54 questions. Try each one, then open the answer to check yourself and read the explanation.
Solve x2 − 2x − 8 > 0
Answer: A (x < −2 or x > 4)
x2 − 2x − 8 = (x + 2)(x − 4). The product is positive outside the roots, so x < −2 or x > 4.
If (x + 3) is a factor of the polynomial x3 + 3x2 + nx − 12, where n is a constant, find the value of n.
Answer: D (−4)
By the factor theorem f(−3) = 0: −27 + 27 − 3n − 12 = 0, so −3n = 12 and n = −4.
The line y = mx − 3 is a tangent to the curve y = 1 − 3x + 2x3 at the point (1, 0). Find the value of the constant m.
Answer: C (3)
dy/dx = −3 + 6x2. At x = 1 the gradient is 3, so m = 3 (and y = 3x − 3 does pass through (1, 0)).
Given that sin θ = √3/2 where θ is acute, find the value of tan 2θ, leaving your answer in surd form.
Answer: A (−√3)
θ = 60°, so tan 2θ = tan 120° = −tan 60° = −√3.
Find the coefficient of x4 in the binomial expansion of (2 + x)6.
Answer: C (60)
The x4 term is 6C4(2)2x4 = 15 × 4x4 = 60x4.
Which of the following binary operations is not commutative?
Answer: D (a * b = a − b + ab)
Swapping a and b changes a − b + ab into b − a + ab, which is different. The other rules stay the same when a and b swap.
Express 2/(3 − √7) in the form a + √b, where a and b are integers.
Answer: B (3 + √7)
Multiply top and bottom by 3 + √7: 2(3 + √7)/(9 − 7) = 3 + √7.
The roots of the quadratic equation 2x2 − 5x + m = 0 are α and β, where m is a constant. Find (α2 + β2) in terms of m.
Answer: A (25/4 − m)
α + β = 5/2 and αβ = m/2. α2 + β2 = (α + β)2 − 2αβ = 25/4 − m.
Given that 2x = 0.125, find the value of x.
Answer: D (−3)
0.125 = 1/8 = 2−3, so x = −3.
The gradient of point P on the curve y = 3x2 − x + 3 is 5. Find the coordinates of P
Answer: A ((1, 5))
dy/dx = 6x − 1 = 5 gives x = 1. Then y = 3 − 1 + 3 = 5, so P is (1, 5).
An arc of length 10.8cm subtends an angle of 1.2 radians at the centre of a circle. Calculate the radius of the circle.
Answer: D (9.0cm)
Arc length l = rθ, so r = 10.8 ÷ 1.2 = 9.0 cm.
The first term of a geometric series is 350. If the sum to infinity is 250, Find the common ratio.
Answer: B (−2/5)
S∞ = a/(1 − r): 250 = 350/(1 − r), so 1 − r = 7/5 and r = −2/5.
p and q are statements such that p ⇒ q. Which of the following is a valid conclusion from the implication?
Answer: C (~q ⇒ ~p)
An implication is always equivalent to its contrapositive: if p ⇒ q then ~q ⇒ ~p.
The roots of a quadratic equation are −3 and 1. Find its equation.
Answer: C (x2 + 2x − 3 = 0)
(x + 3)(x − 1) = x2 + 2x − 3 = 0. Sum of roots −2, product −3.
Given that log3(x − y) = 1 and log3(2x + y) = 2, find the value of x.
Answer: D (4)
x − y = 3 and 2x + y = 9. Adding gives 3x = 12, so x = 4.
If 8Px/8Cx = 6, find the value of x
Answer: C (3)
nPr = r! × nCr, so the ratio is x! = 6 and x = 3.
Evaluate ∫12[(x3 − 1)/x2]dx.
Answer: B (1.0)
The integrand is x − x−2, with antiderivative x2/2 + 1/x. (2 + 1/2) − (1/2 + 1) = 1.
In the diagram, a ladder PS leaning against a vertical wall PR makes angle x° with the horizontal floor. The ladder slides down to a point QT such that angle QTR = 30° and angle SNT = y°. Find the relation between x and y
Answer: D (y = x − 30)
In triangle NST, angle NST = 180° − x and angle NTS = 30°, so y = 180° − (180° − x) − 30° = x − 30°.
If P = (1 1; 2 1), find (P2 + P)
Answer: B ((4 3; 6 4))
P2 = (3 2; 4 3). Adding P gives (4 3; 6 4).
Simplify 2log38 − 3log32.
