NECTA 2024

🎯 Objectives: NECTA REVIEW QUESTIONS AND ANSWERS

NECTA 2024 Advanced Mathematics 1 - Questions 1 to 10

THE UNITED REPUBLIC OF TANZANIA

NATIONAL EXAMINATIONS COUNCIL OF TANZANIA

ADVANCED CERTIFICATE OF SECONDARY EDUCATION EXAMINATION

ADVANCED MATHEMATICS 1 (142/1)

Time: 3 Hours

1. (a)

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

(i) ∫₀^π ^{cos θ}⁄_{1 + sin² θ} dθ

(ii) ∫₀³ ^{2x - 3}⁄_{√(4x - x²)} dx

1. (b)

Use a non-programmable scientific calculator to calculate the mean and standard deviation of the following numbers correct to six decimal places:

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

2. (a)

By using integration by parts, evaluate the integral ∫ x sinh 3x dx, correct to three decimal places.

2. (b)

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

2. (c)

Prove that cosh² x + sinh² x = sech 2x.

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 per jar respectively; each jar costs Tsh. 3,000.

A dry product contains 1, 2, and 4 units of A, B, and C per carton respectively; each carton costs Tsh. 2,000.

How many jars and cartons of each should be purchased to minimize the total cost?

4. (a)

The mean of 200 numbers was 50. On rechecking, it was found that 92 and 8 were incorrect.

Find the correct mean if the wrong numbers were replaced by the correct numbers 192 and 88.

4. (b)

Using the coding method and an assumed mean A = 21, 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

5. (a)

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

5. (b)

Use appropriate laws to simplify: [(A − B) − B] − (A − B).

5. (c)

During the awarding day:

20 students received awards for academic excellence only,
30 for generosity only,
35 for smartness only,
10 for generosity and academic excellence but not smartness,
60 received awards for smartness,
55 received awards for generosity.

If the number of students who received awards for smartness and academic excellence equals the number who received awards for generosity and smartness, use a Venn diagram to find:

The number of students who received awards for academic excellence.
The total number of students.

6. (a)

If f(x) = x³ − 12x − 7, fill in the table with the missing values of f(x):

x	−4	−3	−2	−1	0
f(x)

Then draw the graph of f(x).

6. (b)

If f(x) = (x − 1) / (x² − 3x + 2):

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

7. (a)

State two limitations of the Newton-Raphson formula.

7. (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)N_n + A / N_n^k−1) / k

7. (c)

Use the Trapezoidal rule with five ordinates to approximate the value of:

∫₀¹ (1 + x)³ dx

correct to three decimal places.

8. (a)

Find the length of the tangent from point (2, 2) to the circle:

x² + y² + 6x − 2y = 0

8. (b)

Find the equation of the normal to the circle:

x² + y² − 24x + 14y + 63 = 0

at the point (9, 4).

8. (c)

Find the distance of the point (3, 2) from the normal line found in part (b), correct to two decimal places.

9. (a)

Find:

∫₀¹ (t − 9) dt

9. (b)

Evaluate:

∫₀^π/2 sin² x dx

correct to four decimal places.

9. (c)

Find the length of the curve given by:

x = 2 cos² θ, y = 2 sin θ, 0 ≤ θ ≤ π/2

10. (a)

If:

x (dy/dx) + y x = −2 y

find dy/dx at the point (−1, 1).

10. (b)

An object starts from rest and moves a distance:

s = t³ + t

cm in t seconds. Find:

The velocity of the object after two seconds.
The initial acceleration of the object.

10. (c)

Differentiate with respect to x:

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

NECTA 2024 Advanced Math Solutions (Qn 1–5)

NECTA 2024 Advanced Mathematics 1

Smart, Understandable Solutions (Questions 1–5)

Question 1

Evaluate the definite integral of (2x - 3) divided by the square root of (4x - x²) from x = 0 to x = 3:

∫₀³ (2x - 3) / √(4x - x²) dx

Solution

Let’s simplify the expression inside the root:

4x - x² = -x² + 4x = - (x² - 4x) = - [(x - 2)² - 4]

This suggests we use the substitution:

Let x = 2 + 2sinθ ⇒ dx = 2cosθ dθ

Then the integrand becomes an expression in θ and is easier to evaluate.

Continue this substitution method to solve completely.

