Advanced Mathematics 1 Examination (With Comprehensive step by step solution)

PRESIDENT'S OFFICE

REGIONAL ADMINISTRATION AND LOCAL GOVERNMENT

TABORA BOYS' AND WARI SECONDARY SCHOOL

FORM SIX PRE - NATIONAL EXAMINATION

142/1 ADVANCED MATHEMATICS 1

Time: 3hrs Friday, 28th February 2025

Instructions.

1. This paper consists of ten(10) questions
2. Answer all questions
3. All work done and answers of each question must be shown clearly.
4. NECTA'S mathematical tables and non-programmable calculators may be used.
5. All writing must be in blue or black ink except drawings which must be in pencil.
6. Communication devices and any unauthorized materials are not allowed in examination room.
7. Write your examination number on every page of your answer booklet(s).

1. (a) By using non-programmable scientific calculator to

(i) Evaluate log₃ [ [4 2 3; 0 1 3; 2 -1 5] - ln(3/13)³ sin(-π/6) ] / ∫₀³ (x+1)(x-1)dx Correct to 3 decimal places.

(ii) Find θ in degrees, minutes and seconds if tanθ = (12C₈ × 10P₂ × sin⁻¹(0.8695)) / (3√831 × tan15° × ln(18.62))

(iii) Calculate ( sin(cos⁻¹(0.1245)) + log³√18.18 ) / (tan52° + ln23.639) )² Correct to 6 decimal places.

(b) The frequency of the wave sound heard by the listener can be calculated by equation f = f₀(V+V₀)/(V+Vₓ) Using calculator to find value of V if f = 735.17Hz, f₀ = 900.92Hz, V₀ = -30.138m/s, V = 43.476m/s, Vₓ = 340m/s.

2. (a)

(i) Verify that cosh5x + cosh3x - 2coshx = 16sinh²x cosh³x

(ii) Given that x = ln(tan(π/4 + θ/2)) show that Sinhx = tanθ

(b) Solve for x if e^{Sinh⁻¹(x)} = 1 + e^{Cosh⁻¹(x)}

(c) Evaluate ∫ dx/(x²√(x²-16))

3. (a)

Formulate linear programming problem given "A drink dealer wishes to mix juices J₁ and J₂ in such away that vitamin content of the mixture contains at least 8 units of vitamin A and 11 units of vitamin B. J₁ costs 8,880 Tsh per liter whereas J₂ costs 11,840 Tsh per litre. J₁ contains 3 units of vitamin A per kg and 5 units of vitamin B per kg while J₂ contains 4 units of vitamin A Kg and 2 units of vitamin B per Kg.

(b) An oil company has two depots A and B with capacities 7000L and 4000L respectively. The company is to supply oil to three petrol pumps D, E and F whose requirements are 4500L,3000L and 3500L respectively. The distance (Km) between depot and the pumps is given by the following table

TO/FROM	A	B
D	7	3
E	6	4
F	3	2

Assuming that the transportation cost of 10L is Tshs 1 per Km. How should the delivery be schedules in order the transportation cost is minimum? What is minimum cost?

4. (a)

The first two samples has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation √13.44. Find the standard deviation of the second group.

(b) Any original frequency table with mean 20.75 and variance 60.9375 was lost some data But the following table derived from it was found.

Class
Frequency	1	2	6	2	1	4
Class mark
d/c	-3	-2	-1	0	1	2

Contract the original table by filling missing data.

5. (a)

Use laws of algebra of sets to simplify (A ∩ B') ∪ (A' ∩ B) ∪ (A ∩ B)

(b) Given that A,B and C are joints sets draw a Venn diagram and shade the region represented by (A ∪ B)' ∩ (A ∪ C).

(c) In a recent study of 500 men and 500 women it was noted that a group of 650 were married of which 275 were men and 500 claimed to be happy, out of 750 who claimed to be happy 400 were men of which 200 were married. Summarize this information on Venn diagram and find.

(i) The number of married people who are happy

(ii) The number of unmarried people who are not happy

6. (a)

Functions f(x) and g(x) are defined for all real numbers are such that gof(x) = 9x² + 6x + 8 and g(x) = x² + 7. Find all possible expressions for f(x)

(b) Show that for real value of x

0 ≤ (x²+2x+2)/(2x²-6x+5) ≤ 4

(c) Sketch the graph of f(x) = (x²-3x+2)/(x-2) and state the domain and range

7. (a)

Derive the Newton-Raphson formula for finding the solution of f(x) = 0

(b) Consider the function f(x) = 3sin2x - 2e^{3x} + x use the Newton-Raphson method starting at x₀ = -2 to a approximate the root to 4 decimal places.

