QUESTIONS
MAT 001: ADVANCED PURE MATHEMATICS
1. (a) Given that C = (2k −2; 3 k), where k is a real number. Find the determinant of C, hence find C⁻¹, leaving your answer in terms of k. [5 Marks]
(b) Given the matrix A = (6 −2; −4 1), and I is a 2×2 identity matrix. Prove that A² = 7A + 2I. [5 Marks]
(c) The Cubic polynomial f(x) is defined by f(x) = 4x³ − 7x − 3. Use the theory of quadratic equation to obtain the minimal value of the quotient q(x) when f(x) is divided by (x − 2). [5 Marks]
[TOTAL = 15 Marks]
2. (a) The sum of the first ten terms of an arithmetic progression is 400 and the sum of the next ten terms is 1000. What is the first term and the common difference. [5 marks]
(b) Find the term independent of x in the expansion of (2x² + 1/x)⁹. [5 Marks]
(c) Solve the equation 4sin²x + 8cos x − 7 = 0 for 0° ≤ x ≤ 360°. [5 Marks]
[TOTAL = 15 Marks]
MAT 002: CALCULUS
3. (a) Evaluate the following:
i. lim(x→0) x(1+x)/(1−x²) [3 Marks]
ii. lim(x→∞) (2x²+1)/((2−x)(2+x)) [2 Marks]
(b) If ∫₋₁ᵐ (6x−5)dx = 4, find the positive value of m. [5 Marks]
(c) Solve dy/dx + (1/x)y = sin x [5 Marks]
[TOTAL = 15 Marks]
4. (a) Evaluate ∫ (3x+1)/((x+1)²(x−2)) dx [4 Marks]
(b) Expand sin 2x as a series as far as the 7th term. [7 Marks]
(c) Solve y'' + 2y' + 4y = 0, y = y(t). [4 Marks]
[TOTAL = 15 Marks]
MAT 003: STATISTICS
5. (a) Simplify ᵖ⁺¹C₄ − ᵖ⁻¹C₄ [5 Marks]
(b) The deviation of 85 from some numbers are −6, −4, −2, 1, 2, 3, 4, 5 and 8.
Find and correct to 2 decimal places the:
(i) mean [4 Marks]
(ii) Variance. [6 Marks]
[TOTAL = 15 Marks]
6. (a)(i) Use Spearman's correlation coefficient to calculate the relation between the scores of 5 students in mathematics and Physics test. [5 Marks]
| Maths (x) | 2 | 3 | 6 | 4 | 7 |
|---|---|---|---|---|---|
| Physics (y) | 9 | 6 | 8 | 2 | 5 |
(ii) Give interpretation of your result. [1 mark]
(b) Different plastic clothes pegs are made in various colours. The number of white pegs may be modeled by a binomial distribution with parameter p equal to 0.2. The contents of packets of 50 pegs of mixed colours may be considered to be random samples. Determine the probability that a packet contains:
(i) Less than or equal to 15 white pegs; [3 Marks]
(ii) Exactly 7 white pegs. [3 Marks]
(c) Samsung and Johnson regularly play each other at darts. The probability that Johnson wins any game is 0.3, and the outcome of each game is independent of the outcome of every other game. Find the probability that, in a match of 15 games, Samsung win exactly 8 games. [3 Marks]
[TOTAL = 15 Marks]
MAT 004: APPLIED MATHEMATICS
6. (b) [previous part not shown] ...car to increase from 5 ms⁻¹ to 9 ms⁻¹ [5 marks]
(c) A particle moves towards centre of attraction starting from rest at a distance α from the centre. If its velocity at any distance r from the centre varies as √((α²−r²)/r²), show that its acceleration is −k²α²/r³, where k is the proportionality constant. [5 marks]
[TOTAL = 15 Marks]
7. (a) Define the term moment of force. [2 Marks]
(b) A uniform rod AB of mass 10kg is hinged on the wall at end A. Find:
(i) the force that must be applied perpendicular to the rod at end B which will maintain the rod horizontal, [5 Marks]
(ii) the force that must be applied perpendicular to the rod at end B which will maintain the rod at angle 60° to the downward vertical through A. [4 Marks]
(c) Find the vector equation of a line joining the point with position vectors 5i − 2j + 3k and 2i − 7j + 4k [4 Marks]
[TOTAL = 15 Marks]
8. A body moves in a straight line so that the distance s metres travelled after time t secs is given by s = t³ − 2t² + t.
