WAEC Further Mathematics 2016 past questions

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

Ten questions from the 2016 WAEC Further Mathematics paper, with answers and explanations. We hold 42 questions from this paper, and the Skolra app has all of them.

WAEC Further Mathematics 2016 questions and answers

1. 1.When f(x) = 2x³+mx²+nx+11 is divided by x²+5x+1, the quotient is 2x-5 and the remainder is 30x+16. Find the values of m and n.
   1. A
      m = 5, n = 7
      Correct
   2. B
      m = 15, n = 7
   3. C
      m = -5, n = 7
   4. D
      m = 5, n = -23
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   Answer: A

   Key idea: Dividend = Divisor × Quotient + Remainder.

   So f(x) = (x²+5x+1)(2x-5) + (30x+16).

   Expanding the product term by term:
   x²×2x = 2x³, x²×(-5) = -5x²
   5x×2x = 10x², 5x×(-5) = -25x
   1×2x = 2x, 1×(-5) = -5

   Collecting like terms: the x² terms give -5x²+10x² = 5x², and the x terms give -25x+2x = -23x. So the product is 2x³ + 5x² - 23x - 5.

   Now add the remainder (30x+16): the x-terms become -23x+30x = 7x, and the constants become -5+16 = 11.

   This gives f(x) = 2x³ + 5x² + 7x + 11, which matches the given form 2x³+mx²+nx+11 — notice the constant term 11 matches exactly, confirming the working is correct.

   So m = 5 and n = 7 is correct.

2. 2.The remainder when x³-2x+m is divided by x-1 is equal to the remainder when 2x³+x-m is divided by 2x+1. Find the value of m.
   1. A
      7/8
   2. B
      1/8
      Correct
   3. C
      -1/8
   4. D
      5/8
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   Answer: B

   By the remainder theorem, when a polynomial is divided by (x - a), the remainder is the value of the polynomial at x = a. When it is divided by (bx - a), the remainder is the value at x = a/b.

   First remainder: f(x) = x³ - 2x + m divided by (x - 1). Here a = 1, so the remainder is
   f(1) = 1³ - 2(1) + m = 1 - 2 + m = m - 1.

   Second remainder: g(x) = 2x³ + x - m divided by (2x + 1). Setting 2x + 1 = 0 gives x = -1/2 (not +1/2 — this is the trap that leads to 7/8), so the remainder is
   g(-1/2) = 2(-1/2)³ + (-1/2) - m = 2(-1/8) - 1/2 - m = -1/4 - 1/2 - m = -3/4 - m.

   Equating the two remainders (as the question states they are equal):
   m - 1 = -3/4 - m
   2m = -3/4 + 1 = 1/4
   m = 1/8.

3. 3.If (2t-3s)(t-s) = 0, find t/s.
   1. A
      t/s = 2/3 or 1
   2. B
      t/s = 1 only
   3. C
      t/s = -1 or -3/2
   4. D
      t/s = 1 or 3/2
      Correct
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   Answer: D

   When a product of two factors equals zero, at least one of the factors must be zero. So we treat each bracket separately.

   First factor: 2t - 3s = 0 → 2t = 3s → t/s = 3/2.

   Second factor: t - s = 0 → t = s → t/s = 1.

   Since either factor can be the one that vanishes, both values satisfy the original equation, so t/s = 1 or 3/2 is correct.

   Picking only one of the two cases (t/s = 1 only) ignores that the other factor can also be zero. Writing the ratio upside down as s/t instead of t/s gives the tempting but wrong 2/3 or 1. Introducing a sign error while solving the first factor (as if it were 2t + 3s = 0) leads to the negative pair, -1 or -3/2, which does not match either factor correctly.

4. 4.Find the coefficient of x³ in the expansion of [(1/3)(2+x)]⁶.
   1. A
      160/27
   2. B
      160
   3. C
      80/729
   4. D
      160/729
      Correct
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   Answer: D

   Rewrite the expression as a single bracket raised to a power: [(1/3)(2+x)]⁶ = (1/3)⁶ × (2+x)⁶. The factor 1/3 applies to the whole bracket (2+x), not just to x, so it must be raised to the power 6 along with everything else.

   Expand (2+x)⁶ using the binomial theorem. The general term is C(6,k)·2^(6−k)·xᵏ. For the x³ term, set k = 3:
   C(6,3) = 20, and 2^(6−3) = 2³ = 8.
   So the x³ term inside (2+x)⁶ is 20 × 8 × x³ = 160x³.

   Now bring back the outer factor (1/3)⁶ = 1/729. Multiplying through gives the coefficient of x³ as 160 × 1/729 = 160/729, which is the correct result.

5. 5.If the solution set of x²+kx-5=0 is (-1,5), find the value of k.
   1. A
      6
   2. B
      -4
      Correct
   3. C
      4
   4. D
      -5
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   Answer: B

   The solution set (-1, 5) means the roots of the equation are x = -1 and x = 5.

