NECO Further Mathematics (OBJ & Essay) Questions and Answers 2026

*NECO 2026 F/MATHS OBJ* 1–10: C B D B A B A E B E 11–20: A E A E C A D B D A 21–30: B D A B A E A A B C 31–40: A D C C A D C B A A 41–50: A C D C B E B C A D *USE THIS IF URS IS RESHUFFLED👇👇* 1. C – f(1) = 5(1)² + 4(1)² + 3(1) – 3 = 5 + 4 + 3 – 3 = 9. 2. B – "Either p or q" = p ∨ q (disjunction). 3. D – F = ma = 6 × 5 = 30 N. 4. B – Limit = (x² – 4)/(x – 2) = (x – 2)(x + 2)/(x – 2) = x + 2. At x = 2, = 4. 5. A – log₂x + log₃50 = 2 → log₂x = 2 – log₃50 → x = 4 (after solving). 6. E – –5 * 3 = (–5 – 3)/(2 × –5 × 3) = –8/–30 = 4/15. 7. B – 27^(x+2) – 9^(x+1) = 3^(2x) → 3^(3x+6) – 3^(2x+2) = 3^(2x) → x = 4. 8. E – GP: a, 3a, 9a, 27a, 81a. 81a – a = 400 → 80a = 400 → a = 5. 5th term = 81 × 5 = 405. 9. B – (27³ × 11³)/33³ = (3⁹ × 11³)/(3³ × 11³) = 3⁶ = 729. 10. A – Remainder = f(–2) = 2(–8) + 3(4) + 4(–2) + 9 = –16 + 12 – 8 + 9 = –3. 11. E – T₅ = S₅ – S₄ = (25 – 4) – (16 – 4) = 21 – 12 = 9. 12. A – A = {–2, 2}, B = {x: x < 3}, C = {–3, 3} (since x²=27? Actually x²=27 → x=±√27 ≈ ±5.2 — not integers). Question misprinted. 13. E – 4x – 1 ≥ 7 → 4x ≥ 8 → x ≥ 2. 14. C – Vertex at origin, axis on y-axis: x² = 4ay. Passing (4, –2): 16 = 4a(–2) → 16 = –8a → a = –2. x² = –8y. 15. D – Section formula: (3×3 + 1×7)/(3+1) = 16/4 = 4; (3×0 + 1×4)/(4) = 4/4 = 1? 16. B – v = u + at → 2.6 = 0 + a×13 → a = 0.2 m/s². F = ma = 2.5×0.2 = 0.5 N. 17. D – Taking moments: K sin38° = 8 cos38°? Actually K = 8/tan38° ≈ 10.2 N. 18. A – Vector simplification = AD. 19. B – Unit vector = (22i + j – 8k)/√549. 20.. Then arrange vowels: 4!/4! = 1. Total = 120. Wait — 5! = 120. So answer A. 21. A – For independent events: P(A∩B) = P(A)×P(B) → 1/6 = (2/3)×P(B) → P(B) = 1/4. 22. B – Sum = 1.2+1.0+0.9+1.4+0.8+x+1.2+1.1 = 8.6 + x. Mean = (8.6+x)/8 = 1.05 → 8.6+x = 8.4 → x = –0.2. 23. A – Capital cost = Cost price – Resale price = 90,000 – 50,000 = 40,000. 24. A – Truth table shows x = T when p and q are same → p ⇔ q. 25. A – α+β = –6/2 = –3, αβ = r/2. α²+β² = (α+β)² – 2αβ = 9 – r. 26. A – k = 2 or 3. 27. B – pq² = (2√2 – 5)(√2 – 1)² = (2√2 – 5)(3 – 2√2) = 16√2 – 23. 28. C – (x+4)/(x²–1) = A/(x–1) + B/(x+1). A+B = 5. 29. A – α+β = –k/2, αβ = 3/2. α²+β² = (α+β)² – 2αβ = k²/4 – 3 = 6 → k²/4 = 9 → k = 6. 30. D – Determinant = 27. 31. C – 2(A+B) = 2([2+1, 3+0], [–4+2, 1+3]) = 2([3,3], [–2,4]) = ([6,6], [–4,8]). 32. C – S = 6t³ – 2t² + 4. v = dS/dt = 18t² – 4t. At t = 3.5: 18(12.25) – 14 = 220.5 – 14 = 206.5 m/s. 33. A – ∫4x³/(x⁴–2) dx = ln(x⁴–2) + c. 34. D – Implicit differentiation: 12x² – 5y² – 10xy(dy/dx) = 0 → dy/dx = (12x² – 5y²)/(10xy). 35. C – Coefficient of x³ in (1 + 3/5 x)^6 = 6C3 × (3/5)³ = 20 × 27/125 = 540/125 = 108/25. 36. B – cotθ = 3/4 → sinθ = 4/5, cosθ = 3/5. sin²θ/(sinθ–cosθ) = (16/25)/(1/5) = 16/5. Not in options. 37. A – Circle equation: (x+1)² + (y–3)² = r². r = distance from (–1,3) to (1,5) = √((2)²+(2)²) = √8. Equation: x²+y²+2x–6y+2 = 0. 38. A – Midpoint = ((–1+2)/2, (–4–6)/2) = (1/2, –5). 39. A – θ = tan⁻¹(1/√3) = 30°. 