WAEC Further Mathematics 2024 past questions and answers

5/√2 = 5√2/2 and √8/8 = 2√2/8 = √2/4. So 5√2/2 − √2/4 = 9√2/4 and m = 9/4.

g(3) = 25 − 9 = 16, then f(16) = √16 = 4.

Let u = √x: u − 6/u = 1 gives u² − u − 6 = 0, (u − 3)(u + 2) = 0. Since u = √x > 0, u = 3 and x = 9.

h(−3) = 9 − 3p + 2 = 5, so 3p = 6 and p = 2.

First term 3, twelfth term 58. S₁₂ = 12/2 × (3 + 58) = 366.

x = 2² = 4. log₄ 128 = log₂ 128 / log₂ 4 = 7/2.

cos 300° = ½, sin 120° = √3/2, tan 135° = −1. (¼ − 4 × ¾)/1 = ¼ − 3 = −2¾.

y = (2 − x)/x gives xy = 2 − x, x(y + 1) = 2, x = 2/(y + 1). So f⁻¹(x) = 2/(x + 1), x ≠ −1.

Let p = 2ˣ: p² − 5p + 4 = 0, (p − 1)(p − 4) = 0. 2ˣ = 1 gives x = 0 and 2ˣ = 4 gives x = 2.

3r = 2p − q = (4 − 10, 8 + 1) = (−6, 9), so r = (−2, 3).

2n − r = (−2 − 1)i + (6 − 2)j = −3i + 4j, and |−3i + 4j| = √(9 + 16) = 5.0.

dy/dx = 2ux + 3. At x = −1: −2u + 3 = 9, so u = −3.

The x² term is ⁴C₂(−2x)² = 6 × 4x² = 24x², so k = 24.

Possible sums that divide 12: 2 (1 way), 3 (2), 4 (3), 6 (5), 12 (1) — 12 of 36 outcomes. P = 12/36 = 1/3.

a = (43 − 15)/12 = 7/3 ms⁻², so F = ma = 42 × 7/3 = 98 N.

By De Morgan's law, the complement of an intersection is the union of the complements: (M ∩ N)′ = M′ ∪ N′.

s = ut + ½at² = 0 + ½ × 4 × 16 = 32 m.

Rearrange: 14x² − 9x + 1 < 0, i.e. (7x − 1)(2x − 1) < 0. This holds between the roots: 1/7 < x < 1/2.

The force acts along the plane, so the normal reaction balances only mg cos 30° = 400 × 0.866 ≈ 346.4 N. The nearest option is 346.0 N.

The given line has gradient ½, so the perpendicular has gradient −2. y − 5 = −2(x + 3) gives y + 2x + 1 = 0.

The y² term is ⁵C₃ y²(−2x)³ = 10 × (−8x³)y², so its coefficient is −80x³.

f′(x) = 2x + 3 = 0 gives x = −1½.

7 choices for the first digit, 6 for the second and 5 for the third: 7 × 6 × 5 = 210.

The top row gives 12 + 4m = 24, so m = 3 (check: 14 + 5(3) = 29 = 2 × 14.5).

Momentum: 80(25) + m(−10) = (80 + m)(12.8). So 2000 − 10m = 1024 + 12.8m, 976 = 22.8m and m ≈ 43 kg.

Order on the bench matters: ¹²P₅ = 12 × 11 × 10 × 9 × 8 = 95040.

Angle XYZ is at Y, between XY and YZ, opposite XZ. Cosine rule: cos α = (10² + 9² − 7²)/(2 × 10 × 9) = 132/180 = 11/15.

Differentiate implicitly: 2y dy/dx + 2y + 2x dy/dx = 0, so dy/dx(2y + 2x) = −2y and dy/dx = −y/(x + y).

Total = 4 × 5 + 3 × 12 = 56, and 56 ÷ 7 = 8.

Gradients are 3 and 7. tan θ = (7 − 3)/(1 + 21) = 4/22, so θ ≈ 10.3° ≈ 10°.

