WAEC Further Mathematics 2014 past questions

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

Ten questions from the 2014 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 2014 questions and answers

1.Integrate (x - 1/x)² with respect to x.
1. A
  x³/3 - 2x + 1/x + C
2. B
  x³/3 + 2x - 1/x + C
3. C
  x³/3 - 2x - 1/(2x) + C
4. D
  x³/3 - 2x - 1/x + C
  Correct
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Answer: D
Step 1: Expand the bracket first.
(x - 1/x)² = x² - 2(x)(1/x) + (1/x)² = x² - 2 + 1/x²
Step 2: Integrate term by term.
∫x² dx = x³/3
∫(-2) dx = -2x
∫x⁻² dx = x⁻¹/(-1) = -1/x
Step 3: Combine, with the constant of integration.
∫(x - 1/x)² dx = x³/3 - 2x - 1/x + C
2.Find the derivative of ∛(3x³+1) with respect to x.
1. A
  3x²/∛(3x³+1)
2. B
  9x²/∛((3x³+1)²)
3. C
  3x²/∛((3x³+1)²)
  Correct
4. D
  x²/∛((3x³+1)²)
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Answer: C
Let y = ∛(3x³+1) = (3x³+1)^(1/3). This is a chain-rule problem: differentiate the outer power function, then multiply by the derivative of the inner bracket.
Outer derivative: bring down the power and reduce it by 1,
d/dx[(3x³+1)^(1/3)] = (1/3)(3x³+1)^(-2/3) × d/dx(3x³+1)
Inner derivative: d/dx(3x³+1) = 9x²
Multiplying: (1/3) × 9x² × (3x³+1)^(-2/3) = 3x²(3x³+1)^(-2/3) = 3x²/∛((3x³+1)²). This is the correct derivative.
The option with 9x² on top skips the (1/3) that comes from the power rule on the outer function, so it is 3 times too large.
3.Find the coefficient of x³ in the binomial expansion of (3x+4)⁴ in ascending powers of x.
1. A
  108
2. B
  432
  Correct
3. C
  864
4. D
  768
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Answer: B
Expand (3x + 4)⁴ using the binomial theorem, writing the terms in ascending powers of x:
General term: C(4,k) × (3x)^k × 4^(4−k)
Since we want the term in x³, set k = 3:
C(4,3) = 4
(3x)³ = 27x³
4^(4−3) = 4¹ = 4
So the term is 4 × 27x³ × 4 = 432x³, giving a coefficient of 432.
As a check, the full expansion is 256 + 768x + 864x² + 432x³ + 81x⁴, and adding all the coefficients gives 2401, which equals 7⁴ — matching (3+4)⁴. This confirms the expansion is correct.
4.Given that 3x+4y+6=0 and 4x-by+3=0 are perpendicular, find the value of b.
1. A
  -16/3
2. B
  3
  Correct
3. C
  -3
4. D
  16/3
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Answer: B
For a line written as Ax + By + C = 0, the gradient is -A/B.
Line 1: 3x + 4y + 6 = 0 → gradient m₁ = -3/4.
Line 2: 4x - by + 3 = 0 → gradient m₂ = -4/(-b) = 4/b.
Two lines are perpendicular when the product of their gradients is -1:
m₁ × m₂ = -1
(-3/4) × (4/b) = -1
-3/b = -1
b = 3
Check: with b = 3, line 2 becomes 4x - 3y + 3 = 0, with gradient 4/3. Multiplying, (-3/4) × (4/3) = -1, which confirms the lines are indeed perpendicular. So 3 is correct.
The value -3 is what you get if you slip on the gradient formula and use +A/B instead of -A/B for one of the lines, turning the equation into (3/4) × (4/b) = -1.
5.Express (14N, 240°) as a column vector.
1. A
  (-7, -12.12)
  Correct
2. B
  (-12.12, -7)
3. C
  (-7, 12.12)
4. D
  (7, -12.12)
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Answer: A
A vector given as (magnitude, direction) is converted to column vector form using x = r cosθ and y = r sinθ, where r = 14 N and θ = 240°.
240° lies in the third quadrant (between 180° and 270°), where both cosine and sine are negative. Using the reference angle of 60° (since 240° = 180° + 60°):
cos240° = -cos60° = -0.5
sin240° = -sin60° = -0.8660 (i.e. -√3/2)
So:
x = 14 × (-0.5) = -7
y = 14 × (-0.8660) = -7√3 ≈ -12.12
The column vector is therefore (-7, -12.12).
6.Solve (log2 m)² - log2(m³) = 10.
1. A
  m = 5 or m = -2
2. B
  m = 1024 or m = 1/16
3. C
  m = 32 or m = 1/4
