64 questions. Try each one, then open the answer to check yourself and read the explanation.
Evaluate (0.13)3 correct to three significant figures.
Answer: B (0.00220)
0.13 × 0.13 × 0.13 = 0.002197. The first three significant figures are 2, 1, 9 and the next digit 7 rounds the 9 up, giving 0.00220.
Simplify: 11011two − 1101two
Answer: C (1110two)
11011two = 27 and 1101two = 13. 27 − 13 = 14 = 8 + 4 + 2 = 1110two.
Simplify: (252/3 ÷ 251/6) / ((1/5)−7/6 × (1/5)1/6)
Answer: B (1)
Numerator: 252/3 − 1/6 = 251/2 = 5. Denominator: (1/5)−7/6 + 1/6 = (1/5)−1 = 5. So the value is 5 ÷ 5 = 1.
Simplify: (x − 4)/4 − (x − 3)/6
Answer: B ((x − 6)/12)
Use the common denominator 12: [3(x − 4) − 2(x − 3)]/12 = (3x − 12 − 2x + 6)/12 = (x − 6)/12.
Given that y = 1 − 2x/(4x − 3), find the value of x for which y is undefined.
Answer: B (3/4)
y is undefined when the denominator is zero: 4x − 3 = 0, so x = 3/4.
P is a point on the same plane with a fixed point A. If P moves such that it is always equidistant from A, the locus of P is
Answer: C (a circle with centre A)
All points at a fixed distance from one point form a circle with that point as centre.
A fair coin is tossed three times. Find the probability of getting two heads and one tail.
Answer: B (3/8)
There are 2 × 2 × 2 = 8 equally likely outcomes. HHT, HTH and THH give two heads and one tail, so the probability is 3/8.
If 30% of y is equal to x, what, in terms of x, is 30% of 3y?
Answer: D (3x)
30% of 3y = 3 × (30% of y) = 3x.
A baker used 40% of a 50 kg bag of flour. If 1/8 of the amount used was for cake, how many kilograms of flour was used for cake?
Answer: A (2½)
Flour used = 40% of 50 kg = 20 kg. Flour for cake = 1/8 × 20 kg = 2½ kg.
If tan y = 0.404, where y is acute, find cos 2y.
Answer: B (0.719)
y = tan−1(0.404) ≈ 22°, so 2y ≈ 44° and cos 44° ≈ 0.719.
In the diagram, PQUV, PQTU, QRTU and QRST are parallelograms. |UV| = 4.8 cm and the perpendicular distance between PR and VS is 5 cm. Calculate the area of quadrilateral PRSV.
Answer: C (60 cm2)
Opposite sides of parallelograms are equal, so PQ = QR = 4.8 cm and VU = UT = TS = 4.8 cm. PRSV is a trapezium with parallel sides 9.6 cm and 14.4 cm: area = ½(9.6 + 14.4) × 5 = 60 cm2.
Which of these diagrams show(s) the correct method of constructing a perpendicular line to another line PQ at a given point X on PQ?
Answer: C (I and II only)
In I, arcs from P and Q (equidistant from X) meet above X, and the line through X and that point is perpendicular. In II, the 60° and 120° marks on the semicircle are used and the angle between them is bisected, which also gives 90°. III shows no construction at all.
In the diagram, |QR| = 5 cm, PQ̂R = 60° and PŜR = 45°. Find |PS|, leaving your answer in surd form.
Answer: D (5√6 cm)
In right triangle PQR, |PR| = 5 tan 60° = 5√3. In right triangle PRS, |PS| = |PR| ÷ sin 45° = 5√3 × √2 = 5√6 cm.
Represent the solution of the inequality (x − 2)/2 ≥ (x + 3)/4 on the number line.
Answer: D (Diagram D)
Multiply both sides by 4: 2(x − 2) ≥ x + 3, so 2x − 4 ≥ x + 3 and x ≥ 7. This is a closed (filled) dot at 7 with the arrow pointing right, as in D.