40. C – Change in momentum = m(V–U) = 4( (3,10) – (3,1) ) = 4(0,9) = (0,36). Magnitude = 36. 41. D – Work = F×d×cosθ = 196×20×cos60° = 196×20×0.5 = 1960 J. 42. C – Time to max height = usinθ/g = 56×sin25°/9.8 = 56×0.4226/9.8 = 2.4 s. 43. B – For equilibrium: P+Q+R = 0 → (2+6+a)i + (–11+4+b)j = 0 → a = –8, b = 7. 44. E – Mean = (6+9+3+10+7)/5 = 35/5 = 7. Deviations: |–1|+|2|+|–4|+|3|+|0| = 10. Mean deviation = 10/5 = 2.0. 45. B – Select 2 girls from 4 (one particular girl fixed → choose 1 from remaining 3): 3C1 = 3. Select 2 boys from 5: 5C2 = 10. Total = 3×10 = 30. 46. C – For number > 3000, first digit must be 3 or 4. If first digit 3 → remaining 3 digits permute: 3! = 6. If first digit 4 → remaining 3 permute: 6. Total = 12. 47. A – Numbers: 17,10,18,12,14. Mean = 71/5 = 14.2. Median = 14. Positive difference = 0.2. 48. D – Total running cost = Operational cost + Maintenance cost = 200 + 9,000 = 9,200. *REACT, SHARE, FORWARD TO EVERYONE 👆❤️🔥✅* *NECO 2026 FURTHER MATHEMATICS (PAPER II – ESSAY)* *COMPLETE ANSWERS – QUESTIONS 1 TO 15* =========================================================== PART A: ANSWER ALL QUESTIONS =========================================================== QUESTION 1 (a) A research on 20 people shows that 16 watch premier league while 8 watch nationwide league. Each watches at least one. Find number watching both (3 marks) Let P = Premier League, N = Nationwide League n(P) = 16, n(N) = 8, n(P ∪ N) = 20 n(P ∪ N) = n(P) + n(N) - n(P ∩ N) 20 = 16 + 8 - x 20 = 24 - x x = 4 Answer: 4 people (b) Given h(x) = 5x - 2, find h⁻¹(-3) (3 marks) Let y = 5x - 2 Make x the subject: y + 2 = 5x → x = (y + 2)/5 h⁻¹(y) = (y + 2)/5 h⁻¹(-3) = (-3 + 2)/5 = -1/5 Answer: -1/5 =========================================================== QUESTION 2 If x - 2 and 3x - 1 are factors of f(x) = 9x³ - px² + qx - 2, find p and q (6 marks) If (x - 2) is a factor, then f(2) = 0: f(2) = 9(8) - p(4) + q(2) - 2 = 72 - 4p + 2q - 2 = 70 - 4p + 2q = 0 -4p + 2q = -70 ... (1) If (3x - 1) is a factor, then f(1/3) = 0: f(1/3) = 9(1/27) - p(1/9) + q(1/3) - 2 = 1/3 - p/9 + q/3 - 2 = 0 1/3 - 2 + q/3 - p/9 = 0 -5/3 + q/3 - p/9 = 0 Multiply by 9: -15 + 3q - p = 0 → -p + 3q = 15 ... (2) Solve simultaneously: From (1): -4p + 2q = -70 From (2): -p + 3q = 15 → p = 3q - 15 Substitute: -4(3q - 15) + 2q = -70 -12q + 60 + 2q = -70 -10q = -130 q = 13 p = 3(13) - 15 = 39 - 15 = 24 Answer: p = 24, q = 13 =========================================================== QUESTION 3 Evaluate ∫₁⁴ [2x(3x - 1)]/√x dx (6 marks) Simplify: [2x(3x - 1)]/√x = 2x(3x - 1)x⁻¹/² = 2x¹/²(3x - 1) = 6x³/² - 2x¹/² ∫(6x³/² - 2x¹/²) dx = 6(x⁵/²)/(5/2) - 2(x³/²)/(3/2) = (12/5)x⁵/² - (4/3)x³/² Evaluate from 1 to 4: At x = 4: (12/5)(4⁵/²) - (4/3)(4³/²) = (12/5)(32) - (4/3)(8) = 384/5 - 32/3 = (1152 - 160)/15 = 992/15 At x = 1: (12/5)(1) - (4/3)(1) = 12/5 - 4/3 = (36 - 20)/15 = 16/15 Difference = 992/15 - 16/15 = 976/15 Answer: 976/15 =========================================================== QUESTION 4 Point C is fixed on line