   For a quadratic x² + kx - 5 = 0, comparing with x² - (sum of roots)x + (product of roots) = 0:

   Sum of roots = -k, so -1 + 5 = -k → 4 = -k → k = -4.

   Check with product of roots = -5 (the constant term, since the coefficient of x² is 1): (-1)(5) = -5. This matches the equation exactly, confirming k = -4 is correct.

   Substituting back: x² - 4x - 5 = (x - 5)(x + 1), which indeed gives roots 5 and -1.

6. 6.The radius of a circle increases at a rate of 0.5 cms⁻¹. Find the rate of change in the area of the circle with radius 7cm. [π = 22/7]
   1. A
      11 cm² s⁻¹
   2. B
      77 cm² s⁻¹
   3. C
      22 cm² s⁻¹
      Correct
   4. D
      44 cm² s⁻¹
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   Answer: C

   The area of a circle is A = πr². Since both the area and the radius are changing with time, we differentiate with respect to time t using the chain rule:

   dA/dt = 2πr × (dr/dt)

   We are given r = 7 cm, dr/dt = 0.5 cm s⁻¹, and π = 22/7. Substituting:

   dA/dt = 2 × (22/7) × 7 × 0.5
   dA/dt = 2 × 22 × 0.5
   dA/dt = 22 cm² s⁻¹

   So the rate of change of the area is 22 cm² s⁻¹.

7. 7.Consider the statements: p: Musa is short; q: Musa is brilliant. Which of the following represents the statement 'Musa is short but not brilliant'?
   1. A
      ~p ∧ q
   2. B
      p ∧ q
   3. C
      p ∨ ~q
   4. D
      p ∧ ~q
      Correct
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   Answer: D

   p: Musa is short. q: Musa is brilliant.

   The statement "Musa is short but not brilliant" joins two conditions with the word "but", which in logic behaves exactly like "and" (conjunction, ∧): Musa is short, AND Musa is not brilliant. "Not brilliant" is the negation of q, written ~q.

   So the statement translates as p ∧ ~q.

   The other options go wrong in specific ways:

   p ∧ q says Musa is short AND brilliant — it drops the negation on q entirely, missing the "not" in "not brilliant".

   ~p ∧ q swaps which part gets negated — it says Musa is NOT short but IS brilliant, the opposite pairing from what the statement says.

8. 8.Find the minimum value of y = 3x²-x-6.
   1. A
      -73/12
      Correct
   2. B
      -17/3
   3. C
      -37/6
   4. D
      -23/4
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   Answer: A

   Since the coefficient of x² is 3 (positive), the curve y = 3x² - x - 6 opens upward, so it has a minimum point.

   To locate the turning point, differentiate and set the gradient to zero:

   dy/dx = 6x - 1 = 0 → x = 1/6.

   Since d²y/dx² = 6 > 0, this confirms a minimum (not a maximum) at x = 1/6.

   Substitute x = 1/6 back into the equation:

   y = 3(1/6)² - (1/6) - 6
   = 3(1/36) - 1/6 - 6
   = 1/12 - 2/12 - 72/12
   = -73/12

   So the minimum value of y is -73/12 (equivalently -6 1/12).

9. 9.Given that r = 3i+4j and t = -5i+12j, find the acute angle between them.
   1. A
      120.5°
   2. B
      30.5°
   3. C
      14.3°
   4. D
      59.5°
      Correct
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   Answer: D

   To find the angle between two vectors r = 3i + 4j and t = -5i + 12j, we use the dot product formula:

   cosθ = (r·t) / (|r||t|)

   Step 1: Find the magnitudes.
   |r| = √(3² + 4²) = √25 = 5
   |t| = √((-5)² + 12²) = √169 = 13

   Step 2: Find the dot product.
   r·t = (3 × -5) + (4 × 12) = -15 + 48 = 33

   Step 3: Find cosθ and θ.
   cosθ = 33 / (5 × 13) = 33/65 = 0.5077
   θ = cos⁻¹(0.5077) ≈ 59.5°

   Since this value is already less than 90°, it is the acute angle required — no further adjustment is needed.

10. 10.Find an expression for y given that dy/dx = x²√x
    1. A
       y = (2/7)x³√x
    2. B
       y = (2/7)x³√x + C
       Correct
    3. C
       y = ¼x⁴ + C
    4. D
       y = (2/5)x³√x + C
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    Answer: B

    First combine the powers of x on the right-hand side: x²√x = x²·x^(1/2) = x^(5/2).

    To get y from dy/dx, integrate with respect to x: raise the power by 1 and divide by the new power.

    New power = 5/2 + 1 = 7/2, so:
    y = x^(7/2) ÷ (7/2) + C = (2/7)x^(7/2) + C.

    Since x^(7/2) = x³ × x^(1/2) = x³√x, the answer is written as y = (2/7)x³√x + C.

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