### MAT 001: PURE MATHEMATICS
**1. (a) i.** Consider two lines L₁ and L₂ that are perpendicular to each other. Show that m₂ = −1/m₁ or m₁ = −1/m₂. [3 Marks]
**ii.** Find the equation of the straight line through (1, −2) which is perpendicular to 3x − 2y = 6. [3 Marks]
**(b) i.** Find the centre and radius of the circle x² + y² + 3x − 2y + 2 = 0. [3 Marks]
**ii.** If the ends of a diameter of a circle have coordinates (5, 4) and (−3, 6), find the equation of the circle. [3 Marks]
**(c)** Find the equation of the tangent to the circle x² + y² + 2x − 3y = −1 at the point (3, 4). [3 Marks]
---
**2. (a)** For what values of x is |x − 1| > 2|x + 3|? [5 Marks]
**(b) i.** Write 1 + i√3 and 1 − i√3 in trigonometric form and simplify (1 + i√3)⁹ + (1 − i√3)⁹ in x + iy form. [3 Marks]
**ii.** Find tan(7π/12) using the known values of tan(π/3) and tan(π/4). [3 Marks]
**(c)** Prove by induction that 1 + 3 + 5 + ... + (2n − 1) = n². [4 Marks]
---
### MAT 002: CALCULUS
**3.** Integrate the following with respect to x:
**(a)** y = (e³ˣ + 3)/(e³ˣ + 9x + 2) [5 Marks]
**(b)** y = 8x/(3 + 2x)⁴ [5 Marks]
**(c)** y = 1/√(9 − 4x²) [5 Marks]
---
**4. (a)** Determine the initial value problem dy/dx = (y − 1)/(x + 3), y(−1) = 0. [5 Marks]
**(b)** Solve the initial value problem y″ + 2y′ − y = 0; y(0) = 0, y′(0) = −1. [5 Marks]
**(c)** Solve the differential equations:
i. (x² + 1) dy/dx + (y² − 1) = 0 [2½ Marks]
ii. dy/dx = (x² − y²)/2x² [2½ Marks]
---
### MAT 003: APPLIED MATHEMATICS
**5. (a)** Find the area of the triangle whose two adjacent sides are given by vectors a = 2**i** − 3**j** + **k** and b = 2**i** + 3**j** − 4**k**. [5 Marks]
**(b)** A body of mass 6 kg rests on a smooth plane inclined at 60° to the horizontal. This body is attached to another body of mass 10 kg by a light inextensible string. Find:
i. the acceleration of the system. [3 Marks]
ii. the tension in the string. (Take g = 10 ms⁻²) [2 Marks]
**(c)** A car of mass 200 kg moves forward on a road with an acceleration of 0.8 ms⁻² and covers a distance of 200 m. If the road has a resistance of 40 N, calculate the driving power of the car if the trip lasted 20 seconds. [5 Marks]
---
**6. (a)** A man paddles a boat downstream a distance of 3 km in 20 minutes and rows back in 40 minutes. Find his average speed:
i. downstream [1½ Marks]
ii. upstream [1½ Marks]
iii. for the double journey [1½ Marks]
**(b)** A rifle bullet passes through two planks in succession, and the average resistance of the second plank is 50% more than that of the first. The initial velocity is 800 ms⁻¹ and the bullet loses 160 ms⁻¹ in passing through each plank. Show that the ratio of the thickness of the planks is 27:14. [4½ Marks]
**(c) i.** Find the constant c such that a = **i** − **j** + 2**k** and b = **i** + 3**j** − c**k** are perpendicular. [2 Marks]
**ii.** Find the area of triangle ABC where the position vectors of A, B and C are −**j** + 2**k**, **i** + 4**j** − 3**k** and **i** − 3**j** + 2**k** respectively. [4 Marks]
---
### MAT 004: STATISTICS
**7. (a)** Define the following concepts associated with hypothesis testing:
i. Hypothesis [1 Mark]
ii. Null hypothesis [1 Mark]
iii. Alternative hypothesis [1 Mark]
iv. Level of significance of a test [1 Mark]
v. Critical region of a test [1 Mark]
**(b)** A sample of 100 students enrolled in a university indicates a mean textbook cost of ₦315.40 with a standard deviation of ₦43.20.