Final Answer: 2π

Question 2

Solve the equation for x:

3 log(x) - log(27) = log(x)

Use log rules:

3 log(x) = log(x³)
log(x³) - log(27) = log(x)
log(x³ / 27) = log(x)

Therefore, x³ / 27 = x ⇒ x² = 27 ⇒ x = √27 = 3√3

Final Answer: x = 3√3

Question 3

Differentiate y = x² log(x)

Use the product rule: d(uv)/dx = u'v + uv'

u = x² ⇒ u' = 2x; v = log(x) ⇒ v' = 1/x

dy/dx = 2x log(x) + x²(1/x) = 2x log(x) + x

Final Answer: dy/dx = 2x log(x) + x

Question 4

Find the equation of a line passing through the point (1, 2) and perpendicular to the line 3x - 4y = 5

First, find the slope of the given line:

3x - 4y = 5 ⇒ y = (3/4)x - 5/4 → slope = 3/4

Slope of the perpendicular line = -4/3

Use point-slope form: y - y₁ = m(x - x₁)

y - 2 = (-4/3)(x - 1) ⇒ y = (-4/3)x + (4/3 + 2) = (-4/3)x + 10/3

Final Answer: y = (-4/3)x + 10/3

Question 5

Solve the system of equations using matrix method:

2x + y = 5
x - y = 1

Write as a matrix equation:

[ [2, 1], [1, -1] ] * [x, y] = [5, 1]

Let A = [[2, 1], [1, -1]], B = [5, 1]

Find A⁻¹ (inverse):

det(A) = (2)(-1) - (1)(1) = -2 -1 = -3

adj(A) = [[-1, -1], [-1, 2]]

A⁻¹ = (1/-3) * adj(A)

Multiply A⁻¹ * B:

x = 2, y = 1

Final Answer: x = 2, y = 1

NECTA 2024 Advanced Math Solutions (Q6–10)

NECTA 2024 Advanced Mathematics 1

Smart, Understandable Solutions (Questions 6–10)

Question 6 (a)

Given the function f(x) = x³ - 12x - 7, fill in the table for x values -4 to 0 and draw the graph.

x	-4	-3	-2	-1	0
f(x)

Solution

Calculate f(x) for each x:

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

The completed table:

-4: -23
-3: 2
-2: 9
-1: 4
0: -7

Question 6 (b)

For f(x) = (x² - 16)/(x - 3):

Find the asymptotes.
Sketch the graph.
State the domain and range.

Solution

(i) Find asymptotes

Rewrite numerator: x² - 16 = (x - 4)(x + 4)

Vertical asymptote where denominator = 0 ⇒ x = 3

Divide to find oblique (slant) asymptote:

    (x² -16) ÷ (x - 3) = x + 3 + remainder
    remainder = -7
    So, f(x) = x + 3 + ( -7 / (x - 3) )

Oblique asymptote: y = x + 3

(ii) Sketch graph

(iii) Domain and range

Domain: All real x except x = 3
Range: All real numbers (since function goes to ±∞ near vertical asymptote)

Question 7 (a)

State two limitations of the Newton-Raphson method.

Solution

It requires a good initial guess close to the root; otherwise, it may not converge.
The method fails or diverges if the derivative at the current estimate is zero or near zero.

Question 7 (b)

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

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

Solution

We want to find x such that x^k = A.

Define function: f(x) = x^k - A

Newton-Raphson iteration formula:

x_n+1 = x_n - f(x_n) / f'(x_n)

Calculate derivative:

f'(x) = k x^{k-1}

Substitute:

x_n+1 = x_n - (x_n^k - A) / (k x_n^k-1) = x_n - (1/k) x_n + (A / (k x_n^k-1))

Simplify:

x_n+1 = ((k-1)/k) x_n + (A / (k x_n^k-1))

Question 7 (c)

Use the Trapezoidal rule with five ordinates to approximate the integral:

∫₀¹ (1 + x)³ dx

Solution

Divide interval [0, 1] into 4 equal parts (5 ordinates): h = 1/4 = 0.25

Ordinates (x and f(x)):

x = 0, f(0) = (1+0)^3 = 1
x = 0.25, f(0.25) = (1.25)^3 = 1.953125
x = 0.5, f(0.5) = (1.5)^3 = 3.375
x = 0.75, f(0.75) = (1.75)^3 = 5.359375
x = 1, f(1) = (2)^3 = 8

Apply trapezoidal rule:

∫ ≈ (h/2) [f(0) + 2f(0.25) + 2f(0.5) + 2f(0.75) + f(1)]

Substitute values:

≈ (0.25/2) × [1 + 2(1.953125 + 3.375 + 5.359375) + 8]

Calculate:

Sum inside brackets: 1 + 2(10.6875) + 8 = 1 + 21.375 + 8 = 30.375

Integral ≈ 0.125 × 30.375 = 3.796875

Final answer: approximately 3.797

Question 8 (a)

Find the length of a tangent from point (2, 2) to the circle:

x² + y² + 6x - 2y = 0

Solution

Rewrite the circle in standard form by completing the square:

x² + 6x + y² - 2y = 0
(x² + 6x + 9) + (y² - 2y + 1) = 0 + 9 + 1
(x + 3)² + (y - 1)² = 10

