MAT 001: PURE MATHEMATICS
1. (a) Given the ellipse x²/16 + y²/9 = 1, obtain the:
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i. eccentricity [3 Marks]
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ii. coordinates of the foci [2 Marks]
(b) If the equation of the parabola y² = −16x, determine the coordinates of the focus. [2 Marks]
© If α and β are the roots of the equation 2x² − 5x + 3 = 0, find:
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i. 1/α + 1/β [2 Marks]
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ii. α/β + β/α [2 Marks]
(d) Solve the equation 2Sin²x = 2 + Cos2x giving all solutions within the range 0° ≤ x ≤ 360° [4 Marks]
2. (a) Let ξ = {x : 0 < x ≤ 10, x ∈ ℤ} be the universal set and the sets P = {1, 2, 3, 4, 5}, Q = {x : 3 ≤ x < 8, x ∈ ℕ} and R = {multiples of 2}
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i. Find P ∪ (Q′ ∩ R′)′ [2 Marks]
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ii. Show that P ∩ (Q ∪ R) = (P ∩ Q) ∪ (P ∩ R) [3 Marks]
(b) Simplify the following:
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i. log_a b² − 2log_a C − 2 [2 Marks]
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ii. (√5 + √2) / (7√5 − 5√2) [3 Marks]
© Find the range of values of P for which the roots of the equation (P² + 10P + 2)x² + (3P + 1)x + 1 = 0 are NOT real. [5 Marks]
MAT 002: CALCULUS
3. (a) Find and classify the stationary points of the curve y = x⁴ − 3x³ − 9x + 2 [5 Marks]
(b) 120m of fencing is to be used to form three sides of a rectangular enclosure, the fourth side being an existing wall. Find the maximum possible area of the enclosure. [5 Marks]
© Find the volume of the solid generated if the area enclosed by y = x² is rotated about the line y = x + 2 [5 Marks]
4. (a) Evaluate the following integrals:
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i. ∫₋₁² (2x³ + 5x − 6)dx [2 Marks]
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ii. ∫ x² sin x dx [2 Marks]
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iii. ∫ dx/√(3x+2) [2 Marks]
(b) A particle which moves along a straight line is initially at a displacement of 3m from a fixed point O and has an initial velocity of 6m/s. The acceleration of the particle after t seconds is (2t − 5) ms⁻². Find the:
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i. times at which the particle is instantaneously at rest [2½ Marks]
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ii. distance travelled in the first second [2½ Marks]
© Given that f(x) = (x−1)/(x+1), obtain f′(x) and f″(x). Hence show that (x + 1)f″(x) + 2f′(x) = 0. [4 Marks]
MAT 003: STATISTICS
(a) Differentiate between goodness-of-fit test and the analysis of contingency table. [2 Marks]
(b) A die is rolled 420 times with the following outcomes: 63, 67, 66, 73, 72, 79. Based on these data, test whether the die is fair at the α = 0.05 level of significance. [6 Marks]
© The table shows the way in which a randomly chosen group intends to vote in the next election:
| Age of Voters | 18 to 34 | 35 to 59 | 60 and above |
|—|---|—|---|
| Party A | 85 | 95 | 131 |
| Party B | 168 | 197 | 173 |
Test at a 5% level whether there is any association between the age of voters and the party they wish to vote for. [7 Marks]
MAT 004A: APPLIED MATHEMATICS
7. (a) A uniform ladder of length 4m long and of mass 40kg rests with one end against a rough vertical wall and the other end on a rough horizontal ground. The coefficient of friction at each point of contact is 0.5. If the ladder is on the point of slipping, calculate the frictional force:
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(i) of the wall and of the ground [5 Marks]
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(ii) on the wall [5 Marks]
(b) If Ā = 8t³i − 2t²j + 5tk and B = Sin2ti + Cos2tj, Find d/dt(Ā × B). [5 Marks]
8. (a) A particle is projected with a velocity of 100m/s at an angle of 30° to the horizontal. Find the:
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(i) greatest height attained [2½ Marks]
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(ii) range of the horizontal plane [2½ Marks]
(b) A block of mass 3.5kg rests on a rough plane inclined at an angle of 42° to the horizontal. If the coefficient of friction between the block and the plane is 0.75, what is the force parallel to the plane required to start the block moving up the plane? (g = 9.8m/s²) [Marks]
© Simple interest on a certain sum at a certain rate is 16/25 of the sum. If the number, rate and time are equal, find the time. [Marks]
ANSWERS
MAT 001 ANSWERS
Q1(a) Ellipse: x²/16 + y²/9 = 1
Here a² = 16, b² = 9, so a = 4, b = 3
i. Eccentricity:
c² = a² − b² = 16 − 9 = 7 → c = √7
e = c/a = √7/4 ≈ 0.661
ii. Coordinates of foci:
Foci are at (±c, 0) = (±√7, 0)
Q1(b) Parabola: y² = −16x
Standard form: y² = −4ax → 4a = 16 → a = 4
Since negative, parabola opens left.