(a) Find the expression for the speed of the body at time t = 2s. [4 Marks]
(b) Find the times at which the body is at rest. [5 Marks]
(c) What is the acceleration of the body at these times? [6 Marks]
[TOTAL = 15 Marks]
Answers
MAT 001: Advanced Pure Mathematics
1(a) C = (2k −2; 3 k)
det(C) = (2k)(k) − (−2)(3) = 2k² + 6
C⁻¹ = 1/det(C) × (k 2; −3 2k) = 1/(2k²+6) · (k 2; −3 2k)
1(b) A = (6 −2; −4 1)
A² = A·A:
- Row 1: [6(6)+(−2)(−4), 6(−2)+(−2)(1)] = [36+8, −12−2] = [44, −14]
- Row 2: [(−4)(6)+1(−4), (−4)(−2)+1(1)] = [−24−4, 8+1] = [−28, 9]
So A² = (44 −14; −28 9)
7A + 2I = (42 −14; −28 7) + (2 0; 0 2) = (44 −14; −28 9)
Since A² = 7A + 2I ✓ (proved)
1(c) Divide f(x) = 4x³ − 7x − 3 by (x−2) using synthetic division (root = 2):
2 | 4 0 -7 -3
| 8 16 18
| 4 8 9 15
q(x) = 4x² + 8x + 9, remainder = 15
Minimum of q(x): vertex at x = −8/(2·4) = −1
q(−1) = 4(1) − 8 + 9 = 5 (minimal value)
2(a) S₁₀ = (10/2)(2a+9d) = 400 → 2a+9d = 80 ...(i)
Sum of next 10 terms = S₂₀ − S₁₀ = 1000 → S₂₀ = 1400
S₂₀ = (20/2)(2a+19d) = 1400 → 2a+19d = 140 ...(ii)
(ii) − (i): 10d = 60 → d = 6
Sub into (i): 2a + 54 = 80 → 2a = 26 → a = 13
2(b) General term of (2x²+1/x)⁹:
Tᵣ = C(9,r)(2x²)⁹⁻ʳ(1/x)ʳ = C(9,r)·2⁹⁻ʳ·x^(18−3r)
Set 18−3r = 0 → r = 6
Term = C(9,6)·2³ = 84 × 8 = 672
2(c) 4sin²x + 8cosx − 7 = 0
Using sin²x = 1−cos²x:
4(1−cos²x) + 8cosx − 7 = 0
−4cos²x + 8cosx − 3 = 0 → 4cos²x − 8cosx + 3 = 0
cosx = [8 ± √(64−48)]/8 = [8±4]/8 → cosx = 1.5 (rejected) or cosx = 0.5
x = 60°, 300°
MAT 002: Calculus
3(a)(i) lim(x→0) x(1+x)/(1−x²) = lim x(1+x)/[(1−x)(1+x)] = lim x/(1−x) = 0/1 = 0
3(a)(ii) lim(x→∞) (2x²+1)/(4−x²); divide num & denom by x²:
= (2+1/x²)/(4/x² −1) → −2
3(b) ∫₋₁ᵐ(6x−5)dx = [3x²−5x]₋₁ᵐ = (3m²−5m) − (3+5) = 3m²−5m−8
Set = 4: 3m²−5m−12 = 0
Discriminant = 25+144 = 169, √169=13
m = (5±13)/6 → m=3 or m=−4/3
m = 3 (positive value)
3(c) dy/dx + y/x = sinx (linear ODE, standard form)
Integrating factor: μ = e^∫(1/x)dx = e^(lnx) = x
d/dx(xy) = x sinx
Integrate RHS by parts: ∫x sinx dx = −xcosx + sinx + C
xy = −xcosx + sinx + C
y = −cosx + sinx/x + C/x
4(a) (3x+1)/[(x+1)²(x−2)] = A/(x+1) + B/(x+1)² + C/(x−2)
3x+1 = A(x+1)(x−2) + B(x−2) + C(x+1)²
- x=−1: −2 = B(−3) → B = 2/3
- x=2: 7 = 9C → C = 7/9
- Compare x² coeff: 0 = A+C → A = −7/9
∫ = A ln|x+1| + B(−1/(x+1)) + C ln|x−2| + K
= (7/9)ln|x−2| − (7/9)ln|x+1| − 2/[3(x+1)] + K
4(b) sinθ = θ − θ³/3! + θ⁵/5! − θ⁷/7! + θ⁹/9! − θ¹¹/11! + θ¹³/13! ...; put θ=2x:
- Term 1: 2x
- Term 2: −(2x)³/6 = −4x³/3
- Term 3: (2x)⁵/120 = 4x⁵/15
- Term 4: −(2x)⁷/5040 = −8x⁷/315
- Term 5: (2x)⁹/362880 = 4x⁹/2835
- Term 6: −(2x)¹¹/39916800 = −8x¹¹/155925
- Term 7: (2x)¹³/6227020800 = 8x¹³/6081075
sin2x = 2x − 4x³/3 + 4x⁵/15 − 8x⁷/315 + 4x⁹/2835 − 8x¹¹/155925 + 8x¹³/6081075 − ...
