2020 JUPEB Mathematics

MAT 001: 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]

© 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]

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]

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]

MAT 003: APPLIED MATHEMATICS

5. (a) Given φ(t) = eᵗ; u(t) = sin t i + cos t j + 3k; v(t) = ti − 2k.

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]

© 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 = 4i − 5j + 3k, B = 2i − 10j − 7k and C = 5i + 7j − 4k, 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]

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]

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]

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©(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©(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©

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©

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©

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 + 3k; v = ti + 0j − 2k

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©

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 = 4i − 5j + 3k, B = 2i − 10j − 7k, C = 5i + 7j − 4k

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]

= 65i + 34j − 30k

(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]

= 89i − 27j + 64k

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©

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©

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:

|—|---|—|---|

| 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©

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

2020 JUPEB Mathematics

MAT 001: PURE MATHEMATICS

MAT 002: CALCULUS

MAT 003: APPLIED MATHEMATICS

MAT 004: STATISTICS

ANSWERS

MAT 001: ADVANCED PURE MATHEMATICS

MAT 002: CALCULUS

MAT 003: APPLIED MATHEMATICS

MAT 004: STATISTICS

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2020 JUPEB Mathematics

MAT 001: PURE MATHEMATICS

MAT 002: CALCULUS

MAT 003: APPLIED MATHEMATICS

MAT 004: STATISTICS

ANSWERS

MAT 001: ADVANCED PURE MATHEMATICS

MAT 002: CALCULUS

MAT 003: APPLIED MATHEMATICS

MAT 004: STATISTICS

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