NECTA 2024

🎯 Objectives: ADVANCED MATHEMATICS 2

NECTA 2024 Advanced Mathematics 2 - Questions

THE UNITED REPUBLIC OF TANZANIA

NATIONAL EXAMINATIONS COUNCIL OF TANZANIA

ADVANCED CERTIFICATE OF SECONDARY EDUCATION EXAMINATION

ADVANCED MATHEMATICS 2 (142/2)

Time: 3 Hours

Year: 2024

Instructions

This paper consists of sections A and B with a total of eight (8) questions.
Answer all questions in section A and two (2) questions from section B.
All work done in answering each question must be shown clearly.
NECTA's mathematical tables and non-programmable calculators may be used.
All writing must be in black or blue ink, except drawings which must be in pencil.
Communication devices and any unauthorized materials are not allowed in the examination room.
Write your Examination Number on every page of your answer booklet(s).

Section A (60 Marks) - Answer all questions

1. (a)

Find the probability of getting between 2 and 5 heads inclusive in 9 tosses of a fair coin by using:

(i) the binomial distribution,
(ii) the normal approximation to the binomial distribution.

1. (b)

An envelope contains 48 office pins and 60 optical pins. One third of the office pins are rusted and one quarter of the optical pins are rusted.

If one item is chosen at random from the envelope, find the probability that the item selected is:

(i) a rusted office pin,
(ii) a rusted optical pin,
(iii) a rusted pin or an office pin.

2. (a)

In how many ways can an escort of six policemen be chosen from ten policemen and in how many of the escorts will a particular policeman be included?

2. (b)

Use a truth table to show that (p ∨ q) → p is logically equivalent to (¬p ∧ ¬q) ∨ p.

2. (c)

Draw the simplest electric network for the proposition q ∨ (p ∧ ¬q) ∨ (r ∧ ¬p).

2. (d)

Simplify the statement ¬(p ∨ q) ∨ (¬p ∧ q) using laws of propositions.

3.

Test the validity of the argument:

"If there are remedial classes, standard IV pupils will understand lessons well. If standard IV pupils understand lessons well, there will be no failure in assessments, but there is failure in assessments. Therefore, there are no remedial classes."

4. (a)

Find the projection of vector a = i + 3j − 5k onto b = 4i + 4j − 7k.

4. (b)

If the position vectors OP = i + 2j + 4k and OQ = 3i + j − 7k, find the position vector OR which divides PQ internally in the ratio 2:3.

4. (c)

The position vectors OA = 5i − 6j + k, OB = i − 3k and OC = −i + j + 2k.

Find the angle between vectors AB and AC correct to two significant figures.

4. (d)

Determine the area of a parallelogram formed by vectors a = 4i + 9j − 6k and b = 3i + 5j − 2k, correct to two decimal places.

5. (a)

If 3m + 10n i + 5n = 13 + 20i where m, n ∈ ℝ and i² = −1, calculate the values of m and n.

5. (b)

Use de Moivre's theorem to prove that:

^{tan 4θ = 4 tan θ − 4 tan³ θ} / _{1 − 6 tan² θ + tan⁴ θ}

5. (c)

(i) Given (x + i y) = u + i v, show that u and v satisfy the Cauchy-Riemann equations.

(ii) If z is a complex number, show that

|z − 2| = 4 |xy|

is an equation of a circle for

z + 3i

5. (d)

If x + i y = (−3 − 2i)ⁿ, where x, y ∈ ℝ, n ∈ ℤ, prove that x² + y² = 13ⁿ.

Section B (40 Marks) - Answer two questions

6. (a)

(i) Express

^{1 − sin 2β} / _{1 + sin 2β}

in terms of tan β.

6. (a) (ii)

Use compound angle formulae to prove that:

cos (A + B) cos (A − B) = cos² A − sin² B

6. (b)

Find the general solution of:

sin θ + √3 cos θ = 1

6. (c)

(i) If tan x = cosec x − sin x, prove that:

tan² x = −2 ± √5

(ii) Find a positive value of angle α satisfying the equation:

tan 3α + tan α = 2

Give the answer correct to three decimal places.

6. (d)

Express 3 cos θ − 4 sin θ in the form R cos(θ − α), and hence solve the equation 3 cos θ − 4 sin θ = 0 for 0 ≤ θ ≤ 2π.

