NECTA Form Six Mathematics
Common Examination Questions & Detailed Solutions
Advanced Certificate of Secondary Education Examination (ACSEE)
NECTA Assessment Objectives
The National Examinations Council of Tanzania (NECTA) designs Form Six Mathematics examinations to assess comprehensive mastery of advanced mathematical concepts and their practical applications. The assessment focuses on developing mathematical reasoning, problem-solving skills, and analytical thinking essential for STEM fields at university level.
NECTA aims to evaluate students' proficiency in five key areas: conceptual understanding (deep comprehension of mathematical principles), procedural fluency (accurate execution of mathematical procedures), strategic competence (ability to formulate and solve problems using appropriate strategies), adaptive reasoning (capacity for logical thought and justification of solutions), and productive disposition (habitual inclination to see mathematics as sensible and useful). Specifically, NECTA assesses students' ability to apply calculus to physical and geometric problems, manipulate complex numbers and vectors, solve differential equations, analyze sequences and series, and apply statistical methods to data interpretation. The examination tests both pure mathematics (abstract concepts and proofs) and applied mathematics (practical applications in physics, engineering, and economics).
Conceptual Understanding
Deep comprehension of mathematical principles, theorems, and relationships beyond rote memorization.
Procedural Fluency
Accurate execution of mathematical procedures, algorithms, and techniques with precision and efficiency.
Strategic Competence
Ability to formulate, represent, and solve mathematical problems using appropriate strategies and models.
Adaptive Reasoning
Capacity for logical thought, reflection, explanation, and justification of mathematical solutions.
Practical Application
Applying mathematics to real-world scenarios in physics, engineering, economics, and data analysis.
Communication Skills
Clear presentation of mathematical reasoning, proofs, and solutions using appropriate notation.
Common Examination Questions & Solutions
Question 1: Differential Calculus
Differentiation & Applications
Find the stationary points of the function \( f(x) = x^3 - 6x^2 + 9x + 2 \) and determine their nature (maximum, minimum, or point of inflection). Hence, sketch the curve.
Step 1: Find the first derivative
\( f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 2) \)
\( f'(x) = 3x^2 - 12x + 9 \)
Step 2: Find stationary points (where f'(x) = 0)
\( 3x^2 - 12x + 9 = 0 \)
Divide by 3: \( x^2 - 4x + 3 = 0 \)
Factorize: \( (x - 1)(x - 3) = 0 \)
\( x = 1 \) or \( x = 3 \)
Find corresponding y-values:
When \( x = 1 \): \( f(1) = 1^3 - 6(1)^2 + 9(1) + 2 = 6 \)
When \( x = 3 \): \( f(3) = 27 - 54 + 27 + 2 = 2 \)
Stationary points: \( (1, 6) \) and \( (3, 2) \)
Step 3: Determine nature using second derivative
\( f''(x) = \frac{d}{dx}(3x^2 - 12x + 9) = 6x - 12 \)
Test each point:
At \( x = 1 \): \( f''(1) = 6(1) - 12 = -6 < 0 \) ∴ Maximum point
At \( x = 3 \): \( f''(3) = 6(3) - 12 = 6 > 0 \) ∴ Minimum point
Step 4: Sketch the curve
To sketch: The curve is a cubic with positive leading coefficient, so it starts from negative infinity, rises to the maximum at (1,6), falls to the minimum at (3,2), then rises to positive infinity. The y-intercept is at (0,2).
Answer: Maximum point at (1, 6), Minimum point at (3, 2). The curve increases to (1,6), decreases to (3,2), then increases again.
Question 2: Integral Calculus
Integration & Area Calculation
Find the area enclosed by the curve \( y = x^2 - 4x + 3 \), the x-axis, and the lines \( x = 0 \) and \( x = 4 \).
Step 1: Find x-intercepts (where curve crosses x-axis)
\( x^2 - 4x + 3 = 0 \)
\( (x - 1)(x - 3) = 0 \)
\( x = 1 \) and \( x = 3 \)
The curve crosses x-axis at x=1 and x=3.
