WAEC SSCE - General Mathematics - 2011 (Essay)
Tambaya 1 Rahoto
In a class of 40 students, 18 passed Mathematics, 19 passed Accounts, 16 passed Economics, 5 passed Mathematics and Accounts only, 6 Mathematics only, 9 Accounts only, 2 Accounts and Economics only. If each student offered at least one of the subjects,
(a) how many students failed in all subjects?
(b) find the percentage number that failed in at least one of Economics and Mathematics
(c) calculate the probability that a student picked at random failed in Accounts?
Let the regions of the three-set Venn diagram (M = Mathematics, A = Accounts, E = Economics) be filled from the "only" data. Given: \(M\) only \(=6\), \(A\) only \(=9\), \(M\cap A\) only \(=5\), \(A\cap E\) only \(=2\).
Let the triple region \(M\cap A\cap E=t\).
Accounts total \(=19\): \(9+5+2+t=19\Rightarrow t=3\).
Mathematics total \(=18\): \(6+5+t+(M\cap E\ \text{only})=18\Rightarrow 6+5+3+(M\cap E\ \text{only})=18\Rightarrow M\cap E\ \text{only}=4\).
Economics total \(=16\): \((E\ \text{only})+4+2+3=16\Rightarrow E\ \text{only}=7\).
Sum of all seven regions \(=6+9+7+5+4+2+3=36\).
(a) Since each of the 40 students offered at least one subject, those in no region failed all three:
\[40-36=4\text{ students failed all subjects.}\](b) Failed at least one of Economics and Mathematics. The complement is passing BOTH Mathematics and Economics \(=(M\cap E\ \text{only})+t=4+3=7\). So the number failing at least one of them \(=40-7=33\).
\[\frac{33}{40}\times100\%=82.5\%\](c) Probability of failing Accounts. Number passing Accounts \(=19\), so number failing Accounts \(=40-19=21\).
\[P(\text{failed Accounts})=\frac{21}{40}\]Bayanin Amsa
Let the regions of the three-set Venn diagram (M = Mathematics, A = Accounts, E = Economics) be filled from the "only" data. Given: \(M\) only \(=6\), \(A\) only \(=9\), \(M\cap A\) only \(=5\), \(A\cap E\) only \(=2\).
Let the triple region \(M\cap A\cap E=t\).
Accounts total \(=19\): \(9+5+2+t=19\Rightarrow t=3\).
Mathematics total \(=18\): \(6+5+t+(M\cap E\ \text{only})=18\Rightarrow 6+5+3+(M\cap E\ \text{only})=18\Rightarrow M\cap E\ \text{only}=4\).
Economics total \(=16\): \((E\ \text{only})+4+2+3=16\Rightarrow E\ \text{only}=7\).
Sum of all seven regions \(=6+9+7+5+4+2+3=36\).
(a) Since each of the 40 students offered at least one subject, those in no region failed all three:
\[40-36=4\text{ students failed all subjects.}\](b) Failed at least one of Economics and Mathematics. The complement is passing BOTH Mathematics and Economics \(=(M\cap E\ \text{only})+t=4+3=7\). So the number failing at least one of them \(=40-7=33\).
\[\frac{33}{40}\times100\%=82.5\%\](c) Probability of failing Accounts. Number passing Accounts \(=19\), so number failing Accounts \(=40-19=21\).
\[P(\text{failed Accounts})=\frac{21}{40}\]Tambaya 2 Rahoto
(a) The area of trapezium PQRS is 60\(cm^{2}\). PQ // RS, /PQ/ = 15 cm, /RS/ = 25 cm and < PSR = 60°. Calculate the : (i) perpendicular height of PQRS ; (ii) |PS|.
(b) Ade received \(\frac{3}{5}\) of a sum of money, Nelly \(\frac{1}{3}\) of the remainder while Austin took the rest. If Austin's share is greater than Nelly's share by N3,000, how much did Ade get?
(a)(i) Perpendicular height. For a trapezium, area \(=\tfrac{1}{2}(\text{sum of parallel sides})\times h\).
\[60=\tfrac{1}{2}(15+25)h=20h\Rightarrow h=3\text{ cm}\]Height \(=3\) cm.
(ii) |PS|. \(PS\) is the slant side at \(S\), where \(\angle PSR=60^{\circ}\). The perpendicular height is the vertical component of \(PS\):
\[h=|PS|\sin 60^{\circ}\Rightarrow |PS|=\frac{h}{\sin 60^{\circ}}=\frac{3}{0.8660}\approx 3.46\text{ cm}\]|PS| \(\approx 3.46\) cm.
(b) Let the total sum be \(T\). Ade takes \(\tfrac{3}{5}T\); remainder \(=\tfrac{2}{5}T\).
Nelly takes \(\tfrac{1}{3}\) of the remainder \(=\tfrac{1}{3}\times\tfrac{2}{5}T=\tfrac{2}{15}T\).
Austin takes the rest of the remainder \(=\tfrac{2}{5}T-\tfrac{2}{15}T=\tfrac{6}{15}T-\tfrac{2}{15}T=\tfrac{4}{15}T\).
Austin's share exceeds Nelly's by \(N3{,}000\):
\[\tfrac{4}{15}T-\tfrac{2}{15}T=\tfrac{2}{15}T=3000\Rightarrow T=3000\times\tfrac{15}{2}=22500\]Ade's share \(=\tfrac{3}{5}\times22500=N13{,}500\).
Ade got \(N13{,}500.00\).
Bayanin Amsa
(a)(i) Perpendicular height. For a trapezium, area \(=\tfrac{1}{2}(\text{sum of parallel sides})\times h\).
\[60=\tfrac{1}{2}(15+25)h=20h\Rightarrow h=3\text{ cm}\]Height \(=3\) cm.
(ii) |PS|. \(PS\) is the slant side at \(S\), where \(\angle PSR=60^{\circ}\). The perpendicular height is the vertical component of \(PS\):
\[h=|PS|\sin 60^{\circ}\Rightarrow |PS|=\frac{h}{\sin 60^{\circ}}=\frac{3}{0.8660}\approx 3.46\text{ cm}\]|PS| \(\approx 3.46\) cm.
(b) Let the total sum be \(T\). Ade takes \(\tfrac{3}{5}T\); remainder \(=\tfrac{2}{5}T\).
Nelly takes \(\tfrac{1}{3}\) of the remainder \(=\tfrac{1}{3}\times\tfrac{2}{5}T=\tfrac{2}{15}T\).
Austin takes the rest of the remainder \(=\tfrac{2}{5}T-\tfrac{2}{15}T=\tfrac{6}{15}T-\tfrac{2}{15}T=\tfrac{4}{15}T\).
