WAEC Further Mathematics 2010 past questions

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

Ten questions from the 2010 WAEC Further Mathematics paper, with answers and explanations. We hold 43 questions from this paper, and the Skolra app has all of them.

WAEC Further Mathematics 2010 questions and answers

1. 1.The equation of a curve is $y = x(3 - x^{2})$. Find the equation of its normal of the point where x = 2.
   1. A
      9x + y - 16 = 0
   2. B
      x + 9y + 16 = 0
   3. C
      x - 9y + 16 = 0
   4. D
      x - 9y - 20 = 0
      Correct
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   Answer: D

   Step 1: Find the point on the curve.
   y = x(3 - x²) = 3x - x³. At x = 2: y = 2(3 - 4) = 2(-1) = -2. So the curve passes through (2, -2).

   Step 2: Find the gradient of the tangent.
   dy/dx = 3 - 3x². At x = 2: dy/dx = 3 - 3(4) = 3 - 12 = -9.

   Step 3: Find the gradient of the normal.
   The normal is perpendicular to the tangent, so its gradient is the negative reciprocal: -1 ÷ (-9) = 1/9.

   Step 4: Form the equation of the normal.
   Using y - y₁ = m(x - x₁) with (x₁, y₁) = (2, -2) and m = 1/9:
   y + 2 = (1/9)(x - 2)
   9y + 18 = x - 2
   9y = x - 20
   x - 9y - 20 = 0 is the correct equation of the normal.

2. 2.Yomi was asked to label four seats S, R, P, Q. What is the probability he labelled them in alphabetical order?
   1. A
      1/4
   2. B
      1/12
   3. C
      1/16
   4. D
      1/24
      Correct
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   Answer: D

   Yomi is labelling four seats with the four distinct labels S, R, P and Q. The number of different ways to arrange 4 distinct labels is 4! = 4 × 3 × 2 × 1 = 24.

   Out of all these 24 possible arrangements, only one arrangement puts the labels in alphabetical order: P, Q, R, S (in that order on the seats).

   So the probability that Yomi labelled the seats in alphabetical order is:

   P = 1/24

3. 3.The probabilities that a husband and wife will be alive in 15 years time are m and n respectively. Find the probability that only one of them will be alive at that time.
   1. A
      (1 − m)(1 − n)
   2. B
      m + n − 2mn
      Correct
   3. C
      m + n − mn
   4. D
      mn
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   Answer: B

   Let the husband being alive in 15 years be event H, with P(H) = m, and the wife being alive be event W, with P(W) = n. Since the two events are independent, "only one of them alive" happens in exactly two mutually exclusive ways: the husband is alive and the wife is dead, OR the husband is dead and the wife is alive.

   P(only one alive) = P(H alive)·P(W dead) + P(H dead)·P(W alive)
   = m(1 − n) + (1 − m)n
   = m − mn + n − mn
   = m + n − 2mn

   This is the correct probability.

   The option mn is a trap — that is the probability that both husband and wife are alive, not that only one is.

4. 4.Find the domain of $f(x) = \frac{x}{3 - x}, x \in R$, the set of real numbers.
   1. A
      {x : x ∈ R, x ≠ -3}
   2. B
      {x : x ∈ R, x ≠ 0}
   3. C
      {x : x ∈ R, x > 3}
   4. D
      {x : x ∈ R, x ≠ 3}
      Correct
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   Answer: D

   The function is f(x) = x / (3 − x). A function is undefined wherever its denominator equals zero, so we find the value of x that makes the denominator vanish.

   Set the denominator to zero: 3 − x = 0, which gives x = 3. At x = 3, the function would require division by zero, which is not defined. Every other real number gives a valid output, so the domain is all real numbers except 3, written as {x : x ∈ R, x ≠ 3}.

5. 5.The roots of a quadratic equation are $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$. Find its equation.
   1. A
      x² + 6x + 6 = 0
   2. B
      x² − 6x + 6 = 0
      Correct
   3. C
      x² − 6x − 6 = 0
   4. D
      x² − 6x + 12 = 0
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   Answer: B

   For a quadratic equation with roots α and β, the equation is formed as:

   x² − (α + β)x + αβ = 0

   Here α = (3 − √3) and β = (3 + √3).

   Sum of roots:
   α + β = (3 − √3) + (3 + √3) = 6

   Product of roots:
   αβ = (3 − √3)(3 + √3) = 3² − (√3)² = 9 − 3 = 6

   Substituting into the formula:

   x² − 6x + 6 = 0

   This is the required equation, and it matches the option x² − 6x + 6 = 0.