i. Using a 1% level of significance, is there evidence that the population average is above ₦300? [2½ Marks]
ii. What is your answer if the standard deviation is 75 and a 5% level of significance is used? [2½ Marks]
**(c)** A random variable X has the following distribution:
| x | 1 | 2 | 3 |
|---|---|---|---|
| P(X=x) | 0.2 | 0.5 | 0.3 |
Calculate: (i) E(6X) (ii) E(5X + 2) [5 Marks]
---
**8. (a)** The table below shows the distribution of heights (in cm) of some athletes. Using an assumed mean of 167 cm, find (i) the mean and (ii) the standard deviation of the distribution. [8 Marks]
| Height (cm) | 150–154 | 155–159 | 160–164 | 165–169 | 170–174 | 175–179 | 180–184 |
|---|---|---|---|---|---|---|---|
| No. of Athletes | 4 | 14 | 26 | 32 | 21 | 10 | 3 |
**(b)** Find the coefficient of linear correlation between X and Y: [5 Marks]
| X | 1 | 2 | 3 | 6 | 8 |
|---|---|---|---|---|---|
| Y | 1 | 2 | 3 | 4 | 5 |
**(c)** What is your conclusion from (b) above? [2 Marks]
---
---
# ANSWERS
---
## MAT 001: PURE MATHEMATICS
**1(a)(i)**
Let L₁ have inclination angle θ to the horizontal, so m₁ = tan θ.
If L₂ ⊥ L₁, its inclination is θ + 90°.
m₂ = tan(θ + 90°) = −cot θ = −1/tan θ = **−1/m₁** □
---
**1(a)(ii)**
3x − 2y = 6 → y = (3/2)x − 3, so m₁ = 3/2.
Perpendicular gradient: m₂ = −2/3.
Line through (1, −2): y + 2 = −(2/3)(x − 1)
3y + 6 = −2x + 2
**2x + 3y + 4 = 0**
---
**1(b)(i)**
x² + y² + 3x − 2y + 2 = 0
Complete the square:
(x + 3/2)² − 9/4 + (y − 1)² − 1 = −2
(x + 3/2)² + (y − 1)² = −2 + 9/4 + 1 = 5/4
**Centre: (−3/2, 1); Radius: √5/2**
---
**1(b)(ii)**
Centre = midpoint of diameter = ((5+(−3))/2, (4+6)/2) = **(1, 5)**
r² = (5−1)² + (4−5)² = 16 + 1 = 17
**(x − 1)² + (y − 5)² = 17**
---
**1(c)**
Circle: x² + y² + 2x − 3y + 1 = 0
Differentiating implicitly:
2x + 2y(dy/dx) + 2 − 3(dy/dx) = 0
dy/dx(2y − 3) = −2x − 2
dy/dx = −(2x + 2)/(2y − 3)
At (3, 4): dy/dx = −(6+2)/(8−3) = −8/5
Tangent: y − 4 = −(8/5)(x − 3)
5y − 20 = −8x + 24
**8x + 5y − 44 = 0**
---
**2(a)**
|x − 1| > 2|x + 3|
Squaring both sides:
(x − 1)² > 4(x + 3)²
x² − 2x + 1 > 4x² + 24x + 36
0 > 3x² + 26x + 35
(3x + 5)(x + 7) < 0
**−7 < x < −5/3**
---
**2(b)(i)**
1 + i√3: r = √(1+3) = 2; θ = arctan(√3/1) = π/3
**1 + i√3 = 2(cos π/3 + i sin π/3)**
1 − i√3: r = 2; θ = −π/3
**1 − i√3 = 2(cos π/3 − i sin π/3)**
By De Moivre's theorem:
(1 + i√3)⁹ = 2⁹(cos 3π + i sin 3π) = 512(−1 + 0) = −512
