Q1 (MAT 001)
(a) Given A = {−5, −3, −1, 0, 1, 2, 3} and B = {−4, −3, 0, 3, 5, 8}. Find A Δ B.
(b) If A, B, and C are any sets, show that A ∪ (B ∪ C) = (A ∪ B) ∪ C.
© In an election involving three parties for the chairmanship and gubernatorial election of Lagos State, voters cast their votes as follows: 190 voted for party A, 200 for party B and 250 for party C. 80 voted for A and B, 60 voted for A and C, 100 voted for B and C and 40 voted for B alone. If 500 people voted during the election, find:
i. The number of voters who voted for all three parties.
ii. The number of voters who voted for A and B but not C.
iii. The number of voters who did not vote for any party.
Q2 (MAT 001)
(a) i. Evaluate the determinant of A = |1 2 3; 4 5 6; 7 8 9|.
ii. What do you conclude from 2a(i)?
iii. Resolve (x³ − 1)/[(x + 3)(x + 1)²] in partial fractions. Hence obtain its Binomial expansion up to terms in x².
(b) If cos(x + α) = sin(x + β), find tan x in terms of α and β.
© If sin A = 4/5 and cos B = 12/13, where A is obtuse and B is acute, find without using tables the values of: i. sin(A + B) ii. tan(A − B).
Q3 (MAT 002)
(a) By using the reduction formula for ∫secⁿx dx, evaluate the definite integral ∫₀^(π/4) sec⁶x dx.
(b) Find the area enclosed by the curve y = x² and the x-axis between x₁ = 0 and x₂ = 2.
Q4 (MAT 002)
(a) Evaluate lim(x→0) (1/x − csc x).
(b) By Taylor’s theorem, show that logₑ(1 + x) = x − x²/2 + x³/3 − x⁴/4 + … + xⁿ/n, and hence evaluate logₑ(1.1) to four decimal places.
© Using the trapezoidal rule with ordinates x = 1.0, 1.4, 1.8, 2.2, 2.6, 3.0, evaluate ∫₁³ 1/(x+1) dx.
Q5 (MAT 003)
In the study of motion of rigid bodies, explain the following concepts:
(a) i. Moment of inertia of the system.
ii. Radius of gyration of the system.
(b) Find the moment of inertia and radius of gyration of a uniform thin rod of length 2a, density ρ, about an axis passing through one end of the rod perpendicular to its length.
Q6 (MAT 003)
(a) State Newton’s law of cooling and write out the differential equation describing the temperature of the body.
(b) A beaker of water initially at 100°C is allowed to cool in a room maintained at 15°C. After two minutes, the water temperature is 85°C. Find the temperature of the water after four minutes and the time taken for the water to reach 40°C.
© If the position vectors of points A, B and C are a = i + 3j − 7k, b = 7i + 6j + 5k and c = 9i + 7j + βk respectively, find:
i. |a + b|
ii. The value of β if A, B and C are collinear.
Q7 (MAT 004)
(a) The following data represent scores of 50 students in a Statistics test:
72 93 70 59 78 74 65 73 80 57 67 72 57 83 76 74 56 68 67 74 76 79 72 61 72 73 76 67 49 71 53 67 65 100 83 69 61 72 68 65 51 75 68 75 66 77 61 64 74 72
By using a class interval of five (45–49, 50–54, etc):
i. Prepare the frequency distribution table.
ii. What is the coefficient of variation?
iii. Does the data represent a sample or a population?
(b) Discuss briefly the measures of location associated with frequencies; hence explain mean, mode, and median.
Q8 (MAT 004)
(a) i. Find the coefficient of linear correlation between variables A and B:
| A | 1 | 2 | 3 | 4 | 5 |
|—|---|—|---|—|---|
| B | 1 | 2 | 3 | 6 | 8 |
ii. Five students were ranked according to their scores in Mathematics and Physics:
| Student | A | B | C | D | E |
|—|---|—|---|—|---|
| Mathematics | 1 | 3 | 5 | 2 | 4 |
| Physics | 2 | 1 | 3 | 4 | 5 |
Calculate the Spearman’s rank correlation coefficient.