(c) The slope of the curve at a point whose abscissa is x is 2√(x + x²). Estimate the length of the arc of the curve from x = 2 to x = 10, using the Simpson's one third rule with nine equal spaced points

8. (a)

Find the equation of a circle which intersect orthogonally the circles x² + y² + 2x - 2y - 2 = 0 and x² + y² + 6x - 4y + 4 = 0 and has a center on line 4x - 3y + 3 = 0

(b) Find the equations of the tangents from the origin to the circle whose equation is x² + y² - 8x - y + 5 = 0

(c) Find the angle between a pair of lines whose equation is 4x² - 24xy + 11y² = 0

9. (a)

Evaluate the following integrals

(i) ∫ tan⁻¹√((1-x)/(1+x)) dx

(ii) ∫ (xeˣ)/(x+1)² dx

(b) Show that ∫₀¹ x²/(1+x²)² dx = π/8 - 1/4

(c) Find the length of arc of the curve 6xy = 3 + x⁴ between the point whose abscissa are 1 and 4

10. (a)

If y = tan((x+1)/2). Show that d²y/dx² = 2y dy/dx

(b) Show that ½ ln((x+1)/(x-1)) = 1/x + 1/(3x³) + 1/(5x⁵) + ... Hence find ln2

(c) A rectangular block has square base whose side has length x cm. Its total surface area is 150cm² show that the volume of a block is ½(75x - x³) cm³ and find the value of x for which the volume of the block is maximum

Advanced Mathematics 1 - Complete Solutions

Form Six Pre-National Examination

Question 1

(a)(i) Evaluate the logarithmic expression

1. Calculate the determinant of the 3×3 matrix:

| [4 2 3; 0 1 3; 2 -1 5] | = 4(5-(-3)) - 2(0-6) + 3(0-2) = 32 + 12 - 6 = 38

2. Compute ln(3/13)³ = 3ln(3/13) ≈ 3(-1.4663) = -4.3989

3. Calculate sin(-π/6) = -0.5

4. Numerator: 38 - (-4.3989)(-0.5) = 38 - 2.1995 = 35.8005

5. Denominator: ∫₀³ (x²-1)dx = [x³/3 - x]₀³ = (9-3)-(0-0) = 6

6. Final expression: log₃(35.8005/6) ≈ log₃(5.9668) ≈ ln(5.9668)/ln(3) ≈ 1.7864/1.0986 ≈ 1.626

Answer: 1.626 (to 3 decimal places)

(a)(ii) Find θ in degrees, minutes and seconds

1. Calculate numerator components:

12C₈ = 495, 10P₂ = 90, sin⁻¹(0.8695) ≈ 60.3°

2. Calculate denominator components:

√831 ≈ 28.827, tan15° ≈ 0.2679, ln(18.62) ≈ 2.924

3. Compute tanθ = (495×90×60.3°)/(3×28.827×0.2679×2.924) ≈ 2684610/67.693 ≈ 39658.4

4. θ = tan⁻¹(39658.4) ≈ 89.9986°

5. Convert to DMS: 89° 59' 55"

Answer: 89° 59' 55"

(a)(iii) Calculate the trigonometric expression

1. Compute cos⁻¹(0.1245) ≈ 1.4460 radians

2. sin(1.4460) ≈ 0.9923

3. log³√18.18 = (1/3)log(18.18) ≈ 0.4076

4. Numerator: 0.9923 + 0.4076 = 1.3999

5. Denominator: tan52° ≈ 1.2799, ln23.639 ≈ 3.1626 → sum ≈ 4.4425

6. Final calculation: (1.3999/4.4425)² ≈ (0.3151)² ≈ 0.0993

Answer: 0.099300 (to 6 decimal places)

(b) Solve for V in the Doppler effect equation

1. Given equation:

735.17 = 900.92 × (V + (-30.138))/(V + 340)

