WAEC Further Mathematics 2017 past questions

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

Ten questions from the 2017 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 2017 questions and answers

1. 1.Given that $a = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $b = \begin{pmatrix} -1 \\ 4 \end{pmatrix}$, evaluate $(2a - \frac{1}{4}b)$.
   1. A
      $\begin{pmatrix} -17/4 \\ -5 \end{pmatrix}$
   2. B
      $\begin{pmatrix} 9/4 \\ 2 \end{pmatrix}$
   3. C
      $\begin{pmatrix} 15/4 \\ 7 \end{pmatrix}$
   4. D
      $\begin{pmatrix} 17/4 \\ 5 \end{pmatrix}$
      Correct
   Show answerHide answer

   Answer: D

   We are given a = (2, 3) and b = (-1, 4), and we want 2a − ¼b.

   First scale each vector component by component:
   2a = 2 × (2, 3) = (4, 6)
   ¼b = ¼ × (-1, 4) = (-¼, 1)

   Now subtract, remembering that subtracting a negative gives addition:
   top: 4 − (-¼) = 4 + ¼ = 17/4
   bottom: 6 − 1 = 5

   So 2a − ¼b = (17/4, 5), and that is the correct result.

2. 2.Differentiate $\frac{5x^{3} + x^{2}}{x}, x\neq 0$ with respect to x.
   1. A
      10x
   2. B
      20x + 3
   3. C
      15x² + 1
   4. D
      10x + 1
      Correct
   Show answerHide answer

   Answer: D

   Step 1: Simplify before differentiating. Since x ≠ 0, divide each term in the numerator by x:
   5x³ ÷ x = 5x², and x² ÷ x = x.
   So (5x³ + x²)/x simplifies to 5x² + x.

   Step 2: Differentiate term by term. The derivative of 5x² is 10x, and the derivative of x is 1. Adding these gives 10x + 1, which is the correct result.

   The option 15x² + 1 comes from a wrong simplification — treating 5x³/x as if the power did not reduce (keeping 5x³ instead of 5x²) before differentiating; that mistake is not valid here.

   The option 20x + 3 comes from trying the quotient rule, [u'v − uv']/v², but adding the two products instead of subtracting them, which flips the sign and inflates the result.

3. 3.The velocity, V, of a particle after t seconds, is $V = 3t^{2} + 2t - 1$. Find the acceleration of the particle after 2 seconds.
   1. A
      14 m/s²
      Correct
   2. B
      8 m/s²
   3. C
      15 m/s²
   4. D
      12 m/s²
   Show answerHide answer

   Answer: A

   Acceleration is the rate of change of velocity, so we differentiate V with respect to t.

   Given V = 3t² + 2t - 1,
   differentiating: a = dV/dt = 6t + 2.

   At t = 2 seconds:
   a = 6(2) + 2 = 12 + 2 = 14 m/s².

   14 m/s² is the correct acceleration.

   15 m/s² is a trap for a candidate who forgets to differentiate at all and instead substitutes t = 2 straight into the velocity equation (3(2)² + 2(2) - 1 = 15) — that gives the velocity at t = 2, not the acceleration.

   12 m/s² comes from differentiating correctly to get 6t, but dropping the constant term (+2) that comes from differentiating 2t.

4. 4.Simplify $\frac{1}{(1-\sqrt{3})^{2}}$
   1. A
      (2 + √3)/2
      Correct
   2. B
      −1/2
   3. C
      2 + √3
   4. D
      (2 − √3)/2
   Show answerHide answer

   Answer: A

   Step 1: Expand the denominator.
   (1 − √3)² = 1² − 2(1)(√3) + (√3)² = 1 − 2√3 + 3 = 4 − 2√3.

   So the expression becomes 1/(4 − 2√3), which factorises as 1/[2(2 − √3)].

   Step 2: Rationalize.
   Multiply top and bottom by the conjugate (2 + √3):
   [1 × (2 + √3)] / [2(2 − √3)(2 + √3)] = (2 + √3) / [2(4 − 3)] = (2 + √3)/2.