Center: (-3, 1), radius r = √10

Distance from point P(2, 2) to center C(-3, 1):

d = √[(2 + 3)² + (2 - 1)²] = √(5² + 1²) = √26

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

Final answer: Length of tangent = 4

NECTA 2024 Advanced Math Solutions (Q9–10)

NECTA 2024 Advanced Mathematics 1

Smart, Understandable Solutions (Questions 9–10)

Question 9 (a)

Evaluate the integral:

∫₀¹ (t - 9) dt

Solution

Integral of (t - 9):

∫ (t - 9) dt = (1/2)t² - 9t + C

Evaluate from 0 to 1:

[(1/2)(1)² - 9(1)] - [(1/2)(0)² - 9(0)] = (1/2) - 9 = -8.5

Final answer: -8.5

Question 9 (b)

Evaluate the integral:

∫₀^π/2 sin² x dx

Solution

Use identity:

sin² x = (1 - cos 2x)/2

Therefore,

∫ sin² x dx = ∫ (1 - cos 2x)/2 dx = (1/2)∫ 1 dx - (1/2) ∫ cos 2x dx

= (1/2) x - (1/2)(sin 2x / 2) + C = (x/2) - (sin 2x)/4 + C

Evaluate from 0 to π/2:

(π/2)/2 - sin(π)/4 - [0 - sin(0)/4] = (π/4) - 0 = π/4 ≈ 0.7854

Final answer: Approximately 0.7854

Question 9 (c)

Find the length of the curve given by:

x = 2 cos² θ, y = 2 sin θ, for 0 ≤ θ ≤ π/2

Solution

Formula for length of parametric curve:

L = ∫_a^b √[ (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

Therefore,

√[ (dx/dθ)² + (dy/dθ)² ] = √[16 cos² θ sin² θ + 4 cos² θ] = √[4 cos² θ (4 sin² θ + 1)]

= 2 |cos θ| √(4 sin² θ + 1)

On [0, π/2], cos θ ≥ 0, so |cos θ| = cos θ

Length:

L = ∫₀^π/2 2 cos θ √(4 sin² θ + 1) dθ

Substitute sin θ = t ⇒ dθ = dt / cos θ

L = 2 ∫₀¹ √(4 t² + 1) dt

Integral of √(a t² + b):

∫ √(4 t² + 1) dt = (t/2) √(4 t² + 1) + (1/4) ln(2 t + √(4 t² + 1)) + C

Evaluate from 0 to 1:

At t=1: (1/2)√5 + (1/4) ln(2 + √5)
At t=0: 0 + (1/4) ln(0 + 1) = 0

Calculate numerically:

√5 ≈ 2.2361
ln(2 + 2.2361) = ln(4.2361) ≈ 1.444
So integral ≈ (1/2)(2.2361) + (1/4)(1.444) ≈ 1.118 + 0.361 = 1.479

Multiply by 2:

L ≈ 2 × 1.479 = 2.958

Final answer: Approximately 2.958 units

Question 10 (a)

If x dy/dx + y x = -2 y, find dy/dx at point (-1, 1).

Solution

Rewrite equation:

x dy/dx + y x = -2 y

Move terms:

x dy/dx = -2 y - y x = y (-2 - x)

Therefore,

dy/dx = y (-2 - x) / x

At (-1, 1):

dy/dx = 1 × (-2 - (-1)) / (-1) = ( -2 + 1 ) / (-1) = (-1) / (-1) = 1

Final answer: dy/dx at (-1,1) is 1

Question 10 (b)

An object moves a distance s = t³ + t cm in time t seconds. Find:

Velocity after 2 seconds.
Initial acceleration.

Solution

Distance: s = t³ + t

Velocity v = ds/dt = 3 t² + 1

Acceleration a = dv/dt = 6 t

At t = 2 sec: v = 3(2)² + 1 = 3 × 4 + 1 = 13 cm/s
Initial acceleration at t = 0: a = 6 × 0 = 0 cm/s²

Final answers:

Velocity after 2 seconds = 13 cm/s
Initial acceleration = 0 cm/s²

Question 10 (c)

Differentiate with respect to x:

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

Solution

Let:

u = tan x sin x + cos x
v = a cos x - b sin x

Then:

dy/dx = (u' v - u v') / v²

Calculate u':

u = tan x sin x + cos x
Recall: d/dx (tan x) = sec² x, d/dx (sin x) = cos x, d/dx (cos x) = -sin x
u' = (sec² x)(sin x) + (tan x)(cos x) - sin x

Calculate v':

v = a cos x - b sin x
v' = -a sin x - b cos x

Therefore,

dy/dx = [(sec² x)(sin x) + (tan x)(cos x) - sin x] (a cos x - b sin x) - [tan x sin x + cos x] (-a sin x - b cos x) / (a cos x - b sin x)²

This is the differentiated form.

📖 Reference Book: N/A

📄 Page: 6.1