Answer: C (3log32)
2log38 = log364 and 3log32 = log38. log3(64/8) = log38 = 3log32.
Which of the semi-interquartile range of a distribution?
Answer: D (½(Upper Quartile − Lower Quartile))
The semi-interquartile range is half the interquartile range: ½(Q3 − Q1).
A stone is projected vertically upwards with a speed of 10ms−1 from a point 8 metres above the ground. Find the maximum height reached. [Take g = 10ms−2]
Answer: A (13 metres)
It rises a further u2/2g = 100/20 = 5 m above the point, so the greatest height is 8 + 5 = 13 m above the ground.
The velocity, v ms−1 of a particle moving in a straight line given by v = 3t2 − 2t + 1 at time t seconds. Find the acceleration of the particle after 3 seconds.
Answer: D (16ms−2)
a = dv/dt = 6t − 2. At t = 3, a = 18 − 2 = 16 ms−2.
Three men, P, Q and R aim at a target the probabilities that P, Q and R hit the target are ½, ⅓ and ¾ respectively. Find the probability that exactly two of them hit the target.
Answer: C (5/12)
Add the three cases with one miss: (½)(⅓)(¼) + (½)(⅔)(¾) + (½)(⅓)(¾) = 1/24 + 6/24 + 3/24 = 10/24 = 5/12.
The position vectors of A and B are (2i + j) and (−i + 4j) respectively. Find |AB|.
Answer: A (3√2)
AB = (−i + 4j) − (2i + j) = −3i + 3j, so |AB| = √(9 + 9) = √18 = 3√2.
Two fair dice, each numbered 1, 2, …, 6, are tossed together. Find the probability that they both show even numbers.
Answer: B (¼)
Each die shows an even number with probability ½, so both do with probability ½ × ½ = ¼.
Calculate, correct to the nearest degree, the angle between the vectors (13, 1) and (1, 4)
Answer: B (72°)
Dot product = 13 + 4 = 17. cos θ = 17/(√170 × √17) = 1/√10 ≈ 0.316, so θ ≈ 72°.
Evaluate (2 3; 4 1)(2, 3)
Answer: C (column vector (13, 11))
Multiply rows by the column: 2 × 2 + 3 × 3 = 13 and 4 × 2 + 1 × 3 = 11, giving a column vector with entries 13 and 11.
If the mean of −1, 0, 9, 3, k, 5 is 2, where k is constant, find the median of the set of numbers.
Answer: B (3/2)
(16 + k)/6 = 2 gives k = −4. In order: −4, −1, 0, 3, 5, 9, so the median is (0 + 3)/2 = 3/2.
Eight football clubs are to play in a league on home and away basis. How many matches are possible?
Answer: C (56)
Each ordered pair (home, away) is a match: 8P2 = 8 × 7 = 56.
Two balls are drawn from a bag containing 3 red, 4 white and 5 black identical balls. Find the probability that they are the entire same colour.
Answer: D (19/66)
Ways to pick two of one colour: 3C2 + 4C2 + 5C2 = 3 + 6 + 10 = 19, out of 12C2 = 66. Probability 19/66.
A force F acting on a body of mass 12kg increases its speed from 5ms−1 to 35ms−1 in 5 seconds. Find the value of F.
Answer: C (72N)
a = (35 − 5)/5 = 6 ms−2, so F = ma = 12 × 6 = 72 N.
Express the force F = (8N, 150°) in the form (ai + bj) where a and b are constants.
Answer: D (−4√3i + 4j)
Measuring 150° anticlockwise from the positive x-axis: a = 8 cos 150° = −4√3 and b = 8 sin 150° = 4, so F = −4√3i + 4j.
Three defective bulbs got mixed up with 7 good ones. If two bulbs are selected at random, what is the probability that both are good?
Answer: C (7/15)
P = (7/10) × (6/9) = 42/90 = 7/15 (selection without replacement).
The ages, in years, of 5 boys are 5, 6, 6, 8 and 10. Calculate, correct to one decimal place, the standard deviation of their ages.
Answer: D (1.8 years)
Mean = 35/5 = 7. Squared deviations 4, 1, 1, 1, 9 add to 16, so variance = 16/5 = 3.2 and s.d. = √3.2 ≈ 1.8.
A body is acted upon by forces F1 = (10 N, 090°) and F2 = (6N, 180°). Find the magnitude of the resultant force, correct to one decimal place.
Answer: B (11.7N)
The forces are at right angles, so the resultant is √(102 + 62) = √136 ≈ 11.7 N.
