62 questions. Try each one, then open the answer to check yourself and read the explanation.
If ₦112.00 exchanges for D14.94, calculate the value of D1.00 in naira.
Answer: B (7.49)
D1.00 = 112 ÷ 14.94 ≈ ₦7.50; the nearest option is 7.49.
Solve for x in the equation (3/5)(2x − 1) = (1/4)(5x − 3).
Answer: D (3)
Multiply by 20: 12(2x − 1) = 5(5x − 3), so 24x − 12 = 25x − 15 and x = 3.
Given that cos x° = 1/r, express tan x° in terms of r.
Answer: D (√(r2 − 1))
Draw a right triangle with adjacent side 1 and hypotenuse r. The opposite side is √(r2 − 1), so tan x° = √(r2 − 1)/1.
The diagram shows a cyclic quadrilateral PQRS with its diagonals intersecting at K. Which of the following triangles is similar to triangle QKR?
Answer: B (ΔPKS)
∠QKR = ∠PKS (vertically opposite) and ∠KQR = ∠SQR = ∠SPR = ∠KPS (angles in the same segment). So ΔQKR ~ ΔPKS.
In the diagram, OP and OR are radii, |PQ| = |QR| and reflex ∠POR is 240°. Calculate the value of x.
Answer: A (60°)
∠POR = 360° − 240° = 120° and ∠PQR = ½ × 240° = 120°. OPQR is a kite, so the angles at P and R are equal: x = (360° − 120° − 120°)/2 = 60°.
If a number is chosen at random from the set {x: 4 ≤ x ≤ 15}, find the probability that it is a multiple of 3 or a multiple of 4.
Answer: C (1/2)
There are 12 numbers (4 to 15). Multiples of 3 or 4: 4, 6, 8, 9, 12, 15, which is 6 numbers. Probability = 6/12 = 1/2.
Solve the equations: 3x − 2y = 7, x + 2y = −3
Answer: A (x = 1, y = −2)
Adding the equations: 4x = 4, so x = 1. Then 1 + 2y = −3 gives y = −2.
One of the factors of (mn − nq − n2 + mq) is (m − n). The other factor is:
Answer: C ((n + q))
Rearrange: mn + mq − n2 − nq = m(n + q) − n(n + q) = (m − n)(n + q).
A cylindrical container has a base radius of 14 cm and height 18 cm. How many litres, correct to the nearest litre, of liquid can it hold? [Take π = 22/7]
Answer: A (11)
V = 22/7 × 142 × 18 = 11,088 cm3 = 11.088 litres ≈ 11 litres.
Find the size of the angle marked x in the diagram.
Answer: C (128°)
Both slanting lines are perpendicular to the same line, so they are parallel. The slanting line makes 52° with the crossing line at the bottom, so it makes 52° with it at the top (alternate angles). x is on a straight line with that angle: x = 180° − 52° = 128°.
A regular polygon of n sides has each exterior angle equal to 45°. Find the value of n.
Answer: B (8)
n = 360° ÷ 45° = 8.
Esther was facing S 20°W. She turned 90° in the clockwise direction. What direction is she facing?
Answer: A (N 70° W)
S 20°W is a bearing of 200°. Turning 90° clockwise gives 290°, which is 70° west of north: N 70° W.
The histogram shows the age distribution of members of a club. What is their modal age?
Answer: A (44.5)
The tallest bar is the 40.5–45.5 class. The crossed lines drawn from its top corners to the neighbouring bars meet near 44.5 years, which is the estimated mode.
The cross section of a uniform prism is a right-angled triangle with sides 3 cm, 4 and 5 cm. If its length is 10 cm, calculate the total surface area.
Answer: B (132 cm2)
Two triangles: 2 × ½ × 3 × 4 = 12 cm2. Three rectangles: 10 × 3 + 10 × 4 + 10 × 5 = 120 cm2. Total = 132 cm2.
Form the equation whose roots are x = ½ and −⅔.
Answer: D (6x2 + x − 2 = 0)
(2x − 1)(3x + 2) = 0 gives 6x2 + x − 2 = 0.