In the diagram, PQ//RS, find x° + y°.
Answer: D (180°)
Because PQ is parallel to RS, the angle between RQ produced and the line through Q towards S equals y° (corresponding angles). That angle and x° lie on a straight line, so x° + y° = 180°.
The ratio of boys to girls in a class is 5:3. Find the probability of selecting at random, a girl from the class.
Answer: C (3/8)
Girls make up 3 out of every 5 + 3 = 8 pupils, so the probability is 3/8.
In the diagram, PR is a diameter of the circle centre O. RS is a tangent at R and QP̂R = 58°. Find ∠QRS.
Answer: C (122°)
The angle in a semicircle is 90°, so ∠PQR = 90° and ∠QRP = 180° − 90° − 58° = 32°. The tangent is perpendicular to the diameter, so ∠PRS = 90°. ∠QRS = 32° + 90° = 122°.
The table below satisfies the relation y = k√x, where k is a positive constant.
x: 1, 4, P
y: 0.5, 1, 2.5
Find the value of k.
Answer: A (0.5)
When x = 1, y = 0.5, so 0.5 = k × √1 and k = 0.5 (x = 4, y = 1 agrees: 0.5 × 2 = 1).
The table below satisfies the relation y = k√x, where k is a positive constant.
x: 1, 4, P
y: 0.5, 1, 2.5
Find the value of P.
Answer: D (25)
With k = 0.5: 2.5 = 0.5√P, so √P = 5 and P = 25.
What must be added to x2 − 3x to make it a perfect square?
Answer: A (9/4)
Add the square of half the coefficient of x: (3/2)2 = 9/4, giving (x − 3/2)2.
In the diagram P, Q, R, S are points on the circle, RQ̂S = 30°, PR̂S = 50° and PŜQ = 20°. What is the value of x° + y°?
Answer: B (130°)
Angles in the same segment: y = ∠PQS = ∠PRS = 50° and ∠SPR = ∠SQR = 30°. In triangle PSR, ∠PSR = 180° − 50° − 30° = 100°, so x = 100° − 20° = 80°. x + y = 130°.
Given that (2x − 1)(x + 5) = 2x2 − mx − 5, what is the value of m?
Answer: C (−9)
(2x − 1)(x + 5) = 2x2 + 9x − 5. Comparing, −m = 9, so m = −9.
What is the place value of 9 in the number 3.0492?
Answer: B (9/1000)
After the decimal point the digits are tenths (0), hundredths (4) and thousandths (9), so the 9 stands for 9/1000.
If the simple interest on a sum of money invested at 3% per annum for 2½ years is ₦123, find the principal.
Answer: C (₦1,640)
P = (100 × I)/(R × T) = (100 × 123)/(3 × 2.5) = 12300/7.5 = ₦1,640.
A machine valued at ₦20,000 depreciates by 10% every year. What will be the value of the machine at the end of two years?
Answer: A (₦16,200)
Each year the value is multiplied by 0.9: 20,000 × 0.9 × 0.9 = ₦16,200.
If 2x + y = 10, and y ≠ 0, which of the following is not a possible value of x?
Answer: B (5)
If x = 5, then y = 10 − 2(5) = 0, which is not allowed. Every other value gives a non-zero y.
In the diagram, P, Q and R are three points in a plane such that the bearing of R from Q is 110° and the bearing of Q from P is 050°. Find angle PQR.
Answer: C (120°)
The north lines are parallel, so the angle between QP and north at Q (measured on the south side) is 50° (alternate angles). ∠PQR = 50° + (180° − 110°) = 50° + 70° = 120°.
If x + y = 12 and 3x − y = 20, find the value of 2x − y.
Answer: C (12)
Adding the equations: 4x = 32, so x = 8 and y = 4. Then 2x − y = 16 − 4 = 12.
Use the graphs to answer this question. What is the equation of the curve?