(a) Solve 2x2 + x − 6 < 0
(b) Express (5 − 2√10)/(3√5 + √2) in the form m√2 + n√5 where m and n are rational numbers.
(a) (2x − 3)(x + 2) < 0
Negative between the roots: −2 < x < 3/2
(b) Multiply by (3√5 − √2)/(3√5 − √2); denominator 45 − 2 = 43
Numerator: 15√5 − 5√2 − 6√50 + 2√20 = 19√5 − 35√2
= −(35/43)√2 + (19/43)√5, so m = −35/43, n = 19/43
Rationalise with the conjugate 3√5 − √2 and collect √2 and √5 terms.
Solve the simultaneous equations log2x − log2y = 2, log2(x − 2y) = 3
log2(x/y) = 2, so x = 4y
x − 2y = 23 = 8
4y − 2y = 8, y = 4
x = 16
Combine the logs into a single log, then remove the logs to get two ordinary equations.
If the quadratic equation (x + 1)(x + 2) = k(3x + 7) has equal roots, find the possible values of the constant k.
x2 + (3 − 3k)x + (2 − 7k) = 0
Equal roots: (3 − 3k)2 = 4(2 − 7k)
9k2 + 10k + 1 = 0, (9k + 1)(k + 1) = 0
k = −1 or k = −1/9
Equal roots means b² − 4ac = 0.
Given that tan 2A = 2 tan A/(1 − tan2A), evaluate tan 15° leaving your answer in surd form.
Put A = 15°: tan 30° = 1/√3 = 2t/(1 − t2), t = tan 15°
t2 + 2√3t − 1 = 0
t = −√3 ± 2
tan 15° > 0, so tan 15° = 2 − √3
Use the double-angle formula with 2A = 30° and keep the positive root.
The probabilities of Rotey obtaining the highest mark in Mathematics, Physics and Biology tests are 0.9, 0.75 and 0.8 respectively. Calculate the probability of getting the highest marks in at least two subjects.
All three: 0.9 × 0.75 × 0.8 = 0.54
M, P not B: 0.9 × 0.75 × 0.2 = 0.135
M, B not P: 0.9 × 0.25 × 0.8 = 0.18
P, B not M: 0.1 × 0.75 × 0.8 = 0.06
At least two = 0.915
'At least two' means exactly two or all three.
A student representative council consist of 8 girls and 6 boys. If an editorial board consisting of 5 persons is to be formed, what is the probability that the board consist of (a) 3 girls and 2 boys; (b) either all girls or all boys?
Total boards: 14C5 = 2002
(a) 8C3 × 6C2 = 56 × 15 = 840; probability 840/2002 = 60/143 (≈ 0.420)
(b) 8C5 + 6C5 = 56 + 6 = 62; probability 62/2002 = 31/1001 (≈ 0.031)
Count favourable committees with combinations and divide by the total number of committees.
The coordinates of the vertices of triangle ABC are A(−2, 1), B(4, −2), and C(1, 8) respectively. If D(x, y) is the foot of the perpendicular from A to BC, find (A) an equation connecting x and y; (b) the unit vector in the direction of BC.
(a) Gradient of BC = (8 + 2)/(1 − 4) = −10/3
D lies on BC: y + 2 = −10/3(x − 4), i.e. 10x + 3y − 34 = 0
(Also AD ⟂ BC gives 3x − 10y + 16 = 0; together D = (292/109, 262/109))
(b) BC = −3i + 10j, |BC| = √109
Unit vector = (−3i + 10j)/√109
D is on line BC, and AD is perpendicular to BC; divide a vector by its length to get a unit vector.
The position vector of a body, with respect to the origin, is given by r = 4t i + (12 − 3t) j at anytime t seconds. (a) Find the velocity of the body (b) Calculate the magnitude of the displacement between t = 0 and t = 5.
(a) v = dr/dt = 4i − 3j (speed 5 ms−1)
(b) r(0) = 12j, r(5) = 20i − 3j
Displacement = 20i − 15j
Magnitude = √(400 + 225) = 25 m
Displacement is the change in position vector, not the difference of the two distances from O.
(a) The 3rd and 6th terms of a geometric progression (G.P.) are 2 and 54 respectively. Find the (i) common ratio (ii) first term (iii) sum of the first ten terms, correct the nearest whole number.
(b) The ratio of the coefficient of x4 to that of x3 in the binomial expansion of (1 + 2x)n is 3 : 1. Find the value of n.