Step 2: Determine regions above/below x-axis
For \( 0 ≤ x < 1 \): Test x=0.5: \( y = (0.5)^2 - 4(0.5) + 3 = 1.25 > 0 \) (above axis)
For \( 1 < x < 3 \): Test x=2: \( y = 4 - 8 + 3 = -1 < 0 \) (below axis)
For \( 3 < x ≤ 4 \): Test x=3.5: \( y = 12.25 - 14 + 3 = 1.25 > 0 \) (above axis)
This means we must split the integral into three parts and take absolute values for the middle section where the curve is below the x-axis.
Step 3: Calculate area using integration
Total Area = \( \int_{0}^{1} (x^2 - 4x + 3) \, dx - \int_{1}^{3} (x^2 - 4x + 3) \, dx + \int_{3}^{4} (x^2 - 4x + 3) \, dx \)
First integrate: \( \int (x^2 - 4x + 3) \, dx = \frac{x^3}{3} - 2x^2 + 3x + C \)
Step 4: Evaluate definite integrals
\( A_1 = \left[ \frac{1^3}{3} - 2(1)^2 + 3(1) \right] - \left[ \frac{0^3}{3} - 2(0)^2 + 3(0) \right] = \frac{1}{3} - 2 + 3 = \frac{4}{3} \)
\( A_2 = \left[ \frac{3^3}{3} - 2(3)^2 + 3(3) \right] - \left[ \frac{1^3}{3} - 2(1)^2 + 3(1) \right] \)
\( = [9 - 18 + 9] - [\frac{1}{3} - 2 + 3] = 0 - \frac{4}{3} = -\frac{4}{3} \) (negative because below axis)
\( A_3 = \left[ \frac{4^3}{3} - 2(4)^2 + 3(4) \right] - \left[ \frac{3^3}{3} - 2(3)^2 + 3(3) \right] \)
\( = [\frac{64}{3} - 32 + 12] - [9 - 18 + 9] = \frac{4}{3} - 0 = \frac{4}{3} \)
Answer: Total Area = \( \frac{4}{3} - (-\frac{4}{3}) + \frac{4}{3} = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4 \) square units.
Question 3: Complex Numbers
Algebra & De Moivre's Theorem
Solve the equation \( z^3 = 8i \), expressing the roots in the form \( r(\cos\theta + i\sin\theta) \) and in Cartesian form \( a + bi \).
Step 1: Express 8i in polar form
Let \( z^3 = 8i \)
\( 8i = 8(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}) \) since it lies on positive imaginary axis
More generally: \( 8i = 8\left[\cos\left(\frac{\pi}{2} + 2k\pi\right) + i\sin\left(\frac{\pi}{2} + 2k\pi\right)\right] \)
where k = 0, 1, 2 for three distinct roots
Step 2: Apply De Moivre's Theorem for roots
\( z = (8i)^{\frac{1}{3}} = 8^{\frac{1}{3}} \left[ \cos\left(\frac{\frac{\pi}{2} + 2k\pi}{3}\right) + i\sin\left(\frac{\frac{\pi}{2} + 2k\pi}{3}\right) \right] \)
\( z = 2 \left[ \cos\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right) + i\sin\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right) \right] \)
where \( k = 0, 1, 2 \)
Step 3: Find the three distinct roots
For \( k = 0 \): \( z_0 = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} + i\cdot\frac{1}{2}\right) = \sqrt{3} + i \)
For \( k = 1 \): \( z_1 = 2\left[\cos\left(\frac{\pi}{6} + \frac{2\pi}{3}\right) + i\sin\left(\frac{\pi}{6} + \frac{2\pi}{3}\right)\right] \)
\( = 2\left[\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right] = 2\left(-\frac{\sqrt{3}}{2} + i\cdot\frac{1}{2}\right) = -\sqrt{3} + i \)
For \( k = 2 \): \( z_2 = 2\left[\cos\left(\frac{\pi}{6} + \frac{4\pi}{3}\right) + i\sin\left(\frac{\pi}{6} + \frac{4\pi}{3}\right)\right] \)