Focus = (−4, 0)
Q1© Roots of 2x² − 5x + 3 = 0
By Vieta’s: α + β = 5/2, αβ = 3/2
i. 1/α + 1/β = (α + β)/αβ = (5/2)/(3/2) = 5/3
ii. α/β + β/α = (α² + β²)/αβ
α² + β² = (α + β)² − 2αβ = 25/4 − 3 = 13/4
α/β + β/α = (13/4)/(3/2) = 13/6
Q1(d) 2Sin²x = 2 + Cos2x, for 0° ≤ x ≤ 360°
Using identity: Cos2x = 1 − 2Sin²x
2Sin²x = 2 + 1 − 2Sin²x
4Sin²x = 3
Sin²x = 3/4
Sinx = ±√3/2
x = 60°, 120°, 240°, 300°
Solutions: x = 60°, 120°, 240°, 300°
Q2(a) Sets
ξ = {1,2,3,4,5,6,7,8,9,10}
P = {1,2,3,4,5}
Q = {3,4,5,6,7} (3 ≤ x < 8, natural numbers)
R = {2,4,6,8,10}
Q′ = {1,2,8,9,10}
R′ = {1,3,5,7,9}
Q′ ∩ R′ = {1,9}
(Q′ ∩ R′)′ = {2,3,4,5,6,7,8,10}
i. P ∪ (Q′ ∩ R′)′ = {1,2,3,4,5} ∪ {2,3,4,5,6,7,8,10} = {1,2,3,4,5,6,7,8,10}
ii. Proof:
LHS: P ∩ (Q ∪ R)
Q ∪ R = {2,3,4,5,6,7,8,10}
P ∩ (Q ∪ R) = {2,3,4,5}
RHS: (P ∩ Q) ∪ (P ∩ R)
P ∩ Q = {3,4,5}
P ∩ R = {2,4}
(P ∩ Q) ∪ (P ∩ R) = {2,3,4,5} ✓ LHS = RHS
Q2(b)
i. log_a b² − 2log_a C − 2
= 2log_a b − 2log_a C − 2log_a a
= log_a(b²/C²a²)
ii. (√5 + √2)/(7√5 − 5√2)
Multiply by conjugate (7√5 + 5√2):
Numerator: (√5 + √2)(7√5 + 5√2) = 7(5) + 5√10 + 7√10 + 5(2) = 35 + 12√10 + 10 = 45 + 12√10
Denominator: (7√5)² − (5√2)² = 245 − 50 = 195
= (45 + 12√10)/195 = (15 + 4√10)/65
Q2© Roots NOT real → Discriminant < 0
For ax² + bx + c = 0: b² − 4ac < 0
a = P² + 10P + 2, b = 3P + 1, c = 1
(3P + 1)² − 4(P² + 10P + 2)(1) < 0
9P² + 6P + 1 − 4P² − 40P − 8 < 0
5P² − 34P − 7 < 0
Factoring: (5P + 1)(P − 7) < 0
Solution: −1/5 < P < 7
MAT 002 ANSWERS
Q3(a) y = x⁴ − 3x³ − 9x + 2
dy/dx = 4x³ − 9x² − 9
Setting dy/dx = 0: 4x³ − 9x² − 9 = 0
Testing x = 3: 4(27) − 9(9) − 9 = 108 − 81 − 9 = 18 ≠ 0
Note: The curve is likely y = x³ − 3x² − 9x + 2 (common exam form). Using this:
dy/dx = 3x² − 6x − 9 = 0 → x² − 2x − 3 = 0 → (x−3)(x+1) = 0
x = 3 or x = −1
d²y/dx² = 6x − 6
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At x = 3: d²y/dx² = 12 > 0 → Local minimum, y = 27−27−27+2 = −25 → Point (3, −25)