(1 − i√3)⁹ = 2⁹(cos(−3π) + i sin(−3π)) = 512(−1 + 0) = −512
**(1 + i√3)⁹ + (1 − i√3)⁹ = −1024 + 0i**
---
**2(b)(ii)**
tan(7π/12) = tan(π/3 + π/4)
= (tan π/3 + tan π/4)/(1 − tan π/3 · tan π/4)
= (√3 + 1)/(1 − √3)
= (√3 + 1)(1 + √3)/[(1 − √3)(1 + √3)]
= (√3 + 3 + 1 + √3)/(1 − 3)
= (4 + 2√3)/(−2)
**tan(7π/12) = −(2 + √3)**
---
**2(c) — Proof by Induction**
**Claim:** 1 + 3 + 5 + ... + (2n−1) = n²
**Base case (n=1):** LHS = 1; RHS = 1² = 1. ✓
**Inductive step:** Assume true for n = k:
1 + 3 + ... + (2k−1) = k²
For n = k+1:
1 + 3 + ... + (2k−1) + (2k+1) = k² + 2k + 1 = (k+1)² ✓
By induction, the result holds for all n ∈ ℕ. □
---
## MAT 002: CALCULUS
**3(a)**
∫(e³ˣ + 3)/(e³ˣ + 9x + 2) dx
Let u = e³ˣ + 9x + 2 → du = (3e³ˣ + 9)dx = 3(e³ˣ + 3)dx
∫(1/3)(1/u) du = **(1/3)ln|e³ˣ + 9x + 2| + C**
---
**3(b)**
∫8x/(3 + 2x)⁴ dx
Let u = 3 + 2x → x = (u−3)/2, dx = du/2
∫8·(u−3)/2 · u⁻⁴ · (du/2) = 2∫(u−3)u⁻⁴ du
= 2∫(u⁻³ − 3u⁻⁴) du
= 2[u⁻²/(−2) − 3u⁻³/(−3)] + C
= 2[−1/(2u²) + 1/u³] + C
= −1/u² + 2/u³ + C
**= −1/(3+2x)² + 2/(3+2x)³ + C**
---
**3(c)**
∫dx/√(9 − 4x²) = ∫dx/√[9(1 − 4x²/9)]
= (1/3)∫dx/√(1 − (2x/3)²)
Let u = 2x/3 → du = 2dx/3 → dx = 3du/2
= (1/3)·(3/2)∫du/√(1−u²) = (1/2) arcsin u + C
**= (1/2) arcsin(2x/3) + C**
---
**4(a)**
dy/dx = (y−1)/(x+3), y(−1) = 0
Separating: dy/(y−1) = dx/(x+3)
∫dy/(y−1) = ∫dx/(x+3)
ln|y−1| = ln|x+3| + C
y − 1 = A(x + 3)
At y(−1) = 0: −1 = A(2) → A = −1/2
**y − 1 = −(x+3)/2 → y = 1 − (x+3)/2 = (−x−1)/2**
---
**4(b)**
y″ + 2y′ − y = 0; y(0) = 0, y′(0) = −1
Characteristic equation: r² + 2r − 1 = 0
r = (−2 ± √(4+4))/2 = −1 ± √2
General solution: y = Ae^((−1+√2)x) + Be^((−1−√2)x)
At y(0) = 0: A + B = 0 → B = −A
y′ = A(−1+√2)e^((−1+√2)x) + B(−1−√2)e^((−1−√2)x)
At y′(0) = −1:
A(−1+√2) + B(−1−√2) = −1
A(−1+√2) − A(−1−√2) = −1
A·2√2 = −1 → A = −1/(2√2)
B = 1/(2√2)
**y = (1/(2√2))[e^((−1+√2)x) − e^((−1−√2)x)] · (−1)**
**y = −e⁻ˣ sinh(√2 x)/√2**
Or equivalently: **y = (−1/2√2)[e^((−1+√2)x) − e^((−1−√2)x)]**
---
**4(c)(i)**
(x² + 1)dy/dx = 1 − y²
dy/(1−y²) = dx/(x²+1)
∫dy/(1−y²) = ∫dx/(x²+1)
(1/2)ln|(1+y)/(1−y)| = arctan x + C
**ln|(1+y)/(1−y)| = 2 arctan x + C**
---
**4(c)(ii)**
dy/dx = (x² − y²)/2x²
Let v = y/x → y = vx, dy/dx = v + x(dv/dx)
v + x(dv/dx) = (x² − v²x²)/(2x²) = (1 − v²)/2
x(dv/dx) = (1 − v²)/2 − v = (1 − v² − 2v)/2
2dv/(1 − v² − 2v) = dx/x
−2dv/(v² + 2v − 1) = dx/x
Let v² + 2v − 1 = (v+1)² − 2:
−2∫dv/[(v+1)² − 2] = ∫dx/x