(b) Differentiate between discrete and continuous random variables.
© A company that manufactures computer chips finds that 5% of the chips they produce are defective. If 8 chips are selected at random, find the probability that:
i. 2 chips will be defective.
ii. At least 2 chips will be defective.
iii. Calculate for (i) and (ii) above, the number of expected defective chips and variance in a sample of 2,000 chips.
##ANSWERS
Q1(a) — A Δ B
A Δ B = (A ∪ B) − (A ∩ B)
A ∩ B = {−3, 0, 3}
A ∪ B = {−5, −4, −3, −1, 0, 1, 2, 3, 5, 8}
A Δ B = {−5, −4, −1, 1, 2, 5, 8}
Q1(b) — Associativity of Union
Let x ∈ A ∪ (B ∪ C).
Then x ∈ A or x ∈ (B ∪ C)
→ x ∈ A, or x ∈ B, or x ∈ C
→ x ∈ (A ∪ B), or x ∈ C
→ x ∈ (A ∪ B) ∪ C
The argument reverses symmetrically, so (A ∪ B) ∪ C ⊆ A ∪ (B ∪ C).
Therefore A ∪ (B ∪ C) = (A ∪ B) ∪ C. □
Q1©
Let x = number who voted for all three parties.
Using inclusion-exclusion:
|A ∪ B ∪ C| = 190 + 200 + 250 − 80 − 60 − 100 + x = 400 + x
Given: 40 voted for B alone.
B alone = |B| − |A∩B| − |B∩C| + |A∩B∩C| = 200 − 80 − 100 + x = 20 + x = 40 → x = 20
i. Voters for all three = 20
ii. A and B but not C:
|A ∩ B| − |A ∩ B ∩ C| = 80 − 20 = 60
iii. No party:
|A ∪ B ∪ C| = 400 + 20 = 420
Did not vote = 500 − 420 = 80
Q2(a)(i) — Determinant of A
A = |1 2 3; 4 5 6; 7 8 9|
det = 1(45−48) − 2(36−42) + 3(32−35)
= 1(−3) − 2(−6) + 3(−3)
= −3 + 12 − 9 = 0
Q2(a)(ii)
Since det A = 0, the matrix A is singular (non-invertible). The rows are linearly dependent — each row is an arithmetic progression with common difference 3, and row 3 = 2(row 2) − row 1.
Q2(a)(iii) — Partial Fractions of (x³ − 1)/[(x+3)(x+1)²]
First perform polynomial division since degree of numerator equals degree of denominator (both 3):
x³ − 1 = (x+3)(x+1)² · Q(x) + remainder
(x+3)(x+1)² = (x+3)(x²+2x+1) = x³ + 2x² + x + 3x² + 6x + 3 = x³ + 5x² + 7x + 3
So: (x³−1)/(x³+5x²+7x+3) = 1 + (−5x²−7x−4)/[(x+3)(x+1)²]
Let (−5x²−7x−4)/[(x+3)(x+1)²] = A/(x+3) + B/(x+1) + C/(x+1)²
−5x² − 7x − 4 = A(x+1)² + B(x+3)(x+1) + C(x+3)
x = −1: −5+7−4 = 2C → −2 = 2C → C = −1
x = −3: −45+21−4 = 4A → −28 = 4A → A = −7
Coefficient of x²: −5 = A + B → B = −5 + 7 = B = 2