Answer: B (y = 3 − 2x − x2)
The curve cuts the x-axis at x = −3 and x = 1 and turns at the top, so y = −(x + 3)(x − 1) = 3 − 2x − x2. Check: at x = −1, y = 3 + 2 − 1 = 4, the highest point on the graph.
Use the graphs to answer this question. Which of these equations best describes the points of intersection of the curve and the line?
Answer: C (x2 + x − 6 = 0)
The curve and the line meet at x = −3 and x = 2, so the equation is (x + 3)(x − 2) = 0, i.e. x2 + x − 6 = 0. (Line y = −x − 3 and curve y = 3 − 2x − x2 give the same equation.)
If loga270 − loga10 + loga(1/3) = 2, what is the value of a?
Answer: A (3)
Combine the logs: loga(270 ÷ 10 × 1/3) = loga9 = 2, so a2 = 9 and a = 3.
A conical water-jug is 7 cm in diameter and 6 cm deep. Find the volume of water it can hold. (Take π = 22/7)
Answer: C (77 cm3)
V = ⅓πr2h = ⅓ × 22/7 × 3.5 × 3.5 × 6 = 77 cm3.
In the diagram, the two circles have a common centre O. If the area of the larger circle is 100π and that of the smaller circle is 49π, find x.
Answer: B (3)
πR2 = 100π gives R = 10 and πr2 = 49π gives r = 7. x is the gap between the circles: 10 − 7 = 3.
If P = [Q(R − T)/15]1/3, make T the subject of the relation.
Answer: C (T = R − 15P3/Q)
Cube both sides: P3 = Q(R − T)/15, so R − T = 15P3/Q and T = R − 15P3/Q.
In the diagram, PQR is a straight line, QRST is a parallelogram, ∠TPQ = 72° and ∠RST = 126°. What type of triangle is ΔPQT?
Answer: C (An isosceles triangle)
In the parallelogram, ∠QRS = 180° − 126° = 54°, and TQ ∥ SR gives ∠PQT = 54° (corresponding angles). Then ∠PTQ = 180° − 72° − 54° = 54°. Two equal angles make ΔPQT isosceles.
A seller allows 20% discount for cash payment on the marked price of his goods. What is the ratio of the cash payment to the marked price?
Answer: D (4:5)
With 20% off, the cash price is 80% of the marked price: 80:100 = 4:5.
The histogram shows the marks scored by a group of pupils in a test. How many pupils took the test?
Answer: D (40)
Add the bar heights for marks 1 to 9: 1 + 3 + 4 + 7 + 8 + 6 + 6 + 4 + 1 = 40.
The histogram shows the marks scored by a group of pupils in a test. If the pass mark was 4, what percentage of the pupils failed the test?
Answer: A (20.0%)
Pupils who scored below 4 (marks 1, 2, 3): 1 + 3 + 4 = 8. 8/40 × 100% = 20%.
Find the nth term of the sequence 5, 10, 20, 40, 80, …
Answer: C (5 × 2n−1)
It is a GP with first term 5 and common ratio 2, so Tn = 5 × 2n−1.
The diagram shows an arc MN of a circle centre O, with radius 10 cm. If ∠MON = 72°, calculate the length of the arc, correct to three significant figures. (Take π = 22/7)
Answer: B (12.6 cm)
Arc length = 72/360 × 2 × 22/7 × 10 = 88/7 ≈ 12.6 cm.
The wheel of a tractor has diameter 1.4 m. What distance does it cover in 100 complete revolutions? (Take π = 22/7)
Answer: D (440 m)
One revolution = πd = 22/7 × 1.4 = 4.4 m. 100 revolutions = 440 m.
If P = {1, 3, 5, 7, 9} and Q = {2, 4, 6, 8, 10} are subsets of a universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, what are the elements of P′ ∩ Q′?
