46 questions. Try each one, then open the answer to check yourself and read the explanation.
If X = (2 3 / 1 4), Y = (1 5 / 0 3) and Z = (6 −1 / 1 2), find 5X − 2Y + 2Z. (Each matrix is written row by row; / separates the rows.)
Answer: C ((20 3 / 7 18))
5X = (10 15 / 5 20), 2Y = (2 10 / 0 6), 2Z = (12 −2 / 2 4). So 5X − 2Y + 2Z = (10 − 2 + 12, 15 − 10 − 2 / 5 − 0 + 2, 20 − 6 + 4) = (20 3 / 7 18).
If 2cos2θ = 1 + sin θ, 0° ≤ θ ≤ 90°, find the value of θ.
Answer: A (30°)
Replace cos²θ by 1 − sin²θ: 2 − 2sin²θ = 1 + sin θ, so 2sin²θ + sin θ − 1 = 0, i.e. (2sin θ − 1)(sin θ + 1) = 0. In the range, sin θ = ½, so θ = 30°.
The function f: x → √(x2 + 5x − 6) is defined on the set of real numbers R. Find the domain of f.
Answer: B ({x : x ∈ R, x ≤ −6, x ≥ 1})
The square root needs x² + 5x − 6 ≥ 0, i.e. (x + 6)(x − 1) ≥ 0. This holds when x ≤ −6 or x ≥ 1.
Given that P = (3 5 −4 / 6 −3 −5 / −2 2 1) and Q = (−3 −2 1 / 1 2 3 / 3 1 2), find |P + Q|. (Each matrix is written row by row; / separates the rows.)
Answer: C (−135)
P + Q = (0 3 −3 / 7 −1 −2 / 1 3 3). Expanding along the first row: 0 − 3(7×3 − (−2)×1) + (−3)(7×3 − (−1)×1) = −3(23) − 3(22) = −69 − 66 = −135.
If √(2(x − 1)3) − 4 = 0, find the value of x.
Answer: A (3)
√(2(x − 1)³) = 4, so 2(x − 1)³ = 16 and (x − 1)³ = 8. Then x − 1 = 2, so x = 3.
Given that (2x + 3)/(x2 − 3x + 2) ≡ A/(x − 1) + B/(x − 2), find the values of A and B.
Answer: A (A = −5, B = 7)
x² − 3x + 2 = (x − 1)(x − 2), so 2x + 3 = A(x − 2) + B(x − 1). Put x = 1: 5 = −A, so A = −5. Put x = 2: 7 = B. Check: −5(x − 2) + 7(x − 1) = 2x + 3.
The coordinates of the centre of a circle is (−2, 5) and its area is 64π cm2. Find the equation of the circle.
Answer: C (x2 + y2 + 4x − 10y − 35 = 0)
πr² = 64π gives r = 8. The circle is (x + 2)² + (y − 5)² = 64, i.e. x² + 4x + 4 + y² − 10y + 25 = 64, so x² + y² + 4x − 10y − 35 = 0.
A straight line makes an angle of 60° with the positive x-axis and cuts the y-axis at y = 7, find the equation of the line.
Answer: A (y = √3x + 7)
The gradient is tan 60° = √3 and the y-intercept is 7, so y = √3x + 7.
If M = {x: −2 ≤ x ≤ 4} and N = {x: 0 < x < 6} where x is a real number, find M ∩ N.
Answer: B ({x: 0 < x ≤ 4})
The numbers in both sets must be greater than 0 (from N) and at most 4 (from M), so M ∩ N = {x: 0 < x ≤ 4}.
If √3 − 2/√3 = x√3, find the value of x.
Answer: C (1/3)
2/√3 = 2√3/3, so √3 − 2√3/3 = √3/3 = (1/3)√3. Hence x = 1/3.
Given that the first term of an exponential sequence (G.P.) is (√3 − 2) and the second term is (7 − 4√3), find the common ratio.
Answer: D (√3 − 2)
r = (7 − 4√3)/(√3 − 2). Since (√3 − 2)² = 3 − 4√3 + 4 = 7 − 4√3, r = √3 − 2.
If 5 + log2 y = 2log2 x, express y in terms of x.
Answer: D (y = x2/32)
5 = log₂ 32, so log₂ y = log₂ x² − log₂ 32 = log₂ (x²/32). Hence y = x²/32.