Simplify: log√27/log√81
Answer: D (3/4)
log√27 = (3/2) log 3 and log√81 = 2 log 3, so the ratio is (3/2) ÷ 2 = 3/4.
Which of these angles can be constructed using ruler and a pair of compasses only?
Answer: C (135°)
135° = 90° + 45°, and both 90° and 45° can be constructed by drawing perpendiculars and bisecting.
The perimeter of a sector of a circle of radius 4 cm is (π + 8) cm. Calculate the angle of the sector.
Answer: A (45°)
Perimeter = arc + 2 × 4, so the arc is π cm. θ/360 × 2π × 4 = π gives θ = 45°.
The length of a piece of stick is 1.75 m. A girl measured it as 1.80 m. Find the percentage error.
Answer: D (20/7 %)
Error = 0.05 m. Percentage error = 0.05/1.75 × 100% = 20/7 %.
What is the value of 3 in the number 42.7531?
Answer: B (3/1000)
After the point: 7 tenths, 5 hundredths, 3 thousandths. So the 3 is worth 3/1000.
The height of a cylinder is equal to its radius. If the volume is 0.216π m3, calculate the radius.
Answer: B (0.60m)
V = πr2h = πr3 = 0.216π, so r3 = 0.216 and r = 0.6 m.
In the diagram, MN//OP, ∠NMQ = 65° and ∠QOP = 125°. What is the size of ∠MQR?
Answer: B (120°)
∠NOP = 180° − 125° = 55°, and MN ∥ OP gives ∠MNQ = 55° (alternate). In ΔMNQ, ∠MQN = 180° − 65° − 55° = 60°. NQR is a straight line, so ∠MQR = 180° − 60° = 120°.
A circle is divided into two sectors in the ratio 3:7. If the radius of the circle is 7 cm, calculate the length of the minor arc of the circle.
Answer: B (13.20 cm)
The minor sector has 3/10 of 360° = 108°. Arc = 108/360 × 2 × 22/7 × 7 = 13.20 cm.
From the diagram, which of the following statements are true?
I. m = q
II. n = q
III. n + p = 180°
IV. p + m = 180°
Answer: B (I and IV)
MNQP is a parallelogram (both pairs of sides parallel). Opposite angles are equal, so m = q (I), and neighbouring angles add to 180°, so p + m = 180° (IV).
Factorize the expression am + bn − an − bm.
Answer: B ((a − b)(m − n))
Group: am − an − bm + bn = a(m − n) − b(m − n) = (a − b)(m − n).
The graph represents the relation y = x2 − 3x − 3. Find the values of x for which x2 − 3x = 7.
Answer: A (−1.55, 4.55)
x2 − 3x = 7 means x2 − 3x − 3 = 4, so read where the curve meets y = 4: x ≈ −1.55 and 4.55 (exactly (3 ± √37)/2).
The graph represents the relation y = x2 − 3x − 3. What is the equation of the line of symmetry of the graph?
Answer: C (x = 1.5)
The line of symmetry passes through the lowest point, x = −b/2a = 3/2 = 1.5.
Simplify: m/n + (m − 1)/5n − (m − 2)/10n where n ≠ 0.
Answer: B (11m/10n)
Over 10n: [10m + 2(m − 1) − (m − 2)]/10n = (10m + 2m − 2 − m + 2)/10n = 11m/10n.
If √72 + √32 − 3√18 = x√8, find the value of x.
Answer: C (1/2)
6√2 + 4√2 − 9√2 = √2, and √8 = 2√2. So √2 = 2x√2 and x = ½.
G varies directly as the square of H. If G is 4 when H is 3, find H when G = 100.
Answer: A (15)
G = kH2: 4 = 9k, so k = 4/9. 100 = (4/9)H2 gives H2 = 225 and H = 15.
Given that n(P) = 19, n(P∪Q) = 38 and n(P∩Q) = 7, find n(Q).
Answer: A (26)
n(P∪Q) = n(P) + n(Q) − n(P∩Q): 38 = 19 + n(Q) − 7, so n(Q) = 26.
What must be added to (2x − 3y) to get (x − 2y)?