   So (2 + √3)/2 is the simplified form.

   The option (2 − √3)/2 is what you get if you mistakenly keep the same sign as the original bracket instead of switching to the conjugate when rationalizing — a very common slip.

   −1/2 comes from wrongly expanding (1 − √3)² as 1² − (√3)² = 1 − 3 = −2 (forgetting the middle term −2√3 entirely), then taking 1/(−2).

5. 5.Evaluate : $\int_{1}^{3} (\frac{x - 1}{(x + 1)^{2}}) \mathrm {d} x$.
   1. A
      ln 2
   2. B
      ln 2 − ½
      Correct
   3. C
      2 ln 2 − ½
   4. D
      ln 2 + ½
   Show answerHide answer

   Answer: B

   Substitute u = x + 1, so x − 1 = u − 2 and dx = du. When x = 1, u = 2; when x = 3, u = 4.

   The integral becomes:
   ∫ (u − 2)/u² du = ∫ (1/u − 2/u²) du = ln u + 2/u

   Evaluating from u = 2 to u = 4:
   At u = 4: ln 4 + 2/4 = ln 4 + ½
   At u = 2: ln 2 + 2/2 = ln 2 + 1

   Subtracting: (ln 4 + ½) − (ln 2 + 1) = ln 4 − ln 2 − ½ = ln 2 − ½ (since ln 4 − ln 2 = ln 2). This is approximately 0.193, and it is the correct value of the integral.

6. 6.Find the magnitude and direction of the vector $p = (5i - 12j)$
   1. A
      17 units, direction 292.6° from the positive x-axis
   2. B
      13 units, direction 292.6° from the positive x-axis
      Correct
   3. C
      13 units, direction 67.4° from the positive x-axis
   4. D
      13 units, direction 112.6° from the positive x-axis
   Show answerHide answer

   Answer: B

   Magnitude: For p = 5i − 12j, the components are x = 5 and y = −12.

   |p| = √(x² + y²) = √(5² + (−12)²) = √(25 + 144) = √169 = 13 units.

   Direction: First find the reference angle: tan θ = |y|/|x| = 12/5 = 2.4, so θ = tan⁻¹(2.4) ≈ 67.4°.

   Since x is positive and y is negative, the vector points into the fourth quadrant (right of the origin, below the horizontal). Measuring direction the standard way — anticlockwise from the positive x-axis, giving a value between 0° and 360° — a fourth-quadrant angle is found as 360° − 67.4° = 292.6°.

   So the vector has magnitude 13 units and direction 292.6°.

7. 7.A force (10i + 4j)N acts on a body of mass 2kg which is at rest. Find the velocity after 3 seconds.
   1. A
      (15i + 6j) ms⁻¹
      Correct
   2. B
      (12i + 30j) ms⁻¹
   3. C
      (30i + 12j) ms⁻¹
   4. D
      (5i + 2j) ms⁻¹
   Show answerHide answer

   Answer: A

   The body starts at rest, so the initial velocity u = 0.

   First find the acceleration using Newton's second law, a = F/m:

   a = (10i + 4j) / 2 = (5i + 2j) m/s²

   Then use v = u + at, with t = 3 seconds:

   v = 0 + (5i + 2j)(3) = (15i + 6j) ms⁻¹

   That is the velocity after 3 seconds.

   A candidate who forgets to divide the force by the mass before multiplying by time gets (30i + 12j) ms⁻¹ — that is F×t, not a×t.
   A candidate who finds the acceleration correctly but forgets to multiply by the time stops at (5i + 2j) ms⁻¹, which is only the acceleration, not the velocity.
   Swapping the i and j components by mistake gives (12i + 30j) ms⁻¹, a simple transposition error.

