### MAT 001: ADVANCED PURE MATHEMATICS
**1. (a) i.** Find the determinant of the matrix $\begin{bmatrix} 2x & 1 \\ y & 2 \end{bmatrix}$ given that 2x + y = 5 and x − 2y = −5.
**ii.** Given that A and B are acute angles with sin A = 2/5 and cos B = 7/25, find without using a calculator the value of cos(A + B). [5 Marks]
**(b)** In a secondary school, 80 students played Hockey or Football. The number that played Football is 5 more than twice the number that played Hockey. If 15 students played both games and every student plays at least one game, find:
i. the number of students that play Football.
ii. the number of students that play Football but not Hockey.
iii. the number of students that play Hockey but not Football. [5 Marks]
**(c) i.** Obtain the binomial expansion of (x + 3y)⁶.
**ii.** Using the expansion obtained in (i), evaluate (1.03)⁶ correct to 4 significant figures. [5 Marks]
---
**2. (a) i.** Express (2 + 3i)² in the form a + bi. Hence evaluate (2b − 3a)/4ab.
**ii.** Given that a complex number Z = 2 + i, express 1 + Z on an Argand diagram. [5 Marks]
**(b)** If (x + 1) is a factor of P(x) = x³ + kx² + x + 6:
i. Compute the value of k.
ii. Hence factorise P(x) completely. [5 Marks]
**(c)** Calculate the length of the tangent to the circle x² + y² + 6x − 4y + 12 = 0 from the external point (3, 5). [5 Marks]
---
### MAT 002: CALCULUS
**3. (a)** Evaluate:
i. $\lim_{x \to 1} \dfrac{\sqrt{5x-4} - \sqrt{x}}{x-1}$
ii. $\lim_{x \to -2} \dfrac{x^2 + 2x + 1}{x^2 + 3x + 2}$ [5 Marks]
**(b)** Evaluate ∫√(a² − x²) dx. [5 Marks]
**(c)** Given Iₙ = ∫tanⁿx dx, show that Iₙ = tanⁿ⁻¹x/(n−1) − Iₙ₋₂. Hence obtain ∫tan⁵x dx. [5 Marks]
---
**4. (a) i.** Solve the differential equation (y² + 2)(2x + 1) dy/dx = 1, y(0) = 1.
**ii.** Expand f(x) = sin²x as a Maclaurin's series in ascending powers of x in the first three terms. [6 Marks]
**(b)** 60 m of fencing is to be used to form three sides of a rectangular enclosure, the fourth side being an existing wall. Compute:
i. the length of the rectangular enclosure in terms of the breadth.
ii. the maximum possible area of the enclosure. [5 Marks]
**(c)** Find the volume generated when the plane figure bounded by y = 2cos 2x, the x-axis and ordinates at x = 0 and x = π/4 rotates about the x-axis through a complete revolution. [4 Marks]
---
### MAT 003: APPLIED MATHEMATICS
**5. (a)** Given φ(t) = eᵗ; u(t) = sin t **i** + cos t **j** + 3**k**; v(t) = t**i** − 2**k**.
i. Find d/dt (u · v).
ii. Evaluate d/dt (u × v). [5 Marks]
**(b)** A load of mass 50 kg is placed in a lift. Calculate the reaction between the floor of the lift and the load when the lift:
i. is moving at constant speed.
ii. moves upward with an acceleration of 3 ms⁻². [5 Marks]
**(c)** A body of mass 8.2 kg is supported by two light inextensible strings attached to it. The other ends are attached to two fixed points in a ceiling 10 m apart. One string is 6 m long and the other is 8 m long. Assuming equilibrium, calculate:
i. the angle made by each string to the horizontal.
ii. the tension in each string. (Take g = 10 ms⁻²) [5 Marks]
---
**6. (a)** A 4 kg ball moving at 10 ms⁻¹ collides with a 16 kg ball moving at 4 ms⁻¹ in the opposite direction. Calculate:
i. the velocity of the balls if they coalesce on impact.