Integrating: y = x³ − 4x² + 2x + c. At (0, 4), c = 4.

y changes by 2 for every unit change in x, so Δy = 2Δx = 2 × 0.1 = 0.20.

Mean = 35/7 = 5. Absolute deviations 1, 1, 4, 3, 4, 0, 3 add to 16. Mean deviation = 16/7 ≈ 2.3.

The gradient is −1 = tan θ, and the angle a line makes with the positive x-axis (0° to 180°) is θ = 135°.

Binomial with n = 12, p = 0.7: P(8) = ¹²C₈(0.7)⁸(0.3)⁴ = 495 × 0.05765 × 0.0081 ≈ 0.2311.

x is in the third quadrant, so sin x and cos x are both negative: sin x = −1/√10, cos x = −3/√10.
sin 2x = 2 sin x cos x = 6/10 = 3/5.
Numerator: 3/5 − (−3/√10) = 3/5 + 3/√10. Denominator: 2/3 + 3/5 = 19/15.
Value = (15/19)(3/5 + 3/√10) = 9/19 + 45/(19√10) = (18 + 9√10)/38 = 9(2 + √10)/38.

Use a right triangle with sides 1, 3, √10 and take the signs for the third quadrant.

First multiple of 7 above 403 is 406 (7 × 58); last below 603 is 602 (7 × 86).
Number of terms n = 86 − 58 + 1 = 29.
Sum = 29/2 × (406 + 602) = 29 × 504 = 14 616.

The multiples form an A.P. with common difference 7.

Seven ordinates means 6 strips: h = (36 − 24)/6 = 2.
x: 24, 26, 28, 30, 32, 34, 36; y = 1/√(x² − 2): 0.04174, 0.03852, 0.03576, 0.03337, 0.03128, 0.02944, 0.02780.
Area ≈ (h/2)[y₀ + y₆ + 2(y₁ + … + y₅)] = 1 × [0.06954 + 2(0.16837)] ≈ 0.406.
(If the limits are 2.4 to 3.6, as the book's working uses: h = 0.2, y = 0.51571, 0.45835, 0.41380, 0.37796, 0.34837, 0.32342, 0.30206, and the integral ≈ 0.466.)

Seven ordinates give six equal strips.

At x = 2: y = 14 − 20 = −6.
dy/dx = 7 − 10x = −13 at x = 2, so the normal has gradient 1/13.
y + 6 = (1/13)(x − 2), i.e. x − 13y − 80 = 0.

The normal is perpendicular to the tangent: m(normal) = −1/m(tangent).

(a) P(M ∩ N′) = P(M) − P(M ∩ N) = 1/2 − 11/50 = 14/50 = 7/25.
(b) P(M′ ∩ N) = P(N) − P(M ∩ N) = 9/20 − 11/50 = 23/100.

A Venn diagram shows 'M only' and 'N only' clearly.

Class boundaries and widths: −0.5–19.5 (20), 19.5–29.5 (10), 29.5–39.5 (10), 39.5–49.5 (10), 49.5–69.5 (20), 69.5–99.5 (30), 99.5–149.5 (50).
Widths differ, so plot frequency density (frequency ÷ width): 0.6, 0.7, 0.6, 0.8, 0.25, 0.3, 0.2 (or per 10 km: 6, 7, 6, 8, 2.5, 3, 2).
Draw touching bars on the class boundaries with these heights; label both axes and state the scale.

With unequal class widths the bar areas, not heights, must match the frequencies.

|u + v|² = |u|² + |v|² + 2|u||v| cos θ.
225 = 100 + 144 + 240 cos θ, so cos θ = −19/240 = −0.0792.
θ ≈ 94.5°.

Use the parallelogram (cosine) rule for the resultant.