Answer: C (∅)
P′ = {2, 4, 6, 8, 10} and Q′ = {1, 3, 5, 7, 9}. They have no common element, so P′ ∩ Q′ = ∅.
In the diagram, what is x + y in terms of z?
Answer: A (180° − z)
The angle at the top of the triangle is vertically opposite z, so it equals z. Angles in a triangle add to 180°, so x + y = 180° − z.
The values of three angles at a point are 3y − 45°, y + 25° and y°. Find the value of y.
Answer: D (76°)
Angles at a point add to 360°: 5y − 20 = 360, so 5y = 380 and y = 76°.
A rectangular carpet 2.5 m long and 2.4 m wide covers 5% of a rectangular floor. Calculate the area of the floor.
Answer: C (120 m2)
Carpet area = 2.5 × 2.4 = 6 m2, which is 5% of the floor. Floor = 6 × 100/5 = 120 m2.
In the diagram, |XY| = 12 cm, |XZ| = 9 cm, |ZN| = 3 cm and ZY||NM. Calculate |MY|.
Answer: B (4 cm)
Since ZY ∥ NM, triangles XZY and XNM are similar: XZ/XN = XY/XM, so 9/12 = 12/XM and XM = 16 cm. |MY| = 16 − 12 = 4 cm.
Which of the following is not a measure of central tendency?
Answer: C (range)
Mean, median and mode describe the centre of the data; the range measures spread (dispersion).
The temperature in a chemical plant was −5°C at 2.00 a.m. The temperature fell by 6°C and then rose again by 7°C. What was the final temperature?
Answer: B (−4°C)
−5 − 6 = −11°C, then −11 + 7 = −4°C.
Simplify: 20/(5√28 − 2√63)
Answer: A (5√7/7)
5√28 = 10√7 and 2√63 = 6√7, so the denominator is 4√7. 20/(4√7) = 5/√7 = 5√7/7.
A ladder 16 m long leans against an electric pole. If the ladder makes an angle of 65° with the ground, how far up the electric pole does its top reach?
Answer: B (14.5 m)
Height = 16 × sin 65° = 16 × 0.9063 ≈ 14.5 m.
(a) Simplify: (4 2/9 − 1 13/15) / (2 1/5 + 4/7 × 2 1/3)
(b) By rationalizing the denominator, simplify: 7√5/√7 (Leave your answer in surd form)
(a) Numerator: 38/9 − 28/15 = (190 − 84)/45 = 106/45.
Denominator: do the multiplication first: 4/7 × 7/3 = 4/3; 11/5 + 4/3 = 53/15.
106/45 ÷ 53/15 = 106/45 × 15/53 = 2/3.
(b) 7√5/√7 × √7/√7 = 7√35/7 = √35.
Change mixed numbers to improper fractions and do × before + in the denominator.
(a) A dealer sold a car to a man and made a profit of 15%. The man then sold it to a woman for ₦120,175.00 at a loss of 5%. How much did the dealer buy the car?
(b) The diameter of the wheel of a car is 36 cm. How many revolutions, correct to three significant figures, will it make to cover a distance of 1.05 km? (Take π = 22/7)
(a) The man sold at 95% of his cost: man's cost = 120,175 × 100/95 = ₦126,500.
This is 115% of the dealer's cost: dealer's cost = 126,500 × 100/115 = ₦110,000.00.
(b) Circumference = 22/7 × 36 = 792/7 cm ≈ 113.14 cm.
1.05 km = 105,000 cm. Revolutions = 105,000 ÷ 792/7 = 928.03 ≈ 928 (3 s.f.).
Work backwards: a 5% loss means selling price = 95% of cost; a 15% profit means selling price = 115% of cost.
(a) Solve for x and y in the following equations: 2x − y = 9/2; x + 4y = 0
(b) In the diagram, TA is a tangent to the circle at A. If BĈA = 40° and DÂT = 52°, find BÂD.
