WAEC Further Mathematics 2006 past questions

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

Ten questions from the 2006 WAEC Further Mathematics paper, with answers and explanations. We hold 47 questions from this paper, and the Skolra app has all of them.

WAEC Further Mathematics 2006 questions and answers

1. 1.An object is projected vertically upwards with a velocity of 80 m/s. Find the : (a) Maximum height reached (b) Time taken to return to the point of projection. [Take g = $10 ms^{-2}$].
   1. A
      Maximum height = 640 m; Time taken to return = 16 s
   2. B
      Maximum height = 160 m; Time taken to return = 8 s
   3. C
      Maximum height = 320 m; Time taken to return = 8 s
   4. D
      Maximum height = 320 m; Time taken to return = 16 s
      Correct
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   Answer: D

   This is a vertical projectile (motion under gravity) problem. Initial velocity u = 80 m/s (upward), g = 10 m/s², and we take upward as positive, so the deceleration due to gravity is -g.

   (b) Maximum height

   At the maximum height, the velocity becomes zero (v = 0). Using v² = u² - 2gh:

   0 = 80² - 2(10)h
   2000h/2000... let's do it cleanly: 0 = 6400 - 20h
   20h = 6400
   h = 320 m

   (a) Time taken to return to the point of projection

   Using v = u - gt, time to reach the maximum height (where v = 0) is:
   0 = 80 - 10t
   t = 8 s

   By symmetry of projectile motion, the time to fall back down equals the time taken to go up. So the total time to return to the point of projection is 2 × 8 s = 16 s.

2. 2.If $(x - 3)$ is a factor of $2x^{3} + 3x^{2} - 17x - 30$, find the remaining factors.
   1. A
      (x + 5)(2x + 2)
   2. B
      (x − 2)(2x + 5)
   3. C
      (x + 2)(2x + 5)
      Correct
   4. D
      (x + 2)(2x − 5)
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   Answer: C

   Step 1: Divide out the known factor. Since (x − 3) is a factor of 2x³ + 3x² − 17x − 30, dividing the cubic by (x − 3) using synthetic division (root 3, coefficients 2, 3, −17, −30) gives a remainder of 0 and a quotient of 2x² + 9x + 10.

   Step 2: Factorise the quotient. For 2x² + 9x + 10, we need two numbers that multiply to 2 × 10 = 20 and add to 9 — these are 4 and 5. So:
   2x² + 4x + 5x + 10 = 2x(x + 2) + 5(x + 2) = (x + 2)(2x + 5).

   Step 3: Confirm. (x + 2)(2x + 5) expands to 2x² + 5x + 4x + 10 = 2x² + 9x + 10, which matches the quotient. So the full factorisation is (x − 3)(x + 2)(2x + 5), and the remaining factors are (x + 2) and (2x + 5).

3. 3.ABCD is a square. Forces of magnitude 14N, 4N, 2N and $2\sqrt{2} N$ act along the sides AB, BC, CD and DA respectively. Find in Newtons, the magnitude of the resultant of the forces.
   1. A
      13.81 N
   2. B
      16.04 N
   3. C
      22.83 N
   4. D
      12.06 N
      Correct
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   Answer: D

   Setting up the square: Let AB and BC be two adjacent (perpendicular) sides of the square. Going round the square A → B → C → D → A, the side CD is parallel to AB but points the opposite way, and DA is parallel to BC but also points the opposite way.

   So we can resolve all four forces onto just two perpendicular directions — along AB and along BC.

   Along the AB direction: the 14N force (along AB) and the 2N force (along CD, which opposes AB) combine to give
   14N − 2N = 12N.

   Along the BC direction: the 4N force (along BC) and the $2\sqrt{2}$N force (along DA, which opposes BC) combine to give
   4N − 2√2N ≈ 4 − 2.83 = 1.17N.

4. 4.Evaluate $\log_{0.25} 8$
   1. A
      -2/3
   2. B
      -3/2
      Correct
   3. C
      3/2
   4. D
      2/3
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   Answer: B

   Evaluate log_0.25 8.

   Write both numbers as powers of the same base, 2:

   0.25 = 1/4 = 2^-2
   8 = 2³

   So log_0.25 8 = log_2^-2 2³.

   Using the rule log_aⁿ(a^m) = m/n, this becomes 3 ÷ (-2) = -3/2.

   A candidate who forgets that the base 0.25 is a negative power of 2 (treating it as 2² instead of 2^-2) gets 3/2, losing the sign. Flipping the fraction the wrong way round (dividing the base's exponent by the number's exponent instead of the other way) gives -2/3 or 2/3 — both mistakes come from mixing up which exponent goes on top.

   Check: 0.25^-3/2 = (1/4)^-3/2 = 4^3/2 = (4^1/2)³ = 2³ = 8. This confirms -3/2 is correct.

5. 5.[TABLE] Age(in years) | 1 - 5 | 6 - 10 | 11 - 15 Frequency | 3 | 5 | 2 [/TABLE] Calculate the standard deviation of the distribution.
   1. A
      12.25 years
   2. B
      7.5 years
   3. C
      3.7 years
   4. D
      3.5 years
      Correct
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   Answer: D

   Step 1: Find the midpoints (x) of each class.
   1–5 → 3, 6–10 → 8, 11–15 → 13.