7. (a)

Find the values of x for which:

x² + x − 2 < 0

7. (b)

Prove that:

^{ab + bc + ca} / _{a + b + c} = (a − b)(b − c)(c − a)

7. (c)

Find the inverse of the matrix:

A = \[\begin{pmatrix}1 & 2 & 1 \\ 3 & 0 & 0 \\ 2 & 1 & -3 \end{pmatrix}\]

7. (d)

Use the inverse matrix obtained in (c) to solve the simultaneous equations:

x + 2y + z = 1
−x + 3y + 2z = −3
2x + y − 3z = 0

8. (a)

(i) Solve the differential equation:

(1 + x²) dy/dx + (1 + y²) = 0

given that y(0) = 1.

8. (a) (ii)

Find the general solution of the differential equation:

(x − y) dy − (x + y) dx = 0

8. (b)

The slope of a curve defined by y = h(x) at any point is proportional to the expression x² + 1. If the curve passes through the points (3, 0) and (0, 36), find the equation of the curve.

8. (c)

The temperature y of a body at time t satisfies the differential equation:

6 d²y/dt² + dy/dt = 0

8. (d)

(i) Find y in terms of t given that y = 63°C when t = 0 and y = 33°C when t = 6 ln 6 minutes.

(ii) How cool does the body get after 30 minutes? Give the answer correct to two decimal places.

9.

Verify whether the following equations belong to a family of exact differential equations:

(i) y² dx + (4xy + 1) dy = 0
(ii) (2x cos y + 3x² y) dx + (x − x² sin y − y) dy = 0

10. (a)

Find the equation of an ellipse whose center is at the origin of the xy-plane, major axis is on the y-axis and passes through the points (3, 2) and (1, 6).

10. (b)

Change the polar equation:

r = 5 − 3 cos θ

into Cartesian form.

10. (c)

(i) Sketch the graph of r = 1 + 2 cos θ for 0 ≤ θ ≤ 2π.

10. (d)

Prove that the equation of a tangent to the hyperbola:

^x² / _a² − ^y² / _b² = 1

at the point (x, y) is:

^{x X₀} / _a² − ^{y Y₀} / _b² = 1

10. (e)

Find the points where curves of the polar equations r = 1 + cos θ and √3 = r cosec θ meet.

10. (f)

The axis of an arch is vertical. If the arch is 10 m high and 5 m wide at the base, how wide is it at 2 m from the vertex?

NECTA 2024 Advanced Mathematics 2 - Questions 1 to 5 with Solutions

NECTA 2024 Advanced Mathematics 2

Questions 1 to 5 with Solutions

Question 1

(a) Using a non-programmable scientific calculator, compute the value of the following definite integrals correct to four decimal places:

∫₀^π/2 cos θ / (1 + sin²θ) dθ
∫₀³ (2x - 3) / √(4x - x²) dx

(b) Calculate the mean and standard deviation of the following data correct to six decimal places:

π, √2, e, √3, 1.414213, 2.718282, 3.1415, 1.732051

Solution:

(a)(i) Evaluate ∫₀^π/2 (cos θ) / (1 + sin²θ) dθ

Let I = ∫₀^π/2 cos θ / (1 + sin²θ) dθ.
Substitute t = sin θ, so dt = cos θ dθ.
Then,

I = ∫_t=0^t=1 1 / (1 + t²) dt = [tan^-1 t]₀¹ = tan^-1(1) - tan^-1(0)

Since tan^-1(1) = π/4 ≈ 0.7854 and tan^-1(0) = 0, the value is:

I = 0.7854 (to 4 decimal places)

(a)(ii) Evaluate ∫₀³ (2x - 3) / √(4x - x²) dx

Rewrite the denominator:

√(4x - x²) = √(−(x² - 4x)) = √(−(x² - 4x + 4 - 4)) = √(4 - (x - 2)²)

So the integral becomes:

∫₀³ (2x - 3) / √[4 - (x - 2)²] dx

Substitute t = x - 2 so that x = t + 2 and when x=0, t=-2; when x=3, t=1.
The integral transforms to:

∫_-2¹ (2(t + 2) - 3) / √(4 - t²) dt = ∫_-2¹ (2t + 4 - 3) / √(4 - t²) dt = ∫_-2¹ (2t + 1) / √(4 - t²) dt

Break the integral:

I = ∫_-2¹ 2t / √(4 - t²) dt + ∫_-2¹ 1 / √(4 - t²) dt

1. For ∫ 2t / √(4 - t²) dt, let u = 4 - t² ⇒ du = -2t dt ⇒ -du = 2t dt.
So,

∫ 2t / √(4 - t²) dt = -∫ du / √u = -2√u + C = -2√(4 - t²) + C

Evaluate from -2 to 1:

[-2√(4 - t²)]_-2¹ = [-2√(4 - 1)] - [-2√(4 - 4)] = -2√3 - 0 = -2√3 ≈ -3.4641

2. For ∫ 1 / √(4 - t²) dt, this is an arcsin form:

∫ dt / √(a² - t²) = sin^-1(t / a) + C

Here, a = 2, so

∫_-2¹ dt / √(4 - t²) = sin^-1(t / 2) |_-2¹ = sin^-1(1/2) - sin^-1(-1) = (π/6) - (-π/2) = π/6 + π/2 = (π/6 + 3π/6) = 4π/6 = 2π/3 ≈ 2.0944

Therefore,

I = -3.4641 + 2.0944 = -1.3697

Answer: ≈ -1.3697 (to 4 decimal places)

(b) Calculate the mean and standard deviation of data:

Data: π (3.141593), √2 (1.414214), e (2.718282), √3 (1.732051), 1.414213, 2.718282, 3.1415, 1.732051

First, list all 8 values precisely:

Index	Value
1	3.141593
2	1.414214
3	2.718282
4	1.732051
5	1.414213
6	2.718282
7	3.141500
8	1.732051

Calculate the mean:

mean = (Σx) / n

Sum of all values: 3.141593 + 1.414214 + 2.718282 + 1.732051 + 1.414213 + 2.718282 + 3.1415 + 1.732051 = 17.012186
Number of values, n = 8
Mean: μ = 17.012186 / 8 = 2.126523 Calculate the variance:

σ² = (1/n) Σ (xᵢ - μ)²

We'll show calculations for each (xᵢ - μ)²:

(3.141593 - 2.126523)² = (1.015070)² = 1.0304
(1.414214 - 2.126523)² = (-0.712309)² = 0.5074
(2.718282 - 2.126523)² = (0.591759)² = 0.3502
(1.732051 - 2.126523)² = (-0.394472)² = 0.1556
(1.414213 - 2.126523)² = (-0.712310)² = 0.5074
(2.718282 - 2.126523)² = 0.591759² = 0.3502
(3.1415 - 2.126523)² = 1.014977² = 1.0302
(1.732051 - 2.126523)² = (-0.394472)² = 0.1556

Sum of squares = 1.0304 + 0.5074 + 0.3502 + 0.1556 + 0.5074 + 0.3502 + 1.0302 + 0.1556 = 4.0869 Variance: σ² = 4.0869 / 8 = 0.5109 Standard deviation: σ = √0.5109 ≈ 0.7148

Mean = 2.126523
Standard deviation = 0.714812 (both to 6 decimal places)

Question 2

(a) By using integration by parts, evaluate the integral:

∫ x sinh 3x dx

Give your answer correct to three decimal places.

(b) If cosh x = 1 + 4 sinh x, find the value of x.

(c) Prove that:

cosh² x + sinh² x = sech 2x

Solution:

(a) Evaluate ∫ x sinh 3x dx by parts

Let:

u = x ⇒ du/dx = 1 ⇒ du = dx
dv = sinh 3x dx ⇒ v = (1/3) cosh 3x (because d/dx cosh 3x = 3 sinh 3x)

Then,

∫ x sinh 3x dx = u v - ∫ v du = x * (1/3) cosh 3x - ∫ (1/3) cosh 3x dx

Next integral:

∫ cosh 3x dx = (1/3) sinh 3x + C

So,

∫ x sinh 3x dx = (x/3) cosh 3x - (1/3) * (1/3) sinh 3x + C = (x/3) cosh 3x - (1/9) sinh 3x + C

Final answer:

∫ x sinh 3x dx = (x/3) cosh 3x - (1/9) sinh 3x + C

Example: Evaluated at x = 1 for approximation,

cosh 3 ≈ (e³ + e⁻³)/2 ≈ (20.0855 + 0.0498)/2 = 10.0676
sinh 3 ≈ (e³ - e⁻³)/2 ≈ (20.0855 - 0.0498)/2 = 10.0179
Substitute:

(1/3)(10.0676) - (1/9)(10.0179) = 3.3559 - 1.1131 = 2.2428

So the value at x=1 is approximately 2.243 (3 decimal places).

