WAEC Further Mathematics 2012 past questions

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

Ten questions from the 2012 WAEC Further Mathematics paper, with answers and explanations. We hold 40 questions from this paper, and the Skolra app has all of them.

WAEC Further Mathematics 2012 questions and answers

1. 1.Find the constant term in the binomial expansion of (2x - 3/x)⁸.
   1. A
      72576
   2. B
      90720
      Correct
   3. C
      1451520
   4. D
      -90720
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   Answer: B

   In the expansion of (2x - 3/x)⁹, the general term is:

   T(k+1) = C(8,k) × (2x)^(8-k) × (-3/x)^k = C(8,k) × 2^(8-k) × (-3)^k × x^(8-2k)

   The constant term occurs when the power of x is zero, i.e. 8 - 2k = 0, so k = 4.

   Substituting k = 4:

   C(8,4) = 70
   2^(8-4) = 2^4 = 16
   (-3)^4 = 81

   Term = 70 × 16 × 81 = 70 × 1296 = 90720

   This is the correct constant term.

   The negative figure comes from wrongly treating (-3)^4 as still negative, forgetting that a negative number raised to an even power becomes positive.

   The smaller figure comes from using the wrong binomial coefficient (mixing up C(8,4) = 70 with C(8,3) = 56), which is a common slip when reading Pascal's triangle or the combination formula.

2. 2.Given that nPr = 90 and nCr = 15, find the value of r.
   1. A
      6
   2. B
      3
      Correct
   3. C
      2
   4. D
      4
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   Answer: B

   There is a direct link between the permutation and combination formulas: nPr = nCr × r!. This is because nCr counts the number of ways to choose r objects (order does not matter), while nPr counts the number of ways to choose AND arrange them (order matters), so nPr is nCr multiplied by the r! ways of arranging the chosen r objects.

   Dividing the given values: r! = nPr ÷ nCr = 90 ÷ 15 = 6.

   Since 3! = 3 × 2 × 1 = 6, it follows that r = 3 is correct.

3. 3.QRS is a triangle such that QR = (3i+2j) and SR = (-5i+3j), find SQ.
   1. A
      SQ = −8i − j
   2. B
      SQ = −2i + 5j
   3. C
      SQ = −8i + j
      Correct
   4. D
      SQ = 8i − j
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   Answer: C

   Let the position vectors of Q, R and S be q, r and s.

   From QR = r − q = 3i + 2j, we get r = q + 3i + 2j.

   From SR = r − s = −5i + 3j, we get s = r − (−5i + 3j) = r + 5i − 3j.

   Substituting for r: s = (q + 3i + 2j) + 5i − 3j = q + 8i − j.

   So SQ = q − s = −8i + j, which is the correct result.

   −2i + 5j comes from wrongly adding QR and SR directly (3i+2j + (−5i+3j)) instead of going Q → R → S → Q correctly — that gives QS, not SQ, and even then the route is set up wrong.
   8i − j and −8i − j both come from sign slips: flipping the overall sign of the whole vector, or getting the j-component sign wrong while the i-component is correct.

4. 4.If y = x³ - x² - x + 6, find the values of x at the turning point.
   1. A
      x = 1 or x = -1/3
      Correct
   2. B
      x = 0 or x = 2/3
   3. C
      x = -1 or x = 1/3
   4. D
      x = 1 or x = 1/3
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   Answer: A

   Differentiate first. For y = x³ - x² - x + 6, dy/dx = 3x² - 2x - 1.

   At a turning point, dy/dx = 0, so 3x² - 2x - 1 = 0. Factorising gives (3x + 1)(x - 1) = 0, so x = 1 or x = -1/3.

   Checking with the quadratic formula: discriminant = (-2)² - 4(3)(-1) = 16, so x = (2 ± 4)/6, which again gives x = 1 or x = -1/3. The pair x = 1 or x = -1/3 is correct.

   The pair x = 0 or x = 2/3 comes from a slip where the candidate drops the constant -1 from the derivative, leaving 3x² - 2x = 0, which factors as x(3x - 2) = 0.

   The pair x = -1 or x = 1/3 swaps the signs on both roots, a common error when the factorisation (3x + 1)(x - 1) is written as (3x - 1)(x + 1) instead.

5. 5.Differentiate x/(x+1) with respect to x.
   1. A
      x/(x+1)²
   2. B
      1/(x+1)
   3. C
      1/(x+1)²
      Correct
   4. D
      -1/(x+1)²
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   Answer: C

   We are asked to differentiate y = x/(x+1), a quotient, so we apply the quotient rule: if y = u/v, then dy/dx = (v·u′ − u·v′)/v².

   Here u = x (so u′ = 1), and v = x + 1 (so v′ = 1). Substituting:
   dy/dx = [(x+1)(1) − x(1)] / (x+1)² = (x + 1 − x) / (x+1)² = 1/(x+1)².

   This is the correct result: the numerator simplifies because the x terms cancel, leaving just 1 on top, while the denominator must be squared as the quotient rule demands.