Find the coefficient of x3 in the binomial expansion of (1 − 3x)5.
Answer: D (−270)
The x³ term is ⁵C₃(−3x)³ = 10 × (−27)x³ = −270x³.
Find the sum of the first 30 terms of the sequence: −5, −3, −1, …
Answer: B (720)
This is an A.P. with a = −5, d = 2. S₃₀ = 30/2 × [2(−5) + 29(2)] = 15 × 48 = 720.
Find the coefficient of x3 in the binomial expansion of [½(3 + x)]5.
Answer: C (45/16)
[½(3 + x)]⁵ = (1/32)(3 + x)⁵. The x³ term of (3 + x)⁵ is ⁵C₃ × 3² x³ = 90x³, so the coefficient is 90/32 = 45/16.
A bag contains 21 apples. If 10 apples are green and the rest are red, how many red apples must be added to the bag to make the probability of selecting a red apple 3/5?
Answer: C (4)
There are 11 red apples. Adding n red: (11 + n)/(21 + n) = 3/5, so 55 + 5n = 63 + 3n and n = 4.
In a horse racing sport, a triple is a type of game. To win this, you need to specify the horses that will finish in the top 3 spots in the order in which they finish. If 7 horses took part in the race, in how many ways can they finish in the top 3 spots?
Answer: C (210 ways)
Order matters, so this is a permutation: ⁷P₃ = 7 × 6 × 5 = 210 ways.
Find the maximum value of the curve y = 2x3 + 4x2 − 8x − 4
Answer: B (12)
dy/dx = 6x² + 8x − 8 = 0 gives 3x² + 4x − 4 = 0, (3x − 2)(x + 2) = 0. d²y/dx² = 12x + 8 is negative at x = −2, so the maximum is y = −16 + 16 + 16 − 4 = 12.
Find the unit vector in the direction of r = 5i − 2j
Answer: C ((5i − 2j)/√29)
|r| = √(25 + 4) = √29, so the unit vector is r/|r| = (5i − 2j)/√29.
How many ways can 15 people be divided into three groups of 3, 8 and 4 in that order?
Answer: D (225,225)
15!/(3! 8! 4!) = ¹⁵C₃ × ¹²C₈ = 455 × 495 = 225,225.
If r denotes a correlation coefficient, which of the following expressions is false?
Answer: D (r = 1.05)
A correlation coefficient always lies between −1 and 1, so r = 1.05 is impossible.
Given that f(x) = 5x2 + px and f(−1) = 2, find the value of p
Answer: C (3)
f(−1) = 5 − p = 2, so p = 3.
Evaluate ∫02 (7x3 − 2x) dx.
Answer: C (24)
∫(7x³ − 2x) dx = 7x⁴/4 − x². From 0 to 2: 7(16)/4 − 4 = 28 − 4 = 24.
Evaluate (4/(3√2)) cos 75° in surd form (radicals)
Answer: A (⅓(√3 − 1))
cos 75° = cos(45° + 30°) = (√6 − √2)/4. So (4/(3√2)) × (√6 − √2)/4 = (√6 − √2)/(3√2) = (√3 − 1)/3.
If 3x2 − 12x + k = 0 has real and unequal roots, find the range of values of k
Answer: A (k < 12)
Real, unequal roots need b² − 4ac > 0: 144 − 12k > 0, so k < 12.
A particle moves in a plane such that its displacement, s, is given by s = (t3 + t2)i + (3t − 2)j. Find the velocity of the particle at t = 3 seconds.
Answer: C (33i + 3j)
v = ds/dt = (3t² + 2t)i + 3j. At t = 3: (27 + 6)i + 3j = 33i + 3j.
An operation * defined on the set of positive real numbers is given by x * y = yx/7, find the inverse of x.
Answer: A (49/x)
Identity: x * e = xe/7 = x gives e = 7. Inverse: x × x⁻¹/7 = 7, so x⁻¹ = 49/x.
A boy of weight 78N sits on a bag of cement weighing 50 kg. Find the reaction of the ground on the bag of cement and the boy. [Take g = 10 ms−2]
Answer: B (578 N)
The ground supports both: the bag's weight is 50 × 10 = 500 N, plus the boy's 78 N, so the reaction is 578 N.
A body of mass 21 kg increases its speed from 15 ms−1 to 43 ms−1 in 12 seconds. Calculate its change in momentum.
