# MAT 001: ADVANCED PURE MATHEMATICS
## Question 1
### (a)(i) Sum of three consecutive terms of a GP = 14, product = 64
Let the three terms be **a/r, a, ar**
**Product:**
(a/r)(a)(ar) = a³ = 64
∴ **a = 4**
**Sum:**
a/r + a + ar = 14
4/r + 4 + 4r = 14
4/r + 4r = 10
Multiply through by r:
4 + 4r² = 10r
4r² − 10r + 4 = 0
2r² − 5r + 2 = 0
(2r − 1)(r − 2) = 0
∴ r = 1/2 or r = 2
**When r = 2:** terms are 4/2, 4, 4(2) = **2, 4, 8**
**When r = 1/2:** terms are 4/(1/2), 4, 4(1/2) = **8, 4, 2**
∴ The three terms are **2, 4, 8**
---
### (a)(ii) Solve: 4ˣ + 2(2ˣ) = 8
Let u = 2ˣ, so 4ˣ = (2²)ˣ = u²
u² + 2u = 8
u² + 2u − 8 = 0
(u + 4)(u − 2) = 0
u = −4 or u = 2
Since 2ˣ > 0, u = 2
2ˣ = 2¹
∴ **x = 1**
---
### (b) Identify conic: 4x² − 25y² + 16x + 50y − 109 = 0
**Complete the square:**
4(x² + 4x) − 25(y² − 2y) = 109
4(x² + 4x + 4) − 25(y² − 2y + 1) = 109 + 16 − 25
4(x + 2)² − 25(y − 1)² = 100
Divide by 100:
**(x + 2)²/25 − (y − 1)²/4 = 1**
This is a **HYPERBOLA** (difference of squares form)
**Centre:** (−2, 1)
Here a² = 25, b² = 4 → a = 5, b = 2
**c² = a² + b² = 25 + 4 = 29 → c = √29**
**Vertices:** (−2 ± 5, 1) = **(3, 1) and (−7, 1)**
**Foci:** (−2 ± √29, 1)
= **(−2 + √29, 1) and (−2 − √29, 1)**
**Asymptotes:**
y − 1 = ±(b/a)(x + 2)
y − 1 = ±(2/5)(x + 2)
**y = 1 + (2/5)(x + 2) and y = 1 − (2/5)(x + 2)**
---
### (c) Express f(x) = 3x³ + 4x² + 2x + 1 as product of three linear factors
**Try rational roots** (factors of 1 over factors of 3): ±1, ±1/3
Test x = −1:
3(−1) + 4(1) + 2(−1) + 1 = −3 + 4 − 2 + 1 = 0 ✓
So (x + 1) is a factor. Divide:
3x³ + 4x² + 2x + 1 ÷ (x + 1):
```
3x³ + 4x² + 2x + 1 = (x+1)(3x² + x + 1)
```
Check discriminant of 3x² + x + 1:
Δ = 1 − 12 = −11 < 0 (no real roots)
∴ Over the reals:
**f(x) = (x + 1)(3x² + x + 1)**
Over complex numbers, factoring 3x² + x + 1:
x = (−1 ± i√11)/6
∴ **f(x) = (x+1)(x − (−1+i√11)/6)(x − (−1−i√11)/6)**
---
## Question 2
### (a) Partial fractions: (2x² + 10x + 4) / [(x+2)(2x² + x − 2)]
First factorise 2x² + x − 2:
Δ = 1 + 16 = 17
x = (−1 ± √17)/4 → **does not factor nicely**, so keep as irreducible quadratic.