ii. the loss of energy resulting from the impact. [6 Marks]
**(b)** If A = 4**i** − 5**j** + 3**k**, B = 2**i** − 10**j** − 7**k** and C = 5**i** + 7**j** − 4**k**, deduce:
i. (A × B)·C and A × (B × C).
ii. unit vectors perpendicular to A and lying in the plane of B and C. [5 Marks]
**(c)** A particle of mass 15 kg is placed on a smooth plane inclined at 30° to the horizontal. Find:
i. the acceleration of the object as it moves down the plane.
ii. the velocity attained after 3 seconds if it starts with an initial velocity of 5 ms⁻¹. (Take g = 10 ms⁻²) [4 Marks]
---
### MAT 004: STATISTICS
**7. (a)** The weekly wages of 1000 workmen are normally distributed with mean ₦70 and standard deviation ₦5. Estimate the number of workers whose weekly wages will be:
i. between ₦69 and ₦72.
ii. more than ₦75.
iii. less than ₦63. [6 Marks]
**(b)** The sugar content of five cups of peak milk was measured yielding: 14.5, 14.2, 14.4, 14.3 and 14.6 mg/cup. At α = 0.05, test H₀: μ = 14.0 against H₁: μ ≠ 14.0. [5 Marks]
**(c)** The scores of students in an Economics test have a mean of 8 and standard deviation of 0.6.
i. Find the Z-scores for students whose scores are 7.5, 8.5, 6.8 and 9.0.
ii. Find the probability of selecting at random a student whose score is between 7 and 9. [4 Marks]
---
**8. (a)** A village is classified into two groups X and Y of 300 people each. A drug is administered to group X only. In groups X and Y, 250 and 200 people respectively recovered. Verify that the drug helps in curing the disease at 5% significance level. (Use χ²₀.₀₅;₁ = 3.841) [6 Marks]
**(b)** X is a random variable representing the number of heads in a fair toss of 4 coins.
i. Find the probability mass function (pmf).
ii. Evaluate: a. P(X = 2) b. P(X ≥ 1) [4 Marks]
**(c)** A bag contains 9 balls: 2 red, 3 blue and 4 black. Three balls are drawn at random. Find the probability that:
i. the three balls are of different colours.
ii. the three balls are of the same colour. [5 Marks]
---
---
# ANSWERS
---
## MAT 001: ADVANCED PURE MATHEMATICS
**1(a)(i)**
Solving the simultaneous equations:
2x + y = 5 ... (1)
x − 2y = −5 → x = 2y − 5 ... (2)
Sub (2) into (1): 2(2y−5) + y = 5 → 5y = 15 → y = 3, x = 1
Matrix: $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$
Determinant = (2×2) − (1×3) = 4 − 3 = **1**
---
**1(a)(ii)**
sin A = 2/5 → cos A = √(1 − 4/25) = √(21)/5
cos B = 7/25 → sin B = √(1 − 49/625) = √576/25 = 24/25
cos(A+B) = cos A cos B − sin A sin B
= (√21/5)(7/25) − (2/5)(24/25)
= 7√21/125 − 48/125
= **(7√21 − 48)/125**
---
**1(b)**
Let H = number playing Hockey.
Football F = 2H + 5.
|H ∪ F| = |H| + |F| − |H ∩ F|
80 = H + (2H+5) − 15
80 = 3H − 10 → 3H = 90 → **H = 30**
**i.** F = 2(30) + 5 = **65**
**ii.** Football only = F − both = 65 − 15 = **50**
**iii.** Hockey only = H − both = 30 − 15 = **15**
---
**1(c)(i)**
(x + 3y)⁶ = Σ C(6,r) x^(6−r)(3y)^r
= x⁶ + 6x⁵(3y) + 15x⁴(9y²) + 20x³(27y³) + 15x²(81y⁴) + 6x(243y⁵) + 729y⁶
**= x⁶ + 18x⁵y + 135x⁴y² + 540x³y³ + 1215x²y⁴ + 1458xy⁵ + 729y⁶**
---
**1(c)(ii)**
(1.03)⁶ = (1 + 0.03)⁶
Set x = 1, 3y = 0.03 → y = 0.01:
= 1 + 18(0.01) + 135(0.0001) + 540(0.000001) + ...