(a)(i) ar2 = 2, ar5 = 54, so r3 = 27 and r = 3
(ii) a = 2/9
(iii) S10 = (2/9)(310 − 1)/(3 − 1) = 59048/9 ≈ 6561
(b) nC4(16) : nC3(8) = 3 : 1
2(n − 3)/4 = 3, so n − 3 = 6 and n = 9
Divide the two GP terms to find r; for (b) use ⁿC₄/ⁿC₃ = (n − 3)/4.
(a) Using the same axes, sketch the curves y = 6 − x − x2 and y = 3x2 − 2x + 3.
(b) Find the x-coordinates of the points of intercession of the two curves in (a)
(c) Calculate the area of the finite region bounded by the two curves (a).
(a) y = 6 − x − x2: ∩-shaped, cuts x-axis at −3 and 2, y-intercept 6, maximum at (−½, 6¼)
y = 3x2 − 2x + 3: ∪-shaped, y-intercept 3, minimum at (⅓, 2⅔), does not cut the x-axis
(b) 6 − x − x2 = 3x2 − 2x + 3 → 4x2 − x − 3 = 0 → (4x + 3)(x − 1) = 0
x = −3/4 or x = 1
(c) Area = ∫ from −3/4 to 1 of (3 + x − 4x2)dx = [3x + x2/2 − 4x3/3]
= 13/6 − (−45/32) = 343/96 ≈ 3.57 square units
Area between curves = ∫(upper − lower)dx between the intersection points.
(a) Evaluate ∫12 x/√(5 − x2) dx
(b) (i) Evaluate |2 −3 1|
|0 1 −2|
|1 2 −3|
(ii) Using your answer in (b)(i), solve the simultaneous equations
2x − 3y + z = 10
y − 2z = −7
x + 2y − 3z = −9
(a) Let u = 5 − x2: integral = [−√(5 − x2)] from 1 to 2 = −1 + 2 = 1
(b)(i) 2(−3 + 4) + 3(0 + 2) + 1(0 − 1) = 7
(ii) Cramer's rule: Δx = 14, Δy = −7, Δz = 21
x = 2, y = −1, z = 3
Use the substitution u = 5 − x²; for the equations use Cramer's rule with the determinant from (b)(i).
(a) Use the trapezium rule with five ordinates to evaluate ∫01 3dx/(1 + x2) correct to four significant figures.
(b) If A = (2 1; 3 2) find the image of the point (1, 2) under the linear transformation A2 + A + 2I, where I is the 2 x 2 unit matrix.
(a) h = 0.25; ordinates at x = 0, 0.25, 0.5, 0.75, 1: 3, 2.8235, 2.4, 1.92, 1.5
Area ≈ ½(0.25)[3 + 1.5 + 2(2.8235 + 2.4 + 1.92)] = 2.348
(b) A2 = (7 4; 12 7)
A2 + A + 2I = (11 5; 15 11)
Image of (1, 2) = (11 + 10, 15 + 22) = (21, 37)
Five ordinates means four strips of width 0.25. Matrices are written (row 1; row 2).
(a) Simplify n+1C3 − n−1C3.
(b) A fair die is thrown five times. Calculate, correct to three decimal places, the probability of containing (i) at most two sixes; (ii) exactly three sixes.
(a) [(n + 1)n(n − 1) − (n − 1)(n − 2)(n − 3)]/6
= (n − 1)(6n − 6)/6 = (n − 1)2
(b) p = 1/6, q = 5/6, n = 5
(i) P(0) + P(1) + P(2) = 0.4019 + 0.4019 + 0.1608 = 0.965
(ii) 5C3(1/6)3(5/6)2 = 250/7776 = 0.032
Write each combination as a product and take out the common factor (n − 1).
The distribution of the lives (in days) of 40 transistor batteries is shown in the table.
Battery life (in days): 26-30, 31-35, 36-40, 41-45, 46-50, 51-55
Frequency: 4, 7, 13, 8, 6, 2
(a) Draw a histogram for the distribution (b) Use your graph in (a) to determine the mode for the distribution. (c) Using an assumed mean of 43 days, calculate the mean of the distribution. (d) What percentage number of batteries will live for less than 31 days or more than 45 days?