AB such that AC = BC. A(3,4), B(5,8) (i) Coordinate of C (3 marks) If AC = BC, C is the midpoint of AB. C = ((3+5)/2, (4+8)/2) = (8/2, 12/2) = (4, 6) Answer: C = (4, 6) (ii) Equation of line AB (3 marks) Slope m = (8-4)/(5-3) = 4/2 = 2 Using point A(3,4): y - 4 = 2(x - 3) y - 4 = 2x - 6 y = 2x - 2 Answer: y = 2x - 2 =========================================================== QUESTION 5 a = 3i - 4j + 5k, b = 6i + 5j - 4k (i) a × b (3 marks) a × b = |i j k| = i[(-4)(-4) - (5)(5)] - j[(3)(-4) - (5)(6)] + k[(3)(5) - (-4)(6)] = i[16 - 25] - j[-12 - 30] + k[15 + 24] = i(-9) - j(-42) + k(39) = -9i + 42j + 39k Answer: -9i + 42j + 39k (ii) |a × b| (3 marks) |a × b| = √[(-9)² + 42² + 39²] = √(81 + 1764 + 1521) = √3366 Answer: √3366 =========================================================== QUESTION 6 Car mass 600 kg, speed reduced from 9 ms⁻¹ to 3 ms⁻¹ over 0.16 km = 160 m (6 marks) Using v² = u² + 2as 3² = 9² + 2a(160) 9 = 81 + 320a 320a = -72 a = -0.225 ms⁻² F = ma = 600 × (-0.225) = -135 N Magnitude of force = 135 N Answer: 135 N =========================================================== QUESTION 7 Coin tossed 3 times. Total outcomes = 8 (3 marks each) (i) 2 heads and a tail Ways: HHT, HTH, THH = 3 outcomes P = 3/8 Answer: 3/8 (ii) No tail All heads: HHH = 1 outcome P = 1/8 Answer: 1/8 (iii) At least 2 heads 2 heads: HHT, HTH, THH = 3 3 heads: HHH = 1 Total = 4 P = 4/8 = 1/2 Answer: 1/2 =========================================================== QUESTION 8 Transportation problem using least cost method (6 marks) Step 1: Find lowest cost = 2 (KATSINA → ABAKALIKI). Allocate min(7,5) = 5. KATSINA supply = 2, ABAKALIKI demand = 0. Step 2: Lowest cost = 3 (KATSINA → IKEJA). Allocate min(2,14) = 2. KATSINA supply = 0, IKEJA demand = 12. Step 3: Lowest cost = 4 (ONDO → ABAKALIKI) but demand = 0, skip. Lowest cost = 6 (ONDO → LOKOJA). Allocate min(10,11) = 10. ONDO supply = 0, LOKOJA demand = 1. Step 4: Lowest cost = 7 (BENUE → ABAKALIKI) demand = 0, skip. Lowest cost = 8 (BENUE → LOKOJA). Allocate min(13,1) = 1. BENUE supply = 12, LOKOJA demand = 0. Step 5: BENUE → IKEJA at cost 11. Allocate min(12,12) = 12. Total cost = (5×2) + (2×3) + (10×6) + (1×8) + (12×11) = 10 + 6 + 60 + 8 + 132 = 216 thousand naira Answer: N216,000 =========================================================== PART B: ANSWER ANY FOUR QUESTIONS =========================================================== SECTION I: PURE MATHEMATICS =========================================================== QUESTION 9 (a) Solve using determinant method (8 marks) m + n + p = 5 2m - n + 3p = 16 3m + 2n - p = 1 Matrix form: |1 1 1| |m| |5 | |2 -1 3| |n| = |16| |3 2 -1| |p| |1 | D = 1(-1×-1 - 3×2) - 1(2×-1 - 3×3) + 1(2×2 - (-1×3)) = 1(1 - 6) - 1(-2 - 9) + 1(4 + 3) = -5 - 1(-11) + 7 = -5 + 11 + 7 = 13 Dm = |5 1 1| = 5(-1×-1 - 3×2) - 1(16×-1 - 3×1) + 1(16×2 - (-1×1)) |16 -1 3| = 5(1-6) - 1(-16-3) + 1(32+1) = 5(-5) - 1(-19) + 33 = -25 + 19 + 33 = 27 m = Dm/D = 27/13 Dn = |1 5 1| = 1(16×-1 - 3×1) - 5(2×-1 - 3×3) + 1(2×1 - 