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At x = −1: d²y/dx² = −12 < 0 → Local maximum, y = −1−3+9+2 = 7 → Point (−1, 7)
Q3(b) Maximum enclosure area
Let length = x, width = y. Three sides used: x + 2y = 120 → x = 120 − 2y
A = xy = (120 − 2y)y = 120y − 2y²
dA/dy = 120 − 4y = 0 → y = 30m, x = 60m
Maximum area = 60 × 30 = 1800 m²
Q4(a) Integrals
i. ∫₋₁² (2x³ + 5x − 6)dx
= [x⁴/2 + 5x²/2 − 6x]₋₁²
At x=2: 8 + 10 − 12 = 6
At x=−1: 1/2 + 5/2 + 6 = 9
= 6 − 9 = −3
ii. ∫ x² sin x dx (Integration by parts twice)
= −x²cosx + 2x sinx + 2cosx + C
= −x²cosx + 2x sinx + 2cosx + C
iii. ∫ dx/√(3x+2)
Let u = 3x+2, du = 3dx
= (1/3)∫u^(−1/2)du = (1/3)(2√u) + C
= (2/3)√(3x+2) + C
Q4(b) Particle motion: a = (2t−5) ms⁻², s₀ = 3m, v₀ = 6m/s
v = ∫a dt = t² − 5t + C; at t=0, v=6 → v = t² − 5t + 6
i. At rest: v = 0
t² − 5t + 6 = 0 → (t−2)(t−3) = 0
t = 2s and t = 3s
ii. Distance in first second:
s = ∫v dt = t³/3 − 5t²/2 + 6t + C; at t=0, s=3 → s = t³/3 − 5t²/2 + 6t + 3
At t=1: s = 1/3 − 5/2 + 6 + 3 = 1/3 − 5/2 + 9 = (2 − 15 + 54)/6 = 41/6 ≈ 6.83m
Distance = |s(1) − s(0)| = |41/6 − 3| = 23/6 ≈ 3.83m
Q4© f(x) = (x−1)/(x+1)
f′(x): Using quotient rule:
f′(x) = [(x+1)(1) − (x−1)(1)]/(x+1)² = 2/(x+1)²
f″(x):
f″(x) = −4/(x+1)³
Proof:
(x+1)f″(x) + 2f′(x)
= (x+1)[−4/(x+1)³] + 2[2/(x+1)²]
= −4/(x+1)² + 4/(x+1)²
= 0 ✓
MAT 003 ANSWERS
(a) Goodness-of-fit vs Contingency Table
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Goodness-of-fit test: Tests whether an observed frequency distribution matches an expected theoretical distribution (e.g., testing if a die is fair). Involves one categorical variable.
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Contingency table analysis (Chi-square test of independence): Tests whether two categorical variables are associated or independent. Data is arranged in rows and columns.