  Correct
4. D
  m = 4 or m = 1/32
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Answer: C
Let x = log₂ m. Since log₂(m³) = 3 log₂ m = 3x, the equation becomes:
x² - 3x = 10 ⇒ x² - 3x - 10 = 0
This factorises as (x - 5)(x + 2) = 0, giving x = 5 or x = -2.
Since x = log₂ m, we convert back: when x = 5, m = 2⁵ = 32; when x = -2, m = 2⁻² = 1/4.
So m = 32 or m = 1/4 is correct.
The option "m = 5 or m = -2" stops at the values of x and forgets the final step of converting from log back to m — a very common slip.
The option "m = 4 or m = 1/32" comes from mixing up the signs of the factors, as if (x + 5)(x - 2) = 0 had been used instead of the correct (x - 5)(x + 2) = 0.
7.For what values of m is 9y² + my + 4 a perfect square?
1. A
  m = ±36
2. B
  m = ±4
3. C
  m = ±12
  Correct
4. D
  m = ±6
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Answer: C
9y² + my + 4 is a perfect square when it matches the pattern (3y + k)², because 9y² = (3y)² and 4 = 2², so k must be ±2.
Expanding: (3y + 2)² = 9y² + 12y + 4, and (3y − 2)² = 9y² − 12y + 4. Matching the middle term to "my" gives m = ±12.
You can confirm this with the perfect-square (equal-roots) condition for a quadratic, b² = 4ac: here a = 9, b = m, c = 4, so m² = 4(9)(4) = 144, giving m = ±12.
8.Find the angle between (5i+3j) and (3i-5j).
1. A
  180°
2. B
  45°
3. C
  90°
  Correct
4. D
  0°
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Answer: C
To find the angle between two vectors a = 5i + 3j and b = 3i - 5j, we use the dot product formula:
a · b = |a||b| cos θ
First find the dot product:
a · b = (5 × 3) + (3 × -5) = 15 - 15 = 0
Since the dot product is zero, cos θ = 0, which means θ = 90°. The two vectors are perpendicular to each other. You can sense-check this quickly: whenever a vector (p, q) is paired with (q, -p) or (-q, p), the dot product always collapses to zero, so they are at right angles.
9.[TABLE] Age | 10-14 | 15-19 | 20-24 | 25-29 | 30-34 Frequency | 6 | 8 | 14 | 10 | 12 [/TABLE] What is the class mark of the median class?
1. A
  22
  Correct
2. B
  20
3. C
  17
4. D
  27
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Answer: A
Step 1: Find the total frequency.
6 + 8 + 14 + 10 + 12 = 50, so N = 50.
Step 2: Locate the median position.
N/2 = 25.
Step 3: Build the cumulative frequency and find the median class.
10-14: 6
15-19: 6+8 = 14
20-24: 14+14 = 28
25-29: 28+10 = 38
30-34: 38+12 = 50
The median class is the first one whose cumulative frequency reaches or passes 25. That happens at 20-24 (cumulative frequency 28), since the class before it (15-19) only reaches 14.
Step 4: Find the class mark of that class.
Class mark = (lower limit + upper limit) / 2 = (20 + 24) / 2 = 22.
10.A box contains 5 red and k blue balls. A ball is selected at random from the box. If the probability of selecting a blue ball is 2/3, find the value of k.
1. A
  3
2. B
  10
  Correct
3. C
  5
4. D
  15
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Answer: B
The box has 5 red balls and k blue balls, so the total number of balls is 5 + k.
Since the probability of picking a blue ball is 2/3, we write:
k / (5 + k) = 2/3
Cross-multiplying: 3k = 2(5 + k) = 10 + 2k.
So 3k - 2k = 10, which gives k = 10.
Check it: total balls = 5 + 10 = 15, and 10/15 = 2/3, which matches the given probability, so k = 10 is correct.
Where the other answers go wrong:
Getting 5 happens when a student multiplies out 2(5 + k) incorrectly as 2k + 5 instead of 2k + 10, which wrongly gives 3k = 2k + 5 and k = 5.

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WAEC Further Mathematics 2014 past questions

WAEC Further Mathematics 2014 questions and answers

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