Paper: Kenya National Examinations Council (KNEC), Diploma in Electrical and Electronic Engineering, Module III — Engineering Mathematics III, June/July 2024. Paper codes: 2521/303, 2601/303, 2602/303 and 2603/303.

This revision lesson includes all eight questions from the supplied four-page paper with independently prepared worked solutions. Candidates are instructed to answer any five questions; all questions carry equal marks. These explanations are study notes, not an official KNEC marking scheme.

Question 1: Eigenvalues and state-transition matrices

1(a) Eigenvalues and eigenvectors (12 marks)

Question. Determine the eigenvalues and corresponding eigenvectors of A = [[2, 3], [3, 2]].

Worked answer.

det(A − λI) = (2 − λ)2 − 9 = (5 − λ)(−1 − λ), so λ = 5 or −1.

  • For λ = 5, an eigenvector is (1,1)T.
  • For λ = −1, an eigenvector is (1,−1)T.

1(b) State-transition matrix for a repeated eigenvalue (8 marks)

Question. For dx/dt = Ax, with A = [[0, 1], [−1, −2]], determine the state-transition matrix Φ(t).

Worked answer.

Write A = −I + N, where N = A + I = [[1,1],[−1,−1]].
Since N2 = 0, the matrix exponential is eAt = e−teNt = e−t(I + tN).
Hence

Φ(t) = e−t[[1 + t, t], [−t, 1 − t]].

This satisfies Φ(0) = I and dΦ/dt = AΦ(t).

Question 2: Double integrals

2(a) Integral over a square (9 marks)

Question. Evaluate ∫01∫011/(xy + 1)2 dy dx.

Worked answer.

For x >

0, ∫01(1 + xy)−2dy = 1/(1 + x);

the value at x = 0 follows by continuity.

Therefore the double integral is ∫01dx/(1 + x) = ln 2.

2(b) Region bounded by two curves (11 marks)

Question. The curves are y = √(1 − x2) and y = √(x − x2). Evaluate ∬Rxy/√(x2 + y2) dxdy and determine the area of R.

Worked answer.

The region lies in the first quadrant.

In polar coordinates the circle is r = 1 and the curve y2 = x − x2 becomes r = cos θ.
Thus 0 ≤ θ ≤ π/2 and cos θ ≤ r ≤ 1.

Since xy/√(x2 + y2) dA = r2cos θ sin θ drdθ, the integral equals (1/3)∫0π/2cos θ sin θ(1 − cos3θ)dθ = 1/10.

The area is (1/2)∫0π/2(1 − cos2θ)dθ = π/8 square units.

Question 3: Line integrals and Green’s theorem

3(a) Path-independent line integral (9 marks)

Question. Show that the integral of x sin y dx + (1/2 x2cos y − 3y2)dy from (0,0) to (1,π/2) is path independent and find its value using a potential function.

Worked answer.

Let P = x sin y and Q = (1/2)x2cos y − 3y2.
Since Py = x cos y = Qx, the field is conservative on the plane.
Integrating P with respect to x gives a potential φ = (1/2)x2sin y − y3 + C.
Therefore the integral is φ(1,π/2) − φ(0,0) = 1/2 − π3/8.

3(b) Verify Green’s theorem on a triangle (11 marks)

Question. Verify Green’s theorem for ∮C[xy dx + (x2 − y)dy], where C is the counter-clockwise boundary of the triangle with vertices (0,0), (1,1) and (0,1).

Worked answer.

With P = xy and Q = x2 − y, Qx − Py = 2x − x = x.
The triangle is 0 ≤ x ≤ y ≤ 1, so ∬Rx dA = ∫01∫0yx dxdy = 1/6.

Directly, the three counter-clockwise sides contribute 1/6, −1/2 and 1/2, respectively, also giving 1/6.

Both sides agree.

Question 4: Newton–Raphson and interpolation

4(a) Newton–Raphson root (8 marks)

Question. Use Newton–Raphson to determine the root of x3 − x2 − 4x + 2 = 0 near x = 2.5.

Worked answer.

With f(x) = x3 − x2 − 4x + 2 and f′(x) = 3x2 − 2x − 4, the iteration is xn+1 = xn − (xn3 − xn2 − 4xn + 2)/(3xn2 − 2xn − 4).

n xn
0 2.500000
1 2.358974
2 2.343119
3 2.342923
4 2.342923

The root near 2.5 is approximately 2.3429231.

4(b) Correct the data error and interpolate (12 marks)

Question. The table gives x = −2, −1, 0, 1, 2, 3, 4, 5 and f(x) = 2, 7, 5, 5, 10, 27, 62, 121. One entry is in error. Use finite differences to locate and correct it, then use Newton–Gregory forward interpolation to determine f(x).

Worked answer.

The data follow a cubic, so the third differences should be constant. The value at x = 0 is the error: it should be 6. The corrected differences are:

x −2 −1 0 1 2 3 4 5
f(x) 2 7 6 5 10 27 62 121
Δf 5 −1 −1 5 17 35 59
Δ2f −6 0 6 12 18 24
Δ3f 6 6 6 6 6

Taking x0 = −2, h = 1 and p = x + 2, Newton–Gregory gives f(x) = 2 + 5p − 3p(p − 1) + p(p − 1)(p − 2).