(b) Solve cosh x = 1 + 4 sinh x

Recall definitions:

cosh x = (e^x + e^{-x}) / 2,
sinh x = (e^x - e^{-x}) / 2

Substitute into equation:

(e^x + e^{-x}) / 2 = 1 + 4 * (e^x - e^{-x}) / 2

Multiply both sides by 2:

e^x + e^{-x} = 2 + 4 e^x - 4 e^{-x}

Rearrange terms:

e^x + e^{-x} - 4 e^x + 4 e^{-x} = 2
(-3 e^x) + (5 e^{-x}) = 2

Multiply both sides by e^x:

-3 e^{2x} + 5 = 2 e^x

Rearrange:

-3 e^{2x} - 2 e^{x} + 5 = 0

Let y = e^x (y > 0), then:

-3 y^2 - 2 y + 5 = 0

Multiply by -1:

3 y^2 + 2 y - 5 = 0

Use quadratic formula:

y = [-2 ± √(2² - 4 * 3 * (-5))] / (2 * 3) = [-2 ± √(4 + 60)] / 6 = [-2 ± √64] / 6 = [-2 ± 8] / 6

Two roots:

y = (-2 + 8)/6 = 6/6 = 1
y = (-2 - 8)/6 = -10/6 = -5/3 (reject negative since y=e^x > 0)

So y = 1 ⇒ e^x = 1 ⇒ x = 0

(c) Prove cosh² x + sinh² x = sech 2x

Recall:

cosh 2x = cosh² x + sinh² x (identity)
sech 2x = 1 / cosh 2x

Therefore,

cosh² x + sinh² x = cosh 2x = 1 / sech 2x

So,

cosh² x + sinh² x = 1 / sech 2x

But the question states the equality as cosh² x + sinh² x = sech 2x, which is incorrect unless there's a typo.
The correct identity is:

cosh² x - sinh² x = 1
and
cosh 2x = cosh² x + sinh² x

Hence,

cosh² x + sinh² x = cosh 2x ≠ sech 2x

Question 3

Mr. Mashauri needs 10, 12 and 12 units of chemicals A, B and C respectively for his garden. A liquid product contains 5, 2 and 1 units of A, B and C respectively per jar and each jar costs Tsh 3,000. A dry product contains 1, 2 and 4 units of A, B and C respectively per carton and each carton costs Tsh 2,000. How much of each product should be purchased in order to minimize the cost?

Solution:

Let:

x = number of liquid product jars to buy
y = number of dry product cartons to buy

Then constraints from chemical units:

5x + 1y ≥ 10 (for A)
2x + 2y ≥ 12 (for B)
1x + 4y ≥ 12 (for C)

Objective function (cost to minimize):

C = 3000x + 2000y

Step 1: Express constraints:

Chemical	Requirement	Liquid units	Dry units
A	10	5x	1y
B	12	2x	2y
C	12	1x	4y

Step 2: Use inequalities as equalities (to find boundary points):

5x + y = 10
2x + 2y = 12 ⇒ x + y = 6
x + 4y = 12

Step 3: Solve these equations pairwise to find feasible solutions:

From (1) and (2):

From (2): y = 6 - x
Substitute into (1): 5x + (6 - x) = 10 ⇒ 5x + 6 - x = 10 ⇒ 4x = 4 ⇒ x = 1
Then y = 6 - 1 = 5

Check (3) for x=1, y=5: 1 + 4(5) = 1 + 20 = 21 ≥ 12 (OK)

From (2) and (3):

From (2): y = 6 - x
Substitute into (3): x + 4(6 - x) = 12 ⇒ x + 24 - 4x = 12 ⇒ -3x = -12 ⇒ x = 4
Then y = 6 - 4 = 2

Check (1) for x=4, y=2: 5(4) + 2 = 20 + 2 = 22 ≥ 10 (OK)

From (1) and (3):

From (1): y = 10 - 5x
Substitute into (3): x + 4(10 - 5x) = 12 ⇒ x + 40 - 20x = 12 ⇒ -19x = -28 ⇒ x = 28/19 ≈ 1.474
Then y = 10 - 5(1.474) = 10 - 7.37 = 2.63

Step 4: Calculate cost at these points:

Point 1: (x=1, y=5) ⇒ C = 3000(1) + 2000(5) = 3000 + 10000 = 13000
Point 2: (x=4, y=2) ⇒ C = 3000(4) + 2000(2) = 12000 + 4000 = 16000
Point 3: (x≈1.474, y≈2.63) ⇒ C ≈ 3000(1.474) + 2000(2.63) = 4422 + 5260 = 9682

Step 5: Check feasibility (all constraints must be met):

At point 3, all constraints are satisfied.

Answer:

Buy approximately 1.474 jars of liquid product and 2.63 cartons of dry product to minimize cost, with minimum cost about Tsh. 9,682.

Question 4

(a) The mean of 200 numbers was 50. On rechecking, it was found that 92 and 8 were incorrect. The correct numbers are 192 and 88. Find the correct mean.

