## MAT 001: PURE MATHEMATICS
### QUESTION 1
**1(a)(i)**
A parabola has equation y² = 4ax. Show that the equation of the tangent at point (at², 2at) is ty = x + at². [3 marks]
**Solution:**
Given parabola: y² = 4ax
Differentiate implicitly:
2y dy/dx = 4a
dy/dx = 2a/y
At point (at², 2at):
Slope m = 2a/(2at) = 1/t
Equation of tangent:
y − 2at = (1/t)(x − at²)
Multiply through by t:
ty − 2at² = x − at²
Therefore: **ty = x + at²** ✓
---
**1(a)(ii)**
Find the focus and directrix of the parabola y² = 12x. [3 marks]
**Solution:**
Compare y² = 12x with y² = 4ax:
4a = 12, so a = 3
**Focus = (3, 0)**
**Directrix: x = −3**
---
**1(b)(i)**
Express the quadratic form 2x² + 5xy + 3y² as a product of two linear factors. [3 marks]
**Solution:**
Try: (2x + 3y)(x + y)
Expanding:
2x² + 2xy + 3xy + 3y² = 2x² + 5xy + 3y² ✓
**Factors: (2x + 3y)(x + y)**
---
**1(b)(ii)**
Determine the angle between the pair of lines represented by x² − 4xy + y² = 0. [3 marks]
**Solution:**
Compare with ax² + 2hxy + by² = 0:
a = 1, h = −2, b = 1
tan θ = 2√(h² − ab) / (a + b)
h² − ab = 4 − 1 = 3
a + b = 1 + 1 = 2
tan θ = 2√3 / 2 = √3
**θ = 60°**
---
**1(c)**
Find the equation of the normal to the ellipse x²/25 + y²/16 = 1 at the point (4, 12/5). [3 marks]
**Solution:**
Differentiate implicitly:
2x/25 + (2y/16)(dy/dx) = 0
dy/dx = −16x/(25y)
At point (4, 12/5):
Slope of tangent = −(16 × 4)/(25 × 12/5) = −64/60 = −16/15
Slope of normal = 15/16
Equation of normal:
y − 12/5 = (15/16)(x − 4)
Multiply through by 80:
80y − 192 = 75x − 300
**75x − 80y − 108 = 0**
---
### QUESTION 2
**2(a)**
Solve the equation z⁴ + 16 = 0, expressing all roots in the form a + bi. [5 marks]
**Solution:**
z⁴ = −16
Write −16 in polar form:
−16 = 16(cos 180° + i sin 180°)
Fourth roots have modulus = 16^(1/4) = 2
Arguments = (180° + 360°k)/4, for k = 0, 1, 2, 3:
- k = 0: θ = 45° → **√2 + √2 i**
- k = 1: θ = 135° → **−√2 + √2 i**
- k = 2: θ = 225° → **−√2 − √2 i**
- k = 3: θ = 315° → **√2 − √2 i**
---
**2(b)(i)**
If α and β are roots of x² − 5x + 3 = 0, find the equation whose roots are (α² + β²) and (αβ). [3 marks]
**Solution:**
By Vieta's formulas:
α + β = 5, αβ = 3
α² + β² = (α + β)² − 2αβ = 25 − 6 = 19
New roots: 19 and 3
Sum = 22, Product = 57
**New equation: x² − 22x + 57 = 0**
---
**2(b)(ii)**
Express (2 − i)/(3 + 4i) + (1 + 2i)/(2 − 3i) in the form x + iy. [3 marks]
**Solution:**
First fraction: (2 − i)/(3 + 4i)
Multiply by (3 − 4i)/(3 − 4i):
= (2 − i)(3 − 4i)/25
= (6 − 8i − 3i + 4i²)/25
= (6 − 11i − 4)/25
= (2 − 11i)/25
Second fraction: (1 + 2i)/(2 − 3i)
Multiply by (2 + 3i)/(2 + 3i):
= (1 + 2i)(2 + 3i)/13
= (2 + 3i + 4i + 6i²)/13
= (2 + 7i − 6)/13
= (−4 + 7i)/13
Adding:
Real part: 2/25 − 4/13 = 26/325 − 100/325 = −74/325
Imaginary part: −11/25 + 7/13 = −143/325 + 175/325 = 32/325
**Answer: −74/325 + (32/325)i**
---
**2(c)**
Use the binomial theorem to expand (2 + 3x)⁵ and hence evaluate (2.03)⁵ correct to 4 decimal places. [4 marks]
**Solution:**
(2 + 3x)⁵ = 32 + 5(16)(3x) + 10(8)(9x²) + 10(4)(27x³) + 5(2)(81x⁴) + 243x⁵
**= 32 + 240x + 720x² + 1080x³ + 810x⁴ + 243x⁵**
For (2.03)⁵: write as (2 + 3x)⁵ where 3x = 0.03, so x = 0.01
= 32 + 240(0.01) + 720(0.0001) + 1080(0.000001) + 810(0.00000001) + ...