Austin's share exceeds Nelly's by \(N3{,}000\):
\[\tfrac{4}{15}T-\tfrac{2}{15}T=\tfrac{2}{15}T=3000\Rightarrow T=3000\times\tfrac{15}{2}=22500\]Ade's share \(=\tfrac{3}{5}\times22500=N13{,}500\).
Ade got \(N13{,}500.00\).
Tambaya 3 Rahoto
(a) Make q the subject of the relation \(t = \sqrt{\frac{pq}{r} - r^{2}q}\).
(b) If \(9^{(1 - x)} = 27^{y}\) and \(x - y = -1\frac{1}{2}\), find the value of x and y.
(a) Square both sides to remove the root:
\[t^2 = \frac{pq}{r} - r^2 q = q\left(\frac{p}{r} - r^2\right) = q\left(\frac{p - r^3}{r}\right).\]
Make q the subject:
\[q = \frac{r\,t^2}{p - r^3}.\]
(b) Write both sides to base 3: \(9^{(1-x)} = 3^{2(1-x)}\) and \(27^{y} = 3^{3y}\). Equating indices,
\[2(1 - x) = 3y \Rightarrow 2 - 2x = 3y. \quad (1)\]
Also \(x - y = -\tfrac{3}{2}\Rightarrow x = y - \tfrac{3}{2}. \quad (2)\)
Substitute (2) into (1):
\[2 - 2\left(y - \tfrac{3}{2}\right) = 3y \Rightarrow 2 - 2y + 3 = 3y \Rightarrow 5 = 5y \Rightarrow y = 1.\]
\[x = 1 - \tfrac{3}{2} = -\tfrac{1}{2}.\]
Answer: \(x = -\tfrac{1}{2},\ y = 1\).
Bayanin Amsa
(a) Square both sides to remove the root:
\[t^2 = \frac{pq}{r} - r^2 q = q\left(\frac{p}{r} - r^2\right) = q\left(\frac{p - r^3}{r}\right).\]
Make q the subject:
\[q = \frac{r\,t^2}{p - r^3}.\]
(b) Write both sides to base 3: \(9^{(1-x)} = 3^{2(1-x)}\) and \(27^{y} = 3^{3y}\). Equating indices,
\[2(1 - x) = 3y \Rightarrow 2 - 2x = 3y. \quad (1)\]
Also \(x - y = -\tfrac{3}{2}\Rightarrow x = y - \tfrac{3}{2}. \quad (2)\)
Substitute (2) into (1):
\[2 - 2\left(y - \tfrac{3}{2}\right) = 3y \Rightarrow 2 - 2y + 3 = 3y \Rightarrow 5 = 5y \Rightarrow y = 1.\]
\[x = 1 - \tfrac{3}{2} = -\tfrac{1}{2}.\]
Answer: \(x = -\tfrac{1}{2},\ y = 1\).
Tambaya 4 Rahoto
(a) 
In the diagram, PQRST is a quadrilateral. PT // QS, < PTQ = 42°, < TSQ = 38° and < QSR = 30°. If < QTS = x and < POT = y, find: (i) x ; (ii) y.
(b) 
In the diagram, PQRS is a circle centre O. If POQ = 150°, < QSR = 40° and < SQP = 45°, calculate < RQS.
(a) In the quadrilateral \(PT \parallel QS\), with \(\angle PTQ = 42^{\circ}\), \(\angle TSQ = 38^{\circ}\) and \(\angle QSR = 30^{\circ}\).
(i) Find \(x = \angle QTS\).
Since \(PT \parallel QS\), \(\angle PTS = \angle TSQ = 38^{\circ}\) (alternate angles). The angles at \(T\) on the straight line give:
\[ 42^{\circ} + x + 38^{\circ} = 180^{\circ}. \]
\[ 80^{\circ} + x = 180^{\circ} \quad\Rightarrow\quad x = 100^{\circ}. \]
(ii) Find \(y = \angle POT\).
Since \(PT \parallel QS\), \(\angle SQT = \angle PTQ = 42^{\circ}\) (alternate angles). In triangle \(QRS\), \(\angle SQR = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}\) (sum of angles in a triangle). The angles on the straight line at \(Q\) give:
\[ y + 42^{\circ} + 60^{\circ} = 180^{\circ}. \]
\[ y + 102^{\circ} = 180^{\circ} \quad\Rightarrow\quad y = 78^{\circ}. \]
(b) \(PQRS\) is a circle centre \(O\), with \(\angle POQ = 150^{\circ}\), \(\angle QSR = 40^{\circ}\) and \(\angle SQP = 45^{\circ}\). Find \(\angle RQS\).
The angle at the centre is twice the angle at the circumference on the same arc \(PQ\):
\[ \angle QSP = \tfrac{1}{2}\times 150^{\circ} = 75^{\circ}. \]
In triangle \(SPQ\), the sum of the angles is \(180^{\circ}\):
\[ \angle QPS + 75^{\circ} + 45^{\circ} = 180^{\circ} \quad\Rightarrow\quad \angle QPS = 60^{\circ}. \]
\(PQRS\) is a cyclic quadrilateral, so opposite angles are supplementary:
\[ \angle QRS = 180^{\circ} - \angle QPS = 180^{\circ} - 60^{\circ} = 120^{\circ}. \]
In triangle \(QRS\), the sum of the angles is \(180^{\circ}\):
\[ \angle RQS = 180^{\circ} - (\angle QRS + \angle QSR) = 180^{\circ} - (120^{\circ} + 40^{\circ}) = 20^{\circ}. \]
\(\angle RQS = 20^{\circ}.\)
Bayanin Amsa
(a) In the quadrilateral \(PT \parallel QS\), with \(\angle PTQ = 42^{\circ}\), \(\angle TSQ = 38^{\circ}\) and \(\angle QSR = 30^{\circ}\).
(i) Find \(x = \angle QTS\).
Since \(PT \parallel QS\), \(\angle PTS = \angle TSQ = 38^{\circ}\) (alternate angles). The angles at \(T\) on the straight line give:
\[ 42^{\circ} + x + 38^{\circ} = 180^{\circ}. \]
\[ 80^{\circ} + x = 180^{\circ} \quad\Rightarrow\quad x = 100^{\circ}. \]
(ii) Find \(y = \angle POT\).
Since \(PT \parallel QS\), \(\angle SQT = \angle PTQ = 42^{\circ}\) (alternate angles). In triangle \(QRS\), \(\angle SQR = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}\) (sum of angles in a triangle). The angles on the straight line at \(Q\) give:
\[ y + 42^{\circ} + 60^{\circ} = 180^{\circ}. \]
\[ y + 102^{\circ} = 180^{\circ} \quad\Rightarrow\quad y = 78^{\circ}. \]
(b) \(PQRS\) is a circle centre \(O\), with \(\angle POQ = 150^{\circ}\), \(\angle QSR = 40^{\circ}\) and \(\angle SQP = 45^{\circ}\). Find \(\angle RQS\).