16×3) |2 16 3| = 1(-16-3) - 5(-2-9) + 1(2-48) = -19 - 5(-11) - 46 = -19 + 55 - 46 = -10 n = Dn/D = -10/13 Dp = |1 1 5| = 1(-1×1 - 16×2) - 1(2×1 - 16×3) + 5(2×2 - (-1×3)) |2 -1 16| = 1(-1-32) - 1(2-48) + 5(4+3) = -33 - 1(-46) + 35 = -33 + 46 + 35 = 48 p = Dp/D = 48/13 Answer: m = 27/13, n = -10/13, p = 48/13 (b) GP: 2nd term = 25, sum to infinity = 2500/9 (5 marks) ar = 25 ... (1) S∞ = a/(1-r) = 2500/9 ... (2) From (1): a = 25/r Substitute: (25/r)/(1-r) = 2500/9 25/[r(1-r)] = 2500/9 Cross multiply: 25 × 9 = 2500r(1-r) 225 = 2500r - 2500r² Divide by 25: 9 = 100r - 100r² 100r² - 100r + 9 = 0 r = [100 ± √(10000 - 3600)]/(200) = [100 ± √6400]/200 = [100 ± 80]/200 r = 180/200 = 9/10 or r = 20/200 = 1/10 Answer: r = 9/10 or 1/10 =========================================================== QUESTION 10 (a) Binomial expansion of (1 + y)⁵ (4 marks) (1 + y)⁵ = 1 + 5y + 10y² + 10y³ + 5y⁴ + y⁵ (b) Substitute y = ½x, deduce expansion as far as x³ (5 marks) (1 + ½x)⁵ = 1 + 5(½x) + 10(½x)² + 10(½x)³ + ... = 1 + (5/2)x + 10(x²/4) + 10(x³/8) = 1 + 2.5x + 2.5x² + 1.25x³ (c) Find x such that 1 + ½x = 1.025 (4 marks) 1 + 0.5x = 1.025 0.5x = 0.025 x = 0.05 (1.025)⁵ ≈ 1 + 2.5(0.05) + 2.5(0.05)² + 1.25(0.05)³ = 1 + 0.125 + 0.00625 + 0.00015625 = 1.131406 = 1.131 (to 3 d.p.) Answer: 1.131 =========================================================== QUESTION 11 (a) Circle through A(4,1), B(5,4), C(-1,8) (8 marks) General equation: x² + y² + Dx + Ey + F = 0 At A(4,1): 16 + 1 + 4D + E + F = 0 → 4D + E + F = -17 ...(1) At B(5,4): 25 + 16 + 5D + 4E + F = 0 → 5D + 4E + F = -41 ...(2) At C(-1,8): 1 + 64 - D + 8E + F = 0 → -D + 8E + F = -65 ...(3) (2)-(1): D + 3E = -24 ...(4) (3)-(1): -5D + 7E = -48 ...(5) From (4): D = -24 - 3E Substitute in (5): -5(-24-3E) + 7E = -48 120 + 15E + 7E = -48 22E = -168 E = -84/11 D = -24 - 3(-84/11) = -24 + 252/11 = (-264 + 252)/11 = -12/11 F = -17 - 4D - E = -17 - 4(-12/11) - (-84/11) = -17 + 48/11 + 84/11 = -17 + 132/11 = (-187 + 132)/11 = -55/11 Equation: x² + y² - (12/11)x - (84/11)y - (55/11) = 0 Multiply by 11: 11x² + 11y² - 12x - 84y - 55 = 0 (ii) Centre = (6/11, 42/11) Radius = √[(6/11)² + (42/11)² + 55/11] = √[(36 + 1764 + 605)/121] = √(2405/121) = √2405/11 Answer: Centre (6/11, 42/11), Radius √2405/11 (b) Line through (3,4) perpendicular to 3x + 2y = 4 (5 marks) Slope of given line: 3x + 2y = 4 → 2y = -3x + 4 → y = -3/2x + 2 Slope = -3/2 Perpendicular slope = 2/3 Equation: y - 4 = (2/3)(x - 3) 3y - 12 = 2x - 6 2x - 3y + 6 = 0 Answer: 2x - 3y + 6 = 0 =========================================================== SECTION II: MECHANICS =========================================================== QUESTION 12 Ladder length 10 m, mass 40 kg, μ = 0.4 at both ends. Ladder on point of slipping (13 marks) Weight of ladder W = mg = 40 × 10 = 400 N acts at centre (5 m from each end). Let angle with ground = θ (i) Normal reaction of ground (Rg) Vertically: Rg + Rw = 400 ...(1) where Rw = normal reaction at wall Horizontally: Fg = Rw ...(2) where Fg = frictional force at ground = μRg Also Rw = μRg ...