= 1 + 0.18 + 0.0135 + 0.00054 + 0.0000001215 + ...
= **1.194** (to 4 s.f.)
More precisely: 1 + 0.18 + 0.0135 + 0.000540 + 0.00001215 + ... ≈ **1.194**
---
**2(a)(i)**
(2 + 3i)² = 4 + 12i + 9i² = 4 + 12i − 9 = −5 + 12i
So a = −5, b = 12.
(2b − 3a)/4ab = (24 − (−15))/(4 × (−5) × 12) = 39/(−240) = **−13/80**
---
**2(a)(ii)**
Z = 2 + i → 1 + Z = 3 + i
Plot point (3, 1) on the Argand diagram with real axis horizontal and imaginary axis vertical. The point 1 + Z lies at coordinates (3, 1).
---
**2(b)(i)**
P(−1) = 0:
(−1)³ + k(−1)² + (−1) + 6 = 0
−1 + k − 1 + 6 = 0
k + 4 = 0 → **k = −4**
---
**2(b)(ii)**
P(x) = x³ − 4x² + x + 6
Dividing by (x + 1):
x³ − 4x² + x + 6 = (x + 1)(x² − 5x + 6) = **(x + 1)(x − 2)(x − 3)**
---
**2(c)**
Length of tangent from external point (x₁, y₁) to circle x² + y² + 2gx + 2fy + c = 0:
L = √(x₁² + y₁² + 2gx₁ + 2fy₁ + c)
Circle: x² + y² + 6x − 4y + 12 = 0, g = 3, f = −2, c = 12
Point: (3, 5)
L = √(9 + 25 + 18 − 20 + 12) = √44 = **2√11 ≈ 6.63 units**
---
## MAT 002: CALCULUS
**3(a)(i)**
$\lim_{x \to 1} \dfrac{\sqrt{5x-4} - \sqrt{x}}{x-1}$ → 0/0 form. Apply L'Hôpital:
Numerator derivative: $\dfrac{5}{2\sqrt{5x-4}} - \dfrac{1}{2\sqrt{x}}$
At x = 1: 5/2 − 1/2 = **2**
---
**3(a)(ii)**
$\lim_{x \to -2} \dfrac{x^2 + 2x + 1}{x^2 + 3x + 2} = \lim_{x \to -2} \dfrac{(x+1)^2}{(x+1)(x+2)} = \lim_{x \to -2} \dfrac{x+1}{x+2} = \dfrac{-1}{0}$
The limit does not exist (tends to −∞ from left, +∞ from right). **The limit is undefined.**
---
**3(b)**
∫√(a² − x²) dx
Let x = a sin θ → dx = a cos θ dθ, √(a²−x²) = a cos θ
∫a cos θ · a cos θ dθ = a²∫cos²θ dθ = a²∫(1 + cos 2θ)/2 dθ
= a²/2 [θ + sin 2θ/2] + C
= a²/2 [θ + sin θ cos θ] + C
Back-substituting (θ = arcsin(x/a), sin θ = x/a, cos θ = √(a²−x²)/a):
**= (a²/2) arcsin(x/a) + (x/2)√(a²−x²) + C**
---
**3(c)**
Iₙ = ∫tanⁿx dx = ∫tanⁿ⁻²x · tan²x dx = ∫tanⁿ⁻²x(sec²x − 1) dx
= ∫tanⁿ⁻²x sec²x dx − ∫tanⁿ⁻²x dx
= tanⁿ⁻¹x/(n−1) − Iₙ₋₂ □
**∫tan⁵x dx:**
I₅ = tan⁴x/4 − I₃
I₃ = tan²x/2 − I₁
I₁ = ∫tan x dx = ln|sec x|
I₃ = tan²x/2 − ln|sec x|
**I₅ = tan⁴x/4 − tan²x/2 + ln|sec x| + C**
---
**4(a)(i)**
(y² + 2)(2x + 1) dy/dx = 1
Separating: (y² + 2) dy = dx/(2x + 1)
∫(y² + 2) dy = ∫dx/(2x + 1)
y³/3 + 2y = (1/2)ln|2x + 1| + C
At y(0) = 1: 1/3 + 2 = (1/2)ln1 + C → C = 7/3
**y³/3 + 2y = (1/2)ln|2x + 1| + 7/3**
---
**4(a)(ii) — Maclaurin series of sin²x**
Using identity: sin²x = (1 − cos 2x)/2
cos 2x = 1 − (2x)²/2! + (2x)⁴/4! − ... = 1 − 2x² + 2x⁴/3 − ...