Answer: D (588 kg ms−1)
Change in momentum = m(v − u) = 21 × (43 − 15) = 21 × 28 = 588 kg ms⁻¹ (the time is not needed).
Simplify: 8! + 6!
Answer: B (6! × 57)
8! = 8 × 7 × 6! = 56 × 6!, so 8! + 6! = 6!(56 + 1) = 6! × 57.
A uniform beam PQ of length 5m and weight 100N rests horizontally on supports P and Q. Find the reaction at P
Answer: A (50 N)
The weight acts at the middle, 2.5 m from each end. Taking moments about Q: R_P × 5 = 100 × 2.5, so R_P = 50 N.
Given that dy/dx = 3x2 − 5, and y = 12 when x = 4, find y
Answer: C (y = x3 − 5x − 32)
Integrating: y = x³ − 5x + k. With x = 4, y = 12: 12 = 64 − 20 + k, so k = −32 and y = x³ − 5x − 32.
A binary operation Δ is defined on the set of real numbers, R, by m Δ n = 1/m3 − 1/n3. Evaluate (√3 − √2) Δ (√3 + √2)
Let m = √3 − √2, n = √3 + √2; mn = 3 − 2 = 1, so 1/m = n and 1/n = m.
m Δ n = 1/m³ − 1/n³ = n³ − m³.
n³ = (√3 + √2)³ = 3√3 + 9√2 + 6√3 + 2√2 = 9√3 + 11√2.
m³ = 9√3 − 11√2.
(√3 − √2) Δ (√3 + √2) = (9√3 + 11√2) − (9√3 − 11√2) = 22√2.
Notice that the two surds multiply to 1, so each is the reciprocal of the other.
A box contains 4 chocolate, 2 oatmeal and 14 ginger identical biscuits. If 5 biscuits are selected at random from the box, find the number of ways this can be done, if 2 must be oatmeal
Both oatmeal biscuits are chosen: ²C₂ = 1 way.
The other 3 come from the 18 non-oatmeal biscuits: ¹⁸C₃ = 816 ways.
Number of ways = 1 × 816 = 816.
Fix the required items first, then choose the rest from what is left.
Given the function f(x) = (2x3 − x)/(x2 − 1), x ≠ ±1, express :
(a) f(x) in the form g(x) + h(x); (b) h(x) in partial fractions
(a) 2x³ − x = 2x(x² − 1) + x, so f(x) = 2x + x/(x² − 1): g(x) = 2x, h(x) = x/(x² − 1).
(b) x/((x − 1)(x + 1)) = A/(x + 1) + B/(x − 1); x = A(x − 1) + B(x + 1).
x = 1: B = ½; x = −1: A = ½.
h(x) = 1/(2(x + 1)) + 1/(2(x − 1)).
Divide first (the top has a higher degree), then split the proper fraction.
Solve: |−2n + 3, n + 1 / 2n, n| = |−2n + 5, 4n / −n + 2, 3n| (determinants written row by row; / separates the rows)
Left: n(−2n + 3) − 2n(n + 1) = −4n² + n.
Right: 3n(−2n + 5) − 4n(−n + 2) = −2n² + 7n.
−4n² + n = −2n² + 7n gives −2n² − 6n = 0, i.e. −2n(n + 3) = 0.
n = 0 or n = −3.
Expand each 2 × 2 determinant as ad − bc, then solve the quadratic.
A bag contains 3 different marbles x, y and z of the same size and shape. The total number of marbles in the bag is 27 and they are such that x : y = 1 : 2 and y : z = 1 : 3
(a) Find the number of z marbles in the bag
(b) Calculate the probability of selecting two of the y marbles at random from the bag one after the other without replacement.
(a) x : y : z = 1 : 2 : 6 (total 9 parts). z = 6/9 × 27 = 18 (and y = 6, x = 3).
(b) P(two y) = 6/27 × 5/26 = 30/702 = 5/117.
Combine the two ratios so y has the same number of parts in both.
Given that m = 2i + pj, n = −3i + 4j and cos θ = 6/(5√13), where θ is an angle between m and n. Find the values of p.
m·n = −6 + 4p; |m| = √(4 + p²), |n| = 5.
(4p − 6)/(5√(4 + p²)) = 6/(5√13), so √13(4p − 6) = 6√(4 + p²).