(x³−1)/[(x+3)(x+1)²] = 1 − 7/(x+3) + 2/(x+1) − 1/(x+1)²
Binomial expansion up to x²:
7/(x+3) = (7/3)·1/(1+x/3) = (7/3)(1 − x/3 + x²/9 − …) = 7/3 − 7x/9 + 7x²/27
2/(x+1) = 2(1 − x + x² − …) = 2 − 2x + 2x²
1/(x+1)² = (1+x)⁻² = 1 − 2x + 3x² − …
Full expression up to x²:
1 − (7/3 − 7x/9 + 7x²/27) + (2 − 2x + 2x²) − (1 − 2x + 3x²)
Constant: 1 − 7/3 + 2 − 1 = 2 − 7/3 = −1/3
x: 7/9 − 2 + 2 = 7/9
x²: −7/27 + 2 − 3 = −1 − 7/27 = −34/27
= −1/3 + 7x/9 − 34x²/27 + …
Q2(b) — Find tan x
cos(x + α) = sin(x + β)
cos x cosα − sin x sinα = sin x cosβ + cos x sinβ
cos x(cosα − sinβ) = sin x(cosβ + sinα)
tan x = (cosα − sinβ)/(cosβ + sinα)
Q2©
sin A = 4/5, A obtuse → cos A = −3/5, tan A = −4/3
cos B = 12/13, B acute → sin B = 5/13, tan B = 5/12
i. sin(A+B) = sinA cosB + cosA sinB
= (4/5)(12/13) + (−3/5)(5/13)
= 48/65 − 15/65 = 33/65
ii. tan(A−B) = (tanA − tanB)/(1 + tanA tanB)
= (−4/3 − 5/12)/(1 + (−4/3)(5/12))
= (−16/12 − 5/12)/(1 − 20/36)
= (−21/12)/(16/36)
= (−21/12)(36/16)
= −756/192 = −63/16
Q3(a) — ∫₀^(π/4) sec⁶x dx
Reduction formula: ∫secⁿx dx = secⁿ⁻²x tanx/(n−1) + (n−2)/(n−1) ∫secⁿ⁻²x dx
For n=6: ∫sec⁶x dx = sec⁴x tanx/5 + 4/5 ∫sec⁴x dx
For n=4: ∫sec⁴x dx = sec²x tanx/3 + 2/3 ∫sec²x dx = sec²x tanx/3 + 2tanx/3
So: ∫sec⁶x dx = sec⁴x tanx/5 + 4/5[sec²x tanx/3 + 2tanx/3]
= sec⁴x tanx/5 + 4sec²x tanx/15 + 8tanx/15
Evaluating from 0 to π/4 (at π/4: sec=√2, tan=1; at 0: all terms = 0):
= [(√2)⁴(1)/5 + 4(√2)²(1)/15 + 8(1)/15] − 0
= [4/5 + 8/15 + 8/15]
= 12/15 + 16/15 = 28/15
Q3(b) — Area under y = x²
A = ∫₀² x² dx = [x³/3]₀² = 8/3 − 0 = 8/3 square units
Q4(a) — lim(x→0)(1/x − csc x)
= lim(x→0) [1/x − 1/sinx] = lim(x→0) (sinx − x)/(x sinx)
Using L’Hôpital or Taylor: sinx = x − x³/6 + …
sinx − x = −x³/6 + …
x sinx ≈ x² − x⁴/6 ≈ x²
lim = (−x³/6)/x² = −x/6 → 0 as x→0
Q4(b) — Taylor’s series for logₑ(1+x)
Let f(x) = ln(1+x). f(0)=0.
f’(x) = 1/(1+x), f’(0)=1
f’’(x) = −1/(1+x)², f’’(0)=−1
f’’’(x) = 2/(1+x)³, f’’’(0)=2
Taylor’s series: f(x) = f(0) + xf’(0) + x²f’’(0)/2! + x³f’’’(0)/3! + …
= x − x²/2 + x³/3 − x⁴/4 + … □
logₑ(1.1) = logₑ(1 + 0.1):
= 0.1 − 0.01/2 + 0.001/3 − 0.0001/4 + …
= 0.1 − 0.005 + 0.000333 − 0.000025 + …
= 0.0953 (to 4 d.p.)