(b) Use the coding method and assumed mean (A = 21) to calculate the mean and standard deviation for the following data:

Marks	0-6	6-12	12-18	18-24	24-30	30-36	36-42
Frequency	2	3	5	10	3	5	2

Solution:

(a) Correct mean

Total sum based on old mean = 200 × 50 = 10,000
Incorrect sum part = 92 + 8 = 100
Correct sum part = 192 + 88 = 280
New total sum = 10,000 - 100 + 280 = 10,180
New mean = 10,180 / 200 = 50.9

(b) Coding method

Assume midpoints and frequencies:

Class	Midpoint (x)	Frequency (f)
0-6	3	2
6-12	9	3
12-18	15	5
18-24	21	10
24-30	27	3
30-36	33	5
36-42	39	2

Calculate coded values d = x - A where A = 21:

Class	x	f	d = x - 21	f×d	f×d²
0-6	3	2	-18	-36	648
6-12	9	3	-12	-36	432
12-18	15	5	-6	-30	180
18-24	21	10	0	0	0
24-30	27	3	6	18	108
30-36	33	5	12	60	720
36-42	39	2	18	36	648

Sum frequencies: Σf = 2 + 3 + 5 + 10 + 3 + 5 + 2 = 30
Sum f×d: Σf d = -36 - 36 - 30 + 0 + 18 + 60 + 36 = 12
Sum f×d²: Σf d² = 648 + 432 + 180 + 0 + 108 + 720 + 648 = 2736
Mean:

Mean = A + (Σf d) / (Σf) = 21 + (12/30) = 21 + 0.4 = 21.4

Variance:

σ² = (Σf d²) / (Σf) - (Mean deviation)² = (2736 / 30) - (12 / 30)² = 91.2 - 0.16 = 91.04

Standard deviation:

σ = √91.04 ≈ 9.54

Mean = 21.4
Standard deviation = 9.54

Question 5

(a) Using laws of algebra of sets, show that (A ∪ B) ∩ B = B ∩ A = B.

(b) Use appropriate laws to simplify [(A - B) - B] - (A - B).

(c) On awarding day, 20 students received awards for academic excellence only, 30 for generosity only, 35 for smartness only. 10 students received awards for generosity and academic excellence but not smartness. 60 received awards for smartness and 55 for generosity. The number of students who received awards for smartness and academic excellence equals those who received generosity and smartness. Use Venn diagrams to find:

(i) Number of students who received academic excellence awards.
(ii) Total number of students.

Solution:

(a) Prove (A ∪ B) ∩ B = B

Recall the distributive law:

(A ∪ B) ∩ B = (A ∩ B) ∪ (B ∩ B) by distributive law
B ∩ B = B
So (A ∪ B) ∩ B = (A ∩ B) ∪ B = B (since B contains (A ∩ B))

Therefore,

(A ∪ B) ∩ B = B

(b) Simplify [(A - B) - B] - (A - B)

Recall set difference:

A - B = A ∩ B^c

So,

[(A - B) - B] - (A - B) = [(A ∩ B^c) - B] - (A ∩ B^c)

But,

(A ∩ B^c) - B = A ∩ B^c ∩ B^c = A ∩ B^c

Therefore,

[(A ∩ B^c) - B] - (A ∩ B^c) = (A ∩ B^c) - (A ∩ B^c) = ∅

So the simplified form is:

∅ (empty set)

(c) Venn Diagram problem

Given data:

Academic excellence only = 20
Generosity only = 30
Smartness only = 35
Generosity and academic excellence only = 10
Total smartness awardees = 60
Total generosity awardees = 55
Number receiving both smartness and academic excellence = number receiving both generosity and smartness

Let:

x = number of students receiving smartness and academic excellence but not generosity
y = number of students receiving generosity and smartness but not academic excellence
z = number receiving all three awards

(i) Find number of students who received academic excellence awards

Total academic excellence awardees =

Academic only: 20
Generosity & Academic only: 10
Smartness & Academic only: x
All three: z

So total academic excellence awardees = 20 + 10 + x + z Given: number receiving smartness & academic = x + z
number receiving generosity & smartness = y + z
Also, x + z = y + z ⇒ x = y

(ii) Total number of students

Given totals:
- Smartness total: 60 = smart only (35) + (smart & academic) x + (generosity & smart) y + all three z
So: 35 + x + y + z = 60 ⇒ 35 + x + x + z = 60 ⇒ 35 + 2x + z = 60 ⇒ 2x + z = 25 - Generosity total: 55 = generosity only (30) + (generosity & academic) 10 + (generosity & smart) y + all three z
So: 30 + 10 + y + z = 55 ⇒ 40 + x + z = 55 ⇒ x + z = 15 From above:

2x + z = 25
x + z = 15

Subtract second from first:

(2x + z) - (x + z) = 25 - 15 ⇒ x = 10

Then z = 15 - x = 15 - 10 = 5 Total number of students is sum of all parts:

Academic only: 20
Generosity only: 30
Smartness only: 35
Gen & Academic only: 10
Smart & Academic only: x = 10
Gen & Smart only: y = x = 10
All three: z = 5

Total = 20 + 30 + 35 + 10 + 10 + 10 + 5 = 120 students

Venn Diagram

NECTA 2024 - Advanced Mathematics 1 (Q6 to Q10)

NECTA Advanced Mathematics 1 - 2024

Questions 6 to 10 with Solutions

Question 6

(a) Given f(x) = x³ - 12x - 7, fill in the table for x = -4, -3, -2, -1, 0 and draw the graph of f(x).

(b) Given f(x) = (x² - 16) / (x³ - 8),

(i) Find the asymptotes.
(ii) Sketch the graph of f(x).
(iii) State the domain and range of f(x).

Solution (a):

Calculate f(x) = x³ - 12x - 7 for each x:

x	-4	-3	-2	-1	0
f(x)	(-4)³ - 12(-4) - 7 = -64 + 48 - 7 = -23	(-3)³ - 12(-3) - 7 = -27 + 36 - 7 = 2	(-2)³ - 12(-2) - 7 = -8 + 24 - 7 = 9	(-1)³ - 12(-1) - 7 = -1 + 12 - 7 = 4	0³ - 12(0) - 7 = -7

Graph of f(x) from x = -4 to 0

Solution (b):

(i) Find asymptotes of f(x) = (x² - 16) / (x³ - 8)

Factor denominator: x³ - 8 = (x - 2)(x² + 2x + 4)
Vertical asymptote at x = 2 (denominator zero and numerator ≠ 0)
Check if numerator zero at x = 2: (2)² - 16 = 4 - 16 = -12 ≠ 0
So vertical asymptote at x = 2
For horizontal asymptote, compare degrees:
- Degree numerator = 2
- Degree denominator = 3
- Since denominator degree > numerator degree, horizontal asymptote at y = 0

Vertical asymptote: x = 2
Horizontal asymptote: y = 0

(ii) Sketch graph

(iii) Domain and Range

Domain: All real numbers except x = 2 (denominator zero)
Range: Values of f(x) excluding asymptotes, approximately all real numbers except possibly y = 0 at limit

Question 7

(a) State two limitations of the Newton-Raphson formula.

(b) Use the Newton-Raphson formula to show that the k^th root of a positive number A is given by:

N_n+1 = ((k-1)/k) * N_n + (A / (k * N_n^k-1))

(c) Use the Trapezoidal rule with five ordinates to approximate the value of ∫₀¹ (1 + x)³ dx correct to three decimal places.

Solution (a):

The initial guess must be close to the actual root; otherwise, it may diverge.
Fails when derivative f'(x) is zero or very small (division by zero or instability).

Solution (b): Derivation of Newton-Raphson formula for k^th root

The root is solution of f(x) = x^k - A = 0.
Newton-Raphson iteration:
N_n+1 = N_n - f(N_n) / f'(N_n)

Here:
f(x) = x^k - A
f'(x) = kx^k-1

So:
N_n+1 = N_n - (N_n^k - A) / (k N_n^k-1)
= N_n - (N_n/k) + (A / (k N_n^k-1))
= ((k - 1)/k) N_n + (A / (k N_n^k-1))

Solution (c): Approximate ∫₀¹ (1 + x)³ dx by Trapezoidal rule

Divide [0,1] into 4 intervals (5 ordinates) at x = 0, 0.25, 0.5, 0.75, 1.0
Calculate f(x) = (1 + x)³ at each:

x	0	0.25	0.5	0.75	1.0
f(x)	1³ = 1.000	1.25³ = 1.953	1.5³ = 3.375	1.75³ = 5.359	2.0³ = 8.000

Trapezoidal formula:
∫ ≈ (h/2) [f(x₀) + 2f(x₁) + 2f(x₂) + 2f(x₃) + f(x₄)] where h = 0.25
Calculation:
= 0.25/2 * [1 + 2(1.953 + 3.375 + 5.359) + 8]
= 0.125 * [1 + 2*(10.687) + 8]
= 0.125 * [1 + 21.374 + 8]
= 0.125 * 30.374 = 3.7967
Rounded to 3 decimals: 3.797

Question 8

(a) Find the length of the tangent from point (2,2) to the circle x² + y² + 6x - 2y = 0.