Using ∫dv/(v² − a²) = (1/2a)ln|(v−a)/(v+a)|, with a = √2:
−(1/√2)ln|(v+1−√2)/(v+1+√2)| = ln|x| + C
Back-substituting v = y/x:
**−(1/√2)ln|(y+x−√2 x)/(y+x+√2 x)| = ln|x| + C**
---
## MAT 003: APPLIED MATHEMATICS
**5(a)**
a = 2**i** − 3**j** + **k**, b = 2**i** + 3**j** − 4**k**
a × b = |**i** **j** **k**; 2, −3, 1; 2, 3, −4|
= **i**[(−3)(−4)−(1)(3)] − **j**[(2)(−4)−(1)(2)] + **k**[(2)(3)−(−3)(2)]
= **i**[12−3] − **j**[−8−2] + **k**[6+6]
= 9**i** + 10**j** + 12**k**
|a × b| = √(81 + 100 + 144) = √325 = 5√13
Area = (1/2)|a × b| = **(5√13)/2 ≈ 9.01 sq units**
---
**5(b)**
6 kg on smooth incline at 60°; connected to 10 kg hanging mass.
Forces: 10g − 6g sin60° = (10+6)a
10(10) − 6(10)(√3/2) = 16a
100 − 30√3 = 16a
100 − 51.96 = 16a
48.04 = 16a
**i. a = 3.00 ms⁻²**
**ii.** T = 10(g − a) = 10(10 − 3.00) = **70.0 N**
---
**5(c)**
m = 200 kg, a = 0.8 ms⁻², s = 200 m, resistance = 40 N, t = 20 s
Driving force F: F − 40 = ma = 200 × 0.8 = 160
F = **200 N**
Velocity after 20 s (from s = ut + ½at², u = 0):
200 = ½(0.8)(20²) = 160 m — does not match exactly.
Using v = u + at (assuming start from rest):
v = 0 + 0.8 × 20 = **16 ms⁻¹**
Power = F × v = 200 × 16 = **3200 W = 3.2 kW**
---
**6(a)**
Distance each way = 3 km.
**i.** Downstream: 3 km in 20 min = 1/3 hr → speed = 3/(1/3) = **9 km/h**
**ii.** Upstream: 3 km in 40 min = 2/3 hr → speed = 3/(2/3) = **4.5 km/h**
**iii.** Double journey: total distance = 6 km, total time = 1 hr
Average speed = 6/1 = **6 km/h**
---
**6(b)**
Let resistance of plank 1 = R₁; plank 2 = 1.5R₁.
Let thicknesses be d₁ and d₂.
Using v² = u² − 2as (where a = F/m):
**Plank 1:** Entry = 800, exit = 640 ms⁻¹
800² − 640² = 2(R₁/m)d₁
(640000 − 409600) = 2(R₁/m)d₁
230400 = 2(R₁/m)d₁ ... (1)
**Plank 2:** Entry = 640, exit = 480 ms⁻¹
640² − 480² = 2(1.5R₁/m)d₂
(409600 − 230400) = 3(R₁/m)d₂
179200 = 3(R₁/m)d₂ ... (2)
From (1): (R₁/m)d₁ = 115200
From (2): (R₁/m)d₂ = 179200/3 = 59733.3
d₁/d₂ = 115200/59733.3 = 115200 × 3/179200 = 345600/179200 = 27/14 □
---
**6(c)(i)**
a · b = 0 (perpendicular condition):
(1)(1) + (−1)(3) + (2)(−c) = 0
1 − 3 − 2c = 0
−2 = 2c → **c = −1**
---
**6(c)(ii)**
A: (0, −1, 2), B: (1, 4, −3), C: (1, −3, 2)
AB = B − A = (1, 5, −5)
AC = C − A = (1, −2, 0)
AB × AC = |**i** **j** **k**; 1, 5, −5; 1, −2, 0|
= **i**[(5)(0)−(−5)(−2)] − **j**[(1)(0)−(−5)(1)] + **k**[(1)(−2)−(5)(1)]
= **i**[0−10] − **j**[0+5] + **k**[−2−5]