sin²x = (1 − cos 2x)/2 = (2x² − 2x⁴/3 + ...)/2
**sin²x = x² − x⁴/3 + 2x⁶/45 − ...**
---
**4(b)(i)**
Three sides of rectangle: L + 2B = 60
**L = 60 − 2B**
---
**4(b)(ii)**
A = LB = (60 − 2B)B = 60B − 2B²
dA/dB = 60 − 4B = 0 → B = 15 m
L = 60 − 30 = 30 m
**Maximum area = 30 × 15 = 450 m²**
---
**4(c)**
V = π∫₀^{π/4} y² dx = π∫₀^{π/4} (2cos 2x)² dx = π∫₀^{π/4} 4cos²2x dx
= 4π∫₀^{π/4} (1 + cos 4x)/2 dx = 2π∫₀^{π/4} (1 + cos 4x) dx
= 2π[x + sin 4x/4]₀^{π/4} = 2π[(π/4 + sin π/4) − 0]
= 2π[π/4 + 0] = **π²/2 cubic units**
---
## MAT 003: APPLIED MATHEMATICS
**5(a)**
u = sin t **i** + cos t **j** + 3**k**; v = t**i** + 0**j** − 2**k**
u · v = t sin t + 0 − 6 = t sin t − 6
**i.** d/dt(u · v) = sin t + t cos t
**= sin t + t cos t**
---
**ii.** u × v = |**i** **j** **k**; sin t, cos t, 3; t, 0, −2|
= **i**(cos t × (−2) − 3 × 0) − **j**(sin t × (−2) − 3t) + **k**(sin t × 0 − cos t × t)
= −2cos t **i** − (−2sin t − 3t)**j** − t cos t **k**
= −2cos t **i** + (2sin t + 3t)**j** − t cos t **k**
d/dt(u × v) = 2sin t **i** + (2cos t + 3)**j** + (t sin t − cos t)**k**
**= 2sin t i + (2cos t + 3)j + (t sin t − cos t)k**
---
**5(b)**
m = 50 kg, g = 10 ms⁻²
**i.** Constant speed (a = 0): R = mg = 50 × 10 = **500 N**
**ii.** Upward acceleration (a = 3 ms⁻²): R = m(g + a) = 50(13) = **650 N**
---
**5(c)**
Strings: 6 m, 8 m, ceiling span = 10 m.
Check: 6² + 8² = 36 + 64 = 100 = 10² → right angle at the mass.