Weight component down the plane = 400 sin 30° = 200 N; normal reaction R = 400 cos 30° ≈ 346.4 N.
(a) 420 N up is more than 200 N, so the body tends to move up and friction acts down the plane: F = 420 − 200 = 220 N.
(b) μ = F/R = 220/346.4 ≈ 0.635.

Decide which way the body tends to move; friction acts the other way.

(a) y = 8x + (27/2)x⁻², so dy/dx = 8 − 27/x³.
(b) 8 − 27/x³ = 0 gives x³ = 27/8, x = 3/2; y = 12 + 27/(2 × 9/4) = 12 + 6 = 18. Stationary point (3/2, 18).
d²y/dx² = 81/x⁴ > 0, so it is a minimum point.
(c) At x = 2, dy/dx = 8 − 27/8 = 37/8; normal gradient = −8/37.
Through (2, 2) as stated: y − 2 = −(8/37)(x − 2), i.e. 8x + 37y − 90 = 0.
(Note: the curve point at x = 2 is (2, 155/8); the normal there is 64x + 296y − 5863 = 0.)

Set dy/dx = 0 for stationary points and use the second derivative for their nature.

Matrices written row by row:
(a) P = (−4 −1 / 2 0), Q = (0 1 / 6 −9), R = (1 −2 / 3 5).
(b)(i) 2P − 3R + Q = (−8 −2 / 4 0) − (3 −6 / 9 15) + (0 1 / 6 −9) = (−11 5 / 1 −24).
(ii) QR = (0 1 / 6 −9)(1 −2 / 3 5) = (3 5 / −21 −57).
(iii) |R| = 5 + 6 = 11; R⁻¹ = (1/11)(5 2 / −3 1).

Each row of the matrix holds the coefficients of x and y in one output coordinate.

(a) 9x/((2x + 1)(x² + 1)) = A/(2x + 1) + (Bx + C)/(x² + 1); 9x = A(x² + 1) + (Bx + C)(2x + 1).
x = −½: −9/2 = (5/4)A, A = −18/5. x² terms: A + 2B = 0, B = 9/5. Constants: A + C = 0, C = 18/5.
Answer: −18/(5(2x + 1)) + (9x + 18)/(5(x² + 1)).
(b) 2m(2m − 1) − 10 = m(m − 1), so 3m² − m − 10 = 0, (3m + 5)(m − 2) = 0. m = 2.

A quadratic factor that doesn't factorise needs a linear numerator Bx + C.

(a) Six marks: median = ½[(2x + y) + (x + 2y)] = 13½, so x + y = 9. With y = x + 1: x = 4, y = 5.
(b) Marks: 11, 12, 13, 14, 14, 17. Mean = 81/6 = 13.5.
Squared deviations: 6.25, 2.25, 0.25, 0.25, 0.25, 12.25; sum 21.5.
SD = √(21.5/6) = √3.583 ≈ 1.89.

With an even number of values, the median is the mean of the middle two.

Bearing θ gives (F sin θ, F cos θ) (east, north):
(a) F₁ = (10, 0); F₂ = (0, −16); F₃ = (3.50, 6.06); F₄ = (−10.39, 6.00).
(b)(i) Resultant = (3.11, −3.94); magnitude = √(3.108² + 3.938²) = √25.16 ≈ 5.02 N.
(ii) Mass = 12/10 = 1.2 kg; a = 5.016/1.2 ≈ 4.18 ms⁻².

Convert each bearing into east and north components, then add.

(a) X = (4t² + t³)/20, so v = (8t + 3t²)/20 and a = (8 + 6t)/20.
(i) t = 15: v = (120 + 675)/20 = 39.75 ms⁻¹.
(ii) Initial acceleration = 8/20 = 0.4; 8 × 0.4 = 3.2 = (8 + 6t)/20, so 6t = 56 and t = 9⅓ s.
(b) 6(7) + m(−5) = (6 + m)(4), so 42 − 5m = 24 + 4m and m = 2 kg.

Differentiate displacement for velocity, and again for acceleration.