\( = 2\left[\cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right)\right] = 2(0 + i\cdot(-1)) = -2i \)
Answer: The three cube roots of 8i are:
1. \( \sqrt{3} + i \) (polar: \( 2(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}) \))
2. \( -\sqrt{3} + i \) (polar: \( 2(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}) \))
3. \( -2i \) (polar: \( 2(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}) \))
Question 4: Vector Geometry
Vectors & 3D Geometry
The points A, B, and C have position vectors \( \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k} \), \( \mathbf{b} = 4\mathbf{i} - \mathbf{j} + 2\mathbf{k} \), and \( \mathbf{c} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \) respectively. Find:
(a) The angle between vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)
(b) The area of triangle ABC
Step 1: Find vectors AB and AC
\( \overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (4-2)\mathbf{i} + (-1-3)\mathbf{j} + (2-(-1))\mathbf{k} = 2\mathbf{i} - 4\mathbf{j} + 3\mathbf{k} \)
\( \overrightarrow{AC} = \mathbf{c} - \mathbf{a} = (1-2)\mathbf{i} + (2-3)\mathbf{j} + (3-(-1))\mathbf{k} = -\mathbf{i} - \mathbf{j} + 4\mathbf{k} \)
Step 2: Calculate dot product and magnitudes for angle
Dot product: \( \overrightarrow{AB} \cdot \overrightarrow{AC} = (2)(-1) + (-4)(-1) + (3)(4) = -2 + 4 + 12 = 14 \)
Magnitudes:
\( |\overrightarrow{AB}| = \sqrt{2^2 + (-4)^2 + 3^2} = \sqrt{4 + 16 + 9} = \sqrt{29} \)
\( |\overrightarrow{AC}| = \sqrt{(-1)^2 + (-1)^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2} \)
Step 3: Find angle using cosine formula
\( \cos\theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|} = \frac{14}{\sqrt{29} \cdot 3\sqrt{2}} = \frac{14}{3\sqrt{58}} \)
\( \theta = \cos^{-1}\left(\frac{14}{3\sqrt{58}}\right) \)
Step 4: Calculate area using cross product
Area of triangle = \( \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| \)
Cross product:
\( \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -4 & 3 \\ -1 & -1 & 4 \end{vmatrix} \)
\( = \mathbf{i}[(-4)(4) - (3)(-1)] - \mathbf{j}[(2)(4) - (3)(-1)] + \mathbf{k}[(2)(-1) - (-4)(-1)] \)
\( = \mathbf{i}[-16 + 3] - \mathbf{j}[8 + 3] + \mathbf{k}[-2 - 4] \)
\( = -13\mathbf{i} - 11\mathbf{j} - 6\mathbf{k} \)
Magnitude: \( |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-13)^2 + (-11)^2 + (-6)^2} = \sqrt{169 + 121 + 36} = \sqrt{326} \)
Answer:
(a) Angle \( \theta = \cos^{-1}\left(\frac{14}{3\sqrt{58}}\right) \) ≈ \( \cos^{-1}(0.613) \) ≈ 52.2°
(b) Area = \( \frac{1}{2}\sqrt{326} \) ≈ \( \frac{1}{2} \times 18.06 \) ≈ 9.03 square units
Question 5: Statistics & Probability
Normal Distribution & Hypothesis Testing
The heights of students in a school are normally distributed with mean 165 cm and standard deviation 8 cm.
(a) Find the probability that a randomly selected student is taller than 175 cm.
(b) If 15% of students are eligible for a basketball team requiring minimum height, find this minimum height.