The angle at the centre is twice the angle at the circumference on the same arc \(PQ\):
\[ \angle QSP = \tfrac{1}{2}\times 150^{\circ} = 75^{\circ}. \]
In triangle \(SPQ\), the sum of the angles is \(180^{\circ}\):
\[ \angle QPS + 75^{\circ} + 45^{\circ} = 180^{\circ} \quad\Rightarrow\quad \angle QPS = 60^{\circ}. \]
\(PQRS\) is a cyclic quadrilateral, so opposite angles are supplementary:
\[ \angle QRS = 180^{\circ} - \angle QPS = 180^{\circ} - 60^{\circ} = 120^{\circ}. \]
In triangle \(QRS\), the sum of the angles is \(180^{\circ}\):
\[ \angle RQS = 180^{\circ} - (\angle QRS + \angle QSR) = 180^{\circ} - (120^{\circ} + 40^{\circ}) = 20^{\circ}. \]
\(\angle RQS = 20^{\circ}.\)
Tambaya 5 Rahoto
(a) Given that \(\sin x = 0.6, 0° \leq x \leq 90°\), evaluate \(2\cos x + 3\sin x\), leaving your answer in the form \(\frac{m}{n}\), where m and n are integers.
(b) 
In the diagram, a semi-circle WXYZ with centre O is inscribed in an isosceles triangle ABC. If /AC/ = /BC/, |OC| = 30 cm and < ACB = 130°, calculate, correct to one decimal place, the (i) radius of the circle ; (ii) area oc the shaded portion. [Take \(\pi = \frac{22}{7}\)].
(a) Evaluating \(2\cos x + 3\sin x\)
Given \(\sin x = 0.6 = \dfrac{3}{5}\) with \(0^\circ \le x \le 90^\circ\), x is acute, so \(\cos x\) is positive.
\[\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - (0.6)^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8 = \frac{4}{5}\]
Then
\[2\cos x + 3\sin x = 2\left(\frac{4}{5}\right) + 3\left(\frac{3}{5}\right) = \frac{8}{5} + \frac{9}{5} = \frac{17}{5}\]
(b) Semi-circle inscribed in isosceles triangle ABC
From the diagram, the diameter WZ lies along the top side AB, the centre O is on AB, and the semicircular arc is tangent to the two equal sides. The apex C is below, with \(|OC| = 30\ \text{cm}\) measured along the axis of symmetry, and \(\angle ACB = 130^\circ\).
By symmetry (AC = BC), the line CO bisects angle C, so the half-angle at C is
\[\frac{130^\circ}{2} = 65^\circ\]
(i) Radius of the circle. The radius drawn to the point where the arc touches a side is perpendicular to that side. In the right-angled triangle formed by O, C and the tangent point, OC is the hypotenuse and the radius r is opposite the \(65^\circ\) angle:
\[r = |OC|\sin 65^\circ = 30 \times 0.9063 = 27.19\]
\[r \approx 27.2\ \text{cm}\]
(ii) Area of the shaded portion. The shaded region is the part of triangle ABC lying outside the semicircle:
\[\text{Shaded} = \text{Area of } \triangle ABC - \text{Area of semicircle}\]
The centre O lies on AB, so OC = 30 cm is the height of the triangle from C to AB. The half-base is
\[\frac{1}{2}|AB| = 30\tan 65^\circ = 30 \times 2.1445 = 64.34\ \text{cm}\]
\[|AB| = 128.67\ \text{cm}\]
\[\text{Area of } \triangle ABC = \frac{1}{2}\times |AB| \times \text{height} = \frac{1}{2}\times 128.67 \times 30 = 1930.1\ \text{cm}^2\]
Area of the semicircle, with \(\pi = \dfrac{22}{7}\) and \(r = 27.2\):
\[\frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7}\times (27.2)^2 = \frac{1}{2}\times \frac{22}{7}\times 739.84 = 1162.6\ \text{cm}^2\]
Therefore
\[\text{Shaded area} = 1930.1 - 1162.6 = 767.5\ \text{cm}^2\]
Bayanin Amsa
(a) Evaluating \(2\cos x + 3\sin x\)
Given \(\sin x = 0.6 = \dfrac{3}{5}\) with \(0^\circ \le x \le 90^\circ\), x is acute, so \(\cos x\) is positive.
\[\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - (0.6)^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8 = \frac{4}{5}\]
Then
\[2\cos x + 3\sin x = 2\left(\frac{4}{5}\right) + 3\left(\frac{3}{5}\right) = \frac{8}{5} + \frac{9}{5} = \frac{17}{5}\]
(b) Semi-circle inscribed in isosceles triangle ABC
From the diagram, the diameter WZ lies along the top side AB, the centre O is on AB, and the semicircular arc is tangent to the two equal sides. The apex C is below, with \(|OC| = 30\ \text{cm}\) measured along the axis of symmetry, and \(\angle ACB = 130^\circ\).
By symmetry (AC = BC), the line CO bisects angle C, so the half-angle at C is
\[\frac{130^\circ}{2} = 65^\circ\]
(i) Radius of the circle. The radius drawn to the point where the arc touches a side is perpendicular to that side. In the right-angled triangle formed by O, C and the tangent point, OC is the hypotenuse and the radius r is opposite the \(65^\circ\) angle:
\[r = |OC|\sin 65^\circ = 30 \times 0.9063 = 27.19\]
\[r \approx 27.2\ \text{cm}\]
(ii) Area of the shaded portion. The shaded region is the part of triangle ABC lying outside the semicircle:
\[\text{Shaded} = \text{Area of } \triangle ABC - \text{Area of semicircle}\]
The centre O lies on AB, so OC = 30 cm is the height of the triangle from C to AB. The half-base is
\[\frac{1}{2}|AB| = 30\tan 65^\circ = 30 \times 2.1445 = 64.34\ \text{cm}\]
\[|AB| = 128.67\ \text{cm}\]
\[\text{Area of } \triangle ABC = \frac{1}{2}\times |AB| \times \text{height} = \frac{1}{2}\times 128.67 \times 30 = 1930.1\ \text{cm}^2\]
Area of the semicircle, with \(\pi = \dfrac{22}{7}\) and \(r = 27.2\):
\[\frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7}\times (27.2)^2 = \frac{1}{2}\times \frac{22}{7}\times 739.84 = 1162.6\ \text{cm}^2\]
Therefore
\[\text{Shaded area} = 1930.1 - 1162.6 = 767.5\ \text{cm}^2\]
Tambaya 6 Rahoto
(a) P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R = 2, find the value of Q when P = 3 and R = 5.
