MAT 001: ADVANCED PURE MATHEMATICS
1. (a) (i). The sum of three consecutive terms of a geometric progression is 14 and the product is 64. Find the terms of the geometric progression.
(ii). Solve for x in the equation: 4ˣ + 2(2ˣ) = 8 [3 Marks]
(b) Identify the conic section defined by 4x² − 25y² + 16x + 50y − 109 = 0. Hence obtain the centre, vertex, focus and the asymptote of the conic. [2 Marks] [6 Marks]
© A cubic polynomial is given by f(x) = 3x³ + 4x² + 2x + product of three linear factors [4 Marks]
[TOTAL = 15 Marks]
2. (a) Resolve the following in partial fractions: (2x² + 10x + 4) / [(x + 2)(2x² + x − 2)] [7 Marks]
(b) cos 60° = 2tan15° / (1 + tan²15°). Find tan15° [5 Marks]
© μ = {x: 1 ≤ x < 10, x ∈ Z}, A = {x: x is a prime number} and B = {x: x is odd}. Find AΔB. [3 Marks]
[TOTAL = 15 Marks]
MAT (Calculus Section)
3. (a) Use first principle to differentiate y = −cos7x. [5 Marks]
(b) Solve the differential equation: xy’ − 2y = x² at x = 1/3 [5 Marks]
© Find the Taylor series expansion of f(x) = 64ˣ at x = 1 [5 Marks]
[TOTAL = 15 Marks]
4. (a) Find the general solution to the equation: d²y/dx² − 4(dy/dx) − 12y = 3e^(5t) [6 Marks]
(b) Use Trapezoidal rule to compute ∫₀^π sin x dx for n = 5 [4 Marks] [5 Marks]
[TOTAL = 15 Marks]
6. (a) A guest house has 5 rooms for rent per day and the number of demands per day follows a Poisson distribution with mean 1.2. Calculate the probability that:
(i). All rooms are vacant on a particular day
(ii). Some clients are rejected on a day [4 Marks]
(b) If the function:
f(x) = 0, x < 2
f(x) = k(3 + 2x), 2 ≤ x ≤ 4
f(x) = 0, x > 4
is a density function, then find:
(i). The value of k [2 Marks]
(ii). The probability that the random variable is in the interval 2 ≤ x ≤ 3 [2 Marks]
© (i). Given a binomial random variable X with the usual notations, find P if n = 6 and 9P(X = 4) = P(X = 2). [4 Marks]
(ii). If 40% of the rings manufactured by a machine are defective, determine the probability that out of 3 rings chosen at random at most 2 rings will be defective. [3 Marks]
[TOTAL = 15 Marks]
MAT 004: APPLIED MATHEMATICS
The vectors P̄ = −4i + 2j − 3k, Q̄ = 4i − j − k, and R̄ = 5i + k respectively define the position of the three vertices P, Q and R of a triangle.
Determine the length of the projection of side PQ onto side QR [5 Marks]
MAT 001: ADVANCED PURE MATHEMATICS
Question 1
(a)(i) Sum of three consecutive terms of a GP = 14, product = 64
Let the three terms be a/r, a, ar
Product:
(a/r)(a)(ar) = a³ = 64
∴ a = 4
Sum:
a/r + a + ar = 14
4/r + 4 + 4r = 14
4/r + 4r = 10
Multiply through by r:
4 + 4r² = 10r
4r² − 10r + 4 = 0
2r² − 5r + 2 = 0
(2r − 1)(r − 2) = 0
∴ r = 1/2 or r = 2
When r = 2: terms are 4/2, 4, 4(2) = 2, 4, 8