Answer: B (y − x)
(x − 2y) − (2x − 3y) = −x + y = y − x.
Simplify: 1¾ − (2⅓ + 4).
Answer: C (−4 7/12)
7/4 − 19/3 = (21 − 76)/12 = −55/12 = −4 7/12.
In the diagram, STUV is a straight line. ∠TSY = ∠UXY = 40° and ∠VUW = 110°. Calculate ∠TYW.
Answer: A (150°)
∠TUX = 180° − 110° = 70°. In ΔTUX, ∠UTX = 180° − 70° − 40° = 70°, so ∠STY = 110°. In ΔSTY, ∠SYT = 180° − 40° − 110° = 30°, so ∠TYW = 180° − 30° = 150°.
Given that 124x = 7(14x), find the value of x.
Answer: D (8)
x2 + 2x + 4 = 7(x + 4), so x2 − 5x − 24 = 0 and (x − 8)(x + 3) = 0. The base must be positive: x = 8.
Find the smaller value of x that satisfies the equation: x2 + 7x + 10 = 0.
Answer: A (−5)
(x + 5)(x + 2) = 0 gives x = −5 or −2. The smaller is −5.
The perpendicular bisectors of the sides of an acute-angled triangle are drawn. Which of these statements is correct? They intersect:
Answer: C (inside the triangle.)
The perpendicular bisectors meet at the circumcentre, which lies inside an acute-angled triangle.
A rectangular garden measures 18.6 m by 12.5 m. Calculate, correct to three significant figures, the area of the garden.
Answer: D (233 m2)
18.6 × 12.5 = 232.5 m2, which is 233 m2 to 3 significant figures.
John pours 96 litres of red oil into a rectangular container with length 220 cm and breadth 40 cm. Calculate, correct to the nearest cm, the height of the oil in the container.
Answer: A (11 cm)
96 litres = 96,000 cm3. Height = 96,000 ÷ (220 × 40) ≈ 10.9 ≈ 11 cm.
In a quiz competition, a student answers n questions correctly and was given D(n + 50) for each question correctly answered. If he gets D600.00 altogether, how many questions did he answer correctly?
Answer: D (10)
n(n + 50) = 600, so n2 + 50n − 600 = 0 and (n + 60)(n − 10) = 0. n cannot be negative, so n = 10.
The Venn diagram shows the number of students in a class who like reading (R), dancing (D) and swimming (S). How many students like dancing and swimming?
Answer: A (7)
S has 16 students: 6 + 3 + 2 + (S and D only) = 16, so S and D only = 5 (the 5 in the diagram). Dancing and swimming = 2 + 5 = 7.
A shopkeeper allows a discount of 15% on the marked price of a mobile phone. If a customer paid GH¢170.00 for a mobile phone, what was the marked price of the phone?
Answer: C (GH¢200.00)
After a 15% discount the customer pays 85% of the marked price. Marked price = 170 × 100/85 = GH¢200.00.
If 27x = 9y, find the value of x/y.
Answer: B (2/3)
33x = 32y, so 3x = 2y and x/y = 2/3.
In the diagram, PQ//TS, PR//TU, reflex angle QPS = 245°, angle PST = 115°, ∠STU = 65° and ∠RPS = x. Find the value of x.
Answer: D (50°)
∠QPT (inside) = 360° − 245° = 115°. Since PQ ∥ TS and PR ∥ TU, ∠QPR = ∠STU = 65°. So x = 115° − 65° = 50°.
Illustrate the inequality −1 < 3x + 5 < 14 on a number line.
Answer: B (Diagram B)
Subtract 5: −6 < 3x < 9, then divide by 3: −2 < x < 3. That is open circles at −2 and 3 joined by a line, as in B.
A boy looks through a window of a building and sees a mango fruit on the ground 50 m away from the foot of the building. If the window is 9 m from the ground, calculate, correct to the nearest degree, the angle of depression of the mango from the window.
Answer: B (10°)
tan θ = 9/50 = 0.18, so θ ≈ 10.2° ≈ 10°.