(a) From x + 4y = 0, x = −4y. Substitute: −8y − y = 9/2, so −9y = 9/2 and y = −½. Then x = 2.
(b) Angle in the alternate segment: ∠ACD = ∠DAT = 52°.
∠BCD = 40° + 52° = 92°.
ABCD is cyclic, so ∠BAD = 180° − 92° = 88°.
Use the alternate segment theorem, then opposite angles of a cyclic quadrilateral.
(a) In a class of 45 students, 32 offered Physics (P), 28 offered Government (G) and 12 did not offer any of the two subjects. (i) Draw the Venn diagram to represent the information. (ii) How many students offered both Physics and Government? (iii) What is n(P∪G)?
(b) If p = 2u/(1 − u) and q = (1 + u)/(1 − u), express (p + q)/(p − q) in terms of u.
(a)(i) Two overlapping circles P and G inside U (45): P only = 32 − x, both = x, G only = 28 − x, outside = 12.
(ii) (32 − x) + x + (28 − x) + 12 = 45, so 72 − x = 45 and x = 27.
(iii) n(P∪G) = 45 − 12 = 33.
(b) p + q = (2u + 1 + u)/(1 − u) = (3u + 1)/(1 − u); p − q = (2u − 1 − u)/(1 − u) = (u − 1)/(1 − u).
(p + q)/(p − q) = (3u + 1)/(u − 1).
Put x in the overlap and add all regions to the total.
Two fair dice are tossed together once. (a) Draw the sample space for the possible outcomes. (b) Find the probability of getting a total: (i) of 7 or 8; (ii) less than 4.
(a) A 6 × 6 table of pairs (1,1) to (6,6), or of totals 2 to 12: 36 equally likely outcomes.
(b)(i) Total 7: 6 outcomes; total 8: 5 outcomes. P = 11/36.
(ii) Total less than 4 (2 or 3): (1,1), (1,2), (2,1) = 3 outcomes. P = 3/36 = 1/12.
Count outcomes from the sample space and divide by 36.
(a) A sector of a circle of radius 8 cm subtends an angle 90° at the centre of the circle. If the sector is folded without overlap to form the curved surface of a cone, find the: (i) base radius; (ii) height; (iii) volume, of the cone. (Take π = 22/7)
(b) A map is drawn to a scale of 1:20,000. Use it to calculate the: (i) distance, in kilometres, represented by 4.5 cm on the map; (ii) distance, in metres, on the map, between two towns 16 km apart; (iii) area on the map, of a forest, which covers 85 km2.
(a)(i) Arc length = 90/360 × 2π × 8 = 4π = circumference of base 2πr, so r = 2 cm.
(ii) Slant height 8 cm: h = √(82 − 22) = √60 ≈ 7.75 cm.
(iii) V = ⅓ × 22/7 × 22 × 7.746 ≈ 32.5 cm3.
(b)(i) 4.5 × 20,000 = 90,000 cm = 0.9 km.
(ii) 16 km = 1,600,000 cm; 1,600,000 ÷ 20,000 = 80 cm = 0.8 m.
(iii) 1 cm represents 200 m, so 1 cm2 represents 0.04 km2. 85 ÷ 0.04 = 2125 cm2.
The arc of the sector becomes the circumference of the cone's base; the sector radius becomes the slant height.
(a)(i) If 4x < 2 + 3x and x − 8 < 3x, what range of values of x satisfies both inequalities? (ii) Represent your result in (i) on the number line.
(b) A shop is sending out a bill for an amount less than £100. The accountant interchanges the two digits and so overcharges the customer by £45. Given that the sum of the two digits is 9, find how much the bill should be.
(a)(i) 4x < 2 + 3x gives x < 2. x − 8 < 3x gives −8 < 2x, so x > −4. Together: −4 < x < 2.
(ii) Number line with open circles at −4 and 2 and the segment between them marked.
(b) Let the bill be 10t + u with t + u = 9. (10u + t) − (10t + u) = 45 gives 9(u − t) = 45, so u − t = 5.
