Paper: Kenya National Examinations Council (KNEC), Diploma in Electrical and Electronic Engineering, Module III — Engineering Mathematics III, October/November 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: Numerical methods

1(a) Newton–Raphson method (11 marks)

Question. For x3 − x2 + 3x − 2 = 0, show that Newton–Raphson gives xn+1 = (2xn3 − xn2 + 2)/(3xn2 − 2xn + 3). Starting with x0 = 0.5, determine the root to six decimal places.

Worked answer.

Let f(x) = x3 − x2 + 3x − 2, so f′(x) = 3x2 − 2x + 3.
Substituting into xn+1 = xn − f(xn)/f′(xn) gives the stated iteration.

n xn
0 0.500000
1 0.727273
2 0.715279
3 0.715225
4 0.715225

The root is x ≈ 0.715225238, or 0.715225 to six decimal places.

1(b) Newton–Gregory forward interpolation (9 marks)

Question. The table gives x = −1, 0, 1, 2, 3, 4 and f(x) = −8, −2, 0, 4, 16, 42. Use Newton–Gregory forward interpolation to determine f(x).

Worked answer.

The forward differences are Δf = 6, 2, 4, 12, 26;

Δ2f = −4, 2, 8, 14;
and Δ3f = 6 throughout, confirming a cubic.
With x0 = −1, h = 1 and p = x + 1:

f(x) = −8 + 6p − 2p(p − 1) + p(p − 1)(p − 2).

Simplifying gives f(x) = x3 − 2x2 + 3x − 2, which reproduces all six tabulated values.

Question 2: Eigenvalues and dynamic systems

2(a) Eigenvalues and eigenvectors (11 marks)

Question. Determine the eigenvalues and corresponding eigenvectors of A = [[−1, 2], [2, 2]].

Worked answer.

det(A − λI) = (−1 − λ)(2 − λ) − 4 = λ2 − λ − 6 = (λ − 3)(λ + 2).
Hence λ = 3 or −2.

  • For λ = 3, an eigenvector is (1,2)T.
  • For λ = −2, an eigenvector is (−2,1)T.

2(b) State-transition matrix (9 marks)

Question. For dx/dt = Ax, where A = [[0, 1], [−2, −3]], determine Φ(t) and show that Φ(0) = I.

Worked answer.

The eigenvalues are −1 and −2, with eigenvectors (1,−1)T and (1,−2)T.

Let P = [[1,1],[−1,−2]].
Then Φ(t) = PeDtP−1 for D = diag(−1,−2).
Multiplication gives

Φ(t) = [[2e−t − e−2t, e−t − e−2t], [−2e−t + 2e−2t, −e−t + 2e−2t]].

Setting t = 0 gives Φ(0) = [[1,0],[0,1]] = I.

Question 3: Multiple integrals

3(a) Evaluate a double integral (5 marks)

Question. Show that ∫12∫x2x1/(x2 + y2) dy dx = (ln 2)tan−1(1/3).

Worked answer.

For fixed x, the inner integral is [tan−1(y/x)/x]x2x = [tan−12 − π/4]/x.
Therefore the double integral equals [tan−12 − π/4]∫12dx/x.
Since tan−12 − tan−11 = tan−1(1/3), the result is (ln 2)tan−1(1/3).

3(b) Change the order of integration (7 marks)

Question. Change the order of integration and evaluate ∫01∫√y11/√[y(1 + x2)] dx dy.

Worked answer.

The region √y ≤ x ≤ 1 is equivalently 0 ≤ x ≤ 1 and 0 ≤ y ≤ x2.
Hence the integral becomes ∫01∫0x21/√[y(1 + x2)] dy dx = 2∫01x/√(1 + x2) dx = 2(√2 − 1).

3(c) Volume between paraboloids (8 marks)

Question. Use a triple integral to find the volume enclosed between z = 2 − x2 − y2 and z = x2 + y2.

Worked answer.

In polar coordinates the surfaces meet at r = 1.

For 0 ≤ r ≤ 1, the upper surface is z = 2 − r2 and the lower surface is z = r2.
Thus V = ∫02π∫01(2 − 2r2)r dr dθ = π cubic units.

Question 4: Line integrals and Green’s theorem

4(a) Line integral on a circular arc (7 marks)

Question. Evaluate ∫C(x + y2)dx − xy dy, where C is the arc of x2 + y2 = 4 from (2,0) to (0,2).

Worked answer.

Use x = 2cos t, y = 2sin t, 0 ≤ t ≤ π/2.
Then dx = −2sin t dt and dy = 2cos t dt.

The integrand becomes −4sin t cos t − 8sin t.

Integrating from 0 to π/2 gives −2 − 8 = −10.

4(b) Work along a straight line (7 marks)

Question. The force field F = −y2i + x2j moves an object from (1,0) to (0,1) along the line segment joining the two points. Find the work done.

Worked answer.

Parameterise the segment by r(t) = (1 − t,t), 0 ≤ t ≤ 1.
Then F = (−t2,(1 − t)2) and r′(t) = (−1,1).
The work is ∫01[t2 + (1 − t)2]dt = 2/3.

