the four corners between the square and the circle):

\[\text{Area of square} = 14^2 = 196\text{ cm}^2\]

\[\text{Area of circle} = \frac{22}{7}\times 7^2 = 154\text{ cm}^2\]

\[196 - 154 = 42\text{ cm}^2\]

Total shaded area:

\[42 + 34.22 = 76.22 \approx 76.2\text{ cm}^2\]

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(a) Simplify \(\sqrt{1001_{two}}\), leaving the answer in base two.

First convert \(1001_{two}\) to base ten:

\[1001_{two}=1(2^3)+0(2^2)+0(2^1)+1(2^0)=8+0+0+1=9_{ten}.\]

Then \(\sqrt{9_{ten}}=3_{ten}\). Convert \(3\) back to base two:

\[3_{ten}=1(2^1)+1(2^0)=11_{two}.\]

Therefore \(\sqrt{1001_{two}}=\mathbf{11_{two}}\).

(b) Find \(z\) in terms of \(x\) and \(y\).

In the diagram, \(O\) is the centre, \(|OP|=x\) (a radius), \(|OK|=y\) and \(\angle OKP=90^\circ\). Since \(OK\) is drawn from the centre perpendicular to the chord \(PQ\), it bisects the chord, so \(K\) is the mid-point of \(PQ\):

\[|KP|=\tfrac{1}{2}|PQ|=\tfrac{z}{2}.\]

Applying Pythagoras' theorem to right-angled triangle \(OKP\):

\[|OP|^2=|OK|^2+|KP|^2\]\[x^2=y^2+\left(\frac{z}{2}\right)^2\]\[\left(\frac{z}{2}\right)^2=x^2-y^2\]\[\frac{z}{2}=\sqrt{x^2-y^2}\]

Therefore \(\displaystyle z=2\sqrt{x^{2}-y^{2}}\).

(c) \(P,Q,R,S\) lie on the circle centre \(O\), with \(\angle POQ=160^\circ\), \(\angle QSR=45^\circ\) and \(\angle PQS=40^\circ\).

(i) \(\angle QPS\): The chord \(PQ\) subtends the central angle \(\angle POQ=160^\circ\). The angle it subtends at the circumference (at \(S\)) is half of this:

\[\angle PSQ=\tfrac{1}{2}\times 160^\circ=80^\circ.\]

In triangle \(PQS\), the three angles sum to \(180^\circ\):

\[\angle QPS=180^\circ-\angle PQS-\angle PSQ=180^\circ-40^\circ-80^\circ=\mathbf{60^\circ}.\]

(ii) \(\angle RQS\): Work out the arcs (central angles) from the given inscribed angles.

• Arc \(PQ\) corresponds to \(\angle POQ=160^\circ\).
• \(\angle PQS=40^\circ\) stands on chord \(PS\), so arc \(PS=2\times40^\circ=80^\circ\).
• \(\angle QSR=45^\circ\) stands on chord \(QR\), so arc \(QR=2\times45^\circ=90^\circ\).

The four arcs around the circle sum to \(360^\circ\):

\[\text{arc }PQ+\text{arc }QR+\text{arc }RS+\text{arc }SP=360^\circ\]\[160^\circ+90^\circ+\text{arc }RS+80^\circ=360^\circ\]\[\text{arc }RS=30^\circ.\]

\(\angle RQS\) stands at \(Q\) on chord \(RS\), so it equals half of arc \(RS\):

\[\angle RQS=\tfrac{1}{2}\times30^\circ=\mathbf{15^\circ}.\]

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Construction of Triangle PQR and the Required Loci

Given: Construct a triangle PQR such that

• PQ = 8 cm,
• PR = 7 cm,
• ∠QPR = 105°.

(a)(i) Construction of ΔPQR

1. Using a ruler, draw a line segment PQ = 8 cm.
2. At point P, construct an angle of 105°:
   • Construct a 90° angle at P by drawing a perpendicular to PQ.
   • Construct a 120° ray at P using equal arcs with the compasses.
   • Bisect the angle between the 90° ray and the 120° ray. The bisected angle is 105°.
3. On the 105° ray, use the compasses to mark a point R such that PR = 7 cm.
4. Join R to Q. This gives the required triangle ΔPQR.

(a)(ii) Locus L<sub>1</sub> of points equidistant from P and Q

The locus of points equidistant from P and Q is the perpendicular bisector of PQ.

1. With the compasses opened to more than half of PQ, draw arcs centred at P above and below PQ.
2. Without changing the compass width, draw arcs centred at Q to intersect the first arcs.
3. Join the two arc-intersection points with a ruler.

This straight line is labelled L<sub>1</sub>.

(a)(iii) Locus l<sub>2</sub> of points equidistant from Q and R

The locus of points equidistant from Q and R is the perpendicular bisector of QR.

1. Draw equal arcs centred at Q and R, using a radius greater than half of QR.
2. Join the two points where the arcs intersect.

This straight line is labelled l<sub>2</sub>.

(b)(i) Point of Intersection

Label the point where L<sub>1</sub> and l<sub>2</sub> intersect as T.

Conclusion: The intersection point T is the centre of the circle passing through P, Q and R.

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