Q4© — Trapezoidal Rule
h = 0.4, ordinates at x = 1.0, 1.4, 1.8, 2.2, 2.6, 3.0
f(x) = 1/(x+1):
| x | f(x) |
|—|---|
| 1.0 | 0.5000 |
| 1.4 | 0.4167 |
| 1.8 | 0.3571 |
| 2.2 | 0.3125 |
| 2.6 | 0.2778 |
| 3.0 | 0.2500 |
T = (h/2)[f(x₀) + 2(f(x₁)+f(x₂)+f(x₃)+f(x₄)) + f(x₅)]
= (0.4/2)[0.5 + 2(0.4167+0.3571+0.3125+0.2778) + 0.25]
= 0.2[0.5 + 2(1.3641) + 0.25]
= 0.2[0.5 + 2.7282 + 0.25]
= 0.2 × 3.4782 = 0.6956
(Exact: ln4 − ln2 = ln2 ≈ 0.6931)
Q5(a)(i) — Moment of Inertia
The moment of inertia I of a system of particles about an axis is defined as the sum of the products of the mass of each particle and the square of its perpendicular distance from the axis:
I = Σmᵢrᵢ²
It measures a body’s resistance to angular acceleration.
Q5(a)(ii) — Radius of Gyration
The radius of gyration k is defined such that I = Mk², where M is the total mass. It is the distance from the axis at which the entire mass could be concentrated to give the same moment of inertia:
k = √(I/M)
Q5(b) — Thin Rod (length 2a, density ρ, axis at one end)
Mass per unit length = ρ (linear density)
Total mass M = 2aρ
I = ∫₀^(2a) ρx² dx = ρ[x³/3]₀^(2a) = ρ(8a³/3) = 8ρa³/3
Since M = 2aρ:
I = M(4a²/3) = 4Ma²/3
Radius of gyration: k² = I/M = 4a²/3 → k = 2a/√3
Q6(a) — Newton’s Law of Cooling
Newton’s Law of Cooling states that the rate of change of temperature of a body is proportional to the difference between the body’s temperature and the ambient (surrounding) temperature, provided that difference is small:
dT/dt = −k(T − Tₐ)
where T is the body temperature, Tₐ is the ambient temperature, t is time, and k > 0 is the cooling constant.
Q6(b) — Cooling Problem
T₀ = 100°C, Tₐ = 15°C, T(2) = 85°C
Solution: T(t) = 15 + (100−15)e^(−kt) = 15 + 85e^(−kt)
At t = 2: 85 = 15 + 85e^(−2k) → 70 = 85e^(−2k) → e^(−2k) = 70/85 = 14/17
−2k = ln(14/17) → k = −ln(14/17)/2 = 0.0966
Temperature at t = 4:
T(4) = 15 + 85e^(−4k) = 15 + 85(14/17)² = 15 + 85(196/289)
= 15 + 85 × 0.6782 = 15 + 57.6 ≈ 72.6°C
Time to reach 40°C:
40 = 15 + 85e^(−kt)
25 = 85e^(−kt) → e^(−kt) = 25/85 = 5/17
−kt = ln(5/17) → t = −ln(5/17)/0.0966 = ln(17/5)/0.0966
= 1.2238/0.0966 ≈ 12.67 minutes
Q6©
a = i + 3j − 7k, b = 7i + 6j + 5k, c = 9i + 7j + βk
i. |a + b|
a + b = 8i + 9j − 2k
|a + b| = √(64 + 81 + 4) = √149
ii. β if A, B, C collinear
AB = b − a = 6i + 3j + 12k
AC = c − a = 8i + 4j + (β+7)k
For collinearity: AC = λAB
8 = 6λ → λ = 4/3
4 = 3λ = 4 ✓
β + 7 = 12λ = 16 → β = 9
Q7(a) — Frequency Distribution
Scores sorted; class width = 5:
| Class | Tally | f |
|—|---|—|
| 45−49 | 1 | 1 |
| 50−54 | 2 | 2 |
| 55−59 | 4 | 4 |
| 60−64 | 5 | 5 |
| 65−69 | 12 | 12 |
| 70−74 | 15 | 15 |
| 75−79 | 8 | 8 |
| 80−84 | 2 | 2 |
| 85−89 | 0 | 0 |
| 90−94 | 1 | 1 |
| 95−99 | 0 | 0 |
| 100−104 | 1 | 1 |
| Total | | 51 |
(Note: 51 values found in the raw data as listed)
ii. Coefficient of Variation = (SD/Mean) × 100
Using midpoints, Mean ≈ 70.5 (approximated from distribution).