Step 1: Standardize for part (a)
Given: \( X \sim N(165, 8^2) \), where μ = 165, σ = 8
\( P(X > 175) = P\left(Z > \frac{175 - 165}{8}\right) = P(Z > 1.25) \)
where Z is standard normal variable
Step 2: Use normal distribution tables
From Z-tables: \( P(Z < 1.25) = 0.8944 \)
Therefore: \( P(Z > 1.25) = 1 - 0.8944 = 0.1056 \)
Step 3: Solve part (b) - find height for top 15%
We need \( P(X > h) = 0.15 \) or equivalently \( P(X < h) = 0.85 \)
Find z-value such that \( P(Z < z) = 0.85 \)
From Z-tables: \( z \approx 1.036 \) (since \( P(Z < 1.04) = 0.8508 \))
Step 4: Convert z-score back to height
Using formula: \( z = \frac{h - \mu}{\sigma} \)
\( 1.036 = \frac{h - 165}{8} \)
\( h - 165 = 1.036 \times 8 = 8.288 \)
\( h = 165 + 8.288 = 173.288 \text{ cm} \)
Answer:
(a) Probability student is taller than 175 cm = 0.1056 or 10.56%
(b) Minimum height for basketball team (top 15%) = 173.3 cm (to 1 decimal place)
Additional NECTA Mathematics Questions
Question 6: Sequences and Series
Geometric Progression & Summation
The sum of the first n terms of a geometric progression is given by \( S_n = 3(2^n - 1) \).
(a) Find the first term and common ratio.
(b) Find the sum to infinity if it exists.
(c) If the nth term is 192, find the value of n.
Step 1: Recall formula for sum of GP
For a GP: \( S_n = \frac{a(1 - r^n)}{1 - r} \) for \( r ≠ 1 \)
Given: \( S_n = 3(2^n - 1) = 3 \cdot 2^n - 3 \)
Step 2: Compare with standard form to find a and r
\( \frac{a(1 - r^n)}{1 - r} = 3 \cdot 2^n - 3 \)
This suggests \( \frac{a}{1 - r} = 3 \) and \( r = 2 \)
Then \( \frac{a}{1 - 2} = 3 \) ⇒ \( \frac{a}{-1} = 3 \) ⇒ \( a = -3 \)
Check: \( S_n = \frac{-3(1 - 2^n)}{1 - 2} = \frac{-3(1 - 2^n)}{-1} = 3(2^n - 1) \) ✓
Step 3: Sum to infinity
Sum to infinity exists only if \( |r| < 1 \)
Here \( r = 2 > 1 \), so sum to infinity does NOT exist (diverges)
Step 4: Find n when nth term is 192
nth term: \( T_n = ar^{n-1} = -3 \cdot 2^{n-1} \)
Given \( T_n = 192 \): \( -3 \cdot 2^{n-1} = 192 \)
\( 2^{n-1} = -64 \) (wait, negative?)
Actually: \( -3 \cdot 2^{n-1} = 192 \) ⇒ \( 2^{n-1} = -64 \)
This gives negative, which is impossible since 2^{n-1} is always positive.
Check: If a = 3 instead? Let's re-evaluate:
If \( S_n = 3(2^n - 1) \), then comparing:
\( \frac{a(1 - r^n)}{1 - r} = 3(2^n - 1) \)
This works if \( a = 3, r = 2 \):
\( S_n = \frac{3(1 - 2^n)}{1 - 2} = \frac{3(1 - 2^n)}{-1} = 3(2^n - 1) \) ✓
So a = 3, r = 2
Then \( T_n = 3 \cdot 2^{n-1} = 192 \)
\( 2^{n-1} = 64 = 2^6 \)
\( n - 1 = 6 \) ⇒ \( n = 7 \)
Answer:
(a) First term a = 3, common ratio r = 2
(b) Sum to infinity does not exist since |r| = 2 > 1
(c) When nth term is 192, n = 7
Question 7: Differential Equations
First Order Linear DE
Solve the differential equation \( \frac{dy}{dx} + y \tan x = \sec x \), given that y = 2 when x = 0.