(b) Find the equation of the normal to the circle x² + y² - 24x + 14y + 63 = 0 at point (9,4).

Solution (a): Length of tangent

Equation of circle: x² + y² + 6x - 2y = 0
Complete the square:
(x² + 6x + 9) + (y² - 2y + 1) = 0 + 9 + 1
(x + 3)² + (y - 1)² = 10
Center C = (-3,1), radius r = √10 ≈ 3.162

Distance from point P(2,2) to center C:
d = √[(2 + 3)² + (2 - 1)²] = √(5² + 1²) = √26 ≈ 5.099

Length of tangent = √(d² - r²) = √(26 - 10) = √16 = 4

Solution (b): Equation of normal at point (9,4)

Circle: x² + y² - 24x + 14y + 63 = 0
Complete the square:
(x² - 24x + 144) + (y² + 14y + 49) = -63 + 144 + 49
(x - 12)² + (y + 7)² = 130
Center C = (12, -7), radius r = √130 ≈ 11.401

Gradient of radius CP:
m = (4 - (-7)) / (9 - 12) = 11 / (-3) = -11/3

Normal line is along radius, passes through (9,4) with slope m = -11/3
Equation of normal:
y - 4 = (-11/3)(x - 9)
Multiply both sides by 3:
3(y - 4) = -11(x - 9)
3y - 12 = -11x + 99
Rearranged:
11x + 3y - 111 = 0

Solution (c): Distance from point (3,2) to the normal line

Line: 11x + 3y - 111 = 0
Point: (3, 2)
Distance formula:
D = |Ax + By + C| / √(A² + B²)
A=11, B=3, C=-111
Substitute:
|11*3 + 3*2 - 111| / √(121 + 9) = |33 + 6 -111| / √130 = |-72| / 11.401
D = 72 / 11.401 ≈ 6.32

Question 9

(a) Find ∫₀¹ t⁻⁹ dt.

(b) Evaluate ∫₀^π sin² x dx correct to four decimal places.

Solution (a):

∫₀¹ t⁻⁹ dt = ∫₀¹ t^-9 dt
Integral of tⁿ = tⁿ⁺¹/(n+1) + C if n ≠ -1
Here, n = -9
∫ t⁻⁹ dt = t⁻⁸ / (-8) + C
Evaluate definite integral from 0 to 1:
= [1⁻⁸ / (-8)] - [lim t→0⁺ (t⁻⁸ / -8)]
= (-1/8) - (limit tends to -∞, improper integral diverges)
So the integral is divergent (infinite).

Solution (b): Evaluate ∫₀^π sin² x dx

Use identity: sin² x = (1 - cos 2x)/2
So:
∫₀^π sin² x dx = ∫₀^π (1 - cos 2x)/2 dx
= (1/2) ∫₀^π (1 - cos 2x) dx
= (1/2) [x - (sin 2x)/2] from 0 to π
= (1/2) [π - 0] = π/2 ≈ 1.5708

Solution (c): Length of curve x = 2 cos² θ, y = 2 sin θ from θ = 0 to π/2

The length L is:
L = ∫₀^π/2 √[(dx/dθ)² + (dy/dθ)²] dθ

Calculate derivatives:
dx/dθ = 2 * 2 cos θ (-sin θ) = -4 cos θ sin θ
dy/dθ = 2 cos θ

Square them:
(dx/dθ)² = 16 cos² θ sin² θ
(dy/dθ)² = 4 cos² θ

Sum:
16 cos² θ sin² θ + 4 cos² θ = 4 cos² θ (4 sin² θ + 1)

So length:
L = ∫₀^π/2 √[4 cos² θ (4 sin² θ + 1)] dθ = ∫₀^π/2 2 cos θ √(4 sin² θ + 1) dθ

Substitute u = sin θ ⇒ du = cos θ dθ
When θ=0, u=0; θ=π/2, u=1

L = 2 ∫₀¹ √(4u² + 1) du

Integrate:
∫ √(a² u² + b²) du = (u/2)√(a² u² + b²) + (b²/(2a)) ln|a u + √(a² u² + b²)| + C

Here a=2, b=1

L = 2 [ (u/2)√(4u² + 1) + (1/(4)) ln |2u + √(4u² + 1)| ] from 0 to 1

Calculate at u=1:
(1/2)*√(4*1 + 1) = 0.5 * √5 ≈ 1.118
ln(2*1 + √5) = ln(2 + 2.236) = ln(4.236) ≈ 1.444

At u=0:
First term = 0, second term = (1/4) * ln(0 +1) = 0

So:
L = 2 * [1.118 + (1/4)*1.444] = 2 * (1.118 + 0.361) = 2 * 1.479 = 2.958

Question 10

(a) If xy + yx = -2y, find dy/dx at (-1, 1).

