### MAT 001: PURE MATHEMATICS
**1. (a)** In a class of 100 students, 35 offer History, 43 offer Geography and 50 offer Economics. 14 students offer History and Geography, 13 offer Geography and Economics and 11 offer History and Economics. The number of students that offer none of the subjects is four times the number of those that offer three subjects.
(i) How many students offer three subjects? [4 Marks]
(ii) How many students offer none? [1 Mark]
(iii) How many students offer only one of the subjects? [2 Marks]
(iv) How many students offer History and Geography only? [1 Mark]
**(b)** From the Venn diagram with universal set U, Circle A and Circle B intersecting:
Inside A only: {j, a, c}; Intersection of A and B: {f, d}; Inside B only: {b, h}; Outside both: {e, g, i, k}.
Find:
(i) A ∩ Bᶜ [1 Mark]
(ii) Aᶜ ∪ B [1 Mark]
**(c)** Solve 2√(x + 5) − √(2x + 8) = 2. [5 Marks]
---
**2. (a)** Prove by mathematical induction that Σₖ₌₁ⁿ k² = n(n+1)(2n+1)/6. [5 Marks]
**(b)** Let ℝ be the set of real numbers. If binary operation ⊗ is defined on ℝ by a ⊗ b = (a + 1)(b + 1) where a, b ∈ ℝ. Solve the equation (2 + 4n) ⊗ (4 − n) = 5n ⊗ n. [5 Marks]
**(c)** An A.P. has a common difference of 2 and a G.P. has a common ratio of 3. The corresponding terms are added together to produce a new sequence. If the 1st term of the sequence is 8 and its 3rd term is 52, find:
(i) the A.P. [2 Marks]
(ii) the G.P. [2 Marks]
(iii) the new sequence. [1 Mark]
---
### MAT 002: CALCULUS
**3. (a)** A vehicle starts from rest and its velocity is measured every second for 8 seconds:
| Time t (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| Velocity v (ms⁻¹) | 0 | 0.4 | 1.0 | 1.7 | 2.9 | 4.1 | 6.2 | 8.0 | 9.4 |
The distance travelled in 8 seconds is given by ∫₀⁸ v dt. Estimate this distance using Simpson's rule. [6 Marks]
**(b)** Evaluate ∫ (x + 1)/(x² − 3x + 2) dx. [6 Marks]
**(c)** Find the differential coefficient of y = 3x² sin 2x. [3 Marks]
---
**4. (a)** Obtain the expansion of eˣ sin x in ascending powers of x as far as the term in x⁵. [8 Marks]
**(b)** Evaluate:
(i) ∫_{π/6}^{π/2} cos x dx [3 Marks]
(ii) ∫₀^{π/4} (sec²x + tan²x) dx [4 Marks]
---
### MAT 003: APPLIED MATHEMATICS
**5. (a)** A block of mass 3.5 kg rests on a rough plane inclined at 42° to the horizontal. If the coefficient of friction between the block and the plane is 0.75, find the least force P parallel to the plane required to:
(i) start moving the block up the plane. [5 Marks]
(ii) prevent the block from sliding down the plane. (Take g = 9.8 ms⁻²) [5 Marks]
**(b)** A body of mass 100 kg is placed in a lift. Find the reaction between the floor of the lift and the body when the lift moves upward with an acceleration of 3.5 ms⁻². (Take g = 9.8 ms⁻²) [5 Marks]
---
**6. (a)** Three points A, B and C are uniquely orientated in space with coordinates (α, −1, 2), (5, 2, 4) and (14, β, 10) respectively. Find the values of α and β such that the three points are collinear. [8 Marks]
**(b)** A car of mass 160 kg is placed on a rough horizontal road with coefficient of friction 0.5. Find the driving force when the car moves with:
(i) an acceleration of 5 ms⁻². [4 Marks]
(ii) a uniform speed of 3 ms⁻¹. [3 Marks]
---
### MAT 004: STATISTICS
**7. (a)** Suppose the probability function of a discrete random variable X is given by:
f(x) = 4!/[(4−x)! x!] × 1/16, for x = 0, 1, 2, 3, 4; and 0 elsewhere.