Set up:
(2x² + 10x + 4) / [(x+2)(2x² + x − 2)] = **A/(x+2) + (Bx + C)/(2x² + x − 2)**
Multiply both sides by (x+2)(2x² + x − 2):
2x² + 10x + 4 = A(2x² + x − 2) + (Bx + C)(x + 2)
**x = −2:**
2(4) + 10(−2) + 4 = A(2(4) + (−2) − 2)
8 − 20 + 4 = A(8 − 2 − 2)
−8 = 4A
**A = −2**
**Expand right side:**
A(2x² + x − 2) + (Bx + C)(x + 2)
= 2Ax² + Ax − 2A + Bx² + 2Bx + Cx + 2C
= (2A + B)x² + (A + 2B + C)x + (−2A + 2C)
**Compare coefficients:**
x²: 2 = 2A + B = 2(−2) + B → B = 2 + 4 = **6**
constant: 4 = −2A + 2C = −2(−2) + 2C = 4 + 2C → 2C = 0 → **C = 0**
Check x: 10 = A + 2B + C = −2 + 12 + 0 = 10 ✓
∴ **(2x² + 10x + 4)/[(x+2)(2x²+x−2)] = −2/(x+2) + 6x/(2x²+x−2)**
---
### (b) cos 60° = 2tan15°/(1 + tan²15°), find tan15°
We know the identity: sin 2θ = 2tanθ/(1 + tan²θ)
So 2tan15°/(1 + tan²15°) = sin 30° = 1/2
But the question states this equals cos 60° = 1/2 ✓ (consistent)
So: 2tan15°/(1 + tan²15°) = 1/2
Let t = tan15°:
4t = 1 + t²
t² − 4t + 1 = 0
t = (4 ± √(16−4))/2 = (4 ± √12)/2 = 2 ± √3
Since 15° is acute and tan15° < tan45° = 1:
**tan15° = 2 − √3**
---
### (c) μ = {1,2,3,...,9}, A = {primes}, B = {odd numbers}, Find AΔB
**A (primes in μ):** {2, 3, 5, 7}
**B (odd numbers in μ):** {1, 3, 5, 7, 9}
**AΔB = (A∪B) − (A∩B)** = elements in A or B but NOT both
A∩B = {3, 5, 7}
A∪B = {1, 2, 3, 5, 7, 9}
**AΔB = {1, 2, 9}**
---
# CALCULUS
---
## Question 3
### (a) Differentiate y = −cos7x from first principles
f(x) = −cos7x
f(x+h) = −cos7(x+h) = −cos(7x+7h)
f'(x) = lim[h→0] [f(x+h) − f(x)]/h
= lim[h→0] [−cos(7x+7h) + cos7x]/h
= lim[h→0] [cos7x − cos(7x+7h)]/h
Using: cos A − cos B = −2 sin((A+B)/2) sin((A−B)/2)
= lim[h→0] [−2 sin(7x + 7h/2) · sin(−7h/2)] / h
= lim[h→0] [2 sin(7x + 7h/2) · sin(7h/2)] / h
= lim[h→0] sin(7x + 7h/2) · [sin(7h/2)/(h/2)] · (1/1) · (7/7)
Wait, let me redo cleanly:
= lim[h→0] [2 sin(7x + 7h/2) · sin(7h/2)] / h
Multiply and divide by 7/2:
= lim[h→0] sin(7x + 7h/2) · [sin(7h/2)/(7h/2)] · 7
As h→0: sin(7x + 7h/2) → sin7x, and sin(7h/2)/(7h/2) → 1
∴ **f'(x) = 7sin7x**
---
### (b) Solve: xy' − 2y = x² at x = 1/3...
*(Note: Reading the image, this appears to be a first-order linear ODE: xy' − 2y = x², with initial condition at x = 1/3)*
Rewrite: dy/dx − (2/x)y = x
**Integrating factor:** μ = e^∫(−2/x)dx = e^(−2ln|x|) = x^(−2) = 1/x²
Multiply through:
(1/x²)dy/dx − (2/x³)y = 1/x
d/dx[y/x²] = 1/x
Integrate both sides:
y/x² = ln|x| + C
**y = x²(ln|x| + C)**
*(Apply initial condition if the value of y at x = 1/3 is given — partial data from image)*
---
### (c) Taylor series of f(x) = 64ˣ about x = 1
*(Note: This appears to be f(x) = 6·4ˣ or possibly f(x) = (64)ˣ. Taking f(x) = 64ˣ)*
f(x) = 64ˣ = (2⁶)ˣ = 2^(6x)
Let a = 1
f(1) = 64
f'(x) = 64ˣ · ln64, f'(1) = 64ln64
f''(x) = 64ˣ(ln64)², f''(1) = 64(ln64)²
f'''(1) = 64(ln64)³
Taylor series about x = 1:
**f(x) = 64 + 64ln64·(x−1) + [64(ln64)²/2!](x−1)² + [64(ln64)³/3!](x−1)³ + ...**
**f(x) = 64 · Σ [(ln64)ⁿ/n!](x−1)ⁿ**, n = 0,1,2,...
---
## Question 4
### (a) General solution of: d²y/dx² − 4(dy/dx) − 12y = 3e^(5t)
*(Note: mixed variables — treating as dy²/dx² − 4dy/dx − 12y = 3e^(5x))*
**Auxiliary equation:**
m² − 4m − 12 = 0
(m − 6)(m + 2) = 0
**m = 6 or m = −2**
**Complementary function:**
**y_c = Ae^(6x) + Be^(−2x)**
**Particular integral** — try y_p = Ce^(5x):
y_p' = 5Ce^(5x), y_p'' = 25Ce^(5x)
Substitute:
25Ce^(5x) − 4(5Ce^(5x)) − 12Ce^(5x) = 3e^(5x)
C(25 − 20 − 12)e^(5x) = 3e^(5x)
−7C = 3
**C = −3/7**
**y_p = −(3/7)e^(5x)**
**General solution:**
**y = Ae^(6x) + Be^(−2x) − (3/7)e^(5x)**
---
### (b) Trapezoidal rule for ∫₀^π sin x dx, n = 5
**h = (π − 0)/5 = π/5**
| x | sin x |
|---|-------|
| x₀ = 0 | 0 |
| x₁ = π/5 | sin36° = 0.5878 |
| x₂ = 2π/5 | sin72° = 0.9511 |
| x₃ = 3π/5 | sin108° = 0.9511 |
| x₄ = 4π/5 | sin144° = 0.5878 |
| x₅ = π | 0 |
**Trapezoidal rule:**
∫ ≈ (h/2)[y₀ + 2(y₁+y₂+y₃+y₄) + y₅]
= (π/10)[0 + 2(0.5878 + 0.9511 + 0.9511 + 0.5878) + 0]
= (π/10)[2 × 3.0778]
= (π/10)(6.1556)
= **1.9332**
*(Exact value = 2; trapezoidal approximation ≈ 1.9332)*
---
# STATISTICS (Question 6)
---
### (a) Poisson distribution, λ = 1.2, 5 rooms
P(X = k) = e^(−λ)·λᵏ/k!
**(i) All rooms vacant → X = 0:**
P(X=0) = e^(−1.2) = **0.3012**
**(ii) Some clients rejected → X > 5 (demand exceeds 5 rooms):**
P(X > 5) = 1 − P(X ≤ 5)
P(X=0) = 0.3012
P(X=1) = e^(−1.2)(1.2) = 0.3614
P(X=2) = e^(−1.2)(1.44/2) = 0.2169
P(X=3) = e^(−1.2)(1.728/6) = 0.0867
P(X=4) = e^(−1.2)(2.0736/24) = 0.0260
P(X=5) = e^(−1.2)(2.48832/120) = 0.0062
P(X≤5) = 0.3012+0.3614+0.2169+0.0867+0.0260+0.0062 = **0.9985**
**P(X>5) = 1 − 0.9985 = 0.0015**
---
### (b) Density function:
f(x) = k(3+2x), 2 ≤ x ≤ 4; 0 otherwise
**(i) Find k:**
∫₂⁴ k(3+2x)dx = 1
k[3x + x²]₂⁴ = 1
k[(12+16) − (6+4)] = 1
k[28 − 10] = 1
18k = 1
**k = 1/18**
**(ii) P(2 ≤ X ≤ 3):**
= ∫₂³ (1/18)(3+2x)dx
= (1/18)[3x + x²]₂³
= (1/18)[(9+9) − (6+4)]
= (1/18)[18 − 10]
= 8/18
= **4/9 ≈ 0.4444**
---
### (c)(i) Binomial: n=6, find p if 9P(X=4) = P(X=2)
P(X=k) = C(6,k)·pᵏ·(1−p)^(6−k)
9·C(6,4)·p⁴(1−p)² = C(6,2)·p²(1−p)⁴
9·15·p⁴(1−p)² = 15·p²(1−p)⁴
9p⁴(1−p)² = p²(1−p)⁴
Divide both sides by p²(1−p)²:
9p² = (1−p)²
3p = 1−p *(taking positive root since 0<p<1)*
4p = 1
**p = 1/4**
---
### (c)(ii) 40% defective, n=3, P(at most 2 defective)
p = 0.4, q = 0.6, n = 3
P(X≤2) = 1 − P(X=3)
= 1 − C(3,3)(0.4)³(0.6)⁰
= 1 − (0.064)
= **0.936**
---
# APPLIED MATHEMATICS (MAT 004)
---
P̄ = −4i + 2j − 3k, Q̄ = 4i − j − k, R̄ = 5i + k
### PQ vector:
PQ = Q − P = (4−(−4))i + (−1−2)j + (−1−(−3))k
= **8i − 3j + 2k**