Angle of 6 m string to horizontal:
sin α = 8/10 = 0.8 → α = 53.13° ≈ **53.1°**
Angle of 8 m string to horizontal:
sin β = 6/10 = 0.6 → β = 36.87° ≈ **36.9°**
Resolving forces (W = 8.2 × 10 = 82 N):
T₁ sin 53.1° + T₂ sin 36.9° = 82 (vertical)
T₁ cos 53.1° = T₂ cos 36.9° (horizontal)
T₁(0.6) = T₂(0.8) → T₁ = 4T₂/3
Sub: (4T₂/3)(0.8) + T₂(0.6) = 82
3.2T₂/3 + 0.6T₂ = 82
1.0667T₂ + 0.6T₂ = 82
1.6667T₂ = 82 → **T₂ = 49.2 N** (8 m string)
**T₁ = 65.6 N** (6 m string)
---
**6(a)**
m₁ = 4 kg, u₁ = 10 ms⁻¹; m₂ = 16 kg, u₂ = −4 ms⁻¹
**i.** By conservation of momentum (perfectly inelastic):
(4)(10) + (16)(−4) = (4 + 16)v
40 − 64 = 20v → v = −24/20 = **−1.2 ms⁻¹**
The combined mass moves at 1.2 ms⁻¹ in the direction of the 16 kg ball.
**ii.** Initial KE = ½(4)(10²) + ½(16)(4²) = 200 + 128 = 328 J
Final KE = ½(20)(1.2²) = 10 × 1.44 = 14.4 J
**Energy loss = 328 − 14.4 = 313.6 J**
---
**6(b)(i)**
A = 4**i** − 5**j** + 3**k**, B = 2**i** − 10**j** − 7**k**, C = 5**i** + 7**j** − 4**k**
A × B = |**i** **j** **k**; 4, −5, 3; 2, −10, −7|
= **i**[(−5)(−7)−(3)(−10)] − **j**[(4)(−7)−(3)(2)] + **k**[(4)(−10)−(−5)(2)]
= **i**[35+30] − **j**[−28−6] + **k**[−40+10]
= 65**i** + 34**j** − 30**k**
(A × B)·C = 65(5) + 34(7) + (−30)(−4) = 325 + 238 + 120 = **683**
B × C = |**i** **j** **k**; 2, −10, −7; 5, 7, −4|
= **i**[(−10)(−4)−(−7)(7)] − **j**[(2)(−4)−(−7)(5)] + **k**[(2)(7)−(−10)(5)]
= **i**[40+49] − **j**[−8+35] + **k**[14+50]
= 89**i** − 27**j** + 64**k**
A × (B × C) = |**i** **j** **k**; 4, −5, 3; 89, −27, 64|
= **i**[(−5)(64)−(3)(−27)] − **j**[(4)(64)−(3)(89)] + **k**[(4)(−27)−(−5)(89)]
= **i**[−320+81] − **j**[256−267] + **k**[−108+445]
= **−239i + 11j + 337k**
---
**6(b)(ii)**
Unit vectors perpendicular to A in the plane of B and C:
A vector in the plane of B and C: D = B + λC = (2+5λ)**i** + (−10+7λ)**j** + (−7−4λ)**k**
For D ⊥ A: A · D = 0
4(2+5λ) + (−5)(−10+7λ) + 3(−7−4λ) = 0
8 + 20λ + 50 − 35λ − 21 − 12λ = 0
37 − 27λ = 0 → λ = 37/27
D = (2 + 185/27)**i** + (−10 + 259/27)**j** + (−7 − 148/27)**k**
= (239/27)**i** + (−11/27)**j** + (−337/27)**k**
|D| = (1/27)√(239² + 11² + 337²) = (1/27)√(57121 + 121 + 113569) = (1/27)√170811 ≈ (1/27)(413.3) ≈ 15.31
**Unit vector = (239i − 11j − 337k)/(27 × 15.31) ≈ (239i − 11j − 337k)/413.3**
---
**6(c)**
Smooth plane, θ = 30°, g = 10 ms⁻²
**i.** a = g sin 30° = 10 × 0.5 = **5 ms⁻²**
**ii.** v = u + at = 5 + 5(3) = **20 ms⁻¹**
---
## MAT 004: STATISTICS
**7(a)**
μ = ₦70, σ = ₦5, N = 1000