If E = MN/(S + N) and E = 75, M = 120, N = 5000, find S.
Answer: C (3000)
75 = 600,000/(S + 5000), so S + 5000 = 8000 and S = 3000.
(a) Simplify: (½ of ¼ ÷ ⅓)/(⅙ − ¾ + ½).
(b) Given that √x = 101̄.6741, without using calculators, find the value of x.
(a) Numerator: ½ × ¼ = 1/8; 1/8 ÷ 1/3 = 3/8.
Denominator: (2 − 9 + 6)/12 = −1/12.
3/8 ÷ (−1/12) = 3/8 × (−12) = −9/2 = −4½.
(b) log √x = 1̄.6741, so ½ log x = 1̄.6741 and log x = 2 × 1̄.6741 = 1̄.3482.
x = antilog 1̄.3482 = 0.2229.
'of' means multiply; do it before the division.
(a) Make q the subject of the relation t = √(pq/r − r2q).
(b) If 9(1−x) = 27y and x − y = −1½, find the value of x + y.
(a) Square: t2 = pq/r − r2q. Multiply by r: rt2 = pq − r3q = q(p − r3).
q = rt2/(p − r3).
(b) 32(1 − x) = 33y, so 2 − 2x = 3y, i.e. 2x + 3y = 2.
x − y = −3/2 gives 2x − 2y = −3. Subtracting: 5y = 5, y = 1, then x = −½.
x + y = ½.
Write 9 and 27 as powers of 3.
A sector of a circle with radius 21 cm has an area of 280 cm2. (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 π = 22/7]
(a) Area = ½ × r × arc, so arc = 2 × 280/21 = 26.67 cm (the angle is about 72.7°).
Perimeter = 26.67 + 21 + 21 ≈ 68.7 cm.
(b) The arc becomes the base circumference: 2 × 22/7 × r = 26.67, so r ≈ 4.24 cm. The slant height is 21 cm.
sin(half the vertical angle) = 4.24/21 ≈ 0.202, so half-angle ≈ 11.7°.
Vertical angle ≈ 23°.
The radius of the sector becomes the slant height of the cone.
(a) In the diagram, PQRST is a quadrilateral. PT//QS, ∠PTQ = 42°, ∠TSQ = 38° and ∠QSR = 30°. If ∠QTS = x and ∠PQT = y, find: (i) x; (ii) y.
(b) In the diagram, PQRS is a circle centre O. If PÔQ = 150°, ∠QSR = 40° and ∠SQP = 45°, calculate ∠RQS.
(a)(i) PT ∥ QS, so ∠TQS = ∠PTQ = 42° (alternate). In ΔQTS: x = 180° − 42° − 38° = 100°.
(ii) In ΔQRS (right angle at R): ∠RQS = 180° − 90° − 30° = 60°. P, Q, R are on a straight line, so y = 180° − 60° − 42° = 78°.
(b) ∠PSQ = ½ × 150° = 75° (angle at centre). ∠PSR = 75° + 40° = 115°.
Opposite angles of cyclic PQRS: ∠PQR = 180° − 115° = 65°.
∠RQS = 65° − 45° = 20°.
The angle at the centre is twice the angle at the circumference.
A library received a $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 new books over that of new furniture.
(a) Angles = percentage × 3.6°: magazines 36°, new books 126°, repairs 54°, furniture 108°, training 36°. Draw and label the pie chart.
(b) New books = 35% of $1,300 = $455; furniture = 30% of $1,300 = $390.
Increase = 65/390 × 100% ≈ 16.7% ≈ 17%.
Percentage increase is measured against the smaller amount (furniture).
In a class of 40 students, 18 passed Mathematics, 19 passed Accounts, 16 passed Economics, 5 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 the subjects? (b) find the percentage number who failed in at least one of Economics and Mathematics; (c) calculate the probability that a student at random failed in Accounts.
Let x passed all three. Accounts: 9 + 5 + 2 + x = 19, so x = 3.
Mathematics and Economics only = 18 − 6 − 5 − 3 = 4. Economics only = 16 − 2 − 3 − 4 = 7.