### QR vector:
QR = R − Q = (5−4)i + (0−(−1))j + (1−(−1))k
= **i + j + 2k**
### Length of projection of PQ onto QR:
proj = (PQ · QR)/|QR|
PQ · QR = 8(1) + (−3)(1) + 2(2) = 8 − 3 + 4 = **9**
|QR| = √(1+1+4) = √6
**Projection = 9/√6 = 9√6/6 = 3√6/2 ≈ 3.674**
---
### Area of triangle:
PQ × QR = |i j k |
|8 −3 2 |
|1 1 2 |
= i[(−3)(2)−(2)(1)] − j[(8)(2)−(2)(1)] + k[(8)(1)−(−3)(1)]
= i[−6−2] − j[16−2] + k[8+3]
= **−8i − 14j + 11k**
|PQ × QR| = √(64 + 196 + 121) = √381
**Area = (1/2)√381 ≈ (1/2)(19.52) ≈ 9.76 square units**
# SOLUTIONS
# MAT 001: ADVANCED PURE MATHEMATICS
## Question 1
### (a) Prove by induction: 5ⁿ + 2(4ⁿ⁺¹) − 1 is divisible by 4
Let P(n): 5ⁿ + 2(4ⁿ⁺¹) − 1 = 4M for some integer M
**Step 1 — Base case n = 1:**
5¹ + 2(4²) − 1 = 5 + 32 − 1 = 36 = 4(9) ✓
**Step 2 — Assume true for n = k:**
5ᵏ + 2(4ᵏ⁺¹) − 1 = 4M ... (*)
**Step 3 — Prove true for n = k+1:**
Show: 5ᵏ⁺¹ + 2(4ᵏ⁺²) − 1 is divisible by 4
5ᵏ⁺¹ + 2(4ᵏ⁺²) − 1
= 5·5ᵏ + 2·4·4ᵏ⁺¹ − 1
= 5·5ᵏ + 8·4ᵏ⁺¹ − 1
From (*): 5ᵏ = 4M − 2(4ᵏ⁺¹) + 1
Substitute:
= 5[4M − 2(4ᵏ⁺¹) + 1] + 8(4ᵏ⁺¹) − 1
= 20M − 10(4ᵏ⁺¹) + 5 + 8(4ᵏ⁺¹) − 1
= 20M − 2(4ᵏ⁺¹) + 4
= 20M − 2(4ᵏ⁺¹) + 4
Factor out 4:
= 4[5M − 2(4ᵏ⁻¹)·2 ... ]
Let me redo cleanly:
= 20M + 4 − 2(4ᵏ⁺¹)
= 4(5M + 1) − 2(4ᵏ⁺¹)
= 4(5M + 1) − 2·4·4ᵏ
= 4(5M + 1) − 8·4ᵏ
= 4(5M + 1 − 2·4ᵏ)
Since M and k are integers, this is divisible by 4. ✓
**By the principle of mathematical induction, P(n) is true for all positive integers n.**
---
### (b) (x+5) and (2x−1) are factors of ax³ + 31x² + bx − 10
Since (x+5) is a factor: f(−5) = 0
a(−125) + 31(25) + b(−5) − 10 = 0
−125a + 775 − 5b − 10 = 0
−125a − 5b = −765
**25a + b = 153 ... (1)**
Since (2x−1) is a factor: f(1/2) = 0
a(1/8) + 31(1/4) + b(1/2) − 10 = 0
Multiply through by 8:
a + 62 + 4b − 80 = 0
a + 4b = 18 ... (2)
From (2): a = 18 − 4b
Substitute in (1):
25(18 − 4b) + b = 153
450 − 100b + b = 153
−99b = −297
**b = 3**
a = 18 − 4(3) = 18 − 12 = **a = 6**
So f(x) = 6x³ + 31x² + 3x − 10
**Find remaining factor** by dividing by (x+5)(2x−1) = 2x²+9x−5:
6x³ + 31x² + 3x − 10 ÷ (2x²+9x−5):
6x³ + 31x² + 3x − 10 = (2x²+9x−5)(3x+2)
Check: (2x²+9x−5)(3x+2)
= 6x³+4x²+27x²+18x−15x−10
= 6x³+31x²+3x−10 ✓
**Remaining factor: (3x + 2)**
Full factorisation: **(x+5)(2x−1)(3x+2)**
---
### (c) Solve (4ˣ)(3ˣ) = 12 using logarithms
Take log of both sides:
log(4ˣ · 3ˣ) = log12
x·log4 + x·log3 = log12
x(log4 + log3) = log12
x·log12 = log12
**x = 1**
---
## Question 2
### (a) log₁₀(2x² + 3x + 8) < 1
log₁₀(2x² + 3x + 8) < log₁₀(10)
2x² + 3x + 8 < 10 *(argument must be positive)*
2x² + 3x − 2 < 0
(2x − 1)(x + 2) < 0
Critical values: x = 1/2 and x = −2
Sign analysis:
- x < −2: positive
- −2 < x < 1/2: negative ✓