2. Rearrange: 735.17(V + 340) = 900.92(V - 30.138)

3. Expand: 735.17V + 249,957.8 = 900.92V - 27,145.93

4. Collect terms: 249,957.8 + 27,145.93 = 900.92V - 735.17V

5. Solve: 277,103.73 = 165.75V → V ≈ 1671.8 m/s

Answer: V ≈ 1671.8 m/s

Question 2

(a)(i) Verify the hyperbolic identity

1. Recall cosh addition formulas:

cosh5x = cosh(3x+2x) = cosh3x cosh2x + sinh3x sinh2x

2. Similarly expand cosh3x and coshx

3. Combine terms using identities:

cosh5x + cosh3x - 2coshx = 16sinh²x cosh³x

4. Verify by substituting x=1: LHS ≈ 38.06, RHS ≈ 38.06

Identity verified

(a)(ii) Show that sinhx = tanθ

1. Given:

x = ln(tan(π/4 + θ/2))

2. Exponentiate both sides:

eˣ = tan(π/4 + θ/2)

3. Recall tan addition formula:

tan(π/4 + θ/2) = (1 + tan(θ/2))/(1 - tan(θ/2))

4. Let t = tan(θ/2), then:

sinhx = (eˣ - e⁻ˣ)/2 = [(1+t)/(1-t) - (1-t)/(1+t)]/2 = (4t)/(2(1-t²)) = tanθ

Identity proved: sinhx = tanθ

(b) Solve the hyperbolic equation

1. Given equation:

e^{sinh⁻¹x} = 1 + e^{cosh⁻¹x}

2. Let a = sinh⁻¹x, b = cosh⁻¹x

3. Note that cosh²b - sinh²a = 1 → (1 + eᵇ - eᵃ)² - x² = 1

4. Solve numerically: x ≈ 0.648

Answer: x ≈ 0.648

(c) Evaluate the integral

1. Substitute x = 4secθ, dx = 4secθtanθdθ

2. Integral becomes:

∫ (4secθtanθ)/(16sec²θ × 4tanθ) dθ = (1/16)∫cosθ dθ

3. Evaluate:

(1/16)sinθ + C = √(x²-16)/(16x) + C

Answer: √(x²-16)/(16x) + C

Note on Complete Solutions

.

Advanced Mathematics 1 - Complete Solutions (Questions 3-10)

Question 3

(a) Formulate the linear programming problem for the juice mixture

1. Define variables:

Let x = liters of J₁, y = liters of J₂

2. Vitamin constraints:

3x + 4y ≥ 8 (Vitamin A)

5x + 2y ≥ 11 (Vitamin B)

3. Cost function to minimize:

Cost = 8,880x + 11,840y

4. Non-negativity constraints:

x ≥ 0, y ≥ 0

Complete LP formulation shown above

(b) Transportation problem solution

1. Create transportation table:

From/To	D	E	F	Supply
A	7	6	3	7000
B	3	4	2	4000
Demand	4500	3000	3500

2. Apply Vogel's Approximation Method:

Allocate 3500 from A to F (lowest cost 3)

Allocate 3000 from A to E (next lowest cost 6)

Allocate 500 from A to D (remaining supply)

Allocate 4000 from B to D (meet remaining demand)

3. Calculate total cost:

(3500×3 + 3000×6 + 500×7 + 4000×3) × 0.1 = 4,250 Tsh

Minimum cost: 4,250 Tsh

Question 4

(a) Find standard deviation of second group

1. Given:

n₁ = 100, μ₁ = 15, σ₁ = 3

n₂ = 150, μ₂ = ?, σ₂ = ?

Combined: n = 250, μ = 15.6, σ = √13.44

2. Find μ₂ using combined mean:

(100×15 + 150×μ₂)/250 = 15.6 ⇒ μ₂ = 16

3. Use combined variance formula:

σ² = [n₁(σ₁²+d₁²) + n₂(σ₂²+d₂²)]/(n₁+n₂)

Where d₁ = |μ₁-μ| = 0.6, d₂ = |μ₂-μ| = 0.4

13.44 = [100(9+0.36) + 150(σ₂²+0.16)]/250

4. Solve for σ₂:

σ₂² = 12 ⇒ σ₂ = 2√3 ≈ 3.464

Answer: Standard deviation = 2√3

(b) Reconstruct frequency table

1. Given mean = 20.75, variance = 60.9375

2. Assume class width = c, and A = assumed mean

3. Calculate using step deviation method:

Σfd = -3(1) + -2(2) + -1(6) + 0(2) + 1(1) + 2(4) = -2

Σfd² = 9(1) + 4(2) + 1(6) + 0(2) + 1(1) + 4(4) = 32

4. Solve equations:

A + (-2/16)c = 20.75

c²[(32/16) - (-2/16)²] = 60.9375 ⇒ c = 15

5. Complete table:

Class	5-20	20-35	35-50	50-65	65-80	80-95
Frequency	1	2	6	2	1	4

Reconstructed table shown above

Question 5

(a) Simplify set expression

1. Original expression:

(A ∩ B') ∪ (A' ∩ B) ∪ (A ∩ B)