   Step 2: Find N, Σfx and Σfx².
   N = 3 + 5 + 2 = 10
   Σfx = (3×3) + (5×8) + (2×13) = 9 + 40 + 26 = 75
   Σfx² = (3×9) + (5×64) + (2×169) = 27 + 320 + 338 = 685

   Step 3: Find the mean.
   Mean = Σfx/N = 75/10 = 7.5

   Step 4: Find the variance, then the standard deviation.
   Variance = (Σfx²/N) − mean² = 68.5 − 56.25 = 12.25
   Standard deviation = √12.25 = 3.5 years.

6. 6.Find the least value of the function $f(x) = 3x^{2} + 18x + 32$.
   1. A
      -5
   2. B
      5
      Correct
   3. C
      41
   4. D
      -3
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   Answer: B

   f(x) = 3x² + 18x + 32 is a quadratic (upward-opening, since the coefficient of x² is positive), so it has a least (minimum) value at its turning point.

   Find the x-coordinate of the turning point using x = -b/(2a):
   x = -18/(2 × 3) = -18/6 = -3.

   Substitute x = -3 back into f(x):
   f(-3) = 3(-3)² + 18(-3) + 32 = 3(9) - 54 + 32 = 27 - 54 + 32 = 5.

   Confirming by completing the square: f(x) = 3(x² + 6x) + 32 = 3[(x + 3)² - 9] + 32 = 3(x + 3)² + 5. Since (x + 3)² can never be negative, the smallest f(x) can ever be is 5.

   5 is correct.

7. 7.If $a = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ and $b = \begin{pmatrix} -3 \\ 5 \end{pmatrix}$, find a vector c such that $4a + 3c = b$.
   1. A
      $\begin{pmatrix} 5 \\ 1 \end{pmatrix}$
   2. B
      $\begin{pmatrix} -5 \\ 1 \end{pmatrix}$
   3. C
      $\begin{pmatrix} -2 \\ 1 \end{pmatrix}$
   4. D
      $\begin{pmatrix} -5 \\ -1 \end{pmatrix}$
      Correct
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   Answer: D

   We are given a = (3, 2) and b = (-3, 5), and the equation 4a + 3c = b.

   Make c the subject first: 3c = b - 4a.

   Compute 4a: 4 × (3, 2) = (12, 8).
   Now subtract from b: b - 4a = (-3 - 12, 5 - 8) = (-15, -3).

   Divide by 3 to get c: c = (-15/3, -3/3) = (-5, -1). That is the correct vector.

   The vector (5, 1) comes from working out 4a - b instead of b - 4a, which flips the sign on both components.
   The vector (-2, 1) comes from forgetting to multiply a by 4 before subtracting, i.e. using b - a instead of b - 4a.
   The vector (-5, 1) comes from making a sign slip only on the second (y) component while the x-component is correctly handled.

8. 8.The area of a sector of a circle is 3$cm^{2}$. If the sector subtends an angle of 1.5 radians at the centre, calculate the radius of the circle.
   1. A
      4 cm
   2. B
      √2 cm
   3. C
      2 cm
      Correct
   4. D
      1 cm
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   Answer: C

   Formula: the area of a sector in terms of the radius r and the angle θ (in radians) is A = ½r²θ.

   Here A = 3 cm² and θ = 1.5 radians, so:
   3 = ½ × r² × 1.5
   3 = 0.75r²
   r² = 3 ÷ 0.75 = 4
   r = √4 = 2 cm.

   The radius 2 cm is correct because it comes straight from the standard sector-area formula, squaring and square-rooting correctly.

   4 cm is what you get if you find r² = 2A/θ correctly (r² = 4) but then forget to take the square root, reading r² itself off as the radius.

   1 cm comes from putting the 2 in the wrong place, dividing by 2θ instead of multiplying: r² = A/(2θ) = 3/3 = 1, giving a radius that is too small.

9. 9.Find the coordinates of the point on the curve $y = x^{2} + 4x - 2$, where the gradient is zero.
   1. A
      (-2, -6)
      Correct
   2. B
      (-2, -4)
   3. C
      (2, 10)
   4. D
      (-4, -2)
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   Answer: A

   The gradient of the curve y = x² + 4x - 2 is given by differentiating:

   dy/dx = 2x + 4

   At the point where the gradient is zero, set dy/dx = 0:
   2x + 4 = 0
   2x = -4
   x = -2

   Now substitute x = -2 back into the original equation to find y:
   y = (-2)² + 4(-2) - 2
   y = 4 - 8 - 2
   y = -6

   So the point where the gradient is zero is (-2, -6).

10. 10.Solve the inequality $2x^{2} + 5x - 3 \geq 0$.
    1. A
       x ≤ -1/2 or x ≥ 3
    2. B
       x ≤ -3 or x ≥ 1/2
       Correct
    3. C
       -3 ≤ x ≤ 1/2
    4. D
       -1/2 ≤ x ≤ 3
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    Answer: B

    Step 1: Factorise. 2x² + 5x - 3 = (2x - 1)(x + 3). Check by expanding: 2x·x + 2x·3 - 1·x - 1·3 = 2x² + 6x - x - 3 = 2x² + 5x - 3. Correct.

    Step 2: Find the critical values. 2x - 1 = 0 gives x = 1/2. x + 3 = 0 gives x = -3.

    Step 3: Decide which region satisfies ≥ 0. The curve y = 2x² + 5x - 3 is an upward-opening parabola (positive coefficient of x²), so it is below zero only between its roots and above (or on) zero outside them. So the solution is x ≤ -3 or x ≥ 1/2.

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