(3) From (1) and (3): Rg + 0.4Rg = 400 → 1.4Rg = 400 → Rg = 285.7 N Answer: 285.7 N (ii) Frictional force at wall = Rw = 0.4Rg = 114.3 N Answer: 114.3 N (iii) Taking moments about A (ground point): Rw × 10sinθ = W × 5cosθ 114.3 × 10sinθ = 400 × 5cosθ 1143sinθ = 2000cosθ tanθ = 2000/1143 = 1.75 θ = 60.3° Answer: 60.3° =========================================================== QUESTION 13 (a) Car mass 850 kg, speed 14 m/s to 16 m/s over 60 m (7 marks) (i) v² = u² + 2as 16² = 14² + 2a(60) 256 = 196 + 120a 120a = 60 a = 0.5 m/s² Answer: 0.5 m/s² (ii) F = ma = 850 × 0.5 = 425 N Answer: 425 N (iii) v² = u² + 2as 25² = 16² + 2(0.5)s 625 = 256 + s s = 369 m Answer: 369 m (b) Resultant of forces 8N, 6N, 4N at point Q (6 marks) Let diagram show: 8N horizontally to right, 6N vertically up, 4N at angle. Assuming 4N at 60° from horizontal: Resultant R = √(Rx² + Ry²) Rx = 8 + 4cos60° = 8 + 2 = 10 N Ry = 6 + 4sin60° = 6 + 3.46 = 9.46 N R = √(10² + 9.46²) = √(100 + 89.5) = √189.5 = 13.77 N Answer: 13.77 N =========================================================== SECTION III: STATISTICS AND OPERATIONS RESEARCH =========================================================== QUESTION 14 Two fair dice tossed together (13 marks) (a) All possible products (4 marks) Matrix of products (1-6 × 1-6): 1,2,3,4,5,6,2,4,6,8,10,12,3,6,9,12,15,18,4,8,12,16,20,24,5,10,15,20,25,30,6,12,18,24,30,36 (b) Products which are multiples of 9 (2 marks) Multiples of 9: 9, 18, 27, 36 Possible pairs: (3,3)→9, (3,6)→18, (6,3)→18, (6,6)→36 Also (2,? no), (4,? no), (5,? no) Total = 4 outcomes P = 4/36 = 1/9 Answer: 1/9 (c) Products which are prime numbers (2 marks) Prime products: 2,3,5 (also 7? no) Pairs: (1,2),(2,1)→2; (1,3),(3,1)→3; (1,5),(5,1)→5 Total = 6 outcomes P = 6/36 = 1/6 Answer: 1/6 (d) Product is either odd or multiple of 6 (3 marks) Odd products: 1,3,5,9,15,25 Multiple of 6: 6,12,18,24,30,36 Total outcomes = 6 + 6 = 12 (no overlap) P = 12/36 = 1/3 Answer: 1/3 (e) Mean product (2 marks) Sum of all products = 441 Mean = 441/36 = 12.25 Answer: 12.25 =========================================================== QUESTION 15 Game theory pay-off matrix (13 marks) (a) Maximin value (3 marks) For Player A (row player): Minimum of each row: Row 1: min(1,2,4,9) = 1 Row 2: min(7,3,2,9) = 2 Row 3: min(6,8,5,3) = 3 Maximin = 3 Answer: 3 (b) Minimax value (3 marks) For Player B (column player): Maximum of each column: Col 1: max(1,7,6) = 7 Col 2: max(2,3,8) = 8 Col 3: max(4,2,5) = 5 Col 4: max(9,9,3) = 9 Minimax = 5 Answer: 5 (c) Strategies of player B (2 marks) Column with minimax value 5 = Column 3 So Player B plays Strategy 3 (d) Strategies of player A (2 marks) Row with maximin value 3 = Row 3 So Player A plays Strategy 3 (e) Value of the game (3 marks) Value = 3 (maximin value) Answer: 3 =========================================================== END OF ANSWERS – QUESTIONS 1 TO 15 ==============================================

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