= 32 + 2.4 + 0.0720 + 0.001080 + 0.00000810 + ...
**(2.03)⁵ ≈ 34.4731** (correct to 4 decimal places)
---
## MAT 002: CALCULUS
### QUESTION 3
**3(a)**
Evaluate ∫(x³ + 2x)/(x² + 1)² dx [5 marks]
**Solution:**
Split the numerator:
x³ + 2x = x(x² + 1) + x
So: (x³ + 2x)/(x² + 1)² = x/(x² + 1) + x/(x² + 1)²
∫x/(x² + 1) dx = (1/2)ln(x² + 1)
For ∫x/(x² + 1)² dx, let u = x² + 1:
= (1/2)∫u⁻² du = −1/(2(x² + 1))
**Answer: (1/2)ln(x² + 1) − 1/(2(x² + 1)) + C**
---
**3(b)**
Find ∫tan⁴x sec²x dx [5 marks]
**Solution:**
Let u = tan x, du = sec²x dx
∫tan⁴x sec²x dx = ∫u⁴ du = u⁵/5 + C
**= tan⁵x/5 + C**
---
**3(c)**
Evaluate ∫₀^(π/4) (sin x + cos x)/(3 + sin 2x) dx [5 marks]
**Solution:**
Let t = sin x − cos x, dt = (cos x + sin x) dx
t² = 1 − sin 2x → sin 2x = 1 − t²
3 + sin 2x = 4 − t²
When x = 0: t = −1; when x = π/4: t = 0
∫₋₁⁰ dt/(4 − t²) = (1/4)∫₋₁⁰ [1/(2+t) + 1/(2−t)] dt
= (1/4)[ln|2+t| − ln|2−t|]₋₁⁰
= (1/4)[(ln 2 − ln 2) − (ln 1 − ln 3)]
= (1/4)(ln 3)
**= (ln 3)/4**
---
### QUESTION 4
**4(a)**
Find the general solution of d²y/dx² − 5 dy/dx + 6y = e²ˣ. [5 marks]
**Solution:**
Auxiliary equation: m² − 5m + 6 = 0 → (m − 2)(m − 3) = 0 → m = 2, 3
Complementary function: y_c = C₁e²ˣ + C₂e³ˣ
Since e²ˣ appears in the CF, try y_p = Axe²ˣ:
y_p' = Ae²ˣ(1 + 2x)
y_p'' = Ae²ˣ(4 + 4x)
Substituting:
Ae²ˣ(4 + 4x) − 5Ae²ˣ(1 + 2x) + 6Axe²ˣ = e²ˣ
Ae²ˣ(4 + 4x − 5 − 10x + 6x) = e²ˣ
−Ae²ˣ = e²ˣ
A = −1
**General solution: y = C₁e²ˣ + C₂e³ˣ − xe²ˣ**
---
**4(b)**
Solve (x² − xy + y²)dx − xy dy = 0. [5 marks]
**Solution:**
Homogeneous equation. Let y = vx, dy = v dx + x dv:
(x² − vx² + v²x²)dx − vx²(v dx + x dv) = 0
x²(1 − v + v²)dx − vx²(v dx + x dv) = 0
(1 − v + v² − v²)dx = vx dv
(1 − v)dx = vx dv
dx/x = v/(1 − v) dv = [−1 + 1/(1−v)] dv
Integrating:
ln|x| = −v − ln|1 − v| + C
Substituting back v = y/x:
ln|x| = −y/x − ln|(x − y)/x| + C
**x ln|x − y| + y = Cx**
---
**4(c)**
A curve passes through (2, 3) with dy/dx = x²√(1 + x²). Find the equation of the curve. [5 marks]
**Solution:**
y = ∫x²√(1 + x²) dx
Using integration by parts:
y = (x/8)(2x² + 1)√(1 + x²) − (1/8)ln|x + √(1 + x²)| + C
Using point (2, 3):
3 = (2/8)(9)√5 − (1/8)ln(2 + √5) + C
C = 3 − (9√5/4) + (1/8)ln(2 + √5)
**y = (x/8)(2x² + 1)√(1 + x²) − (1/8)ln|x + √(1 + x²)| + C**
where C = 3 − 9√5/4 + (1/8)ln(2 + √5)