6. 6.The marks obtained by 10 students in a test are as follows: 3, 7, 6, 2, 8, 5, 9, 1, 4 and 10. Find the mean mark.
   1. A
      4.5
   2. B
      5.0
   3. C
      6.0
   4. D
      5.5
      Correct
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   Answer: D

   Step 1: Add up all the marks.
   3 + 7 + 6 + 2 + 8 + 5 + 9 + 1 + 4 + 10 = 55

   Step 2: Divide by the number of students.
   Mean = Total marks ÷ Number of students = 55 ÷ 10 = 5.5

   So the correct mean mark is 5.5.

   A candidate who gets 4.5 has likely forgotten to include one of the ten marks (for example, dropping the 10) before dividing, giving 45 ÷ 10. A candidate who gets 5.0 has made a small slip while adding the figures, landing on a total of 50 instead of 55. A candidate who gets 6.0 has over-added somewhere and arrived at a total of 60 instead of 55. Always re-add the list carefully and remember to divide by the actual number of data items given.

7. 7.Simplify: (cos2θ - 1)/sin2θ
   1. A
      −tan θ
      Correct
   2. B
      cot θ
   3. C
      tan θ
   4. D
      −cot θ
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   Answer: A

   Simplify (cos2θ − 1)/sin2θ.

   Use the double-angle identities: cos2θ = 1 − 2sin²θ and sin2θ = 2sinθcosθ.

   So cos2θ − 1 = (1 − 2sin²θ) − 1 = −2sin²θ.

   Therefore:
   (cos2θ − 1)/sin2θ = (−2sin²θ)/(2sinθcosθ) = −sinθ/cosθ = −tan θ.

   This is the correct simplification.

   A student who forgets the negative sign in cos2θ − 1 = −2sin²θ ends up with plain tan θ — that mistake drops the sign along the way.

   A student who mixes up which trig ratio sits on top after cancelling the sin θ factors gets cot θ instead of tan θ, inverting the fraction sinθ/cosθ to cosθ/sinθ.

   Combining both slips (wrong sign and inverted ratio) gives −cot θ, which looks plausible but is still not the simplified form.

8. 8.Given that P = (2,1; 5,-3) and Q = (4,-8; 1,-2), Find (2P - Q).
   1. A
      (0, -10; -9, 4)
   2. B
      (-2, 9; 4, -1)
   3. C
      (0, 10; 9, -4)
      Correct
   4. D
      (8, -6; 11, -8)
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   Answer: C

   P = (2, 1; 5, -3) and Q = (4, -8; 1, -2).

   First find 2P by doubling every entry of P:
   2P = (4, 2; 10, -6)

   Now subtract Q entry by entry:
   2P - Q = (4 - 4, 2 - (-8); 10 - 1, -6 - (-2))
   = (0, 10; 9, -4)

   So (0, 10; 9, -4) is correct.

   (-2, 9; 4, -1) comes from forgetting to double P first and just computing P - Q — a very common slip.

   (0, -10; -9, 4) is what you get if you reverse the subtraction order and compute Q - 2P instead of 2P - Q; notice every entry is just the sign-flip of the correct matrix.

   (8, -6; 11, -8) comes from adding Q to 2P instead of subtracting it.

9. 9.Two forces 10N and 6N act in the directions 060° and 330° respectively. Find the x-component of their resultant.
   1. A
      10.20 N
   2. B
      8.66 N
   3. C
      5.66 N
      Correct
   4. D
      11.66 N
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   Answer: C

   Setting up the components. Bearings are measured clockwise from North, so for any force acting at a bearing θ, the horizontal (x, East) component is F sin θ and the vertical (y, North) component is F cos θ. This is opposite to the usual maths convention (where x uses cosine), and mixing the two up is the most common slip in this type of question.

   Force 1 (10N at 060°):
   x₁ = 10 sin60° = 10 × 0.8660 = 8.660 N

   Force 2 (6N at 330°):
   x₂ = 6 sin330° = 6 × (-0.5) = -3.000 N, since 330° lies just before North going anticlockwise, so its sine is negative (it pulls slightly to the West).

   Resultant x-component = 8.660 + (-3.000) = 5.66 N, which is the correct answer.

10. 10.Evaluate ∫₋₂³ (3x² - 2x - 12) dx
    1. A
       -30
       Correct
    2. B
       30
    3. C
       -18
    4. D
       12
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    Answer: A

    Finding the antiderivative: integrate each term of 3x² - 2x - 12 separately.

    ∫3x² dx = x³, ∫-2x dx = -x², ∫-12 dx = -12x.

    So F(x) = x³ - x² - 12x.

    Evaluating at the limits:
    F(3) = 3³ - 3² - 12(3) = 27 - 9 - 36 = -18
    F(-2) = (-2)³ - (-2)² - 12(-2) = -8 - 4 + 24 = 12

    Applying the Fundamental Theorem of Calculus: the definite integral equals F(upper) - F(lower) = F(3) - F(-2) = -18 - 12 = -30. This is the correct value.

    A candidate who evaluates only at the upper limit and stops gets -18, forgetting that a definite integral requires subtracting the lower-limit value too.

    A candidate who evaluates only at the lower limit and stops gets 12, the same mistake in the other direction.

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