WAEC Further Mathematics 2008 past questions

Checked by the Skolra question team · Last updated 23 October 2026

Ten questions from the 2008 WAEC Further Mathematics paper, with answers and explanations. We hold 41 questions from this paper, and the Skolra app has all of them.

WAEC Further Mathematics 2008 questions and answers

1. 1.A test consists of 12 questions out of which candidates are to answer 10. If the first 6 are compulsory, in how many ways can each candidate select her questions?
   1. A
      30
   2. B
      360
   3. C
      15
      Correct
   4. D
      66
   Show answerHide answer

   Answer: C

   Setting up the problem: There are 12 questions in all, and a candidate must answer 10 of them. Since the first 6 questions are compulsory, every candidate automatically answers those 6. What is left to decide is how to pick the remaining 10 − 6 = 4 questions from the 12 − 6 = 6 questions that are not compulsory.

   The order in which the questions are answered does not matter, so this is a combination:

   Number of ways = ⁶C₄ = 6!/(4!2!) = (6×5)/(2×1) = 15.

   So the candidate can select her questions in 15 different ways.

2. 2.Express 1/(1 - sin 45°) in surd form.
   1. A
      −(1 + √2)
   2. B
      (2 + √2)/3
   3. C
      2 + √2
      Correct
   4. D
      2 − √2
   Show answerHide answer

   Answer: C

   Step 1: Recall sin 45° = √2/2.

   Step 2: So 1 − sin 45° = 1 − √2/2 = (2 − √2)/2.

   Step 3: Expressing 1/(1 − sin 45°) in surd form means rationalizing the denominator:

   1 ÷ [(2 − √2)/2] = 2/(2 − √2).

   Multiply top and bottom by the conjugate (2 + √2):

   2(2 + √2) / [(2 − √2)(2 + √2)] = 2(2 + √2) / (4 − 2) = 2(2 + √2)/2 = 2 + √2.

   This is the correct surd form.

   Where the wrong answers come from:

   2 − √2 comes from multiplying by the wrong sign, or mixing up (1 + sin 45°) with (1 − sin 45°) during rationalization — the sign on √2 flips.

   −(1 + √2) comes from forgetting to halve √2 (treating sin 45° as √2 instead of √2/2) before rationalizing, which drags a negative sign and a 1 into the final surd.

3. 3.The distance between P(x, 7) and Q(6, 19) is 13 units. Find the values of x.
   1. A
      x = 5 or x = 7
   2. B
      x = 5 or x = -5
   3. C
      x = -1 or x = -11
   4. D
      x = 1 or x = 11
      Correct
   Show answerHide answer

   Answer: D

   Using the distance formula: for P(x, 7) and Q(6, 19), the distance is

   |PQ| = √[(x−6)² + (7−19)²] = 13

   Squaring both sides: (x−6)² + (−12)² = 13², so (x−6)² + 144 = 169, which gives (x−6)² = 25.

   Taking the square root of both sides: x − 6 = ±5. So x = 6 + 5 = 11, or x = 6 − 5 = 1.

   That makes x = 1 or x = 11 the correct answer.

   A student who forgets to add the 6 back after taking the square root of 25 stops at x = 5 or x = −5 — that is only the value of (x−6), not x itself.

   A student who mixes up the sign when reversing the subtraction gets x = −1 or x = −11 instead, by writing x − (−6) = ±5 incorrectly.

4. 4.If p = (2, -2) and q = (3, 4), find |q - (1/2)p|.
   1. A
      5
   2. B
      √37
   3. C
      √29
      Correct
   4. D
      2√10
   Show answerHide answer

   Answer: C

   Step 1: Scale p by 1/2.
   p = (2, -2), so (1/2)p = (1, -1).

   Step 2: Subtract from q.
   q - (1/2)p = (3, 4) - (1, -1) = (3-1, 4-(-1)) = (2, 5).

   Step 3: Find the magnitude.
   |(2, 5)| = √(2² + 5²) = √(4 + 25) = √29. This is the correct result.

   The option √37 comes from forgetting to halve p before subtracting — a student who computes q - p directly gets (1, 6), whose magnitude is √37.

   The option 5 comes from a sign slip: adding (1/2)p to q instead of subtracting gives (4, 3), whose magnitude is a clean 5 — tempting, but it answers the wrong expression.

5. 5.If f(x) = 2x² - 3x - 1, find the value of x for which f(x) is minimum.
   1. A
      x = 3/2
   2. B
      x = 3/4
      Correct
   3. C
      x = -3/4
   4. D
      x = 3/8
   Show answerHide answer

   Answer: B

   f(x) = 2x² - 3x - 1

   Since the coefficient of x² (which is 2) is positive, the curve is a parabola that opens upward, so it has a minimum point. The minimum occurs where the gradient (derivative) is zero.

   Differentiate: f'(x) = 4x - 3.

   Set f'(x) = 0: 4x - 3 = 0, so 4x = 3, giving x = 3/4.

   This matches the vertex formula x = -b/(2a) = -(-3)/(2 × 2) = 3/4, so x = 3/4 is correct.

6. 6.If sin A = 3/5 and cos B = 15/17, where A is an obtuse angle and B is acute, find the value of cos (A + B).
   1. A
      36/85
   2. B
      -84/85
      Correct
   3. C
      84/85
   4. D
      -36/85
   Show answerHide answer

   Answer: B

   Step 1: Find cos A. Since sin A = 3/5 and A is obtuse, A lies in the second quadrant where cosine is negative. Using the 3-4-5 right triangle, cos A = -4/5.

