WAEC Further Mathematics 2011 past questions

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

Ten questions from the 2011 WAEC Further Mathematics paper, with answers and explanations. We hold 41 questions from this paper, and the Skolra app has all of them.

WAEC Further Mathematics 2011 questions and answers

1. 1.$f(x) = p + qx$, where p and q are constants. If f(1) = 7 and f(5) = 19, find f(3).
   1. A
      13
      Correct
   2. B
      10
   3. C
      16
   4. D
      9
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   Answer: A

   Since f(x) = p + qx, substitute the two given values to form simultaneous equations:

   f(1) = p + q = 7
   f(5) = p + 5q = 19

   Subtracting the first equation from the second removes p: (p + 5q) - (p + q) = 19 - 7, which gives 4q = 12, so q = 3. Putting this back into p + q = 7 gives p = 4.

   Now find f(3) = p + 3q = 4 + (3 x 3) = 4 + 9 = 13.

2. 2.If $y^{2} + xy - x = 0$, find $\frac{\mathrm d y}{\mathrm d x}$.
   1. A
      dy/dx = (y − 1) / (x + 2y)
   2. B
      dy/dx = (1 − y) / 2y
   3. C
      dy/dx = (1 − y) / (2y − x)
   4. D
      dy/dx = (1 − y) / (x + 2y)
      Correct
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   Answer: D

   Differentiate each term of y² + xy − x = 0 with respect to x, treating y as a function of x.

   d/dx(y²) = 2y(dy/dx) — chain rule.
   d/dx(xy) = x(dy/dx) + y — product rule, since xy is a product of two functions of x.
   d/dx(x) = 1.

   Putting it together: 2y(dy/dx) + x(dy/dx) + y − 1 = 0.

   Collect the dy/dx terms: (dy/dx)(x + 2y) = 1 − y, so dy/dx = (1 − y)/(x + 2y) is correct.

   The option (1 − y)/2y comes from forgetting that xy needs the product rule — a student who treats d/dx(xy) as just y (dropping the x·dy/dx piece) is left with only 2y in the denominator.

   The option (y − 1)/(x + 2y) is a sign slip: moving the −1 across the equation the wrong way flips the numerator's sign.

3. 3.The equation of a circle is $3x^{2} + 3y^{2} + 6x - 12y + 6 = 0$. Find its radius
   1. A
      √7
   2. B
      √3
      Correct
   3. C
      √5
   4. D
      3√2
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   Answer: B

   Before you can read off the radius of a circle, the equation must first be arranged so that the coefficients of x² and y² are both 1.

   Starting from 3x² + 3y² + 6x - 12y + 6 = 0, divide every term by 3:

   x² + y² + 2x - 4y + 2 = 0

   Now compare this with the general equation of a circle, x² + y² + 2gx + 2fy + c = 0:

   2g = 2, so g = 1
   2f = -4, so f = -2
   c = 2

   The radius is given by r = √(g² + f² - c):

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

   So the radius of the circle is √3 units.

4. 4.Determine the coefficient of $x^{2}$ in the expansion of $(a + 3x)^{6}$.
   1. A
      45a⁴
   2. B
      135a⁴
      Correct
   3. C
      135a³
   4. D
      270a⁴
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   Answer: B

   Using the binomial expansion, the general term of (a + 3x)⁶ is:

   T(k+1) = ⁶C(k) · a^(6−k) · (3x)^k

   The term containing x² is the one where k = 2:

   T₃ = ⁶C₂ · a⁴ · (3x)²

   ⁶C₂ = 15, and (3x)² = 9x², so:

   T₃ = 15 × a⁴ × 9x² = 135a⁴x²

   So the coefficient of x² is 135a⁴.

   A candidate who forgets to square the 3 inside (3x)² — using 3 instead of 9 — wrongly gets 45a⁴. One who doubles the binomial coefficient by mistake lands on 270a⁴. And one who miscounts the power of a (using a³ instead of a⁴, as if k = 3 rather than k = 2) gets 135a³. Always square the whole bracket term (3x), not just the x.

5. 5.Solve $9^{2x + 1} = 81^{3x + 2}$
   1. A
      x = -5/4
   2. B
      x = 3/4
   3. C
      x = 0
   4. D
      x = -3/4
      Correct
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   Answer: D

   Step 1: Express both sides with the same base. Since 9 = 3² and 81 = 3⁴, rewrite the equation as:
   3^2(2x+1) = 3^4(3x+2)

   Step 2: Equate the exponents (since the bases are now equal):
   2(2x + 1) = 4(3x + 2)
   4x + 2 = 12x + 8

   Step 3: Collect like terms.
   2 − 8 = 12x − 4x
   −6 = 8x
   x = −6/8 = −3/4

   This is the correct value, and it is the one that satisfies the original equation.

   The option x = 0 comes from a slip while expanding 4(3x + 2): a candidate who multiplies only the x‑term (getting 12x) but mishandles the constant ends up with 4x + 2 = 12x + 2, which wrongly cancels to x = 0.