---
## MAT 003: APPLIED MATHEMATICS
### QUESTION 5
**5(a)**
Find the scalar triple product [a, b, c] where a = i + 2j − k, b = 2i − j + 3k, c = 3i + j + 2k. Hence determine if the vectors are coplanar. [5 marks]
**Solution:**
[a, b, c] = det:
| 1 2 −1 |
| 2 −1 3 |
| 3 1 2 |
= 1[(−1)(2) − (3)(1)] − 2[(2)(2) − (3)(3)] + (−1)[(2)(1) − (−1)(3)]
= 1(−2 − 3) − 2(4 − 9) − 1(2 + 3)
= −5 + 10 − 5
= **0**
Since [a, b, c] = 0, **the vectors are coplanar.**
---
**5(b)**
s = 2t³ − 9t² + 12t + 4
**(i)** Velocity and acceleration:
v = ds/dt = **6t² − 18t + 12**
a = dv/dt = **12t − 18**
**(ii)** Particle at rest (v = 0):
6t² − 18t + 12 = 0
t² − 3t + 2 = 0
(t − 1)(t − 2) = 0
**t = 1 s and t = 2 s**
**(iii)** Acceleration when v = 0:
At t = 1: a = 12(1) − 18 = **−6 m/s²**
At t = 2: a = 12(2) − 18 = **+6 m/s²**
---
**5(c)**
Masses: m₁ = 8 kg, m₂ = 12 kg, g = 10 m/s²
**(i) Common acceleration:**
a = (m₂ − m₁)g / (m₁ + m₂) = (12 − 8)(10) / (12 + 8) = 40/20 = **2 m/s²**
**(ii) Tension:**
For 12 kg: T = m₂(g − a) = 12(10 − 2) = 12 × 8 = **96 N**
---
### QUESTION 6
**6(a)**
Projectile at 30°, u = 40 m/s, g = 10 m/s²
uₓ = 40 cos 30° = 20√3 m/s
u_y = 40 sin 30° = 20 m/s
**(i) Maximum height:**
H = u_y²/(2g) = 400/20 = **20 m**
**(ii) Time of flight:**
T = 2u_y/g = 2(20)/10 = **4 s**
**(iii) Horizontal range:**
R = uₓ × T = 20√3 × 4 = **80√3 m**
---
**6(b)**
For equilibrium: F₁ + F₂ + F₃ = 0
(3 − 2 + a)i + (4 + 5 + b)j = 0
(1 + a) = 0 and (9 + b) = 0
**a = −1, b = −9**
---
**6(c)**
m = 5 kg, F = 60 N at 30°, μ = 0.4, g = 10 m/s²
**(i) Normal reaction:**
R = mg − F sin 30° = 5(10) − 60(0.5) = 50 − 30 = **20 N**
**(ii) Acceleration:**
Friction force f = μR = 0.4 × 20 = 8 N
Net force = F cos 30° − f = 60(√3/2) − 8 = 30√3 − 8 ≈ 51.96 − 8 = 43.96 N
a = Net force / m = 43.96/5 = **8.79 m/s²**
---
## MAT 004: STATISTICS
### QUESTION 7
**7(a)**
**(i) Type I error:** Rejecting a true null hypothesis (false positive).
**(ii) Type II error:** Failing to reject a false null hypothesis (false negative).
**(iii) Power of a test:** The probability of correctly rejecting a false null hypothesis = 1 − P(Type II error).
**(iv) p-value:** The probability of obtaining a test statistic as extreme as, or more extreme than, the observed value, assuming the null hypothesis is true.