(a) Passed at least one = 6 + 9 + 7 + 5 + 2 + 4 + 3 = 36, so 40 − 36 = 4 failed all.
(b) Passed both Economics and Mathematics = 4 + 3 = 7, so 40 − 7 = 33 failed at least one of them: 33/40 × 100% = 82.5%.
(If read as failing both: 40 − n(M ∪ E) = 40 − 27 = 13, i.e. 32.5%.)
(c) Failed Accounts = 40 − 19 = 21; probability = 21/40.
Fill in the Venn diagram from the 'only' numbers first, then use the totals.
(a) Divide (x2 − 4)/(x2 + x) by (x2 − 4x + 4)/(x + 1).
(b) The diagram shows the graphs of y = ax2 + bx + c and y = mx + k where a, b, c, m and k are constants. Use the graph(s) to: (i) find the roots of the equation ax2 + 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 = ax2 + bx + c.
(a) (x − 2)(x + 2)/[x(x + 1)] × (x + 1)/(x − 2)2 = (x + 2)/[x(x − 2)].
(b)(i) The graphs meet at P and H: x ≈ −1.5 and x ≈ 1.5.
(ii) L(−1, 0), M(0, 2), N(2, 0). M gives c = 2. L: a − b + 2 = 0; N: 4a + 2b + 2 = 0. Solving: a = −1, b = 1. Curve: y = −x2 + x + 2.
(iii) The line of symmetry is halfway between the roots −1 and 2: x = 0.5.
Substitute each point's coordinates into y = ax2 + bx + c.
(a) Given that sin x = 0.6 and 0° ≤ x ≤ 90°, evaluate 2 cos x + 3 sin x, leaving your answer in the form 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, (i) radius of the circle; (ii) area of the shaded portion. (Take π = 22/7)
(a) cos x = √(1 − 0.36) = 0.8. 2(0.8) + 3(0.6) = 3.4 = 17/5.
(b)(i) OC bisects ∠ACB, so ∠OCY = 65°. The radius OY is perpendicular to the tangent CB: r = 30 sin 65° ≈ 27.2 cm.
(ii) OA = 30 tan 65° ≈ 64.3 cm. Area of ΔABC = ½ × 2(64.34) × 30 ≈ 1930.1 cm2.
Semicircle = ½ × 22/7 × 27.192 ≈ 1161.7 cm2.
Shaded area ≈ 1930.1 − 1161.7 ≈ 768.4 cm2 (767.5 cm2 if r = 27.2 is used).
The radius to a point of contact is perpendicular to the tangent.
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|.
(a) Draw QR = 7 cm; construct 60° at Q and mark QP = 7 cm; with centres P and R, radius 7 cm, find S. Join PS and RS.
(b) Bisect ∠PQR (points equidistant from PQ and QR) and draw the perpendicular bisector of QR (points equidistant from Q and R). X is where they meet.
(c) |XR| ≈ 4.0 cm (exactly 3.5/cos 30° ≈ 4.04 cm).
The bisector of the angle at Q makes 30° with QR.
(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 cm3, 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) r2 : R2 = 9 : 49, so r : R = 3 : 7 and R = 12 × 7/3 = 28 cm.
Volume = 4/3 × 22/7 × 283 ≈ 91,989 cm3.
(b)(i) ∠XYZ = 90° + 45° = 135°. XZ2 = 32 + 52 − 2(3)(5) cos 135° = 34 + 21.21 = 55.21, so XZ ≈ 7.43 ≈ 7 km.
(ii) sin ∠YXZ = 5 sin 135°/7.43 ≈ 0.476, so ∠YXZ ≈ 28.4°. Bearing of Z from X = 270° + 28.4° ≈ 298°.
Ratio of surface areas = (ratio of radii)2.
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.
x/(x + 37) = 0.26 → x = 0.26x + 9.62 → 0.74x = 9.62 → x = 13. Total = 50.
(a) Median = average of 25th and 26th scores. Cumulative frequencies: 2, 7, 20, 31, … so both are 4: median = 4.
(b) Mean = (2 + 10 + 39 + 44 + 45 + 60)/50 = 200/50 = 4.