(a) Bars on class boundaries 25.5, 30.5, …, 55.5 with heights 4, 7, 13, 8, 6, 2
(b) Modal class 35.5–40.5; crossing lines give mode ≈ 35.5 + 6/(6 + 5) × 5 ≈ 38.2 days
(c) Mid-points 28, 33, 38, 43, 48, 53; d = −15, −10, −5, 0, 5, 10; Σfd = −145
Mean = 43 − 145/40 = 39.375 ≈ 39.4 days
(d) Less than 31: 4; more than 45: 6 + 2 = 8; 12/40 × 100 = 30%
On the histogram, join the corners of the tallest bar to the neighbouring bars; their crossing gives the mode.
The table below shows the corresponding values of two variables X and Y.
X: 33, 31, 28, 25, 23, 22, 19, 17, 16, 14
Y: 4, 6, 4, 10, 12, 10, 14, 15, 18, 22
(a) Plot a scatter diagram to represent the data. (b) Calculate x̄, the mean of x and ȳ, the mean of y. (c) Draw the line of best fit to pass through (x̄, ȳ) (d) From your graph in (c), determine the (i) relationship between X and Y. (ii) value of Y when X is 24.
(a) Plot the 10 points with X on the horizontal axis
(b) x̄ = 228/10 = 22.8, ȳ = 115/10 = 11.5
(c) Straight line through (22.8, 11.5) following the trend of the points
(d)(i) Negative linear relationship: Y decreases as X increases, roughly Y = 31.5 − 0.88X (gradient read from the graph about −0.9)
(ii) When X = 24, Y ≈ 10.4 (about 10–11 from the graph)
Find the gradient and intercept of your line of best fit to state the relationship.
(a) The position vectors of points L and M are (5i + 6j) and (13i + 4j) respectively. If point K lies on LM such that LK : KM is 2:3, find the position vector of K.
(b) Three poles are situated at points A, B and C on the same horizontal plane such that AB = (8km, 060°) and BC = (12km, 130°). Calculate: (i) |AB|, correct to three decimal places (ii) the bearing of C from A, correct to the nearest degree.
(a) LM = 8i − 2j; LK = (2/5)LM = (16/5)i − (4/5)j
OK = 5i + 6j + (16/5)i − (4/5)j = (41/5)i + (26/5)j
(b) Angle ABC = 180° − 130° + 60° = 110°
(i) |AC|2 = 82 + 122 − 2(8)(12)cos 110° = 273.67, |AC| = 16.543 km (|AB| itself is the given 8 km)
(ii) AC = (8 sin 60° + 12 sin 130°)i + (8 cos 60° + 12 cos 130°)j = 16.121i − 3.714j
Bearing of C from A = 90° + tan−1(3.714/16.121) ≈ 103°
Use the cosine rule in triangle ABC, then resolve the displacements east and north to get the bearing.
Coplanar forces 4N, 8N 6N, 4N and 5N act at a point as shown in the diagram. If the 6N force act in the direction 090°, calculate the (a) magnitude of the resultant force; (b) direction of the resultant force.
Angles measured anticlockwise from the 6N force (east): 6N at 0°, 8N at 55°, 4N at 155°, 4N at −25°, 5N at 205°
East components: 6 + 8 cos 55° − 4 cos 25° + 4 cos 25° − 5 cos 25° = 6.057 N
North components: 8 sin 55° + 4 sin 25° − 4 sin 25° − 5 sin 25° = 4.440 N
(a) |R| = √(6.0572 + 4.4402) ≈ 7.5 N
(b) tan−1(4.440/6.057) ≈ 36.2° north of east, i.e. bearing ≈ 054°
Resolve every force along and perpendicular to the 6N force, add, then use Pythagoras and tan⁻¹.
(a) The position vectors of points A, B and C are i + 5j, 3i + 9j, and −i + j respectively. (i) Show that points A, B and C are collinear. (ii) Determine the ratio |AB| : |BC|
(b) A uniform beam XY of mass 10kg and length 24m is hunged horizontally from a cross bar by two vertical inextensible strings, one attached to X and the other at a point M, 4m away from Y, a mass of 50kg is suspended at a point N which is 8m from X. If the system remains in equilibrium, calculate the tensions in the strings [Take g = 10ms−2]
(a)(i) AB = 2i + 4j, BC = −4i − 8j = −2AB; parallel with common point B, so A, B, C are collinear
(ii) |AB| : |BC| = √20 : √80 = 1 : 2
(b) Weights: beam 100 N at 12 m from X, load 500 N at 8 m from X; M is 20 m from X
Moments about X: 20TM = 100 × 12 + 500 × 8 = 5200, TM = 260 N
Resolving: TX + TM = 600, TX = 340 N
Take moments about one string to remove its tension, then resolve vertically.