Solving: u = 7, t = 2. The bill should be £27.
Use open circles because the inequalities are strict.
In the diagram, ABCDEO is two-thirds of a circle centre O. The radius AO is 7 cm and |AB| = |BC| = |CD| = |DE|. Calculate, correct to the nearest whole number, the area of the shaded portion. (Take π = 22/7)
Two-thirds of a circle is 240°, shared by 4 equal chords: each chord subtends 60° at O.
Area of one sector = 60/360 × 22/7 × 72 = 25.67 cm2.
Area of one triangle = ½ × 7 × 7 × sin 60° = 21.22 cm2.
Shaded area = 4 × (25.67 − 21.22) = 4 × 4.45 ≈ 17.8 cm2 ≈ 18 cm2.
Each shaded segment = sector − triangle.
(a) Using a ruler and a pair of compasses only: (i) construct ΔXYZ such that |XY| = 8 cm and ∠YXZ = ∠ZYX = 45°; (ii) locate a point P inside the triangle equidistant from XY and XZ, and also equidistant from YX and YZ; (iii) construct a circle touching the three sides of the triangle. (iv) measure the radius of the circle.
(b) The length of the sides of a hexagon are x − 5, 2x, 2x + 7, 2x and 2x − 1. If the perimeter is 144 cm, find the value of x.
(a)(i) Draw XY = 8 cm; construct 45° at X and at Y (bisect a constructed 90°); the arms meet at Z.
(ii) Bisect ∠X and ∠Y; P is where the bisectors meet.
(iii) Drop a perpendicular from P to XY; with centre P and that radius draw the inscribed circle.
(iv) Radius ≈ 1.7 cm (exact value 4√2 − 4 ≈ 1.66 cm).
(b) Perimeter: (x − 5) + 2x + 2x + (2x + 7) + 2x + (2x − 1) = 11x + 1 = 144, so x = 13.
The point equidistant from two lines lies on the bisector of the angle between them.
The following table shows the distribution of test scores in a class.
Scores: 1, 2, 3, 4, 5, 7, 8, 9, 10
No. of Pupils: 1, 1, 5, 3, k2 + 1, 6, 2, 3, 4
(a) If the mean score of the class is 6, find the: (i) value of k; (ii) median score.
(b) Draw a bar chart for the distribution.
(c) If a pupil is picked at random from the class, what is the probability that he/she will score less than 6?
(a)(i) Σf = k2 + 26; Σfx = 1 + 2 + 15 + 12 + 5(k2 + 1) + 42 + 16 + 27 + 40 = 5k2 + 160.
(5k2 + 160)/(k2 + 26) = 6 gives k2 = 4, so k = 2.
(ii) 30 pupils; median = average of 15th and 16th scores. Cumulative: 1, 2, 7, 10, 15 (score 5), 21 (score 7). Median = (5 + 7)/2 = 6.
(b) Bars of heights 1, 1, 5, 3, 5, 6, 2, 3, 4 above scores 1, 2, 3, 4, 5, 7, 8, 9, 10, with labelled axes.
(c) Scores below 6: 1 + 1 + 5 + 3 + 5 = 15 pupils. P = 15/30 = ½.
Set the mean formula equal to 6 and solve for k2.
(a) The diagram shows a pyramid standing on a cuboid. The dimensions of the cuboid are 4 m by 3 m by 2 m and the slant edge of the pyramid is 5 m. Calculate the volume of the shape.
(b) The 2nd, 3rd and 4th terms of an AP are x − 2, 5 and x + 2 respectively. Calculate the value of x.
(a) Cuboid volume = 4 × 3 × 2 = 24 m3.
Diagonal of the 4 m × 3 m top = √(16 + 9) = 5 m, half-diagonal = 2.5 m.
Pyramid height = √(52 − 2.52) = √18.75 ≈ 4.33 m.