(i) Find P(X ≤ 3). [3 Marks]
(ii) Obtain the distribution function of X. [1 Mark]
(iii) Find P(X > 0). [2 Marks]
(iv) Find the mean and variance of X. [4 Marks]
**(b)** The following table lists the frequency distribution for 60 rolls of a die:
| Outcome | 1-spot | 2-spot | 3-spot | 4-spot | 5-spot | 6-spot |
|---|---|---|---|---|---|---|
| Frequency | 7 | 12 | 8 | 15 | 11 | 7 |
Test at the 5% significance level whether the null hypothesis that the die is fair is true. [5 Marks]
---
**8. (a)(i)** Distinguish between Type I and Type II errors. [2 Marks]
**(ii)** Past records show that the mean score of Year II Accountancy students in Business Statistics is 56% with a standard deviation of 7%. A random sample of 45 Year II students was tested and the mean score was 52%. Verify at α = 0.05 whether the average score of the students in this course has changed. [9 Marks]
**(b)** An Accountancy student claims he is 90% sure of the postings he made in all transactions. Would you agree with this claim if in a given transaction he makes 20 out of 25 postings correctly? Use α = 0.05. [4 Marks]
---
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# ANSWERS
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## MAT 001: PURE MATHEMATICS
**1(a)**
Let H = History, G = Geography, E = Economics, x = number offering all three.
Given: None = 4x.
By inclusion-exclusion:
100 = 35 + 43 + 50 − (14 + 13 + 11) + x + 4x
100 = 128 − 38 + 5x
100 = 90 + 5x → 5x = 10 → **x = 2**
**(i)** Students offering three subjects = **2**
**(ii)** Students offering none = 4 × 2 = **8**
**(iii)** Only H = 35 − (14 + 11 − 2) − 2 = 35 − 23 = 12
Only G = 43 − (14 + 13 − 2) − 2 = 43 − 25 = 18
Only E = 50 − (13 + 11 − 2) − 2 = 50 − 22 = 28
Total offering only one subject = 12 + 18 + 28 = **58**
**(iv)** History and Geography only = |H ∩ G| − |H ∩ G ∩ E| = 14 − 2 = **12**
---
**1(b)**
**(i)** A ∩ Bᶜ = elements in A but not in B = **{j, a, c}**
**(ii)** Aᶜ ∪ B = elements not in A, or in B = **{b, h, e, g, i, k, f, d}**
---
**1(c)**
2√(x+5) − √(2x+8) = 2
2√(x+5) = 2 + √(2x+8)
Squaring both sides:
4(x+5) = 4 + 4√(2x+8) + (2x+8)
4x + 20 = 2x + 12 + 4√(2x+8)
2x + 8 = 4√(2x+8)
x + 4 = 2√(2x+8)
Squaring again:
x² + 8x + 16 = 4(2x+8)
x² + 8x + 16 = 8x + 32
x² = 16 → x = ±4
Check x = 4: 2√9 − √16 = 6 − 4 = 2 ✓
Check x = −4: 2√1 − √0 = 2 − 0 = 2 ✓
**x = 4 or x = −4**
---
**2(a) — Proof by Mathematical Induction**
**Claim:** Σₖ₌₁ⁿ k² = n(n+1)(2n+1)/6
**Step 1 — Base case (n = 1):**
LHS = 1² = 1
RHS = 1(2)(3)/6 = 1 ✓
**Step 2 — Inductive step:**
Assume true for n = m: Σₖ₌₁ᵐ k² = m(m+1)(2m+1)/6
For n = m+1:
Σₖ₌₁^(m+1) k² = m(m+1)(2m+1)/6 + (m+1)²
= (m+1)[m(2m+1)/6 + (m+1)]
= (m+1)[(2m² + m + 6m + 6)/6]
= (m+1)(2m² + 7m + 6)/6
= (m+1)(m+2)(2m+3)/6
This is exactly the formula with n = m+1: (m+1)(m+2)(2(m+1)+1)/6 ✓
By induction, the formula holds for all n ∈ ℕ. □
---
**2(b)**
a ⊗ b = (a+1)(b+1)
LHS: (2+4n) ⊗ (4−n) = (3+4n)(5−n) = 15 − 3n + 20n − 4n² = 15 + 17n − 4n²
RHS: 5n ⊗ n = (5n+1)(n+1) = 5n² + 5n + n + 1 = 5n² + 6n + 1
Setting LHS = RHS:
15 + 17n − 4n² = 5n² + 6n + 1
9n² − 11n − 14 = 0
n = [11 ± √(121 + 504)]/18 = [11 ± √625]/18 = [11 ± 25]/18
**n = 2 or n = −7/9**
---
**2(c)**
Let AP: a, a+2, a+4, ... and GP: b, 3b, 9b, ...