(b) An aeroplane flies from town A(20°N, 60°E) to town B(20°N, 20°E). (i) if the journey takes 6 hours, calculate, correct to 3 significant figures, the average speed of the aeroplane. (ii) if it then flies due North from town B to town C, 420 km away, calculate correct to the nearest degree, the latitude of town C. [Take radius of the earth = 6400 km and \(\pi\) = 3.142].
(a) \(P\propto\dfrac{Q}{R^{2}}\Rightarrow P=\dfrac{kQ}{R^{2}}\).
Using \(P=1,\ Q=8,\ R=2\): \(1=\dfrac{8k}{4}=2k\Rightarrow k=\tfrac{1}{2}\).
When \(P=3,\ R=5\): \(3=\dfrac{\tfrac{1}{2}\,Q}{25}=\dfrac{Q}{50}\Rightarrow Q=150\).
\(Q=150\)
(b)(i) \(A(20^{\circ}N,60^{\circ}E)\) and \(B(20^{\circ}N,20^{\circ}E)\) are on the same parallel of latitude, so the flight is along the parallel \(20^{\circ}N\). Difference in longitude \(=60-20=40^{\circ}\).
\[\text{Distance}=\frac{40}{360}\times2\pi R\cos20^{\circ}=\frac{40}{360}\times2(3.142)(6400)(0.9397)\approx 4199\text{ km}\]\[\text{Average speed}=\frac{4199}{6}\approx 700\text{ km/h (3 s.f.)}\](ii) Flying due North from \(B\) is along a meridian, where distance \(=\dfrac{\theta}{360}\times2\pi R\), \(\theta\) the change in latitude.
\[420=\frac{\theta}{360}\times2(3.142)(6400)\Rightarrow \theta=\frac{420\times360}{40217.6}\approx 3.76^{\circ}\]Starting at \(20^{\circ}N\) and moving north, latitude of \(C=20+3.76\approx 24^{\circ}N\).
Latitude of C \(\approx 24^{\circ}N\).
Bayanin Amsa
(a) \(P\propto\dfrac{Q}{R^{2}}\Rightarrow P=\dfrac{kQ}{R^{2}}\).
Using \(P=1,\ Q=8,\ R=2\): \(1=\dfrac{8k}{4}=2k\Rightarrow k=\tfrac{1}{2}\).
When \(P=3,\ R=5\): \(3=\dfrac{\tfrac{1}{2}\,Q}{25}=\dfrac{Q}{50}\Rightarrow Q=150\).
\(Q=150\)
(b)(i) \(A(20^{\circ}N,60^{\circ}E)\) and \(B(20^{\circ}N,20^{\circ}E)\) are on the same parallel of latitude, so the flight is along the parallel \(20^{\circ}N\). Difference in longitude \(=60-20=40^{\circ}\).
\[\text{Distance}=\frac{40}{360}\times2\pi R\cos20^{\circ}=\frac{40}{360}\times2(3.142)(6400)(0.9397)\approx 4199\text{ km}\]\[\text{Average speed}=\frac{4199}{6}\approx 700\text{ km/h (3 s.f.)}\](ii) Flying due North from \(B\) is along a meridian, where distance \(=\dfrac{\theta}{360}\times2\pi R\), \(\theta\) the change in latitude.
\[420=\frac{\theta}{360}\times2(3.142)(6400)\Rightarrow \theta=\frac{420\times360}{40217.6}\approx 3.76^{\circ}\]Starting at \(20^{\circ}N\) and moving north, latitude of \(C=20+3.76\approx 24^{\circ}N\).
Latitude of C \(\approx 24^{\circ}N\).
Tambaya 7 Rahoto
Using ruler and a pair of compasses only,
(a) construct a rhombus PQRS of side 7 cm and < PQR = 60°;
(b) locate point X such that X lies on the locus of points equidistant from PQ and QR and also equidistant from Q and R ;
(c) measure |XR|.
Construction and loci
- Draw \(QR=7\text{ cm}\).
- With centres \(Q\) and \(R\), and radius \(7\text{ cm}\), draw arcs intersecting at \(P\). This gives \(QP=QR=7\text{ cm}\) and \(\angle PQR=60^\circ\).
- With centres \(P\) and \(R\), again using radius \(7\text{ cm}\), draw arcs intersecting at \(S\) (choose the intersection other than \(Q\)). Join \(PS\) and \(RS\). All four sides are \(7\text{ cm}\), so \(PQRS\) is a rhombus.
- The locus of points equidistant from the lines \(PQ\) and \(QR\) is the angle bisector of \(\angle PQR\). Construct the internal angle bisector.
- The locus of points equidistant from points \(Q\) and \(R\) is the perpendicular bisector of \(QR\).
- The intersection of these loci is \(X\).
Finding \(|XR|\)
The angle bisector makes an angle of \(30^\circ\) with \(QR\). The perpendicular bisector meets \(QR\) at its midpoint, so the horizontal distance from \(X\) to \(R\) is \(3.5\text{ cm}\).
Using coordinates with \(Q=(0,0)\) and \(R=(7,0)\), the perpendicular bisector is \(x=3.5\). At this point on the \(30^\circ\) angle bisector,
\[ y=3.5\tan 30^\circ=\frac{7}{2\sqrt3}. \] \[ |XR|=\sqrt{3.5^2+\left(\frac{7}{2\sqrt3}\right)^2} =\sqrt{\frac{49}{3}} =\frac{7}{\sqrt3} \approx 4.04\text{ cm}. \]Therefore, \(|XR|\approx 4.0\text{ cm}\) when measured to the nearest millimetre.
The supplied reference answer is inconsistent with the question: it gives \(|XP|=3.8\text{ cm}\), whereas the question asks for \(|XR|\). In the exact construction, \(|XP|=|XR|\approx4.04\text{ cm}\), so \(3.8\text{ cm}\) is not consistent with the stated dimensions.
Examination reminder: “Equidistant from two lines” means an angle bisector; “equidistant from two points” means a perpendicular bisector.
Bayanin Amsa
Construction and loci
- Draw \(QR=7\text{ cm}\).
- With centres \(Q\) and \(R\), and radius \(7\text{ cm}\), draw arcs intersecting at \(P\). This gives \(QP=QR=7\text{ cm}\) and \(\angle PQR=60^\circ\).
- With centres \(P\) and \(R\), again using radius \(7\text{ cm}\), draw arcs intersecting at \(S\) (choose the intersection other than \(Q\)). Join \(PS\) and \(RS\). All four sides are \(7\text{ cm}\), so \(PQRS\) is a rhombus.