Pyramid volume = ⅓ × 12 × 4.33 ≈ 17.32 m3.
Total ≈ 41.3 m3.
(b) Common difference: 5 − (x − 2) = (x + 2) − 5, so 7 − x = x − 3 and x = 5.
The pyramid's vertex is above the centre of the top face, so use half the diagonal with Pythagoras.
(a) Two lines AB and CD intersect at X such that CÂX is equal to BD̂X. If |AX| = 6 cm, |XB| = 4 cm and |CX| = 3 cm, find |XD|.
(b) The diagram shows the positions of three points X, Y and Z on a horizontal plane. The bearing of Y from X is 312° and that of Y from Z is 022°. If |XY| = 32 km and |ZY| = 50 km, calculate, correct to one decimal place: (i) |XZ|; (ii) the bearing of Z from X.
(a) ∠CAX = ∠BDX and ∠AXC = ∠DXB (vertically opposite), so ΔAXC ~ ΔDXB with A↔D, C↔B.
AX/DX = CX/BX: 6/|XD| = 3/4, so |XD| = 8 cm.
(b)(i) At Y: bearing of X from Y = 132°, bearing of Z from Y = 202°, so ∠XYZ = 70°.
|XZ|2 = 322 + 502 − 2(32)(50) cos 70° = 3524 − 1094.5 = 2429.5, so |XZ| ≈ 49.3 km.
(ii) sin ∠YXZ = 50 sin 70°/49.29 ≈ 0.9532, ∠YXZ ≈ 72.4°.
Bearing of Z from X = 312° − 72.4° = 239.6°.
Back bearings differ by 180°: use them to find the angle at Y, then the cosine and sine rules.
(Questions 14 and 15 are for candidates in Ghana, Sierra Leone and The Gambia only.)
(a) Using a scale of 2 cm to 2 units on both axes, draw, on a graph paper, two perpendicular axes Ox and Oy for −8 ≤ x ≤ 8 and −8 ≤ y ≤ 8.
(b) Draw the ΔPQR with vertices P(2, 5), Q(2, −1) and R(8, 4).
(c) Describe fully, two different transformations that map P to Q.
(d) Draw the image ΔP1Q1R1 of ΔPQR under a rotation through 90° anticlockwise about O.
(e) Find the vector PP1.
(a), (b) Axes drawn to scale; triangle with P(2, 5), Q(2, −1), R(8, 4).
(c) Any two of: a reflection in the line y = 2; a translation by the vector (0, −6); a rotation of 180° about (2, 2).
(d) Rotation 90° anticlockwise about O maps (x, y) to (−y, x): P1(−5, 2), Q1(1, 2), R1(−4, 8).
(e) PP1 = (−5 − 2, 2 − 5) = (−7, −3), written as a column vector.
P and Q have the same x-coordinate, so the mirror line is halfway between y = 5 and y = −1.
(a) Under the mapping h(x) = px2 − qx + 2, the image of 3 is 14 and the image of −2 is 24. Find: (i) the values of p and q; (ii) the elements whose image is 4.
(b) In the pentagon ABCDE, AE = (2, 0), DE = (−1, 2), DC = (−2, −5) and CB = (−3, 2) (column vectors). Show that ABCD is a parallelogram.
(a)(i) h(3) = 9p − 3q + 2 = 14 gives 3p − q = 4. h(−2) = 4p + 2q + 2 = 24 gives 2p + q = 11. Adding: 5p = 15, p = 3, q = 5.
(ii) 3x2 − 5x + 2 = 4 → 3x2 − 5x − 2 = 0 → (3x + 1)(x − 2) = 0, so x = 2 or x = −⅓.
(b) AB = AE + ED + DC + CB = (2, 0) + (1, −2) + (−2, −5) + (−3, 2) = (−2, −5).
AB = DC = (−2, −5), so AB and DC are equal and parallel; therefore ABCD is a parallelogram.
A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.