When r = 1/2: terms are 4/(1/2), 4, 4(1/2) = 8, 4, 2
∴ The three terms are 2, 4, 8
(a)(ii) Solve: 4ˣ + 2(2ˣ) = 8
Let u = 2ˣ, so 4ˣ = (2²)ˣ = u²
u² + 2u = 8
u² + 2u − 8 = 0
(u + 4)(u − 2) = 0
u = −4 or u = 2
Since 2ˣ > 0, u = 2
2ˣ = 2¹
∴ x = 1
(b) Identify conic: 4x² − 25y² + 16x + 50y − 109 = 0
Complete the square:
4(x² + 4x) − 25(y² − 2y) = 109
4(x² + 4x + 4) − 25(y² − 2y + 1) = 109 + 16 − 25
4(x + 2)² − 25(y − 1)² = 100
Divide by 100:
(x + 2)²/25 − (y − 1)²/4 = 1
This is a HYPERBOLA (difference of squares form)
Centre: (−2, 1)
Here a² = 25, b² = 4 → a = 5, b = 2
c² = a² + b² = 25 + 4 = 29 → c = √29
Vertices: (−2 ± 5, 1) = (3, 1) and (−7, 1)
Foci: (−2 ± √29, 1)
= (−2 + √29, 1) and (−2 − √29, 1)
Asymptotes:
y − 1 = ±(b/a)(x + 2)
y − 1 = ±(2/5)(x + 2)
y = 1 + (2/5)(x + 2) and y = 1 − (2/5)(x + 2)
© Express f(x) = 3x³ + 4x² + 2x + 1 as product of three linear factors
Try rational roots (factors of 1 over factors of 3): ±1, ±1/3
Test x = −1:
3(−1) + 4(1) + 2(−1) + 1 = −3 + 4 − 2 + 1 = 0
So (x + 1) is a factor. Divide:
3x³ + 4x² + 2x + 1 ÷ (x + 1):
3x³ + 4x² + 2x + 1 = (x+1)(3x² + x + 1)
Check discriminant of 3x² + x + 1:
Δ = 1 − 12 = −11 < 0 (no real roots)
∴ Over the reals:
f(x) = (x + 1)(3x² + x + 1)
Over complex numbers, factoring 3x² + x + 1:
x = (−1 ± i√11)/6
∴ f(x) = (x+1)(x − (−1+i√11)/6)(x − (−1−i√11)/6)
Question 2
(a) Partial fractions: (2x² + 10x + 4) / [(x+2)(2x² + x − 2)]
First factorise 2x² + x − 2:
Δ = 1 + 16 = 17
x = (−1 ± √17)/4 → does not factor nicely, so keep as irreducible quadratic.
Set up:
(2x² + 10x + 4) / [(x+2)(2x² + x − 2)] = A/(x+2) + (Bx + C)/(2x² + x − 2)
Multiply both sides by (x+2)(2x² + x − 2):
2x² + 10x + 4 = A(2x² + x − 2) + (Bx + C)(x + 2)
x = −2:
2(4) + 10(−2) + 4 = A(2(4) + (−2) − 2)
8 − 20 + 4 = A(8 − 2 − 2)
−8 = 4A
A = −2
Expand right side:
A(2x² + x − 2) + (Bx + C)(x + 2)
= 2Ax² + Ax − 2A + Bx² + 2Bx + Cx + 2C
= (2A + B)x² + (A + 2B + C)x + (−2A + 2C)
Compare coefficients:
x²: 2 = 2A + B = 2(−2) + B → B = 2 + 4 = 6
constant: 4 = −2A + 2C = −2(−2) + 2C = 4 + 2C → 2C = 0 → C = 0
Check x: 10 = A + 2B + C = −2 + 12 + 0 = 10 ✓
∴ (2x² + 10x + 4)/[(x+2)(2x²+x−2)] = −2/(x+2) + 6x/(2x²+x−2)
(b) cos 60° = 2tan15°/(1 + tan²15°), find tan15°
We know the identity: sin 2θ = 2tanθ/(1 + tan²θ)
So 2tan15°/(1 + tan²15°) = sin 30° = 1/2
But the question states this equals cos 60° = 1/2 ✓ (consistent)
So: 2tan15°/(1 + tan²15°) = 1/2
Let t = tan15°:
4t = 1 + t²
t² − 4t + 1 = 0
t = (4 ± √(16−4))/2 = (4 ± √12)/2 = 2 ± √3
Since 15° is acute and tan15° < tan45° = 1:
tan15° = 2 − √3
© μ = {1,2,3,…,9}, A = {primes}, B = {odd numbers}, Find AΔB
A (primes in μ): {2, 3, 5, 7}
B (odd numbers in μ): {1, 3, 5, 7, 9}