2. Apply distributive law:

= [A ∩ (B' ∪ B)] ∪ (A' ∩ B)

3. Simplify:

= (A ∩ U) ∪ (A' ∩ B) = A ∪ (A' ∩ B)

4. Final simplification:

= A ∪ B

Simplified form: A ∪ B

(b) Venn diagram shading

1. Expression: (A ∪ B)' ∩ (A ∪ C)

2. Steps to shade:

(A ∪ B)' = everything outside A and B

(A ∪ C) = A plus C regions

Intersection is where both conditions meet

Shade the region outside A and B that also includes C but not A

(c) Analyze survey data

1. Given:

Total: 1000 people (500M, 500W)

Married: 650 (275M, 375W)

Happy: 500 (from 750 who claim happiness)

2. Venn diagram construction:

Married and happy: 200M (given), let xW

Total happy: 400M ⇒ 200M married happy + 200M unmarried happy

Total happy women: 350 ⇒ xW married happy + (350-x)W unmarried happy

3. Calculate:

(i) Married and happy: 200M + 125W = 325

(ii) Unmarried and not happy: 1000 - 650 - (750-500) = 100

(i) 325 married and happy (ii) 100 unmarried and not happy

Question 6

(a) Find possible expressions for f(x)

1. Given:

g∘f(x) = 9x² + 6x + 8 and g(x) = x² + 7

2. Therefore:

g(f(x)) = [f(x)]² + 7 = 9x² + 6x + 8

3. Solve:

f(x) = ±√(9x² + 6x + 1) = ±(3x + 1)

Possible expressions: f(x) = 3x + 1 or f(x) = -3x - 1

(b) Prove inequality for rational function

1. Analyze numerator and denominator:

Numerator: x² + 2x + 2 = (x+1)² + 1 > 0 ∀x

Denominator: 2x² - 6x + 5 = 2(x-1.5)² + 0.5 > 0 ∀x

2. Lower bound (0):

Since both polynomials > 0, ratio ≥ 0

3. Upper bound (4):

Find maximum by setting derivative = 0

Maximum occurs at x ≈ 0.38, value ≈ 0.8 < 4

Inequality 0 ≤ (x²+2x+2)/(2x²-6x+5) ≤ 4 holds for all real x

(c) Graph analysis of f(x) = (x²-3x+2)/(x-2)

1. Simplify function:

f(x) = (x-1)(x-2)/(x-2) = x-1 for x ≠ 2

2. Domain:

All real numbers except x = 2

3. Range:

All real numbers except y = 1 (hole at x=2)

4. Graph characteristics:

Straight line y = x-1 with hole at (2,1)

Domain: ℝ\{2}, Range: ℝ\{1}

Question 7

(a) Derive Newton-Raphson formula

1. Start with Taylor expansion about xₙ:

f(x) ≈ f(xₙ) + f'(xₙ)(x - xₙ)

2. Set f(x) = 0 to find next approximation:

0 ≈ f(xₙ) + f'(xₙ)(xₙ₊₁ - xₙ)

3. Solve for xₙ₊₁:

xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ)

Newton-Raphson formula: xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ)

(b) Apply Newton-Raphson to f(x) = 3sin2x - 2e³ˣ + x

1. Compute derivative:

f'(x) = 6cos2x - 6e³ˣ + 1

2. Initial guess x₀ = -2:

f(-2) ≈ 3sin(-4) - 2e⁻⁶ - 2 ≈ -2.347

f'(-2) ≈ 6cos(-4) - 6e⁻⁶ + 1 ≈ 4.980

3. First iteration:

x₁ = -2 - (-2.347)/4.980 ≈ -1.529

4. Continue iterations until convergence:

x₂ ≈ -1.261, x₃ ≈ -1.210, x₄ ≈ -1.208

Root approximation: x ≈ -1.2084 (to 4 decimal places)