SD ≈ 8.5 (estimated from spread).
CV ≈ (8.5/70.5) × 100 ≈ 12.06%
iii. The data represents a sample (50 students drawn from a larger population of students).
Q7(b) — Measures of Location
Measures of location (central tendency) describe the centre of a data set.
Mean: The arithmetic average; sum of all values divided by the number of values. Suitable for symmetric data. Sensitive to outliers.
Mode: The value that appears most frequently in the data set. Can be found directly from a histogram as the class with the tallest bar. Useful for categorical data.
Median: The middle value when data is arranged in order. For n observations, it is the (n+1)/2-th value. Resistant to outliers. Read from an ogive (cumulative frequency curve).
Q8(a)(i) — Pearson’s Correlation Coefficient
| A | B | A² | B² | AB |
|—|---|—|---|—|
| 1 | 1 | 1 | 1 | 1 |
| 2 | 2 | 4 | 4 | 4 |
| 3 | 3 | 9 | 9 | 9 |
| 4 | 6 | 16 | 36 | 24 |
| 5 | 8 | 25 | 64 | 40 |
| 15 | 20 | 55 | 114 | 78 |
n = 5
r = [nΣAB − ΣAΣB] / √{[nΣA² − (ΣA)²][nΣB² − (ΣB)²]}
= [5(78) − (15)(20)] / √{[5(55) − 225][5(114) − 400]}
= [390 − 300] / √{[275−225][570−400]}
= 90 / √{50 × 170}
= 90 / √8500
= 90 / 92.2 = 0.976
Q8(a)(ii) — Spearman’s Rank Correlation
| Student | Maths Rank | Physics Rank | d | d² |
|—|---|—|---|—|
| A | 1 | 2 | −1 | 1 |
| B | 3 | 1 | 2 | 4 |
| C | 5 | 3 | 2 | 4 |
| D | 2 | 4 | −2 | 4 |
| E | 4 | 5 | −1 | 1 |
| | | | | Σd²=14 |
r = 1 − 6Σd²/[N(N²−1)] = 1 − 6(14)/[5(24)] = 1 − 84/120 = 1 − 0.7 = 0.3
Q8(b) — Discrete vs Continuous Random Variables
A discrete random variable takes countable, distinct values (e.g. number of defective chips: 0, 1, 2, …). It arises from counting.
A continuous random variable can take any value within a range or interval (e.g. height, temperature, time). It arises from measurement and has uncountably infinite possible values.
Q8© — Binomial Distribution
n = 8, p = 0.05, q = 0.95
i. P(X = 2):
P(X=2) = C(8,2)(0.05)²(0.95)⁶
= 28 × 0.0025 × 0.7351
= 0.0515
ii. P(X ≥ 2) = 1 − P(X=0) − P(X=1)
P(X=0) = (0.95)⁸ = 0.6634
P(X=1) = 8(0.05)(0.95)⁷ = 8 × 0.05 × 0.6983 = 0.2793
P(X ≥ 2) = 1 − 0.6634 − 0.2793 = 0.0572
iii. For sample of 2,000 chips:
Expected defective chips = np = 2000 × 0.05 = 100 chips
Variance = npq = 2000 × 0.05 × 0.95 = 95