- x > 1/2: positive
Also check argument > 0: 2x²+3x+8 > 0
Discriminant = 9 − 64 = −55 < 0, always positive ✓
∴ **−2 < x < 1/2**
---
### (b) Conditions for general conic ax²+bxy+cy²+dx+fy+g = 0
Let Δ = b²− 4ac
**(i) Straight line:** a = b = c = 0 (no second-degree terms); degenerate case
**(ii) Circle:** b = 0 and a = c (equal coefficients, no xy term)
**(iii) Parabola:** b² − 4ac = 0
**(iv) Hyperbola:** b² − 4ac > 0
**(v) Ellipse:** b² − 4ac < 0 (and a ≠ c)
---
### (c) Show arg(Z₁Z₂) = arg(Z₁) + arg(Z₂)
Let Z₁ = r₁(cosθ₁ + i sinθ₁) where arg(Z₁) = θ₁
Let Z₂ = r₂(cosθ₂ + i sinθ₂) where arg(Z₂) = θ₂
**Multiply:**
Z₁Z₂ = r₁r₂(cosθ₁ + i sinθ₁)(cosθ₂ + i sinθ₂)
= r₁r₂[(cosθ₁cosθ₂ − sinθ₁sinθ₂) + i(sinθ₁cosθ₂ + cosθ₁sinθ₂)]
= r₁r₂[cos(θ₁+θ₂) + i sin(θ₁+θ₂)]
*(using compound angle formulas)*
The modulus of Z₁Z₂ is r₁r₂, and the argument is (θ₁ + θ₂)
∴ **arg(Z₁Z₂) = θ₁ + θ₂ = arg(Z₁) + arg(Z₂)** ∎
---
# MAT 002: CALCULUS
---
## Question 3
### (a)(i) Evaluate ∫ 16x/(6+4x)³ dx
Let u = 6 + 4x → x = (u−6)/4, dx = du/4
∫ 16·(u−6)/4 / u³ · du/4
= ∫ (u−6)/u³ du
= ∫ (u⁻² − 6u⁻³) du
= −u⁻¹ + 3u⁻² + C
= **−1/(6+4x) + 3/(6+4x)² + C**
---
### (a)(ii) Given f′(x) = 1 + 1/x², f(1) = 0, find f(x)
f(x) = ∫(1 + x⁻²)dx = x − x⁻¹ + C = x − 1/x + C
Apply f(1) = 0:
1 − 1 + C = 0 → **C = 0**
∴ **f(x) = x − 1/x**
---
### (b) ∫₁² (x+1)/(x²+4) dx by Trapezoidal rule, n = 5
h = (2−1)/5 = 0.2
| x | f(x) = (x+1)/(x²+4) |
|---|----------------------|
| 1.0 | 2/5 = 0.4000 |
| 1.2 | 2.2/5.44 = 0.4044 |
| 1.4 | 2.4/5.96 = 0.4027 |
| 1.6 | 2.6/6.56 = 0.3963 |
| 1.8 | 2.8/7.24 = 0.3867 |
| 2.0 | 3/8 = 0.3750 |
Trapezoidal rule:
∫ ≈ (h/2)[f(x₀) + 2(f(x₁)+f(x₂)+f(x₃)+f(x₄)) + f(x₅)]
= (0.2/2)[0.4000 + 2(0.4044+0.4027+0.3963+0.3867) + 0.3750]
= 0.1[0.4000 + 2(1.5901) + 0.3750]
= 0.1[0.4000 + 3.1802 + 0.3750]
= 0.1 × 3.9552
= **≈ 0.3955**
---
### (c) Turning points of y = x³ − 2x² + x + 4
dy/dx = 3x² − 4x + 1 = 0
(3x − 1)(x − 1) = 0
**x = 1/3 or x = 1**
d²y/dx² = 6x − 4
**At x = 1/3:** d²y/dx² = 2 − 4 = −2 < 0 → **LOCAL MAXIMUM**
y = (1/27) − (2/9) + (1/3) + 4 = 1/27 − 6/27 + 9/27 + 108/27 = **112/27 ≈ 4.148**
**At x = 1:** d²y/dx² = 6 − 4 = 2 > 0 → **LOCAL MINIMUM**
y = 1 − 2 + 1 + 4 = **4**
**Maximum value = 112/27 at x = 1/3**
---
## Question 4
### (a) s = t³ − 2t² + t
**(i) Speed (velocity):**
v = ds/dt = **3t² − 4t + 1**
**(ii) Body at rest → v = 0:**
3t² − 4t + 1 = 0
(3t − 1)(t − 1) = 0
**t = 1/3 s or t = 1 s**
**(iii) Acceleration a = dv/dt = 6t − 4**
At t = 1/3: a = 2 − 4 = **−2 ms⁻²**
At t = 1: a = 6 − 4 = **+2 ms⁻²**
---
### (b)(i) lim(x→0) (4x + sin6x)/(x + sin4x)
Divide numerator and denominator by x:
= lim(x→0) [4 + (sin6x)/x] / [1 + (sin4x)/x]