6. 6.A line is perpendicular to $3x - y + 11 = 0$ and passes through the point (1, -5). Find its equation.
   1. A
      3x − y − 8 = 0
   2. B
      x − 3y − 16 = 0
   3. C
      x + 3y + 14 = 0
      Correct
   4. D
      x + 3y − 14 = 0
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   Answer: C

   Step 1: Find the gradient of the given line.
   Rewrite $3x - y + 11 = 0$ as $y = 3x + 11$. Its gradient is $3$.

   Step 2: Find the gradient of the perpendicular line.
   For perpendicular lines, $m_1 \times m_2 = -1$. So the required gradient is $-\dfrac{1}{3}$.

   Step 3: Use the point (1, -5) with gradient $-\dfrac{1}{3}$.
   $y - (-5) = -\dfrac{1}{3}(x - 1)$
   $y + 5 = -\dfrac{1}{3}x + \dfrac{1}{3}$
   Multiply through by 3: $3y + 15 = -x + 1$
   $x + 3y + 14 = 0$

   So the correct equation is x + 3y + 14 = 0. Checking: at $x = 1$, $1 + 3y + 14 = 0 \Rightarrow y = -5$, which matches the given point.

7. 7.Find the remainder when $5x^{3} + 2x^{2} - 7x - 5$ is divided by (x - 2).
   1. A
      37
   2. B
      -23
   3. C
      29
      Correct
   4. D
      39
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   Answer: C

   Using the Remainder Theorem: when a polynomial f(x) is divided by (x - a), the remainder equals f(d). Here the divisor is (x - 2), so a = 2.

   f(2) = 5(2)³ + 2(2)² - 7(2) - 5
   = 5(8) + 2(4) - 14 - 5
   = 40 + 8 - 14 - 5
   = 29

   So 29 is the correct remainder.

   A candidate who mistakenly treats the divisor as (x + 2) and substitutes x = -2 instead of x = 2 gets -40 + 8 + 14 - 5 = -23, which is wrong because it answers a different division problem.

   Another common slip is mishandling the sign of the constant term, -5, flipping it to +5 partway through the working, which gives 40 + 8 - 14 + 5 = 39.

8. 8.A function is defined by $f(x) = \frac{3x + 1}{x^{2} - 1}, x \neq \pm 1$. Find f(-3).
   1. A
      -1
      Correct
   2. B
      4/5
   3. C
      5/4
   4. D
      -4/5
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   Answer: A

   f(x) = (3x + 1)/(x² − 1), so substitute x = −3 directly.

   Numerator: 3(−3) + 1 = −9 + 1 = −8.

   Denominator: (−3)² − 1 = 9 − 1 = 8.

   So f(−3) = −8/8 = −1.

   A candidate who mishandles the denominator sign, taking (−3)² as −9 instead of +9, gets −8/−10, which wrongly simplifies to 4/5. Flipping only the sign of the denominator (treating it as −8) gives −4/5 instead. Inverting the correct fraction by mistake, dividing the denominator by the numerator rather than the other way round, gives 5/4. Squaring correctly and subtracting correctly, as shown above, gives the true value of −1.

9. 9.Find $\lim \limits_{x \to 3} \frac{x + 3}{x^{2} - x - 12}$
   1. A
      1/3
   2. B
      -2/3
   3. C
      1
   4. D
      -1
      Correct
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   Answer: D

   Since the denominator does not vanish at x = 3, this limit is found by direct substitution — there is no need to factor or cancel anything.

   Numerator: 3 + 3 = 6.

   Denominator: 3² − 3 − 12 = 9 − 3 − 12 = −6.

   So the limit is 6 ÷ (−6) = −1.

   A student who drops the negative sign when combining 9 − 3 − 12 ends up with a denominator of 18 instead of −6, giving the wrong value of 1/3.

   Another common slip is forgetting the minus sign in the subtraction altogether and treating the denominator as +6, which wrongly gives 1 instead of −1.

   A third mistake comes from miscomputing 3² as 6 rather than 9, which leads to a denominator of −9 and a wrong result of −2/3.

10. 10.The fourth term of a geometric sequence is 2 and the sixth term is 8. Find the common ratio.
    1. A
       4
    2. B
       ±4
    3. C
       ±½
    4. D
       ±2
       Correct
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    Answer: D

    In a geometric sequence, the nth term is T_n = ar^n-1, where a is the first term and r is the common ratio.

    The fourth term: T₄ = ar³ = 2
    The sixth term: T₆ = ar⁵ = 8

    Divide the sixth term by the fourth term so that "a" cancels out:

    T₆ ÷ T₄ = ar⁵ ÷ ar³ = r²

    8 ÷ 2 = 4, so r² = 4.

    Taking the square root of both sides gives r = ±2. A square root always has two possible values (one positive, one negative), and since r is only ever raised to odd and even powers here, both +2 and −2 genuinely satisfy the sequence, so neither can be dropped.

    So the common ratio is ±2.

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