On simplifying, f(x) = x3 − 2x + 6.

Question 5: Fourier series

5(a) Periodic piecewise-linear function (13 marks)

Question. A periodic function is f(t) = 2 − t for −1 < t < 0, f(t) = t for 0 < t < 1, extended by f(t + 2) = f(t). Sketch it on −1 < t < 3 and determine its Fourier series.

Worked answer.

The two line segments on (−1,1) repeat every 2 units. Over −1 < t < 3, repeat the same pattern on 1 < t < 3. The Fourier interval has L = 1.

The coefficients are a0/2 = 3/2, an = 2[(-1)n − 1]/(n2π2) and bn = 2[(-1)n − 1]/(nπ).
Thus

f(t) = 3/2 − 4Σm=0∞[cos((2m + 1)πt)/( (2m + 1)2π2) + sin((2m + 1)πt)/((2m + 1)π)].

At its jump points, the series converges to the midpoint of the two one-sided limits.

5(b) Half-range Fourier cosine series (7 marks)

Question. Determine the Fourier cosine series of f(t) = (π − t)2, 0 < t < π.

Worked answer.

For L = π, a0/2 = (1/π)∫0π(π − t)2dt = π2/3.
For n ≥ 1, an = (2/π)∫0π(π − t)2cos(nt)dt = 4/n2.
Therefore

f(t) = π2/3 + 4Σn=1∞cos(nt)/n2, for 0 ≤ t ≤ π.

Question 6: Complex variables

6(a) Hyperbolic complex function (9 marks)

Question. Given f(z) = cosh z, with z = x + jy, express f(z) = u + jv and show that u and v satisfy the Cauchy–Riemann equations and that u is harmonic.

Worked answer.

Using the addition formula, cosh(x + jy) = cosh x cos y + j sinh x sin y.
Thus u = cosh x cos y and v = sinh x sin y.
We have ux = sinh x cos y = vy and uy = −cosh x sin y = −vx.
Also uxx = cosh x cos y and uyy = −cosh x cos y, so ∇2u = 0.

6(b) Image of the circle under the printed transformation (11 marks)

Question. The paper gives w = (z + j)/(z + 2j) for |z| = 2, then asks for the centre and radius of an image circle.

Worked answer and source note. The pole z = −2j lies on |z| = 2, so this transformation does not map the full circle to a finite circle. Rearranging gives z = j(1 − 2w)/(w − 1). Therefore |1 − 2w| = 2|w − 1|. For w = u + jv, this simplifies to u = 3/4. The image is the straight line Re(w) = 3/4, so it has no circle centre or radius. The wording in the supplied paper is internally inconsistent here.

Question 7: Surface integrals

7(a) Area of a cone (6 marks)

Question. Determine the area of the surface z2 = 4(x2 + y2) between z = 0 and z = 2.

Worked answer.

The required part is the upper cone z = 2r, 0 ≤ r ≤ 1.
Its base radius is 1 and its slant height is √(12 + 22) = √5.
The curved surface area is πrℓ = π√5 square units.

7(b) Integral over an upper hemisphere (7 marks)

Question. Evaluate ∬Sx dS where S is the upper hemisphere x2 + y2 + z2 = 4, z ≥ 0.

Worked answer.

The upper hemisphere and its surface element are symmetric under x ↦ −x, while the integrand x changes sign. The positive and negative halves cancel, so the integral is 0.

7(c) Divergence theorem on a sphere (7 marks)

Question. Use the divergence theorem for F = 2xi − 3yj + 4zk on the sphere x2 + y2 + z2 = 9.

Worked answer and source note. With the standard outward orientation, ∇·F = 2 − 3 + 4 = 3. The sphere encloses volume (4/3)π(33) = 36π, so the outward flux is 3(36π) = 108π. The scan appears to say “upward unit normals”; the divergence theorem for a closed sphere requires outward normals, so the stated wording is ambiguous if read literally.

Question 8: Jordan form and Fourier series

8(a) Jordan canonical form (15 marks)

Question. Determine the Jordan canonical form of A = [[0, 1], [−1, −2]].

Worked answer.

The characteristic polynomial is (λ + 1)2.

Since A is not equal to −I and has only one independent eigenvector, its Jordan form is J = [[−1, 1], [0, −1]].
For example, an eigenvector is v = (1,−1)T and a generalized eigenvector satisfying (A + I)w = v is w = (1,0)T.

8(b) Fourier sine series (5 marks)

Question. Determine the Fourier sine series of f(t) = t, 0 < t < π.

Worked answer.

The sine coefficients are bn = (2/π)∫0πt sin(nt)dt = 2(−1)n+1/n.
Therefore

t = 2Σn=1∞(−1)n+1sin(nt)/n, for 0 <

t < π.

Revision note: these independently prepared explanations are for study and are not the official KNEC marking scheme. Check the supplied scan for the original notation and instructions.

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