= lim(x→0) [4 + 6·(sin6x)/6x] / [1 + 4·(sin4x)/4x]
= [4 + 6(1)] / [1 + 4(1)]
= 10/5
= **2**
---
### (b)(ii) Show if y = tan r, then d²y/dr² = 2(dy/dr)
y = tan r
dy/dr = sec²r
d²y/dr² = 2 sec r · sec r tan r = 2 sec²r · tan r
Now: 2(dy/dr) = 2sec²r
Hmm — let me re-read. The equation is d²y/dr² = 2(dy/dr).
Actually: dy/dr = sec²r = 1 + tan²r = 1 + y²
d²y/dr² = d/dr(sec²r) = 2secr · secr·tanr = 2sec²r·tanr
2·dy/dr = 2sec²r
These are equal only if tanr = 1, so this holds specifically.
**Alternative reading** — perhaps y = tanr means:
dy/dr = sec²r
d²y/dr² = 2sec²r·tanr = 2·sec²r·y = 2y·(dy/dr)
So the correct identity is:
**d²y/dr² = 2y · dy/dr**
*This is likely the intended statement. Proof:*
y = tan r → dy/dr = sec²r = 1 + tan²r = 1 + y²
d²y/dr² = d/dr(1 + y²) = 2y·(dy/dr) ∎
---
### (c) IVP: d²y/dt² − 4(dy/dt) + 3y = 0, y(1) = 1, y′(1) = 0
**Auxiliary equation:**
m² − 4m + 3 = 0
(m−1)(m−3) = 0
m = 1 or m = 3
**General solution:**
y = Ae^t + Be^(3t)
dy/dt = Ae^t + 3Be^(3t)
**Apply y(1) = 1:**
Ae + Be³ = 1 ... (1)
**Apply y′(1) = 0:**
Ae + 3Be³ = 0 ... (2)
(2) − (1): 2Be³ = −1 → **B = −e⁻³/2**
From (1): Ae = 1 − Be³ = 1 + 1/2 = 3/2 → **A = 3/(2e)**
∴ **y = (3/2e)eᵗ − (e⁻³/2)e^(3t)**
= **(3/2)e^(t−1) − (1/2)e^(3t−3)**
= **(1/2)[3e^(t−1) − e^(3(t−1))]**
---
# MAT 003: STATISTICS
---
## Question 5
### (a) Regression: Y = β₁X + β₀
Data:
| X | 50 | 65 | 70 | 38 | 35 | 52 | 80 |
|---|----|----|----|----|----|----|-----|
| Y | 20 | 27 | 13 | 17 | 9 | 15 | 16 |
**Compute sums (n = 7):**
| X | Y | XY | X² |
|---|---|----|----|
| 50 | 20 | 1000 | 2500 |
| 65 | 27 | 1755 | 4225 |
| 70 | 13 | 910 | 4900 |
| 38 | 17 | 646 | 1444 |
| 35 | 9 | 315 | 1225 |
| 52 | 15 | 780 | 2704 |
| 80 | 16 | 1280 | 6400 |
| **ΣX=390** | **ΣY=117** | **ΣXY=6686** | **ΣX²=23398** |
X̄ = 390/7 = 55.71
Ȳ = 117/7 = 16.71
**β₁ = (ΣXY − nX̄Ȳ)/(ΣX² − nX̄²)**
= (6686 − 7×55.71×16.71)/(23398 − 7×55.71²)
= (6686 − 6512.5)/(23398 − 21724.6)
= 173.5/1673.4
= **β₁ ≈ 0.1037**
**β₀ = Ȳ − β₁X̄**
= 16.71 − 0.1037×55.71
= 16.71 − 5.78
= **β₀ ≈ 10.93**
**Regression equation: Y = 0.1037X + 10.93**
**(ii) When X = 10:**
Y = 0.1037(10) + 10.93 = 1.037 + 10.93 = **≈ 11.97**
---
### (b) Hypothesis test: μ = 10.5, n=15, x̄=10.9, s=0.6, α=1%
**H₀:** μ = 10.5
**H₁:** μ ≠ 10.5 (two-tailed)
**Test statistic (t-test, unknown σ):**
t = (x̄ − μ)/(s/√n)
= (10.9 − 10.5)/(0.6/√15)
= 0.4/(0.6/3.873)
= 0.4/0.1549
= **t = 2.582**
**Critical value:** t(α/2, n−1) = t(0.005, 14) = **2.977**
**Decision:** |t_calc| = 2.582 < 2.977 = t_critical
**Conclusion: Fail to reject H₀.** At 1% significance level, there is insufficient evidence to conclude that the mean age is different from 10.5 days.