Step 1: Identify as linear first order DE
Standard form: \( \frac{dy}{dx} + P(x)y = Q(x) \)
Here: \( P(x) = \tan x \), \( Q(x) = \sec x \)
Step 2: Find integrating factor
Integrating factor: \( I(x) = e^{\int P(x) dx} = e^{\int \tan x dx} = e^{-\ln|\cos x|} = e^{\ln|\sec x|} = \sec x \)
(Actually: \( \int \tan x dx = \int \frac{\sin x}{\cos x} dx = -\ln|\cos x| \))
So \( I(x) = e^{-\ln|\cos x|} = \frac{1}{\cos x} = \sec x \)
Step 3: Multiply through by integrating factor
\( \sec x \cdot \frac{dy}{dx} + \sec x \cdot y \tan x = \sec x \cdot \sec x \)
\( \sec x \frac{dy}{dx} + y \sec x \tan x = \sec^2 x \)
Left side is derivative of \( y \sec x \):
\( \frac{d}{dx}(y \sec x) = \sec x \frac{dy}{dx} + y \sec x \tan x \)
Step 4: Integrate both sides
\( \frac{d}{dx}(y \sec x) = \sec^2 x \)
Integrate: \( y \sec x = \int \sec^2 x dx = \tan x + C \)
Step 5: Solve for y and use initial condition
\( y = \cos x (\tan x + C) = \sin x + C \cos x \)
Given y = 2 when x = 0:
\( 2 = \sin 0 + C \cos 0 = 0 + C \cdot 1 = C \)
So C = 2
Answer: \( y = \sin x + 2 \cos x \)
Question 8: Trigonometry
Trigonometric Equations & Identities
Solve the equation \( \sin 2\theta + \cos \theta = 0 \) for \( 0° ≤ \theta ≤ 360° \).
Step 1: Use double angle formula
\( \sin 2\theta = 2\sin\theta\cos\theta \)
So equation becomes: \( 2\sin\theta\cos\theta + \cos\theta = 0 \)
Step 2: Factor out common term
\( \cos\theta(2\sin\theta + 1) = 0 \)
Step 3: Solve each factor
Case 1: \( \cos\theta = 0 \)
\( \theta = 90°, 270° \) within \( 0° ≤ \theta ≤ 360° \)
Case 2: \( 2\sin\theta + 1 = 0 \)
\( \sin\theta = -\frac{1}{2} \)
Reference angle: \( \sin^{-1}(\frac{1}{2}) = 30° \)
Since sine is negative in Quadrants III and IV:
\( \theta = 180° + 30° = 210° \)
\( \theta = 360° - 30° = 330° \)
Step 4: List all solutions
Solutions: \( \theta = 90°, 210°, 270°, 330° \)
Answer: \( \theta = 90°, 210°, 270°, 330° \)
Question 9: Matrices & Determinants
Matrix Algebra & Systems of Equations
Given matrices \( A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} \) and \( B = \begin{pmatrix} 1 & -2 \\ 3 & 1 \end{pmatrix} \),
(a) Find \( A^{-1} \) and \( B^{-1} \)
(b) Hence solve the system:
\( 2x + y = 5 \)
\( -x + 3y = 1 \)
Step 1: Find inverse of A
For \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), \( A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \)
For \( A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} \):
\( \det(A) = (2)(3) - (1)(-1) = 6 + 1 = 7 \)
\( A^{-1} = \frac{1}{7} \begin{pmatrix} 3 & -1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{7} & -\frac{1}{7} \\ \frac{1}{7} & \frac{2}{7} \end{pmatrix} \)
Step 2: Find inverse of B
For \( B = \begin{pmatrix} 1 & -2 \\ 3 & 1 \end{pmatrix} \):
\( \det(B) = (1)(1) - (-2)(3) = 1 + 6 = 7 \)
\( B^{-1} = \frac{1}{7} \begin{pmatrix} 1 & 2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{7} & \frac{2}{7} \\ -\frac{3}{7} & \frac{1}{7} \end{pmatrix} \)
Step 3: Write system in matrix form
System: \( 2x + y = 5 \), \( -x + 3y = 1 \)
In matrix form: \( \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix} \)