4(c) y″+2y′+4y=0
Auxiliary equation: r² +2r+4 = 0
r = [−2 ± √(4−16)]/2 = −1 ± i√3
y(t) = e⁻ᵗ[C₁cos(√3 t) + C₂sin(√3 t)]
MAT 003: Statistics
5(a) Using ⁿC₄ = n(n−1)(n−2)(n−3)/24:
ᵖ⁺¹C₄ = (p+1)p(p−1)(p−2)/24
ᵖ⁻¹C₄ = (p−1)(p−2)(p−3)(p−4)/24
Difference = (p−1)(p−2)[(p+1)p − (p−3)(p−4)]/24
(p+1)p = p²+p; (p−3)(p−4) = p²−7p+12
Bracket = p²+p − p²+7p−12 = 8p−12
Difference = (p−1)(p−2)(8p−12)/24 = (p−1)(p−2)·4(2p−3)/24
= (p−1)(p−2)(2p−3)/6
5(b) Deviations (n=9): −6,−4,−2,1,2,3,4,5,8
Σd = −6−4−2+1+2+3+4+5+8 = 11
Mean = A + Σd/n = 85 + 11/9 = 86.22
Σd² = 36+16+4+1+4+9+16+25+64 = 175
Variance = Σd²/n − (Σd/n)² = 175/9 − (11/9)² = 19.4444 − 1.4938 = 17.95
6(a)(i) Rank the data:
| x | 2 | 3 | 6 | 4 | 7 | Rank(x) | 1 | 2 | 4 | 3 | 5 |
| y | 9 | 6 | 8 | 2 | 5 | Rank(y) | 5 | 3 | 4 | 1 | 2 |
d = Rx−Ry: −4, −1, 0, 2, 3
d²: 16, 1, 0, 4, 9 → Σd² = 30
ρ = 1 − 6Σd²/[n(n²−1)] = 1 − (6×30)/(5×24) = 1 − 180/120 = −0.5
6(a)(ii) ρ = −0.5 shows a moderate negative correlation between Maths and Physics scores — students scoring higher in Maths tended to score lower in Physics.
6(b) n=50, p=0.2, q=0.8 → mean=10, variance=8, σ=√8≈2.828
(i) Normal approximation with continuity correction:
Z = (15.5−10)/2.828 = 1.945 → P(X≤15) ≈ 0.974
(ii) Exact binomial: P(X=7) = C(50,7)(0.2)⁷(0.8)⁴³
C(50,7) = 99,884,400
(0.2)⁷ = 1.28×10⁻⁵
(0.8)⁴³ ≈ 6.79×10⁻⁵
P(X=7) = 99,884,400 × 1.28×10⁻⁵ × 6.79×10⁻⁵ ≈ 0.087
6(c) P(Johnson wins)=0.3 → P(Samsung wins)=0.7; n=15, want exactly 8 Samsung wins:
P(X=8) = C(15,8)(0.7)⁸(0.3)⁷
C(15,8) = 6435
(0.7)⁸ = 0.05765; (0.3)⁷ = 0.0002187
P = 6435 × 0.05765 × 0.0002187 ≈ 0.081
MAT 004: Applied Mathematics
7(a) Moment of force is the turning effect produced by a force about a point, equal to the product of the force and the perpendicular distance from the point to the line of action of the force: M = F × d.
7(b) Rod AB, mass 10 kg, hinged at A; weight W = mg acts at the midpoint (L/2 from A). Taking g = 9.8 m/s²: W = 98 N.
(i) Rod horizontal, force F applied ⊥ to rod at B — take moments about A:
F·L = W·(L/2) → F = W/2 = 49 N
(ii) Rod at 60° to the downward vertical, force F′ ⊥ to rod at B. Weight's moment arm about A = (L/2)sin60° (component perpendicular to rod); F′'s moment arm = L (already perpendicular):
F′·L = W(L/2)sin60°
F′ = (W/2)sin60° = 49 × 0.866 ≈ 42.4 N
7(c) Points: a = 5i−2j+3k, b = 2i−7j+4k
Direction vector = b − a = (2−5)i + (−7+2)j + (4−3)k = −3i−5j+k
Vector equation: r = (5i−2j+3k) + λ(−3i−5j+k), λ ∈ ℝ
8. s = t³−2t²+t
(a) v = ds/dt = 3t²−4t+1
At t=2: v = 3(4)−4(2)+1 = 12−8+1 = 5 m/s
(b) Rest ⟹ v=0: 3t²−4t+1=0
Discriminant = 16−12 = 4, √4=2
t = (4±2)/6 → t = 1/3 s and t = 1 s
(c) a = dv/dt = 6t−4
At t=1/3: a = 2−4 = −2 m/s²
At t=1: a = 6−4 = 2 m/s²