Squaring: 13(16p² − 48p + 36) = 36(4 + p²), giving 43p² − 156p + 81 = 0.
(43p − 27)(p − 3) = 0, so p = 3 or p = 27/43.
Check: cos θ is positive, so 4p − 6 > 0; only p = 3 satisfies the original equation (p = 27/43 gives a negative cosine, an extra root from squaring).
Use cos θ = m·n/(|m||n|), and check roots after squaring.
A body of mass 12 kg rests on a rough plane inclined at an angle of 30° to the horizontal. The coefficient of friction between the body and the plane is 2/3.
A force of magnitude P Newtons acts on the body along the inclined plane. Find the value of P if the body is at the point of moving: (a) down the plane
(b) up the plane [Take g = 10 ms−2]
R = mg cos 30° = 120 × 0.866 = 103.9 N; limiting friction F = μR = ⅔ × 103.9 = 69.3 N.
Component of weight down the plane = 120 sin 30° = 60 N.
(a) Moving down: friction acts up the plane: P + 60 = 69.3, so P ≈ 9.3 N.
(b) Moving up: friction acts down the plane: P = 60 + 69.3 ≈ 129.3 N.
Friction always opposes the motion that is about to happen.
A committee of 6 people is to be chosen from 7 mid-fielders, 5 attackers and 8 referees. If each discipline must be represented, determine the number of ways of selecting the committee to include: (a) exactly one attacker and one mid-fielder (b) exactly two referees (c) at least two persons from the attackers
(a) 1 mid-fielder, 1 attacker, 4 referees: 7 × 5 × ⁸C₄ = 7 × 5 × 70 = 2450.
(b) 2 referees (⁸C₂ = 28) and 4 from mid-fielders and attackers with at least one of each: ¹²C₄ − ⁷C₄ − ⁵C₄ = 495 − 35 − 5 = 455. Total 28 × 455 = 12 740.
(c) 2 attackers (⁵C₂ = 10): (7×56 + 21×28 + 35×8) = 1260, giving 12 600;
3 attackers (10): (7×28 + 21×8) = 364, giving 3 640;
4 attackers (5): 7×8 = 56, giving 280;
5 attackers leaves one place, so not possible.
Total = 12 600 + 3 640 + 280 = 16 520.
Every committee needs at least one of each group, so list the allowed splits.
(a) Solve: (1/4)(m + 1)/3 + 4 = 5(1/2)(m + 1)/3
(b) Given that p log3 x = log81 x, find the value of p
(a) Let u = (½)^((m + 1)/3); then (¼)^((m + 1)/3) = u².
u² − 5u + 4 = 0, (u − 1)(u − 4) = 0, u = 1 or 4.
u = 1: (m + 1)/3 = 0, m = −1.
u = 4 = (½)⁻²: (m + 1)/3 = −2, m = −7.
m = −1 or m = −7.
(b) log₈₁ x = log₃ x / log₃ 81 = ¼ log₃ x, so p = ¼.
A substitution turns the index equation into a quadratic.
In a study to find the number of languages 50 people can speak, it was revealed that (m2 + 3) of them speak Yoruba, (5m − 3) speak Hausa and (3m + 1) speak Igbo. 4 of them speak all three languages and 14 of them speak exactly two languages. Those who speak Igbo and Yoruba only are half of those who speak Yoruba and Hausa only. Those who speak Hausa and Igbo only are 1 less than those who speak Yoruba and Hausa only. 6 of them could not speak any of the languages. (a) Illustrate the information provided in a Venn Diagram (b) Find the: (i) Value of m
(ii) Number of people who speak one language only
Let Yoruba and Hausa only = a; Igbo and Yoruba only = a/2; Hausa and Igbo only = a − 1.
a + a/2 + a − 1 = 14, so a = 6: Y∩H only 6, Y∩I only 3, H∩I only 5, all three 4, none 6.
(a) Venn diagram: three overlapping circles in a rectangle (50) with these regions; Yoruba only m² − 10, Hausa only 5m − 18, Igbo only 3m − 11, outside 6.
(b)(i) (m² − 10) + (5m − 18) + (3m − 11) + 6 + 3 + 5 + 4 + 6 = 50, so m² + 8m − 65 = 0, (m + 13)(m − 5) = 0; m = 5 (m > 0).
(ii) One language only = 15 + 7 + 4 = 26.
Find the 'exactly two' regions first from the given relations.