New sequence term 1: a + b = 8 → a = 8 − b
New sequence term 3: (a+4) + 9b = 52
(8−b+4) + 9b = 52 → 12 + 8b = 52 → b = 5, a = 3
**(i) AP: 3, 5, 7, ...**
**(ii) GP: 5, 15, 45, ...**
**(iii) New sequence: 8, 20, 52, ...**
---
## MAT 002: CALCULUS
**3(a) — Simpson's Rule**
n = 8 (even), h = 1. Velocities:
y₀=0, y₁=0.4, y₂=1.0, y₃=1.7, y₄=2.9, y₅=4.1, y₆=6.2, y₇=8.0, y₈=9.4
Simpson's Rule: Distance = (h/3)[y₀ + y₈ + 4(y₁+y₃+y₅+y₇) + 2(y₂+y₄+y₆)]
= (1/3)[0 + 9.4 + 4(0.4+1.7+4.1+8.0) + 2(1.0+2.9+6.2)]
= (1/3)[9.4 + 4(14.2) + 2(10.1)]
= (1/3)[9.4 + 56.8 + 20.2]
= (1/3)(86.4)
= **28.8 m**
---
**3(b)**
∫(x+1)/[(x−1)(x−2)] dx
Partial fractions: (x+1)/[(x−1)(x−2)] = A/(x−1) + B/(x−2)
x+1 = A(x−2) + B(x−1)
x = 1: 2 = −A → A = −2
x = 2: 3 = B → B = 3
∫[−2/(x−1) + 3/(x−2)] dx = **−2 ln|x−1| + 3 ln|x−2| + C**
---
**3(c)**
y = 3x² sin 2x
By product rule:
dy/dx = 3x²(2 cos 2x) + 6x(sin 2x)
**dy/dx = 6x² cos 2x + 6x sin 2x**
---
**4(a) — Expansion of eˣ sin x**
eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + ...
sin x = x − x³/3! + x⁵/5! − ...
Multiplying term by term up to x⁵:
eˣ sin x = (1 + x + x²/2 + x³/6 + x⁴/24 + x⁵/120 + ...)(x − x³/6 + x⁵/120 − ...)
Collecting by power:
x¹: 1·x = x
x²: 1·0 + x·x = x²
x³: 1·(−x³/6) + x·x² + (x²/2)·x = −x³/6 + x³ + x³/2 = x³(−1/6 + 1 + 1/2) = x³(4/3) → **x³/3**
Wait — careful collection:
x³ terms: (1)(−1/6) + (1)(1) + (1/2)(1) ← coefficients of x³:
From sin x term (−x³/6): coefficient −1/6
From (x)(x²) via eˣ·sin: 1 × 1 = 1 (eˣ gives x, sin x gives x²? No — sin x has no x² term)
Let me redo cleanly:
eˣ = 1 + x + x²/2 + x³/6 + x⁴/24 + x⁵/120
sin x = x − x³/6 + x⁵/120
Coefficient of xⁿ in product = Σ aₖ·bₙ₋ₖ where aₖ from eˣ, bₙ₋ₖ from sin x:
**x¹:** a₀·b₁ = 1·1 = **1**
**x²:** a₁·b₁ = 1·1 = **1**
**x³:** a₀·b₃ + a₂·b₁ = (1)(−1/6) + (1/2)(1) = −1/6 + 3/6 = **1/3**
**x⁴:** a₁·b₃ + a₃·b₁ = (1)(−1/6) + (1/6)(1) = **0**
**x⁵:** a₀·b₅ + a₂·b₃ + a₄·b₁ = (1)(1/120) + (1/2)(−1/6) + (1/24)(1)
= 1/120 − 1/12 + 1/24 = 1/120 − 10/120 + 5/120 = **−4/120 = −1/30**
**eˣ sin x = x + x² + x³/3 − x⁵/30 + ...**
---
**4(b)(i)**
∫_{π/6}^{π/2} cos x dx = [sin x]_{π/6}^{π/2} = sin(π/2) − sin(π/6) = 1 − 1/2 = **1/2**
---
**4(b)(ii)**
∫₀^{π/4} (sec²x + tan²x) dx
Using identity tan²x = sec²x − 1:
sec²x + tan²x = 2sec²x − 1
∫₀^{π/4} (2sec²x − 1) dx = [2 tan x − x]₀^{π/4}
= (2·1 − π/4) − (0 − 0)