AΔB = (A∪B) − (A∩B) = elements in A or B but NOT both
A∩B = {3, 5, 7}
A∪B = {1, 2, 3, 5, 7, 9}
AΔB = {1, 2, 9}
Question 3
(a) Differentiate y = −cos7x from first principles
f(x) = −cos7x
f(x+h) = −cos7(x+h) = −cos(7x+7h)
f’(x) = lim[h→0] [f(x+h) − f(x)]/h
= lim[h→0] [−cos(7x+7h) + cos7x]/h
= lim[h→0] [cos7x − cos(7x+7h)]/h
Using: cos A − cos B = −2 sin((A+B)/2) sin((A−B)/2)
= lim[h→0] [−2 sin(7x + 7h/2) · sin(−7h/2)] / h
= lim[h→0] [2 sin(7x + 7h/2) · sin(7h/2)] / h
= lim[h→0] sin(7x + 7h/2) · [sin(7h/2)/(h/2)] · (1/1) · (7/7)
Wait, let me redo cleanly:
= lim[h→0] [2 sin(7x + 7h/2) · sin(7h/2)] / h
Multiply and divide by 7/2:
= lim[h→0] sin(7x + 7h/2) · [sin(7h/2)/(7h/2)] · 7
As h→0: sin(7x + 7h/2) → sin7x, and sin(7h/2)/(7h/2) → 1
∴ f’(x) = 7sin7x
(b) Solve: xy’ − 2y = x² at x = 1/3…
(Note: Reading the image, this appears to be a first-order linear ODE: xy’ − 2y = x², with initial condition at x = 1/3)
Rewrite: dy/dx − (2/x)y = x
Integrating factor: μ = e^∫(−2/x)dx = e^(−2ln|x|) = x^(−2) = 1/x²
Multiply through:
(1/x²)dy/dx − (2/x³)y = 1/x
d/dx[y/x²] = 1/x
Integrate both sides:
y/x² = ln|x| + C
y = x²(ln|x| + C)
(Apply initial condition if the value of y at x = 1/3 is given — partial data from image)
© Taylor series of f(x) = 64ˣ about x = 1
(Note: This appears to be f(x) = 6·4ˣ or possibly f(x) = (64)ˣ. Taking f(x) = 64ˣ)
f(x) = 64ˣ = (2⁶)ˣ = 2^(6x)
Let a = 1
f(1) = 64
f’(x) = 64ˣ · ln64, f’(1) = 64ln64
f’’(x) = 64ˣ(ln64)², f’’(1) = 64(ln64)²
f’’’(1) = 64(ln64)³
Taylor series about x = 1:
f(x) = 64 + 64ln64·(x−1) + 64(ln64)²/2!² + 64(ln64)³/3!³ + …
f(x) = 64 · Σ (ln64)ⁿ/n!ⁿ, n = 0,1,2,…
Question 4
(a) General solution of: d²y/dx² − 4(dy/dx) − 12y = 3e^(5t)
(Note: mixed variables — treating as dy²/dx² − 4dy/dx − 12y = 3e^(5x))
Auxiliary equation:
m² − 4m − 12 = 0
(m − 6)(m + 2) = 0
m = 6 or m = −2
Complementary function:
y_c = Ae^(6x) + Be^(−2x)
Particular integral — try y_p = Ce^(5x):
y_p’ = 5Ce^(5x), y_p’’ = 25Ce^(5x)
Substitute:
25Ce^(5x) − 4(5Ce^(5x)) − 12Ce^(5x) = 3e^(5x)
C(25 − 20 − 12)e^(5x) = 3e^(5x)
−7C = 3
C = −3/7
y_p = −(3/7)e^(5x)
General solution:
y = Ae^(6x) + Be^(−2x) − (3/7)e^(5x)
(b) Trapezoidal rule for ∫₀^π sin x dx, n = 5
h = (π − 0)/5 = π/5
| x | sin x |
|—|-------|
| x₀ = 0 | 0 |
| x₁ = π/5 | sin36° = 0.5878 |
| x₂ = 2π/5 | sin72° = 0.9511 |
| x₃ = 3π/5 | sin108° = 0.9511 |
| x₄ = 4π/5 | sin144° = 0.5878 |
| x₅ = π | 0 |
Trapezoidal rule:
∫ ≈ (h/2)[y₀ + 2(y₁+y₂+y₃+y₄) + y₅]
= (π/10)[0 + 2(0.5878 + 0.9511 + 0.9511 + 0.5878) + 0]
= (π/10)[2 × 3.0778]
= (π/10)(6.1556)
= 1.9332
(Exact value = 2; trapezoidal approximation ≈ 1.9332)
STATISTICS (Question 6)
(a) Poisson distribution, λ = 1.2, 5 rooms
P(X = k) = e^(−λ)·λᵏ/k!