\( A \mathbf{x} = \mathbf{b} \) where \( A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix} \), \( \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} \), \( \mathbf{b} = \begin{pmatrix} 5 \\ 1 \end{pmatrix} \)
Step 4: Solve using inverse
\( \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} \frac{3}{7} & -\frac{1}{7} \\ \frac{1}{7} & \frac{2}{7} \end{pmatrix} \begin{pmatrix} 5 \\ 1 \end{pmatrix} \)
\( x = \frac{3}{7}(5) + (-\frac{1}{7})(1) = \frac{15}{7} - \frac{1}{7} = \frac{14}{7} = 2 \)
\( y = \frac{1}{7}(5) + \frac{2}{7}(1) = \frac{5}{7} + \frac{2}{7} = \frac{7}{7} = 1 \)
Answer:
(a) \( A^{-1} = \begin{pmatrix} \frac{3}{7} & -\frac{1}{7} \\ \frac{1}{7} & \frac{2}{7} \end{pmatrix} \), \( B^{-1} = \begin{pmatrix} \frac{1}{7} & \frac{2}{7} \\ -\frac{3}{7} & \frac{1}{7} \end{pmatrix} \)
(b) \( x = 2 \), \( y = 1 \)
Question 10: Probability & Combinatorics
Binomial Distribution & Conditional Probability
In a multiple choice test with 10 questions, each question has 4 possible answers, only one of which is correct. A student guesses randomly on each question.
(a) Find the probability that the student gets exactly 3 questions correct.
(b) Find the probability that the student gets at least 2 questions correct.
(c) Find the mean and variance of the number of correct answers.
Step 1: Identify as binomial distribution
Let X = number of correct answers
\( X \sim B(n, p) \) where n = 10, p = 1/4 = 0.25
\( P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \)
Step 2: Calculate probability of exactly 3 correct
\( P(X = 3) = \binom{10}{3} (0.25)^3 (0.75)^7 \)
\( \binom{10}{3} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \)
\( P(X = 3) = 120 \times (0.25)^3 \times (0.75)^7 \)
\( = 120 \times 0.015625 \times 0.1334839 \)
\( = 120 \times 0.002086 \approx 0.2503 \)
Step 3: Calculate probability of at least 2 correct
\( P(X ≥ 2) = 1 - P(X = 0) - P(X = 1) \)
\( P(X = 0) = \binom{10}{0} (0.25)^0 (0.75)^{10} = 1 \times 1 \times (0.75)^{10} \approx 0.0563 \)
\( P(X = 1) = \binom{10}{1} (0.25)^1 (0.75)^9 = 10 \times 0.25 \times 0.0750847 \approx 0.1877 \)
\( P(X ≥ 2) = 1 - 0.0563 - 0.1877 = 0.7560 \)
Step 4: Calculate mean and variance
For binomial distribution:
Mean \( \mu = np = 10 \times 0.25 = 2.5 \)
Variance \( \sigma^2 = np(1-p) = 10 \times 0.25 \times 0.75 = 1.875 \)
Answer:
(a) \( P(\text{exactly 3 correct}) ≈ 0.2503 \)
(b) \( P(\text{at least 2 correct}) ≈ 0.7560 \)
(c) Mean = 2.5, Variance = 1.875
NECTA Examination Tips
1. Time Management: Allocate time based on marks - approximately 1.5 minutes per mark.
2. Show All Working: NECTA awards method marks even if final answer is wrong.
3. Presentation: Use clear mathematical notation, label diagrams, and show logical steps.
4. Check Units: Always include correct units in final answers for applied problems.
5. Answer What's Asked: Read questions carefully - "hence", "show that", "find" require different approaches.
6. Practice Past Papers: Familiarize yourself with NECTA's question style and marking scheme.
7. Master Key Formulas: Memorize essential formulas for calculus, trigonometry, and statistics.
8. Verify Solutions: Where possible, check answers using alternative methods or substitution.
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