---
### (c) Poisson with standard deviation = 2
For Poisson: Var(X) = λ, so SD = √λ = 2 → **λ = 4**
**(i) E(X) = λ = 4**
**(ii) P(X < 3) = P(X=0) + P(X=1) + P(X=2)**
P(X=k) = e⁻⁴·4ᵏ/k!
P(X=0) = e⁻⁴ = 0.01832
P(X=1) = e⁻⁴·4 = 0.07326
P(X=2) = e⁻⁴·16/2 = 0.14653
**P(X < 3) = e⁻⁴(1 + 4 + 8) = 13e⁻⁴ ≈ 0.2381**
---
## Question 6
### (a) Chi-square test for independence (gender vs education)
**Observed frequencies:**
| | Formal | Informal | Total |
|-------|--------|----------|-------|
| Men | 700 | 390 | 1090 |
| Women | 420 | 490 | 910 |
| **Total** | **1120** | **880** | **2000** |
**Expected frequencies:** E = (Row total × Col total)/Grand total
E(Men, Formal) = (1090×1120)/2000 = **610.4**
E(Men, Informal) = (1090×880)/2000 = **479.6**
E(Women, Formal) = (910×1120)/2000 = **509.6**
E(Women, Informal) = (910×880)/2000 = **400.4**
**χ² = Σ(O−E)²/E:**
= (700−610.4)²/610.4 + (390−479.6)²/479.6 + (420−509.6)²/509.6 + (490−400.4)²/400.4
= (89.6)²/610.4 + (−89.6)²/479.6 + (−89.6)²/509.6 + (89.6)²/400.4
= 8028.16/610.4 + 8028.16/479.6 + 8028.16/509.6 + 8028.16/400.4
= 13.15 + 16.74 + 15.75 + 20.05
= **χ² = 65.69**
**Critical value:** χ²(0.05, 1) = 3.841
**Since 65.69 > 3.841, reject H₀.**
**Conclusion: At 5%, there is significant evidence that men have more formal education than women.**
---
### (b) p(X=x) = rx², x=1,2; r(5−x)², x=3,4
**(i) Find r:**
Sum of all probabilities = 1:
r(1)² + r(2)² + r(5−3)² + r(5−4)² = 1
r(1 + 4 + 4 + 1) = 1
10r = 1
**r = 1/10**
**(ii) E(X) and Var(X):**
| x | p(x) | xp(x) | x²p(x) |
|---|------|-------|--------|
| 1 | 1/10 | 1/10 | 1/10 |
| 2 | 4/10 | 8/10 | 16/10 |
| 3 | 4/10 | 12/10 | 36/10 |
| 4 | 1/10 | 4/10 | 16/10 |
E(X) = (1+8+12+4)/10 = 25/10 = **2.5**
E(X²) = (1+16+36+16)/10 = 69/10 = 6.9
Var(X) = E(X²) − [E(X)]² = 6.9 − 6.25 = **0.65**
---
### (c) Tyre dealer z-test: n=100, x̄=15267, μ=15200, σ=1248, α=5%
**H₀:** μ = 15200
**H₁:** μ ≠ 15200
**Test statistic:**
z = (x̄ − μ)/(σ/√n)
= (15267 − 15200)/(1248/√100)
= 67/124.8
= **z = 0.537**
**Critical value:** z(0.025) = ±1.96
**Since |0.537| < 1.96, fail to reject H₀.**
**Conclusion: At 5%, the claim is valid. The sample is consistent with the population mean of 15200 kms.**
---
# MAT 004: APPLIED MATHEMATICS
---
## Question 7
### (a) v(t) = cos t î + (3−t²) ĵ + 2e^(−3t) k̂
**Displacement** = ∫₀² v(t) dt
∫₀² cos t dt = [sin t]₀² = sin2 − 0 = 0.9093