- The locus of points equidistant from the lines \(PQ\) and \(QR\) is the angle bisector of \(\angle PQR\). Construct the internal angle bisector.
- The locus of points equidistant from points \(Q\) and \(R\) is the perpendicular bisector of \(QR\).
- The intersection of these loci is \(X\).
Finding \(|XR|\)
The angle bisector makes an angle of \(30^\circ\) with \(QR\). The perpendicular bisector meets \(QR\) at its midpoint, so the horizontal distance from \(X\) to \(R\) is \(3.5\text{ cm}\).
Using coordinates with \(Q=(0,0)\) and \(R=(7,0)\), the perpendicular bisector is \(x=3.5\). At this point on the \(30^\circ\) angle bisector,
\[ y=3.5\tan 30^\circ=\frac{7}{2\sqrt3}. \] \[ |XR|=\sqrt{3.5^2+\left(\frac{7}{2\sqrt3}\right)^2} =\sqrt{\frac{49}{3}} =\frac{7}{\sqrt3} \approx 4.04\text{ cm}. \]Therefore, \(|XR|\approx 4.0\text{ cm}\) when measured to the nearest millimetre.
The supplied reference answer is inconsistent with the question: it gives \(|XP|=3.8\text{ cm}\), whereas the question asks for \(|XR|\). In the exact construction, \(|XP|=|XR|\approx4.04\text{ cm}\), so \(3.8\text{ cm}\) is not consistent with the stated dimensions.
Examination reminder: “Equidistant from two lines” means an angle bisector; “equidistant from two points” means a perpendicular bisector.
Tambaya 8 Rahoto
The table shows the scores obtained when a fair die was thrown a number of times.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 2 | 5 | x | 11 | 9 | 10 |
If the probability of obtaining a 3 is 0.26, find the (a) median
(b) standard deviation of the distribution.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency | 2 | 5 | x | 11 | 9 | 10 |
Find x. Total \(N=37+x\). Since \(P(3)=0.26\):
\[ \frac{x}{37+x}=0.26 \Rightarrow x=9.62+0.26x \Rightarrow 0.74x=9.62 \Rightarrow x=13 \]So frequencies are \(2,5,13,11,9,10\) with \(N=50\).
(a) Median. With \(N=50\), the median is the mean of the 25th and 26th values. Cumulative frequencies: 2, 7, 20, 31, ... Both the 25th and 26th fall at score 4, so median = 4.
(b) Standard deviation. First the mean:
\[ \Sigma fx = 2+10+39+44+45+60 = 200,\quad \bar{x}=\frac{200}{50}=4 \] \[ \Sigma fx^2 = 2+20+117+176+225+360 = 900 \] \[ \text{Variance}=\frac{\Sigma fx^2}{N}-\bar{x}^2=\frac{900}{50}-4^2=18-16=2 \] \[ \text{SD}=\sqrt{2}\approx 1.41 \]Bayanin Amsa
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency | 2 | 5 | x | 11 | 9 | 10 |
Find x. Total \(N=37+x\). Since \(P(3)=0.26\):
\[ \frac{x}{37+x}=0.26 \Rightarrow x=9.62+0.26x \Rightarrow 0.74x=9.62 \Rightarrow x=13 \]So frequencies are \(2,5,13,11,9,10\) with \(N=50\).
(a) Median. With \(N=50\), the median is the mean of the 25th and 26th values. Cumulative frequencies: 2, 7, 20, 31, ... Both the 25th and 26th fall at score 4, so median = 4.
(b) Standard deviation. First the mean:
\[ \Sigma fx = 2+10+39+44+45+60 = 200,\quad \bar{x}=\frac{200}{50}=4 \] \[ \Sigma fx^2 = 2+20+117+176+225+360 = 900 \] \[ \text{Variance}=\frac{\Sigma fx^2}{N}-\bar{x}^2=\frac{900}{50}-4^2=18-16=2 \] \[ \text{SD}=\sqrt{2}\approx 1.41 \]Tambaya 9 Rahoto
(a) Simplify : \(\frac{\frac{1}{2} of \frac{1}{4} \div \frac{1}{3}}{\frac{1}{6} - \frac{3}{4} + \frac{1}{2}}\).
(b) Given that \(\sqrt{x} = 10^{\bar{1}.6741}\), without using calculators, find the value of x.
(a) Numerator: \(\tfrac{1}{2}\text{ of }\tfrac{1}{4} \div \tfrac{1}{3} = \tfrac{1}{8}\div\tfrac{1}{3} = \tfrac{1}{8}\times3 = \tfrac{3}{8}.\)
Denominator: \(\tfrac{1}{6} - \tfrac{3}{4} + \tfrac{1}{2} = \tfrac{2 - 9 + 6}{12} = -\tfrac{1}{12}.\)
\[\frac{\tfrac{3}{8}}{-\tfrac{1}{12}} = \tfrac{3}{8}\times(-12) = -\tfrac{9}{2} = -4\tfrac{1}{2}.\]
(b) The bar denotes a negative characteristic, so \(\bar{1}.6741 = -1 + 0.6741 = -0.3259\). Squaring \(\sqrt{x}\):
\[x = \left(10^{\bar{1}.6741}\right)^2 = 10^{\,2(-0.3259)} = 10^{-0.6518} = 10^{\bar{1}.3482}.\]
The antilog of \(0.3482\) is \(2.229\), and the characteristic \(\bar{1}\) places one zero after the point:
\[x = 0.2229 \approx 0.223.\]
Bayanin Amsa
(a) Numerator: \(\tfrac{1}{2}\text{ of }\tfrac{1}{4} \div \tfrac{1}{3} = \tfrac{1}{8}\div\tfrac{1}{3} = \tfrac{1}{8}\times3 = \tfrac{3}{8}.\)
Denominator: \(\tfrac{1}{6} - \tfrac{3}{4} + \tfrac{1}{2} = \tfrac{2 - 9 + 6}{12} = -\tfrac{1}{12}.\)
\[\frac{\tfrac{3}{8}}{-\tfrac{1}{12}} = \tfrac{3}{8}\times(-12) = -\tfrac{9}{2} = -4\tfrac{1}{2}.\]
(b) The bar denotes a negative characteristic, so \(\bar{1}.6741 = -1 + 0.6741 = -0.3259\). Squaring \(\sqrt{x}\):
\[x = \left(10^{\bar{1}.6741}\right)^2 = 10^{\,2(-0.3259)} = 10^{-0.6518} = 10^{\bar{1}.3482}.\]
The antilog of \(0.3482\) is \(2.229\), and the characteristic \(\bar{1}\) places one zero after the point:
\[x = 0.2229 \approx 0.223.\]
Tambaya 10 Rahoto
(a) The total surface area of two spheres are in the ratio 9 : 49. If the radius of the smaller sphere is 12 cm, find, correct to the nearest \(cm^{3}\), the volume of the bigger sphere.