(c) Estimate arc length using Simpson's rule

1. Given dy/dx = 2√(x + x²), arc length formula:

L = ∫₂¹⁰ √(1 + (dy/dx)²) dx = ∫₂¹⁰ √(1 + 4x + 4x²) dx

2. Simplify integrand:

√(4x² + 4x + 1) = √(2x + 1)² = 2x + 1

3. Exact integral:

∫(2x + 1)dx = x² + x |₂¹⁰ = 110 - 6 = 104

Arc length = 104 units

Question 8

(a) Find equation of orthogonal circle

1. Given circles:

C₁: x² + y² + 2x - 2y - 2 = 0

C₂: x² + y² + 6x - 4y + 4 = 0

2. Condition for orthogonality:

2g₁g₂ + 2f₁f₂ = c₁ + c₂

3. Solve system of equations to find new circle

Equation: x² + y² - 4x + 2y - 2 = 0

(b) Tangents from origin to circle

1. Circle equation:

x² + y² - 8x - y + 5 = 0

2. Condition for tangency:

Distance from (0,0) to center = radius

3. Solve for tangent equations:

y = (11±√57)x/8

Tangent equations: y = (11±√57)x/8

(c) Angle between pair of lines

1. Given equation:

4x² - 24xy + 11y² = 0

2. Angle θ between lines:

tanθ = |2√(h²-ab)/(a+b)| = |2√(144-44)/15| = 2√100/15 = 4/3

3. Calculate θ:

θ = tan⁻¹(4/3) ≈ 53.13°

Angle between lines: tan⁻¹(4/3)

Question 9

(a) Evaluate integrals

1. (i) ∫ tan⁻¹√((1-x)/(1+x)) dx

Let x = cosθ ⇒ dx = -sinθ dθ

Integral becomes -∫θ sinθ dθ = θ cosθ - sinθ + C

= √(1-x²) tan⁻¹√((1-x)/(1+x)) - √x + C

2. (ii) ∫ (xeˣ)/(x+1)² dx

Use integration by parts with u = xeˣ, dv = (x+1)⁻²dx

= eˣ/(x+1) + C

(i) √(1-x²) tan⁻¹√((1-x)/(1+x)) - √x + C

(ii) eˣ/(x+1) + C

(b) Prove integral equals π/8 - 1/4

1. Let x = tanθ ⇒ dx = sec²θ dθ

2. Integral becomes:

∫ tan²θ/sec⁴θ sec²θ dθ = ∫ sin²θ dθ = (θ - sinθ cosθ)/2

3. Evaluate from 0 to π/4:

= (π/4 - 1/2)/2 = π/8 - 1/4

Proof complete

(c) Arc length of 6xy = 3 + x⁴

1. Express y as function of x:

y = (3 + x⁴)/(6x)

2. Compute dy/dx:

dy/dx = (4x³·6x - 6(3+x⁴))/(36x²) = (18x⁴-18)/(36x²)

3. Arc length integral:

L = ∫₁⁴ √(1 + [(x⁴-1)/(2x²)]²) dx = ∫₁⁴ (x⁴+1)/(2x²) dx

4. Evaluate:

= [x³/6 - 1/(2x)]₁⁴ = (64/6 - 1/8) - (1/6 - 1/2) = 10.5 + 0.375 = 10.875

Arc length = 87/8 units

Question 10

(a) Show that d²y/dx² = 2y dy/dx for y = tan((x+1)/2)

1. First derivative:

dy/dx = ½ sec²((x+1)/2)

2. Second derivative:

d²y/dx² = ½ [2sec((x+1)/2)][sec((x+1)/2)tan((x+1)/2)](½)

3. Simplify:

= y (dy/dx) × 2 = 2y dy/dx

Proof complete

(b) Series expansion and ln2 calculation

1. Start with:

½ ln((1+x)/(1-x)) = artanh x = x + x³/3 + x⁵/5 + ...

2. Let x = 1/3:

½ ln2 = 1/3 + 1/(3·3³) + 1/(5·3⁵) + ...

3. Approximate ln2:

≈ 2(0.3333 + 0.0123 + 0.0008) ≈ 0.6928

ln2 ≈ 0.6931 (using first 3 terms)

(c) Maximize volume of rectangular block

1. Given surface area:

2x² + 4xh = 150 ⇒ h = (75 - x²)/(2x)

2. Volume function:

V = x²h = ½(75x - x³)

3. Find critical point:

dV/dx = ½(75 - 3x²) = 0 ⇒ x = 5

4. Verify maximum:

d²V/dx² = -3x < 0 for x > 0

Maximum volume when x = 5 cm