4(c) Green’s theorem on an upper semicircle (6 marks)

Question. Evaluate ∮C(ex − y2)dx + (ey + x)dy, where C is the counter-clockwise boundary of the upper semicircle x2 + y2 = 1 together with the x-axis.

Worked answer.

Put P = ex − y2 and Q = ey + x.
Then Qx − Py = 1 + 2y.
Over the upper unit half-disk, ∬1 dA = π/2 and ∬2y dA = 4/3.

Green’s theorem gives the line integral as π/2 + 4/3.

Question 5: Harmonic functions and complex mappings

5(a) Harmonic function and conjugate (11 marks)

Question. Show that u(x,y) = e4xcos 4y − 6x + 2y + 3 is harmonic and determine a conjugate harmonic function v(x,y) such that f(z) = u + jv is analytic.

Worked answer.

The exponential-trigonometric terms give uxx = 16e4xcos 4y and uyy = −16e4xcos 4y;

the linear terms have zero second derivatives.

Thus ∇2u = 0.
From vy = ux = 4e4xcos 4y − 6, we get v = e4xsin 4y − 6y + g(x).
Using vx = −uy gives g′(x) = −2.
Therefore v = e4xsin 4y − 2x − 6y + C.

5(b) Image circle under the printed transformation (9 marks)

Question. The scan’s transformation is read as w = (z − 2j)/(z + j). The circle |z| = 3 is mapped to the w-plane; determine the centre and radius of the image circle.

Worked answer.

Rearranging gives z = −j(w + 2)/(w − 1).
Thus |w + 2| = 3|w − 1|.
Writing w = u + jv and completing the square gives (u − 11/8)2 + v2 = 81/64.

The image circle has centre (11/8,0) and radius 9/8.

Question 6: Fourier series

6(a) Half-range cosine series (7 marks)

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

Worked answer.

For L = π, a0/2 = (1/π)∫0π(π − t)dt = π/2.
For n ≥ 1, an = (2/π)∫0π(π − t)cos(nt)dt = 2[1 − (−1)n]/(πn2).
Only odd terms remain, so

f(t) = π/2 + (4/π)Σm=0∞cos((2m + 1)t)/(2m + 1)2.

6(b) Fourier series for the capacitor charge (13 marks)

Question. From the graph, determine q(t) over one period and find its Fourier series.

Worked answer.

The graph gives q(t) = 2t/π for 0 ≤ t <
π and q(t) = −1 for π ≤ t <

2π, repeated with period 2π. Its mean a0/2 is zero.

The coefficients are an = 2[(-1)n − 1]/(π2n2) and bn = [1 − 3(-1)n]/(πn).
Therefore

q(t) = Σn=1∞{2[(-1)n − 1]cos(nt)/(π2n2) + [1 − 3(-1)n]sin(nt)/(πn)}.

At the jumps t = 0 and t = π, the series converges to the average of the left- and right-hand limits.

Question 7: Surface integrals and Stokes’ theorem

7(a) Flux through a paraboloid (10 marks)

Question. Evaluate ∬SF·n dS for F = zi + 2xj + 3yk, where S is z = 1 − x2 − y2 above the xy-plane.

Worked answer.

The projection is the unit disk. For the upward orientation, n dS = (2x,2y,1)dxdy.

The integrand F·n dS = [2xz + 4xy + 3y]dxdy, with z = 1 − x2 − y2.

Each term is odd in x or y over the symmetric disk, so the flux is 0.

7(b) Stokes’ theorem on a triangular plane (10 marks)

Question. Use Stokes’ theorem for F = 3xi + yj + 2yk, where C is the counter-clockwise boundary of x + 2y + z = 1 in the first octant.

Worked answer.

∇×F = (2,0,0).
Write z = 1 − x − 2y;

the upward oriented surface element is n dS = (1,2,1)dxdy. The projected triangle has vertices (0,0), (1,0), (0,1/2) and area 1/4.

Hence ∮CF·dr = ∬S(∇×F)·n dS = 2(1/4) = 1/2.

Question 8: Symmetric matrices and Fourier series

8(a) Determine a symmetric matrix (11 marks)

Question. A 2 × 2 symmetric matrix A has eigenvalues 3 and −1. Given that an eigenvector for λ = 3 is (−1,1)T, determine an eigenvector for λ = −1 and the matrix A.

Worked answer.

For a symmetric matrix, eigenvectors belonging to distinct eigenvalues are orthogonal.

A vector orthogonal to (−1,1)T is (1,1)T, so this is an eigenvector for λ = −1.
Using the corresponding unit eigenvectors in A = PDPT gives A = [[1,−2],[−2,1]].

8(b) Fourier series of a periodic triangular function (9 marks)

Question. Over one period, f(t) = π − t for −π < t < 0 and f(t) = π + t for 0 < t < π. Sketch the function and determine its Fourier series.

Worked answer.

The function is f(t) = π − |t| on (−π,π), extended periodically with period 2π.
It is even, so bn = 0.
Its mean is a0/2 = π/2, and an = 2[1 − (−1)n]/(πn2).
Hence

f(t) = π/2 + (4/π)Σm=0∞cos((2m + 1)t)/(2m + 1)2.

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.

PreviousNext