**(v) One-tailed test:** A hypothesis test where the critical region lies entirely in one tail of the distribution.
---
**7(b)**
H₀: μ ≥ 2000, H₁: μ < 2000
n = 64, x̄ = 1950, s = 120, α = 5%
**(i) Test statistic:**
z = (x̄ − μ₀)/(s/√n) = (1950 − 2000)/(120/√64) = −50/15 = **−3.33**
Critical value at 5% (one-tailed): z = −1.645
Since −3.33 < −1.645, **reject H₀.** The data does not support the manufacturer's claim.
**(ii) p-value:**
P(Z < −3.33) = **0.0004**
---
**7(c)**
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| P(X=x) | 0.10 | 0.20 | 0.30 | 0.25 | 0.15 |
**(i) E(X):**
E(X) = 0(0.10) + 1(0.20) + 2(0.30) + 3(0.25) + 4(0.15)
= 0 + 0.20 + 0.60 + 0.75 + 0.60 = **2.15**
**E(X²):**
E(X²) = 0²(0.10) + 1²(0.20) + 2²(0.30) + 3²(0.25) + 4²(0.15)
= 0 + 0.20 + 1.20 + 2.25 + 2.40 = **6.05**
Var(X) = E(X²) − [E(X)]² = 6.05 − (2.15)² = 6.05 − 4.6225 = **1.4275**
**(ii) Var(3X − 2):**
Var(3X − 2) = 9 Var(X) = 9 × 1.4275 = **12.8475**
---
### QUESTION 8
**8(a)**
Data:
| Age (X) | 45 | 52 | 38 | 60 | 48 | 55 | 42 | 50 |
|---|---|---|---|---|---|---|---|---|
| BP (Y) | 130 | 145 | 120 | 155 | 135 | 150 | 125 | 140 |
n = 8
ΣX = 390, ΣY = 1100
X̄ = 48.75, Ȳ = 137.5
ΣX² = 45² + 52² + 38² + 60² + 48² + 55² + 42² + 50²
= 2025 + 2704 + 1444 + 3600 + 2304 + 3025 + 1764 + 2500 = **19366**
ΣY² = 130² + 145² + 120² + 155² + 135² + 150² + 125² + 140²
= 16900 + 21025 + 14400 + 24025 + 18225 + 22500 + 15625 + 19600 = **152300**
ΣXY = (45×130)+(52×145)+(38×120)+(60×155)+(48×135)+(55×150)+(42×125)+(50×140)
= 5850 + 7540 + 4560 + 9300 + 6480 + 8250 + 5250 + 7000 = **54230**
Sₓₓ = ΣX² − n X̄² = 19366 − 8(48.75)² = 19366 − 19012.5 = **353.5**
S_yy = ΣY² − n Ȳ² = 152300 − 8(137.5)² = 152300 − 151250 = **1050**
Sₓᵧ = ΣXY − n X̄ Ȳ = 54230 − 8(48.75)(137.5) = 54230 − 53625 = **605**
**(i) Correlation coefficient:**
r = Sₓᵧ / √(Sₓₓ · S_yy) = 605 / √(353.5 × 1050) = 605 / √371175 = 605 / 609.24 ≈ **0.993**
**(ii) Regression line of Y on X:**
b = Sₓᵧ / Sₓₓ = 605 / 353.5 = **1.712**
a = Ȳ − b X̄ = 137.5 − 1.712(48.75) = 137.5 − 83.46 = **54.04**
**Regression line: Y = 1.712X + 54.04**
---
**8(b)**
f(x) = k(4x − x²) for 0 ≤ x ≤ 4, 0 otherwise
**(i) Find k:**
∫₀⁴ k(4x − x²) dx = 1
k[2x² − x³/3]₀⁴ = 1
k[32 − 64/3] = 1
k[96/3 − 64/3] = 1
k(32/3) = 1
**k = 3/32**
**(ii) P(1 ≤ X ≤ 3):**
P = (3/32)∫₁³ (4x − x²) dx
= (3/32)[2x² − x³/3]₁³
= (3/32)[(18 − 9) − (2 − 1/3)]
= (3/32)[9 − 5/3]
= (3/32)(22/3)
= 22/32
**P(1 ≤ X ≤ 3) = 11/16 = 0.6875**