∫₀² (3−t²)dt = [3t − t³/3]₀² = 6 − 8/3 = 10/3 = 3.3333
∫₀² 2e^(−3t)dt = [−(2/3)e^(−3t)]₀² = −(2/3)e⁻⁶ + 2/3 = (2/3)(1 − e⁻⁶) = 0.6616
**s(2) = 0.9093î + 3.3333ĵ + 0.6616k̂**
**|s| = √(0.9093² + 3.3333² + 0.6616²)**
= √(0.827 + 11.111 + 0.438)
= √12.376
= **≈ 3.518 m**
**Acceleration** a(t) = dv/dt:
aₓ = −sin t
aᵧ = −2t
a_z = −6e^(−3t)
At t = 2:
aₓ = −sin2 = −0.9093
aᵧ = −4
a_z = −6e⁻⁶ = −0.01487
**a(2) = −0.9093î − 4ĵ − 0.01487k̂**
**|a| = √(0.9093² + 4² + 0.01487²)**
= √(0.827 + 16 + 0.000221)
= √16.827
= **≈ 4.102 ms⁻²**
---
### (b) Mass 12.5 kg on smooth inclined plane, θ = 42°, g = 9.8 ms⁻²
**(i) Acceleration down the plane:**
a = g sin θ = 9.8 × sin42° = 9.8 × 0.6691
= **6.557 ms⁻²**
**(ii) Velocity after 10s, u = 5 ms⁻¹:**
v = u + at = 5 + 6.557×10 = 5 + 65.57
= **70.57 ms⁻¹**
---
### (c) Projectile: v = 40 ms⁻¹, θ = 55°, g = 9.8 ms⁻²
**(i) Time of flight:**
T = 2u sinθ/g
= 2×40×sin55°/9.8
= 2×40×0.8192/9.8
= 65.536/9.8
= **≈ 6.69 s**
**(ii) Range:**
R = u²sin2θ/g
= 40²×sin110°/9.8
= 1600×0.9397/9.8
= 1503.5/9.8
= **≈ 153.4 m**
---
## Question 8
### (a) Collision: Toyota Camry (500kg, 30ms⁻¹) hits Honda Accord (400kg, 50ms⁻¹) head-on
*(Head-on: take Toyota's direction as positive, Honda is negative)*
u_T = +30, u_H = −50, m_T = 500, m_H = 400
v_T = +40 ms⁻¹ (given)
**By conservation of momentum:**
m_T·u_T + m_H·u_H = m_T·v_T + m_H·v_H
500(30) + 400(−50) = 500(40) + 400·v_H
15000 − 20000 = 20000 + 400v_H
−5000 − 20000 = 400v_H
−25000 = 400v_H
**v_H = −62.5 ms⁻¹**
*(Honda continues in its original direction at 62.5 ms⁻¹)*
**(ii) Loss in KE of Honda Accord:**
KE_initial = ½ × 400 × 50² = 200 × 2500 = 500,000 J
KE_final = ½ × 400 × 62.5² = 200 × 3906.25 = 781,250 J
*(KE increased due to head-on — Honda gained energy from impact)*
**Change in KE of Honda = 781,250 − 500,000 = 281,250 J**
*(The Accord gained 281,250 J — the "loss" refers to the system's total KE loss):*
Total KE before = ½(500)(30²) + ½(400)(50²) = 225,000 + 500,000 = 725,000 J
Total KE after = ½(500)(40²) + ½(400)(62.5²) = 400,000 + 781,250 = 1,181,250 J
*(Note: Since KE after > KE before, this suggests an explosive/energy-releasing collision. The problem likely intends the Accord's velocity to be negative after collision in magnitude sense.)*
**Loss in KE of Accord = |KE_after − KE_before| of Accord = 281,250 J gained, meaning the Accord actually received energy during the collision.**