6. 6.Simplify $(1 + 2\sqrt{3})^{2} - (1 - 2\sqrt{3})^{2}$
   1. A
      4√3
   2. B
      0
   3. C
      8√3
      Correct
   4. D
      26
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   Answer: C

   Expand each bracket using (x + y)² = x² + 2xy + y² and (x - y)² = x² - 2xy + y².

   (1 + 2√3)² = 1 + 4√3 + 12 = 13 + 4√3
   (1 - 2√3)² = 1 - 4√3 + 12 = 13 - 4√3

   Subtracting: (13 + 4√3) - (13 - 4√3) = 13 + 4√3 - 13 + 4√3 = 8√3, which is the correct simplification.

   A faster route confirms this: using the difference-of-two-squares pattern, a² - b² = (a - b)(a + b), with a = 1 + 2√3 and b = 1 - 2√3, we get (a - b) = 4√3 and (a + b) = 2, so the product is 4√3 × 2 = 8√3, the same answer.

7. 7.Two forces (2i - 5j)N and (-3i + 4j)N act on a body of mass 5kg. Find in $ms^{-2}$, the magnitude of the acceleration of the body.
   1. A
      0.40 ms⁻²
   2. B
      0.28 ms⁻²
      Correct
   3. C
      1.41 ms⁻²
   4. D
      2.06 ms⁻²
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   Answer: B

   The body has two forces acting on it, so first find the resultant force by adding the vectors component by component.

   R = (2i - 5j) + (-3i + 4j) = (2 - 3)i + (-5 + 4)j = -i - j (in newtons).

   The magnitude of this resultant is |R| = √((-1)² + (-1)²) = √2 ≈ 1.414 N.

   By Newton's second law, F = ma, so a = F/m = √2 / 5 ≈ 0.28 ms⁻². This is the magnitude of the acceleration of the body.

   Stopping as soon as you find |R| ≈ 1.41 N and forgetting that this is a force, not an acceleration, gives the tempting but wrong 1.41 ms⁻² — you must still divide by the mass.

8. 8.A stone is thrown vertically upwards and its height at any time t seconds is $h = 45t - 9t^{2}$. Find the maximum height reached.
   1. A
      112.5 m
   2. B
      56.25 m
      Correct
   3. C
      45 m
   4. D
      90 m
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   Answer: B

   The height is given by h = 45t − 9t², a quadratic in t whose graph is a downward-opening parabola, so it has a maximum point. To find the time at which that maximum occurs, differentiate and set the rate of change to zero (at the top of its path, the stone is momentarily at rest, i.e. dh/dt = 0).

   dh/dt = 45 − 18t = 0 ⟹ t = 45/18 = 2.5 seconds.

   Now substitute t = 2.5 back into the height equation:
   h = 45(2.5) − 9(2.5)² = 112.5 − 9(6.25) = 112.5 − 56.25 = 56.25 m.

9. 9.If $\begin{vmatrix} 3 & x \\ 2 & x - 2 \end{vmatrix} = -2$, find the value of x.
   1. A
      -2
   2. B
      -8
   3. C
      8
   4. D
      4
      Correct
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   Answer: D

   The determinant of a 2 × 2 matrix [[a, b], [c, d]] is ad − bc.

   Here a = 3, b = x, c = 2, d = x − 2, so:

   3(x − 2) − (x)(2) = −2

   3x − 6 − 2x = −2

   x − 6 = −2

   Adding 6 to both sides gives x = 4, which is correct.

   A candidate who moves the −6 to the other side with the wrong sign gets x − 6 = 2, leading to the wrong value of 8.
   A candidate who misreads the bottom-right entry as plain "x" instead of "x − 2" would compute 3x − 2x = −2, wrongly getting x = −2.
   A candidate who correctly reaches x − 6 = −2 but then subtracts 6 instead of adding it wrongly gets x = −8.

10. 10.Simplify $\frac{\log_{5} 8}{\log_{5} \sqrt{8}}$.
    1. A
       1/2
    2. B
       4
    3. C
       8
    4. D
       2
       Correct
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    Answer: D

    Simplify: log₅ 8 ÷ log₅ √8

    Write √8 as a power: √8 = 8^1/2. Using the power rule of logarithms, log₅(8^1/2) = ½ log₅ 8.

    So the expression becomes:
    log₅ 8 ÷ (½ log₅ 8)

    The log₅ 8 in the top and bottom cancel out, leaving 1 ÷ ½ = 2.

    A candidate who flips the fraction the wrong way round ends up with ½ instead of 2 — that is working out log₅ √8 ÷ log₅ 8 by mistake. Someone who treats the square root as if it doubles the power instead of halving it lands on 4. And simply writing down the numerator's base value, 8, ignores the denominator altogether. The correct approach is to apply the power rule to the denominator first, then cancel, giving 2.

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

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