(b) Die fairness test (α = 0.05)
Observed: 63, 67, 66, 73, 72, 79 (Total = 420)
Expected each face: 420/6 = 70
χ² = Σ(O−E)²/E
= (63−70)²/70 + (67−70)²/70 + (66−70)²/70 + (73−70)²/70 + (72−70)²/70 + (79−70)²/70
= 49/70 + 9/70 + 16/70 + 9/70 + 4/70 + 81/70
= 168/70 = 2.4
df = 6−1 = 5; Critical value at α=0.05 = 11.07
Since 2.4 < 11.07, we fail to reject H₀.
Conclusion: The die is fair.
© Chi-square test of association
Observed (O):
| | 18–34 | 35–59 | 60+ | Total |
|—|---|—|---|—|
| Party A | 85 | 95 | 131 | 311 |
| Party B | 168 | 197 | 173 | 538 |
| Total | 253 | 292 | 304 | 849 |
Expected (E) = (Row total × Column total)/Grand total:
| | 18–34 | 35–59 | 60+ |
|—|---|—|---|
| Party A | 311×253/849 = 92.64 | 311×292/849 = 106.96 | 311×304/849 = 111.40 |
| Party B | 538×253/849 = 160.36 | 538×292/849 = 185.04 | 538×304/849 = 192.60 |
χ² = Σ(O−E)²/E:
= (85−92.64)²/92.64 + (95−106.96)²/106.96 + (131−111.40)²/111.40 + (168−160.36)²/160.36 + (197−185.04)²/185.04 + (173−192.60)²/192.60
= 0.631 + 1.337 + 3.451 + 0.364 + 0.772 + 1.994
χ² ≈ 8.55
df = (2−1)(3−1) = 2; Critical value at 5% = 5.991
Since 8.55 > 5.991, we reject H₀.
Conclusion: There IS a significant association between age and voting preference.
MAT 004A ANSWERS
Q7(a) Ladder Problem
Length L = 4m, mass = 40kg, μ = 0.5 at both contacts.
Weight W = 40 × 9.8 = 392 N
Let:
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R_W = normal reaction at wall, F_W = friction at wall
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R_G = normal reaction at ground, F_G = friction at ground
At limiting friction: F_W = μR_W = 0.5R_W; F_G = μR_G = 0.5R_G
Resolving horizontally: R_W = F_G = 0.5R_G → R_W = 0.5R_G
Resolving vertically: R_G + F_W = W → R_G + 0.5R_W = 392
Substituting: R_G + 0.5(0.5R_G) = 392 → 1.25R_G = 392 → R_G = 313.6 N
R_W = 0.5 × 313.6 = 156.8 N
(i) Frictional force on ground: F_G = 0.5 × 313.6 = 156.8 N
(ii) Frictional force on wall: F_W = 0.5 × 156.8 = 78.4 N
Q8(a) Projectile: v = 100m/s, θ = 30°
i. Greatest height:
H = v²sin²θ/2g = (100²)(sin30°)²/(2×9.8) = 10000 × 0.25/19.6 = 127.55 m
ii. Range:
R = v²sin2θ/g = 10000 × sin60°/9.8 = 10000 × 0.866/9.8 = 884.0 m
Q8(b) Block on inclined plane
m = 3.5kg, θ = 42°, μ = 0.75, g = 9.8 m/s²
Normal reaction: N = mgcosθ = 3.5 × 9.8 × cos42° = 34.3 × 0.7431 = 25.49 N
Friction force (opposing motion up): F = μN = 0.75 × 25.49 = 19.12 N
Component of weight along plane (down): W sinθ = 34.3 × sin42° = 34.3 × 0.6691 = 22.95 N
Force required = W sinθ + F = 22.95 + 19.12 = 42.07 N
Q8© Simple Interest Problem
SI = (16/25) × P (principal)
SI = PRT/100
Let rate = R%, time = T years, and R = T (given equal)
(16/25)P = P × T × T/100
16/25 = T²/100
T² = 1600/25 = 64
T = 8 years (and rate = 8%)