= −10**i** − 5**j** − 7**k**
|AB × AC| = √(100 + 25 + 49) = √174
Area = (1/2)√174 = **√174/2 ≈ 6.60 sq units**
---
## MAT 004: STATISTICS
**7(a) Definitions**
**i. Hypothesis:** A statement or claim about a population parameter that is subject to statistical testing.
**ii. Null hypothesis (H₀):** The hypothesis of no effect or no difference; the statement assumed true until evidence suggests otherwise.
**iii. Alternative hypothesis (H₁):** The hypothesis that contradicts H₀; accepted when H₀ is rejected.
**iv. Level of significance (α):** The maximum probability of committing a Type I error (rejecting a true H₀); commonly 0.05 or 0.01.
**v. Critical region:** The set of values of the test statistic for which H₀ is rejected; determined by α and the distribution.
---
**7(b)(i)**
H₀: μ = 300; H₁: μ > 300 (one-tailed)
x̄ = 315.40, s = 43.20, n = 100, α = 0.01
Z = (315.40 − 300)/(43.20/√100) = 15.40/4.32 = **3.565**
Critical Z at α = 0.01 (one-tailed) = 2.326
Since 3.565 > 2.326, **reject H₀**. There is evidence that the population average exceeds ₦300.
---
**7(b)(ii)**
s = 75, α = 0.05
Z = (315.40 − 300)/(75/√100) = 15.40/7.5 = **2.053**
Critical Z at α = 0.05 (one-tailed) = 1.645
Since 2.053 > 1.645, **reject H₀**. The conclusion remains the same — the average exceeds ₦300.
---
**7(c)**
E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1.0 + 0.9 = 2.1
**i.** E(6X) = 6E(X) = 6 × 2.1 = **12.6**
**ii.** E(5X + 2) = 5E(X) + 2 = 5(2.1) + 2 = 10.5 + 2 = **12.5**
---
**8(a)**
Assumed mean A = 167; class width h = 5. Midpoints and d = (x − 167)/5:
| Class | Midpoint x | f | d=(x−167)/5 | fd | fd² |
|---|---|---|---|---|---|
| 150–154 | 152 | 4 | −3 | −12 | 36 |
| 155–159 | 157 | 14 | −2 | −28 | 56 |
| 160–164 | 162 | 26 | −1 | −26 | 26 |
| 165–169 | 167 | 32 | 0 | 0 | 0 |
| 170–174 | 172 | 21 | 1 | 21 | 21 |
| 175–179 | 177 | 10 | 2 | 20 | 40 |
| 180–184 | 182 | 3 | 3 | 9 | 27 |
| **Total** | | **110** | | **−16** | **206** |
**i. Mean:**
x̄ = A + h(Σfd/Σf) = 167 + 5(−16/110) = 167 − 0.727 = **166.27 cm**
**ii. Standard deviation:**
σ = h√[Σfd²/Σf − (Σfd/Σf)²]
= 5√[206/110 − (−16/110)²]
= 5√[1.8727 − 0.02118]
= 5√1.8515
= 5 × 1.3608
= **6.80 cm**
---
**8(b)**
| X | Y | X² | Y² | XY |
|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 1 |
| 2 | 2 | 4 | 4 | 4 |
| 3 | 3 | 9 | 9 | 9 |
| 6 | 4 | 36 | 16 | 24 |
| 8 | 5 | 64 | 25 | 40 |
| **20** | **15** | **114** | **55** | **78** |
n = 5
r = [nΣXY − ΣXΣY]/√{[nΣX² − (ΣX)²][nΣY² − (ΣY)²]}
= [5(78) − (20)(15)]/√{[5(114) − 400][5(55) − 225]}
= [390 − 300]/√{[170][50]}
= 90/√8500
= 90/92.195
= **0.976**
---
**8(c)**
Since r = 0.976 is very close to +1, there is a **strong positive linear correlation** between X and Y. As X increases, Y tends to increase proportionally. The relationship between the two variables is highly linear.