= **2 − π/4**
---
## MAT 003: APPLIED MATHEMATICS
**5(a)**
m = 3.5 kg, θ = 42°, μ = 0.75, g = 9.8 ms⁻²
Component of weight along plane = mg sin42° = 3.5 × 9.8 × 0.6691 = **22.95 N**
Normal reaction N = mg cos42° = 3.5 × 9.8 × 0.7431 = **25.49 N**
Maximum friction F = μN = 0.75 × 25.49 = **19.12 N**
**(i) Moving up:** Friction acts down the plane.
P = mg sin42° + F = 22.95 + 19.12 = **42.07 N**
**(ii) Preventing sliding down:** Friction acts up the plane.
P = mg sin42° − F = 22.95 − 19.12 = **3.83 N**
---
**5(b)**
m = 100 kg, a = 3.5 ms⁻² (upward), g = 9.8 ms⁻²
Applying Newton's 2nd law (upward positive):
R − mg = ma
R = m(g + a) = 100(9.8 + 3.5) = 100 × 13.3 = **1330 N**
---
**6(a)**
A(α, −1, 2), B(5, 2, 4), C(14, β, 10)
AB = (5−α, 3, 2); BC = (9, β−2, 6)
For collinearity, BC = k·AB:
z: 6 = 2k → **k = 3**
x: 9 = 3(5−α) → 3 = 5−α → **α = 2**
y: β−2 = 3(3) = 9 → **β = 11**
---
**6(b)**
m = 160 kg, μ = 0.5, g = 9.8 ms⁻²
Friction force = μmg = 0.5 × 160 × 9.8 = **784 N**
**(i) With acceleration a = 5 ms⁻²:**
Net force = ma = 160 × 5 = 800 N
Driving force = Net force + Friction = 800 + 784 = **1584 N**
**(ii) Uniform speed (a = 0):**
Net force = 0 → Driving force = Friction = **784 N**
---
## MAT 004: STATISTICS
**7(a)**
This is Binomial: n = 4, p = 1/2, q = 1/2.
f(x) = C(4,x)(1/2)⁴ = C(4,x)/16
| x | f(x) |
|---|---|
| 0 | 1/16 |
| 1 | 4/16 |
| 2 | 6/16 |
| 3 | 4/16 |
| 4 | 1/16 |
**(i)** P(X ≤ 3) = 1 − P(X=4) = 1 − 1/16 = **15/16**
**(ii) Distribution function F(x):**
F(x) = 0 for x < 0
F(0) = 1/16
F(1) = 5/16
F(2) = 11/16
F(3) = 15/16
F(4) = 1
**(iii)** P(X > 0) = 1 − P(X=0) = 1 − 1/16 = **15/16**
**(iv)** Mean = np = 4 × 1/2 = **2**
Variance = npq = 4 × 1/2 × 1/2 = **1**
---
**7(b) — Chi-Square Test**
H₀: The die is fair (each outcome has probability 1/6)
H₁: The die is not fair
Expected frequency for each outcome = 60/6 = 10
| Outcome | O | E | (O−E)² | (O−E)²/E |
|---|---|---|---|---|
| 1 | 7 | 10 | 9 | 0.9 |
| 2 | 12 | 10 | 4 | 0.4 |
| 3 | 8 | 10 | 4 | 0.4 |
| 4 | 15 | 10 | 25 | 2.5 |
| 5 | 11 | 10 | 1 | 0.1 |
| 6 | 7 | 10 | 9 | 0.9 |
| | | | **χ²** | **5.2** |
Degrees of freedom = 6 − 1 = 5
Critical value at α = 0.05, df = 5: χ²_crit = **11.07**
Since 5.2 < 11.07, we **fail to reject H₀**. The die is fair.
---
**8(a)(i) — Type I and Type II Errors**
**Type I Error:** Rejecting a true null hypothesis (false positive). Its probability is α (the significance level).
**Type II Error:** Failing to reject a false null hypothesis (false negative). Its probability is β.
---
**8(a)(ii) — Z-Test for Mean**
H₀: μ = 56%; H₁: μ ≠ 56% (two-tailed)
Given: μ₀ = 56, σ = 7, n = 45, x̄ = 52, α = 0.05
Z = (x̄ − μ₀)/(σ/√n) = (52 − 56)/(7/√45) = −4/1.0435 = **−3.83**
Critical value: Z_{0.025} = ±1.96
Since |−3.83| > 1.96, we **reject H₀**.
**Conclusion:** The average score of the students has changed significantly at the 5% level.
---
**8(b) — Z-Test for Proportion**
H₀: p = 0.90; H₁: p ≠ 0.90 (two-tailed)
Given: p₀ = 0.90, x = 20, n = 25, p̂ = 20/25 = 0.80, α = 0.05
Z = (p̂ − p₀)/√[p₀(1−p₀)/n] = (0.80 − 0.90)/√[(0.90 × 0.10)/25]
= −0.10/√(0.0036) = −0.10/0.06 = **−1.667**
Critical value: Z_{0.025} = ±1.96
Since |−1.667| < 1.96, we **fail to reject H₀**.
**Conclusion:** There is insufficient evidence at α = 0.05 to dispute the student's claim. We agree that he is 90% accurate.