(b) A cyclist starts from a point X and rides 3 km due West to a point Y. At Y, he changes direction and rides 5 km North- West to a point Z.
(i) How far is he from the starting point, correct to the nearest km? ; (ii) Find the bearing of Z from X, to the nearest degree.
(a) Surface areas of similar spheres are in the ratio of the squares of their radii:
\[\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{9}{49}\Rightarrow \frac{r_{1}}{r_{2}}=\frac{3}{7}\]The smaller radius \(r_{1}=12\) cm corresponds to the ratio part 3, so one part \(=\dfrac{12}{3}=4\) cm, and the bigger radius \(r_{2}=7\times4=28\) cm.
\[V=\frac{4}{3}\pi r_{2}^{3}=\frac{4}{3}(3.142)(28)^{3}=\frac{4}{3}(3.142)(21952)\approx 91964\text{ cm}^{3}\]Volume of bigger sphere \(\approx 91{,}964\text{ cm}^{3}\).
(b) Take \(X\) as origin, West as the negative \(x\)-direction. \(Y=(-3,0)\). North-West is the bearing \(315^{\circ}\), i.e. direction \((-\sin45^{\circ},\cos45^{\circ})\), so
\[Z=(-3,0)+5(-\sin45^{\circ},\cos45^{\circ})=(-3-3.536,\ 3.536)=(-6.536,\ 3.536)\](i) Distance from \(X\):
\[XZ=\sqrt{6.536^{2}+3.536^{2}}=\sqrt{42.72+12.50}=\sqrt{55.22}\approx 7.4\ \text{km}\approx 7\text{ km}\](ii) \(Z\) lies to the North and West of \(X\). The angle west of due North is \(\tan^{-1}\dfrac{6.536}{3.536}=\tan^{-1}1.848=61.6^{\circ}\).
\[\text{Bearing of }Z\text{ from }X=360^{\circ}-61.6^{\circ}\approx 298^{\circ}\]Bearing \(\approx 298^{\circ}\).
Bayanin Amsa
(a) Surface areas of similar spheres are in the ratio of the squares of their radii:
\[\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{9}{49}\Rightarrow \frac{r_{1}}{r_{2}}=\frac{3}{7}\]The smaller radius \(r_{1}=12\) cm corresponds to the ratio part 3, so one part \(=\dfrac{12}{3}=4\) cm, and the bigger radius \(r_{2}=7\times4=28\) cm.
\[V=\frac{4}{3}\pi r_{2}^{3}=\frac{4}{3}(3.142)(28)^{3}=\frac{4}{3}(3.142)(21952)\approx 91964\text{ cm}^{3}\]Volume of bigger sphere \(\approx 91{,}964\text{ cm}^{3}\).
(b) Take \(X\) as origin, West as the negative \(x\)-direction. \(Y=(-3,0)\). North-West is the bearing \(315^{\circ}\), i.e. direction \((-\sin45^{\circ},\cos45^{\circ})\), so
\[Z=(-3,0)+5(-\sin45^{\circ},\cos45^{\circ})=(-3-3.536,\ 3.536)=(-6.536,\ 3.536)\](i) Distance from \(X\):
\[XZ=\sqrt{6.536^{2}+3.536^{2}}=\sqrt{42.72+12.50}=\sqrt{55.22}\approx 7.4\ \text{km}\approx 7\text{ km}\](ii) \(Z\) lies to the North and West of \(X\). The angle west of due North is \(\tan^{-1}\dfrac{6.536}{3.536}=\tan^{-1}1.848=61.6^{\circ}\).
\[\text{Bearing of }Z\text{ from }X=360^{\circ}-61.6^{\circ}\approx 298^{\circ}\]Bearing \(\approx 298^{\circ}\).
Tambaya 11 Rahoto
A sector of a circle with radius 21 cm has an area of 280\(cm^{2}\).
(a) Calculate, correct to 1 decimal place, the perimeter of the sector.
(b) If the sector is bent such that its straight edges coincide to form a cone, calculate, correct to the nearest degree, the vertical angle of the cone. [Take \(\pi = \frac{22}{7}\)].
(a) Perimeter of the sector. First find the angle \(\theta\) from the area:
\[\text{Area} = \frac{\theta}{360}\pi r^2 \Rightarrow 280 = \frac{\theta}{360}\times\frac{22}{7}\times21^2.\]
\[\frac{22}{7}\times441 = 1386,\qquad \frac{\theta}{360} = \frac{280}{1386} = 0.2020.\]
Arc length \(= \dfrac{\theta}{360}\times2\pi r = 0.2020\times\left(2\times\tfrac{22}{7}\times21\right) = 0.2020\times132 = 26.7\text{ cm}.\)
\[\text{Perimeter} = \text{arc} + 2r = 26.7 + 42 = 68.7\text{ cm}.\]
(b) Vertical angle of the cone. When bent, the arc becomes the base circumference and the sector radius becomes the slant height \(l = 21\) cm. Base radius R:
\[2\pi R = 26.7 \Rightarrow R = \frac{26.7}{2\times\tfrac{22}{7}} = 4.24\text{ cm}.\]
Half the vertical angle \(\alpha\) satisfies \(\sin\alpha = \dfrac{R}{l} = \dfrac{4.24}{21} = 0.2020\), so \(\alpha = 11.66^\circ\).
\[\text{Vertical angle} = 2\alpha = 23.3^\circ \approx 23^\circ.\]
Bayanin Amsa
(a) Perimeter of the sector. First find the angle \(\theta\) from the area:
\[\text{Area} = \frac{\theta}{360}\pi r^2 \Rightarrow 280 = \frac{\theta}{360}\times\frac{22}{7}\times21^2.\]
\[\frac{22}{7}\times441 = 1386,\qquad \frac{\theta}{360} = \frac{280}{1386} = 0.2020.\]
Arc length \(= \dfrac{\theta}{360}\times2\pi r = 0.2020\times\left(2\times\tfrac{22}{7}\times21\right) = 0.2020\times132 = 26.7\text{ cm}.\)
\[\text{Perimeter} = \text{arc} + 2r = 26.7 + 42 = 68.7\text{ cm}.\]
(b) Vertical angle of the cone. When bent, the arc becomes the base circumference and the sector radius becomes the slant height \(l = 21\) cm. Base radius R:
\[2\pi R = 26.7 \Rightarrow R = \frac{26.7}{2\times\tfrac{22}{7}} = 4.24\text{ cm}.\]
Half the vertical angle \(\alpha\) satisfies \(\sin\alpha = \dfrac{R}{l} = \dfrac{4.24}{21} = 0.2020\), so \(\alpha = 11.66^\circ\).
\[\text{Vertical angle} = 2\alpha = 23.3^\circ \approx 23^\circ.\]
Tambaya 12 Rahoto
A library received $1,300 grant. It spends 10% of the grant on magazine subscriptions, 35% on new books, 15% to repair damaged books, 30% to buy new furniture and 10% to train library staff.