(i) All rooms vacant → X = 0:
P(X=0) = e^(−1.2) = 0.3012
(ii) Some clients rejected → X > 5 (demand exceeds 5 rooms):
P(X > 5) = 1 − P(X ≤ 5)
P(X=0) = 0.3012
P(X=1) = e^(−1.2)(1.2) = 0.3614
P(X=2) = e^(−1.2)(1.44/2) = 0.2169
P(X=3) = e^(−1.2)(1.728/6) = 0.0867
P(X=4) = e^(−1.2)(2.0736/24) = 0.0260
P(X=5) = e^(−1.2)(2.48832/120) = 0.0062
P(X≤5) = 0.3012+0.3614+0.2169+0.0867+0.0260+0.0062 = 0.9985
P(X>5) = 1 − 0.9985 = 0.0015
(b) Density function:
f(x) = k(3+2x), 2 ≤ x ≤ 4; 0 otherwise
(i) Find k:
∫₂⁴ k(3+2x)dx = 1
k[3x + x²]₂⁴ = 1
k[(12+16) − (6+4)] = 1
k[28 − 10] = 1
18k = 1
k = 1/18
(ii) P(2 ≤ X ≤ 3):
= ∫₂³ (1/18)(3+2x)dx
= (1/18)[3x + x²]₂³
= (1/18)[(9+9) − (6+4)]
= (1/18)[18 − 10]
= 8/18
= 4/9 ≈ 0.4444
©(i) Binomial: n=6, find p if 9P(X=4) = P(X=2)
P(X=k) = C(6,k)·pᵏ·(1−p)^(6−k)
9·C(6,4)·p⁴(1−p)² = C(6,2)·p²(1−p)⁴
9·15·p⁴(1−p)² = 15·p²(1−p)⁴
9p⁴(1−p)² = p²(1−p)⁴
Divide both sides by p²(1−p)²:
9p² = (1−p)²
3p = 1−p (taking positive root since 0<p<1)
4p = 1
p = 1/4
©(ii) 40% defective, n=3, P(at most 2 defective)
p = 0.4, q = 0.6, n = 3
P(X≤2) = 1 − P(X=3)
= 1 − C(3,3)(0.4)³(0.6)⁰
= 1 − (0.064)
= 0.936
APPLIED MATHEMATICS (MAT 004)
P̄ = −4i + 2j − 3k, Q̄ = 4i − j − k, R̄ = 5i + k
PQ vector:
PQ = Q − P = (4−(−4))i + (−1−2)j + (−1−(−3))k
= 8i − 3j + 2k
QR vector:
QR = R − Q = (5−4)i + (0−(−1))j + (1−(−1))k
= i + j + 2k
Length of projection of PQ onto QR:
proj = (PQ · QR)/|QR|
PQ · QR = 8(1) + (−3)(1) + 2(2) = 8 − 3 + 4 = 9
|QR| = √(1+1+4) = √6
Projection = 9/√6 = 9√6/6 = 3√6/2 ≈ 3.674
Area of triangle:
PQ × QR = |i j k |
|8 −3 2 |
|1 1 2 |
= i[(−3)(2)−(2)(1)] − j[(8)(2)−(2)(1)] + k[(8)(1)−(−3)(1)]
= i[−6−2] − j[16−2] + k[8+3]
= −8i − 14j + 11k
|PQ × QR| = √(64 + 196 + 121) = √381
Area = (1/2)√381 ≈ (1/2)(19.52) ≈ 9.76 square units