   Step 2: Find sin B. Since cos B = 15/17 and B is acute, sin B is positive. Using the 8-15-17 right triangle, sin B = 8/17.

   Step 3: Apply the compound angle formula.
   cos(A + B) = cos A cos B - sin A sin B
   = (-4/5)(15/17) - (3/5)(8/17)
   = -60/85 - 24/85
   = -84/85

   The value -84/85 is correct because it correctly uses the negative cosine for the obtuse angle A and subtracts the sine product as the compound angle formula requires.

7. 7.The initial and final velocities of an object of mass 5 kg are u = (1, 3) and v = (4, 7) respectively. Find the magnitude of its change in momentum.
   1. A
      25 kg ms⁻¹
      Correct
   2. B
      35 kg ms⁻¹
   3. C
      5 kg ms⁻¹
   4. D
      125 kg ms⁻¹
   Show answerHide answer

   Answer: A

   Step 1: Find the change in velocity vector.
   v - u = (4, 7) - (1, 3) = (4-1, 7-3) = (3, 4)

   Step 2: Multiply by the mass to get the change in momentum vector.
   Δp = m(v - u) = 5 × (3, 4) = (15, 20)

   Step 3: Find the magnitude using Pythagoras.
   |Δp| = √(15² + 20²) = √(225 + 400) = √625 = 25 kg ms⁻¹

   This is the correct answer.

   A candidate who forgets to multiply by the mass and only finds the magnitude of (v - u), which is √(3² + 4²) = 5, lands on the distractor of 5 kg ms⁻¹ — that is only the change in velocity, not the change in momentum.

8. 8.The sum of the first n terms of a linear sequence is Sn = n² + 2n. Determine the general term of the sequence.
   1. A
      Tn = 2n + 2
   2. B
      Tn = 2n − 1
   3. C
      Tn = n² + 2n
   4. D
      Tn = 2n + 1
      Correct
   Show answerHide answer

   Answer: D

   We are given Sn = n² + 2n, the sum of the first n terms. To find the general term Tn, we use the rule:

   Tn = Sn − S(n−1) for n ≥ 2, and Tn = S1 when n = 1.

   First find S(n−1) by replacing n with (n−1):
   S(n−1) = (n−1)² + 2(n−1) = n² − 2n + 1 + 2n − 2 = n² − 1

   Now subtract:
   Tn = Sn − S(n−1) = (n² + 2n) − (n² − 1) = 2n + 1

   Check with n = 1: S1 = 1² + 2(1) = 3, and from the formula Tn = 2(1) + 1 = 3. They match, so Tn = 2n + 1 holds for every n, including the first term.

   The option Tn = n² + 2n simply repeats the formula for Sn itself — a candidate who forgets to subtract consecutive sums falls into this trap, giving the sum formula instead of the term formula.

9. 9.Find the range of values of x for which 2x² + 7x - 15 > 0.
   1. A
      x < -5 or x > 3/2
      Correct
   2. B
      x < -3/2 or x > 5
   3. C
      -3/2 < x < 5
   4. D
      -5 < x < 3/2
   Show answerHide answer

   Answer: A

   We want the values of x for which 2x² + 7x − 15 is greater than zero.

   Factorise the quadratic. We need two numbers that multiply to give (2 × -15) = -30 and add up to 7: these are 10 and -3.

   2x² + 7x − 15 = 2x² + 10x − 3x − 15
   = 2x(x + 5) − 3(x + 5)
   = (2x − 3)(x + 5)

   So (2x − 3)(x + 5) > 0, and the critical values are x = 3/2 and x = −5.

   Since the coefficient of x² is positive, the curve of the expression opens upward (a smile shape). Such a curve is positive outside its two roots and negative between them.

   Therefore 2x² + 7x − 15 > 0 when x < −5 or x > 3/2, which is the correct range.

   A candidate who forgets that an upward-opening curve is positive outside the roots (and wrongly takes the region between them instead) ends up with −5 < x < 3/2.

10. 10.The sum of the first n terms of a linear sequence is Sn = n² + 2n. Find the common difference of the sequence.
    1. A
       2
       Correct
    2. B
       3
    3. C
       1
    4. D
       4
    Show answerHide answer

    Answer: A

    Given: S_n = n² + 2n, the sum of the first n terms of a linear (arithmetic) sequence.

    To find any term, use T_n = S_n − S_n−1.

    S_n−1 = (n − 1)² + 2(n − 1) = n² − 2n + 1 + 2n − 2 = n² − 1.

    So T_n = (n² + 2n) − (n² − 1) = 2n + 1.

    Check it directly: T₁ = S₁ = 1 + 2 = 3, T₂ = S₂ − S₁ = 8 − 3 = 5, T₃ = S₃ − S₂ = 15 − 8 = 7.
    The sequence runs 3, 5, 7, ... so each term is 2 more than the last — the common difference is 2.

    There is a quicker rule worth remembering: whenever S_n = An² + Bn, the sequence is linear with common difference d = 2A. Here A = 1, giving d = 2 directly, which matches the term-by-term check above.

The questions come from past JAMB, WAEC and NECO papers. We use software to help write the answer options and explanations, and every answer is checked twice before it goes on the site. How we check

Spotted a wrong answer? Tell us from the app or on the contact page and we check it.

Practise all 41 questions we hold from the 2008 paper, on a timed CBT screen

Every question has the answer and an explanation, in the Skolra app.

Download for WindowsDownload for Mac

About the app · Skolra for Windows · Skolra for Mac