(a) Represent this information on a pie chart.
(b) Calculate, correct to the nearest whole number, the percentage increase of the amount for buying books over that of new furniture.
(a) Pie chart
The whole grant of \($1{,}300\) is represented by the full circle, \(360^{\circ}\). Each item's sector angle is found from \(\text{angle}=\dfrac{\text{percentage}}{100}\times360^{\circ}\), which is the same as \(\dfrac{\text{amount}}{1300}\times360^{\circ}\).
| Item | Percentage | Amount ($) | Sector angle |
|---|---|---|---|
| Magazine subscriptions | 10% | 130 | \(\frac{130}{1300}\times360^{\circ}=36^{\circ}\) |
| New books | 35% | 455 | \(\frac{455}{1300}\times360^{\circ}=126^{\circ}\) |
| Repair of damaged books | 15% | 195 | \(\frac{195}{1300}\times360^{\circ}=54^{\circ}\) |
| New furniture | 30% | 390 | \(\frac{390}{1300}\times360^{\circ}=108^{\circ}\) |
| Staff training | 10% | 130 | \(\frac{130}{1300}\times360^{\circ}=36^{\circ}\) |
| Total | 100% | 1300 | \(360^{\circ}\) |
Drawing a circle and marking each sector in turn with the angles above gives the following pie chart:
(b) Percentage increase of books over furniture
Amount for new books \(=35\%\) of \(1300 = \$455\).
Amount for new furniture \(=30\%\) of \(1300 = \$390\).
The increase in the amount is \(455-390=\$65\), and this is expressed as a percentage of the furniture amount:
\[\text{Percentage increase}=\frac{455-390}{390}\times100\%=\frac{65}{390}\times100\%=16.67\%\approx 17\%\]The books allocation is about \(17\%\) more than the furniture allocation (to the nearest whole number).
Bayanin Amsa
(a) Pie chart
The whole grant of \($1{,}300\) is represented by the full circle, \(360^{\circ}\). Each item's sector angle is found from \(\text{angle}=\dfrac{\text{percentage}}{100}\times360^{\circ}\), which is the same as \(\dfrac{\text{amount}}{1300}\times360^{\circ}\).
| Item | Percentage | Amount ($) | Sector angle |
|---|---|---|---|
| Magazine subscriptions | 10% | 130 | \(\frac{130}{1300}\times360^{\circ}=36^{\circ}\) |
| New books | 35% | 455 | \(\frac{455}{1300}\times360^{\circ}=126^{\circ}\) |
| Repair of damaged books | 15% | 195 | \(\frac{195}{1300}\times360^{\circ}=54^{\circ}\) |
| New furniture | 30% | 390 | \(\frac{390}{1300}\times360^{\circ}=108^{\circ}\) |
| Staff training | 10% | 130 | \(\frac{130}{1300}\times360^{\circ}=36^{\circ}\) |
| Total | 100% | 1300 | \(360^{\circ}\) |
Drawing a circle and marking each sector in turn with the angles above gives the following pie chart:
(b) Percentage increase of books over furniture
Amount for new books \(=35\%\) of \(1300 = \$455\).
Amount for new furniture \(=30\%\) of \(1300 = \$390\).
The increase in the amount is \(455-390=\$65\), and this is expressed as a percentage of the furniture amount:
\[\text{Percentage increase}=\frac{455-390}{390}\times100\%=\frac{65}{390}\times100\%=16.67\%\approx 17\%\]The books allocation is about \(17\%\) more than the furniture allocation (to the nearest whole number).
Tambaya 13 Rahoto
(a) Divide \(\frac{x^{2} - 4}{x^{2} + x}\) by \(\frac{x^{2} - 4x + 4}{x + 1}\).
(b) The diagram below shows the graphs of \(y = ax^{2} + bx + c\) and \(y = mx + k\) where a, b, c and m are constants. Use the graph(s) to :
(i) find the roots of the equation \(ax^{2} + bx + c = mx + k\);
(ii) determine the values of a, b and c using the coordinates of points L, M and N and hence write down the equation of the curve;
(iii) determine the line of symmetry of the curve \(y = ax^{2} + bx + c\).
(a) Divide the algebraic fractions.
Dividing by a fraction is the same as multiplying by its reciprocal:
\[\frac{x^{2}-4}{x^{2}+x}\div\frac{x^{2}-4x+4}{x+1}=\frac{x^{2}-4}{x^{2}+x}\times\frac{x+1}{x^{2}-4x+4}\]Factorise each expression fully:
\[x^{2}-4=(x-2)(x+2),\qquad x^{2}+x=x(x+1),\qquad x^{2}-4x+4=(x-2)^{2}\]Substitute the factorised forms:
\[=\frac{(x-2)(x+2)}{x(x+1)}\times\frac{x+1}{(x-2)^{2}}\]Cancel the common factors \((x+1)\) and one \((x-2)\):
\[=\frac{x+2}{x(x-2)}=\frac{x+2}{x^{2}-2x}\](b) Using the graph.
Reading the marked points from the diagram, the curve \(y=ax^{2}+bx+c\) cuts the x-axis at \(L(-1,0)\) and \(N(2,0)\) and cuts the y-axis at \(M(0,2)\). The straight line \(y=mx+k\) passes through \(L(-1,0)\) and cuts the curve again at \(H(1.5,\,1.25)\). The reconstructed graph below shows these features.
(ii) Values of \(a\), \(b\), \(c\) and the equation of the curve. (Solved first, as the roots in part (i) depend on this equation.)
Substitute the coordinates of \(L\), \(M\) and \(N\) into \(y=ax^{2}+bx+c\):
\[L(-1,0):\;\;a(-1)^{2}+b(-1)+c=0\;\Rightarrow\;a-b+c=0\quad(1)\]\[M(0,2):\;\;a(0)^{2}+b(0)+c=2\;\Rightarrow\;c=2\quad(2)\]\[N(2,0):\;\;a(2)^{2}+b(2)+c=0\;\Rightarrow\;4a+2b+c=0\quad(3)\]Put \(c=2\) from (2) into (1) and (3):
\[a-b=-2\quad(1a),\qquad 4a+2b=-2\quad(3a)\]From (1a), \(b=a+2\). Substitute into (3a):
\[4a+2(a+2)=-2\;\Rightarrow\;6a+4=-2\;\Rightarrow\;6a=-6\;\Rightarrow\;a=-1\]Then \(b=a+2=1\) and \(c=2\). Therefore
\[a=-1,\qquad b=1,\qquad c=2,\]and the equation of the curve is
\[y=-x^{2}+x+2.\](i) Roots of \(ax^{2}+bx+c=mx+k\).