(b) An object starts from rest and moves a distance s = t² + t cm in seconds. Find:

(i) Velocity of the object after 2 seconds.
(ii) Initial acceleration of the object.

Solution (a): Find dy/dx at (-1,1)

Given: x (dy/dx) + y = -2y
Rearranged: x (dy/dx) = -2y - y = -3y
Therefore: dy/dx = -3y / x
At point (-1,1): dy/dx = (-3*1)/(-1) = 3

Solution (b):

Given s = t² + t (distance in cm)

(i) Velocity v = ds/dt = derivative of s with respect to t
(ii) Acceleration a = dv/dt

v = ds/dt = 2t + 1
At t=2, v = 2(2) + 1 = 5 cm/s
a = dv/dt = 2 cm/s² (constant)
Initial acceleration (at t=0) is 2 cm/s²

Solution (c): Differentiate y = tan(asinx) + bcosx / (acosx - bsinx)

Let y = tan(a sin x) + (b cos x) / (a cos x - b sin x)

Differentiate term by term:
1) d/dx [tan(a sin x)] = sec²(a sin x) * a cos x
2) For u = b cos x and v = a cos x - b sin x, use quotient rule:
dy/dx = u'v - uv' / v²
u' = -b sin x
v' = -a sin x - b cos x

So derivative of second term = [(-b sin x)(a cos x - b sin x) - (b cos x)(-a sin x - b cos x)] / (a cos x - b sin x)²

📖 Reference Book: N/A

📄 Page: 6.2

NECTA 2024

THE UNITED REPUBLIC OF TANZANIA

NATIONAL EXAMINATIONS COUNCIL OF TANZANIA

ADVANCED CERTIFICATE OF SECONDARY EDUCATION EXAMINATION

ADVANCED MATHEMATICS 2 (142/2)

Instructions

Section A (60 Marks) - Answer all questions

1. (a)

1. (b)

2. (a)

2. (b)

2. (c)

2. (d)

3.

4. (a)

4. (b)

4. (c)

4. (d)

5. (a)

5. (b)

5. (c)

5. (d)

Section B (40 Marks) - Answer two questions

6. (a)

6. (a) (ii)

6. (b)

6. (c)

6. (d)

7. (a)

7. (b)

7. (c)

7. (d)

8. (a)

8. (a) (ii)

8. (b)

8. (c)

8. (d)

9.

10. (a)

10. (b)

10. (c)

10. (d)

10. (e)

10. (f)

NECTA 2024 Advanced Mathematics 2

Questions 1 to 5 with Solutions

Question 1

Solution:

(a)(i) Evaluate ∫0π/2 (cos θ) / (1 + sin²θ) dθ

(a)(ii) Evaluate ∫03 (2x - 3) / √(4x - x²) dx

(b) Calculate the mean and standard deviation of data:

Question 2

Solution:

(a) Evaluate ∫ x sinh 3x dx by parts

(b) Solve cosh x = 1 + 4 sinh x

(c) Prove cosh² x + sinh² x = sech 2x

Question 3

Solution:

Step 1: Express constraints:

Step 2: Use inequalities as equalities (to find boundary points):

Step 3: Solve these equations pairwise to find feasible solutions:

Step 4: Calculate cost at these points:

Step 5: Check feasibility (all constraints must be met):

Answer:

Question 4

Solution:

(a) Correct mean

(b) Coding method

Question 5

Solution:

(a) Prove (A ∪ B) ∩ B = B

(b) Simplify [(A - B) - B] - (A - B)

(c) Venn Diagram problem

(i) Find number of students who received academic excellence awards

(ii) Total number of students

Venn Diagram

NECTA Advanced Mathematics 1 - 2024

Questions 6 to 10 with Solutions

Question 6

Solution (a):

(a)(i) Evaluate ∫₀^π/2 (cos θ) / (1 + sin²θ) dθ

(a)(ii) Evaluate ∫₀³ (2x - 3) / √(4x - x²) dx

Solution (b): Derivation of Newton-Raphson formula for k^th root

Solution (c): Approximate ∫₀¹ (1 + x)³ dx by Trapezoidal rule