**i.** P(69 < X < 72):
Z₁ = (69−70)/5 = −0.2; Z₂ = (72−70)/5 = 0.4
P(−0.2 < Z < 0.4) = Φ(0.4) − Φ(−0.2) = 0.6554 − 0.4207 = 0.2347
Number = 0.2347 × 1000 ≈ **235 workers**
**ii.** P(X > 75):
Z = (75−70)/5 = 1.0
P(Z > 1) = 1 − 0.8413 = 0.1587
Number = 0.1587 × 1000 ≈ **159 workers**
**iii.** P(X < 63):
Z = (63−70)/5 = −1.4
P(Z < −1.4) = 1 − Φ(1.4) = 1 − 0.9192 = 0.0808
Number = 0.0808 × 1000 ≈ **81 workers**
---
**7(b) — One-sample t-test**
Data: 14.5, 14.2, 14.4, 14.3, 14.6; n = 5
x̄ = (14.5+14.2+14.4+14.3+14.6)/5 = 72/5 = **14.4**
s² = Σ(xᵢ − x̄)²/(n−1):
Deviations: 0.1, −0.2, 0, −0.1, 0.2
Squared: 0.01, 0.04, 0, 0.01, 0.04 → Σ = 0.10
s² = 0.10/4 = 0.025 → s = 0.158
t = (x̄ − μ₀)/(s/√n) = (14.4 − 14.0)/(0.158/√5) = 0.4/0.0707 = **5.66**
Critical t (df = 4, α = 0.05, two-tailed): t_crit = 2.776
Since 5.66 > 2.776, **reject H₀**. The mean sugar content differs significantly from 14.0.
---
**7(c)**
μ = 8, σ = 0.6
**i. Z-scores:**
Z(7.5) = (7.5−8)/0.6 = **−0.833**
Z(8.5) = (8.5−8)/0.6 = **0.833**
Z(6.8) = (6.8−8)/0.6 = **−2.000**
Z(9.0) = (9.0−8)/0.6 = **1.667**
**ii.** P(7 < X < 9):
Z₁ = (7−8)/0.6 = −1.667; Z₂ = (9−8)/0.6 = 1.667
P(−1.667 < Z < 1.667) = 2Φ(1.667) − 1 = 2(0.9525) − 1 = **0.9050**
---
**8(a) — Chi-Square Test**
Observed table:
| | Recovered | Not Recovered | Total |
|---|---|---|---|
| Group X (drug) | 250 | 50 | 300 |
| Group Y (no drug) | 200 | 100 | 300 |
| **Total** | **450** | **150** | **600** |
Expected values (E = row total × col total / grand total):
E(X, recovered) = 300×450/600 = 225
E(X, not recovered) = 300×150/600 = 75
E(Y, recovered) = 225; E(Y, not recovered) = 75
χ² = (250−225)²/225 + (50−75)²/75 + (200−225)²/225 + (100−75)²/75
= 625/225 + 625/75 + 625/225 + 625/75
= 2.778 + 8.333 + 2.778 + 8.333 = **22.22**
Critical value χ²₀.₀₅;₁ = 3.841
Since 22.22 > 3.841, **reject H₀**. The drug significantly helps in curing the disease.
---
**8(b)**
X ~ B(4, 0.5), total outcomes = 2⁴ = 16
**i. PMF:**
| X | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| P(X) | 1/16 | 4/16 | 6/16 | 4/16 | 1/16 |
**ii. a.** P(X = 2) = 6/16 = **3/8**
**ii. b.** P(X ≥ 1) = 1 − P(X=0) = 1 − 1/16 = **15/16**
---
**8(c)**
Total ways to draw 3 from 9: C(9,3) = **84**
**i. Three different colours (1R, 1B, 1Bk):**
Ways = C(2,1) × C(3,1) × C(4,1) = 2 × 3 × 4 = 24
P = 24/84 = **2/7**
**ii. Three same colour:**
3 blue: C(3,3) = 1
3 black: C(4,3) = 4
(Cannot get 3 red as only 2 exist)
Total = 1 + 4 = 5
P = 5/84 = **5/84**