The marks obtained by 40 students in an examination are as follows:
29 52 37 79 63 48 65 46 54 32 51 66 74 24 35 61 58 39 49 53
45 58 56 38 59 28 50 49 67 56 37 45 79 68 43 66 26 47 55 71
(a) Construct a grouped frequency distribution table using the class interval 20 − 29, 30 − 39, 40 − 49 …
(b) Using an assumed mean of 54.5, calculate, correct to the nearest whole number, the: (i) Mean mark
(ii) Standard deviation of the distribution
(a) 20–29: 4; 30–39: 6; 40–49: 8; 50–59: 11; 60–69: 7; 70–79: 4 (total 40).
(b) Class marks 24.5 … 74.5; d = x − 54.5 = −30, −20, −10, 0, 10, 20.
Σfd = −120 − 120 − 80 + 0 + 70 + 80 = −170; Σfd² = 3600 + 2400 + 800 + 0 + 700 + 1600 = 9100.
(i) Mean = 54.5 + (−170/40) = 50.25 ≈ 50.
(ii) SD = √(9100/40 − (−4.25)²) = √209.44 ≈ 14.5, i.e. 14 to the nearest whole number.
Tally carefully and check the frequencies add up to 40.
(a) 12% of sugar packaged by a machine in a factory are found to be less than 50kg. If 8 of the bags were selected at random, find, correct to 4 significant figures, the probability that: (i) exactly half; (ii) at least 75% of the bags of sugar are greater than or equal to 50kg
(b) There are 35% defective syringes in a large box. If 14 syringes are selected at random from the box, calculate the probability that between three and six are defective.
(a) p(≥ 50 kg) = 0.88, q = 0.12, n = 8.
(i) P(X = 4) = ⁸C₄(0.88)⁴(0.12)⁴ = 70 × 0.59970 × 0.00020736 ≈ 0.008705.
(ii) At least 6 of 8: P(6) + P(7) + P(8) = 28(0.88)⁶(0.12)² + 8(0.88)⁷(0.12) + (0.88)⁸ = 0.18725 + 0.39233 + 0.35963 ≈ 0.9392.
(b) p = 0.35, q = 0.65, n = 14. Between 3 and 6 (that is, 4 or 5): P(4) + P(5) = 1001(0.35)⁴(0.65)¹⁰ + 2002(0.35)⁵(0.65)⁹ = 0.2022 + 0.2178 ≈ 0.4200.
(If 3 and 6 are included: add P(3) = 0.1366 and P(6) = 0.1759, giving 0.7325.)
Use the binomial distribution P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ.
The position vectors of points P and Q are p = 2i + 3j and q = 4i − 7j respectively. Find (a) |4p − 5q|;
(b) The scalars m and n such that mp − nq ≡ 3i − 7j.
(a) 4p − 5q = (8i + 12j) − (20i − 35j) = −12i + 47j; |4p − 5q| = √(144 + 2209) = √2353 ≈ 48.5.
(b) mp − nq = (2m − 4n)i + (3m + 7n)j = 3i − 7j.
2m − 4n = 3 and 3m + 7n = −7.
Solving: 26n = −23, n = −23/26; m = −7/26.
Compare the i and j parts separately to get two equations.
(a) A ball is thrown from the ground upwards with a velocity of 30ms−1. A second later, another ball held at rest is released from a height of 50m to hit the first ball. Find the: (i) time when the balls meet
(ii) height at which the balls meet [Take g = 10ms−2]
(b) A bullet was released from a hunter's gun with an initial velocity of 8ms−1. It then accelerates at 3.5ms−2 for 15 seconds. Find the (i) final velocity of the bullet
(ii) distance travelled by the bullet
(a) Height of first ball after t s: h = 30t − 5t². Second ball (released at t = 1): h = 50 − 5(t − 1)².
(i) 30t − 5t² = 50 − 5t² + 10t − 5, so 20t = 45 and t = 2.25 s after the first ball is thrown (1.25 s after the second is released).
(ii) h = 30(2.25) − 5(2.25)² = 67.5 − 25.3 ≈ 42.2 m.
(b)(i) v = u + at = 8 + 3.5 × 15 = 60.5 ms⁻¹.
(ii) s = ut + ½at² = 120 + ½ × 3.5 × 225 = 513.75 m.
Measure both balls' heights from the ground with the same clock.