The solutions are the x-coordinates of the points where the line meets the curve, namely \(L\) and \(H\). First find the line \(y=mx+k\) through \(L(-1,0)\) and \(H(1.5,1.25)\):
\[m=\frac{1.25-0}{1.5-(-1)}=\frac{1.25}{2.5}=0.5\]Using \(L(-1,0)\): \(0=0.5(-1)+k\Rightarrow k=0.5\), so \(y=0.5x+0.5\).
Now equate the curve and the line:
\[-x^{2}+x+2=0.5x+0.5\]\[-x^{2}+0.5x+1.5=0\;\Rightarrow\;x^{2}-0.5x-1.5=0\]Solve using the formula \(x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a=1,\,b=-0.5,\,c=-1.5\):
\[x=\frac{0.5\pm\sqrt{(-0.5)^{2}-4(1)(-1.5)}}{2}=\frac{0.5\pm\sqrt{0.25+6}}{2}=\frac{0.5\pm\sqrt{6.25}}{2}=\frac{0.5\pm2.5}{2}\]\[x=\frac{0.5+2.5}{2}=1.5\qquad\text{or}\qquad x=\frac{0.5-2.5}{2}=-1\]The roots are \(x=-1\) and \(x=1.5\), which are exactly the x-coordinates of \(L\) and \(H\) read from the graph.
(iii) Line of symmetry of the curve.
The line of symmetry of \(y=ax^{2}+bx+c\) is \(x=-\dfrac{b}{2a}\). With \(a=-1\) and \(b=1\):
\[x=-\frac{1}{2(-1)}=\frac{1}{2}\]The line of symmetry is \(x=0.5\), the vertical line midway between the roots \(-1\) and \(2\) (shown dashed on the graph, passing through the maximum point \((0.5,\,2.25)\)).
Bayanin Amsa
(a) Divide the algebraic fractions.
Dividing by a fraction is the same as multiplying by its reciprocal:
\[\frac{x^{2}-4}{x^{2}+x}\div\frac{x^{2}-4x+4}{x+1}=\frac{x^{2}-4}{x^{2}+x}\times\frac{x+1}{x^{2}-4x+4}\]Factorise each expression fully:
\[x^{2}-4=(x-2)(x+2),\qquad x^{2}+x=x(x+1),\qquad x^{2}-4x+4=(x-2)^{2}\]Substitute the factorised forms:
\[=\frac{(x-2)(x+2)}{x(x+1)}\times\frac{x+1}{(x-2)^{2}}\]Cancel the common factors \((x+1)\) and one \((x-2)\):
\[=\frac{x+2}{x(x-2)}=\frac{x+2}{x^{2}-2x}\](b) Using the graph.
Reading the marked points from the diagram, the curve \(y=ax^{2}+bx+c\) cuts the x-axis at \(L(-1,0)\) and \(N(2,0)\) and cuts the y-axis at \(M(0,2)\). The straight line \(y=mx+k\) passes through \(L(-1,0)\) and cuts the curve again at \(H(1.5,\,1.25)\). The reconstructed graph below shows these features.
(ii) Values of \(a\), \(b\), \(c\) and the equation of the curve. (Solved first, as the roots in part (i) depend on this equation.)
Substitute the coordinates of \(L\), \(M\) and \(N\) into \(y=ax^{2}+bx+c\):
\[L(-1,0):\;\;a(-1)^{2}+b(-1)+c=0\;\Rightarrow\;a-b+c=0\quad(1)\]\[M(0,2):\;\;a(0)^{2}+b(0)+c=2\;\Rightarrow\;c=2\quad(2)\]\[N(2,0):\;\;a(2)^{2}+b(2)+c=0\;\Rightarrow\;4a+2b+c=0\quad(3)\]Put \(c=2\) from (2) into (1) and (3):
\[a-b=-2\quad(1a),\qquad 4a+2b=-2\quad(3a)\]From (1a), \(b=a+2\). Substitute into (3a):
\[4a+2(a+2)=-2\;\Rightarrow\;6a+4=-2\;\Rightarrow\;6a=-6\;\Rightarrow\;a=-1\]Then \(b=a+2=1\) and \(c=2\). Therefore
\[a=-1,\qquad b=1,\qquad c=2,\]and the equation of the curve is
\[y=-x^{2}+x+2.\](i) Roots of \(ax^{2}+bx+c=mx+k\).
The solutions are the x-coordinates of the points where the line meets the curve, namely \(L\) and \(H\). First find the line \(y=mx+k\) through \(L(-1,0)\) and \(H(1.5,1.25)\):
\[m=\frac{1.25-0}{1.5-(-1)}=\frac{1.25}{2.5}=0.5\]Using \(L(-1,0)\): \(0=0.5(-1)+k\Rightarrow k=0.5\), so \(y=0.5x+0.5\).
Now equate the curve and the line:
\[-x^{2}+x+2=0.5x+0.5\]\[-x^{2}+0.5x+1.5=0\;\Rightarrow\;x^{2}-0.5x-1.5=0\]Solve using the formula \(x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a=1,\,b=-0.5,\,c=-1.5\):
\[x=\frac{0.5\pm\sqrt{(-0.5)^{2}-4(1)(-1.5)}}{2}=\frac{0.5\pm\sqrt{0.25+6}}{2}=\frac{0.5\pm\sqrt{6.25}}{2}=\frac{0.5\pm2.5}{2}\]\[x=\frac{0.5+2.5}{2}=1.5\qquad\text{or}\qquad x=\frac{0.5-2.5}{2}=-1\]The roots are \(x=-1\) and \(x=1.5\), which are exactly the x-coordinates of \(L\) and \(H\) read from the graph.
(iii) Line of symmetry of the curve.
The line of symmetry of \(y=ax^{2}+bx+c\) is \(x=-\dfrac{b}{2a}\). With \(a=-1\) and \(b=1\):
\[x=-\frac{1}{2(-1)}=\frac{1}{2}\]The line of symmetry is \(x=0.5\), the vertical line midway between the roots \(-1\) and \(2\) (shown dashed on the graph, passing through the maximum point \((0.5,\,2.25)\)).