8. 8.(a) Given that $\log_{10} p = a$, $\log_{10} q = b$ and $\log_{10} s = c$, express $\log_{10}\left(\frac{p^{\frac{1}{3}}q^{4}}{s^{2}}\right)$ in terms of $a$, $b$ and $c$. (b) The radius of a circle is 6 cm. If the area is increasing at the rate of $20\,\text{cm}^{2}\text{s}^{-1}$, find, leaving the answer in terms of $\pi$, the rate at which the radius is increasing.
   1. A
      (a) a/3 + 4b + 2c  (b) 5/(3π) cm s⁻¹
   2. B
      (a) a/3 + 4b − 2c  (b) 10/(3π) cm s⁻¹
   3. C
      (a) 3a + 4b − 2c  (b) 5/(6π) cm s⁻¹
   4. D
      (a) a/3 + 4b − 2c  (b) 5/(3π) cm s⁻¹
      Correct
   Show answerHide answer

   Answer: D

   (c) Using the laws of logarithms, break the expression into separate terms:

   log₁₀(p^1/3q⁴/s²) = log₁₀(p^1/3) + log₁₀(q⁴) − log₁₀(s²)

   = (1/3)log₁₀p + 4log₁₀q − 2log₁₀s = a/3 + 4b − 2c.

   A candidate who multiplies instead of dividing the power on p (treating the cube root as a cube) lands on 3a + 4b − 2c, which is one of the tempting wrong answers. Another common slip is mishandling the division inside the logarithm and writing +2c instead of −2c, since dividing by s² subtracts its log rather than adding it.

   (a) The area of a circle is A = πr². Differentiating both sides with respect to time t:

   dA/dt = 2πr · dr/dt

   At the instant r = 6 cm, dA/dt = 20 cm²s⁻¹, so:

   20 = 2π(6)(dr/dt) = 12π(dr/dt)

9. 9.Find the coefficient of $x^{4}$ in the expansion of $(1-2x)^{6}$.
   1. A
      -160
   2. B
      -240
   3. C
      240
      Correct
   4. D
      320
   Show answerHide answer

   Answer: C

   We expand (1 − 2x)⁶ using the binomial theorem. The general term is:

   T(r+1) = C(6, r) · (1)^6−r · (−2x)^r = C(6, r) · (−2)^r · x^r

   We want the term in x⁴, so we set r = 4:

   C(6, 4) = 15, and (−2)⁴ = 16 (a negative number raised to an even power becomes positive).

   Coefficient of x⁴ = 15 × 16 = 240.

   Where the other options go wrong:

   −240 comes from wrongly carrying the minus sign through, as if (−2)⁴ were negative. Since the power 4 is even, the negative sign cancels out — only an odd power of x would give a negative coefficient here.

   So the correct coefficient, matching the actual binomial expansion, is 240.

10. 10.If $y = \frac{1+x}{1-x}$, find $\frac{dy}{dx}$.
    1. A
       -2/(1-x)²
    2. B
       2/(1-x)
    3. C
       2/(1-x)²
       Correct
    4. D
       -2x/(1-x)²
    Show answerHide answer

    Answer: C

    We are given y = (1+x)/(1-x) and need to differentiate it using the quotient rule.

    Let u = 1+x and v = 1-x, so u' = 1 and v' = -1.

    The quotient rule says: dy/dx = (v·u' - u·v')/v²

    dy/dx = [(1-x)(1) - (1+x)(-1)] / (1-x)²
    = [(1-x) + (1+x)] / (1-x)²
    = [1 - x + 1 + x] / (1-x)²
    = 2/(1-x)²

    This is the correct derivative.

    A student who forgets that the derivative of (1-x) is -1 (not +1) and fails to flip the sign on that term will subtract instead of add in the numerator, getting -2x/(1-x)² — this is a very common slip with the quotient rule when one of the functions has a negative coefficient on x.

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

Spotted a wrong answer? Tell us from the app or on the contact page and we check it.

Practise all 47 questions we hold from the 2017 paper, on a timed CBT screen

Every question has the answer and an explanation, in the Skolra app.

Download for WindowsDownload for Mac

About the app · Skolra for Windows · Skolra for Mac