Σf(x − 4)2 = 2(9) + 5(4) + 13(1) + 11(0) + 9(1) + 10(4) = 100.
SD = √(100/50) = √2 ≈ 1.41.
Find x first from the probability, then use the full table.
(a) The area of trapezium PQRS is 60 cm2. PQ//RS, |PQ| = 15 cm, |RS| = 25 cm and ∠PSR = 30°. Calculate the (i) perpendicular height of PQRS; (ii) |PS|.
(b) Ade received 3/5 of a sum of money, Nelly 1/3 of the remainder while Austin took the rest. If Austin's share is greater than Nelly's share by ₦3,000, how much did Ade receive?
(a)(i) 60 = ½(15 + 25) × h = 20h, so h = 3 cm.
(ii) sin 30° = h/|PS|, so |PS| = 3/0.5 = 6 cm.
(b) Let the sum be x. Remainder = 2x/5. Nelly = ⅓ × 2x/5 = 2x/15; Austin = 4x/15.
4x/15 − 2x/15 = 2x/15 = 3000, so x = ₦22,500.
Ade received 3/5 × 22,500 = ₦13,500.
Express each share as a fraction of the whole sum.
(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 π = 3.142]
(a) P = kQ/R2: 1 = 8k/4, so k = ½. Q = 2PR2 = 2 × 3 × 25 = 150.
(b)(i) Longitude difference = 40°. Distance = 40/360 × 2 × 3.142 × 6400 × cos 20° ≈ 4199 km.
Speed = 4199 ÷ 6 ≈ 700 km/h.
(ii) Along a meridian: 420 = θ/360 × 2 × 3.142 × 6400, so θ ≈ 3.76°.
Latitude of C ≈ 20° + 3.76° ≈ 24°N.
Along a parallel of latitude the radius is R cos(latitude); along a meridian it is R.
(Questions 14 and 15 are for candidates in Ghana, Sierra Leone and The Gambia only.)
(a) Using a scale of 2cm to 2 units on both axes, draw on a graph sheet two perpendicular axes OX and OY for the interval −8 ≤ x ≤ 8 and −8 ≤ y ≤ 8.
(b) Draw clearly and label the vertices as appropriate: (i) triangle PQR with P(1, 2), Q(5, 3) and RQ = (2, −3); (ii) the image ΔP′Q′R′ of ΔPQR under a rotation of 180° about the origin where P → P′, Q → Q′ and R → R′; (iii) the image ΔP″Q″R″ of ΔP′Q′R′ under a reflection in the line x = 0 where P′ → P″, Q′ → Q″ and R′ → R″.
(c) Describe fully a single transformation that maps ΔPQR onto ΔP″Q″R″.
(b)(i) R = Q − RQ = (5 − 2, 3 + 3) = (3, 6). Plot P(1, 2), Q(5, 3), R(3, 6).
(ii) (x, y) → (−x, −y): P′(−1, −2), Q′(−5, −3), R′(−3, −6).
(iii) Reflection in x = 0 (the y-axis): (x, y) → (−x, y): P″(1, −2), Q″(5, −3), R″(3, −6).
(c) PQR → P″Q″R″ is (x, y) → (x, −y): a reflection in the x-axis (the line y = 0).
Combine the two rules: (x, y) → (−x, −y) → (x, −y).
(a) Find the image of (−2, 4) under the mapping (x, y) → (2x, y − 3x).
(b) Two functions f and g are defined as: f: x → x2/4 − 9, g: x → 1/2x (x ≠ 0). (i) Evaluate f(4) + g(−1/3). (ii) If f × g = 2, solve for x.
(a) (2(−2), 4 − 3(−2)) = (−4, 10).
(b)(i) f(4) = 16/4 − 9 = −5; g(−⅓) = 1/(−⅔) = −3/2. Sum = −6½.
(ii) (x2/4 − 9) × 1/(2x) = 2 → (x2 − 36)/(8x) = 2 → x2 − 16x − 36 = 0 → (x − 18)(x + 2) = 0 → x = 18 or x = −2.
f × g means multiply the two expressions.