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WAEC Further Mathematics 2022 Objective Past Questions

All 40 questions from the West African Examinations Council (WAEC) Further Mathematics 2022 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

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Further Mathematics 2022 Objective — Question 1

A binary operation Δ is defined on the set of real numbers R, by x Δ y = √(x+y−xy/4), where x,y∈R. Find the value of 4Δ3.

  • A. 16
  • B. 8
  • C. 4
  • D. 2Correct

Explanation

4Δ3 = √(4+3−4(3)/4) = √(7−3) = √4 = 2.

Further Mathematics 2022 Objective — Question 2

Simplify [ (3√6+√54) / √5(3√5) ]⁻¹.

  • A. 5√3/6
  • B. 3√15/6
  • C. 5√6/12Correct
  • D. 5√3/12

Explanation

√54=3√6, so the numerator becomes 3√6+3√6=6√6, and the denominator √5(3√5)=15. Inverting (6√6/15)⁻¹ and rationalizing gives 5√6/12.

Further Mathematics 2022 Objective — Question 3

If log₁₀⁷⁷⁷⁽³ˣ⁻¹⁾ + log₁₀⁷⁷⁷⁴ = log₁₀⁷⁷⁷⁽⁹ˣ⁺²⁾ (i.e. log₁₀[10^(3x−1)] + log₁₀[4] = log₁₀[10^(9x+2)]), find x.

  • A. 1/3
  • B. 1
  • C. 2Correct
  • D. 3

Explanation

Combining the logs: log₁₀[4×10^(3x−1)] = log₁₀[10^(9x+2)], so 4(3x−1)=9x+2, giving 12x−4=9x+2, so 3x=6, x=2.

Further Mathematics 2022 Objective — Question 4

Simplify (9×3ⁿ⁺¹ − 3ⁿ⁺²) / (3ⁿ⁺¹ − 3ⁿ).

  • A. 3
  • B. 9Correct
  • C. 27
  • D. 81

Explanation

Factoring 3ⁿ out of numerator and denominator and simplifying the powers of 3 gives a result of 9.

Further Mathematics 2022 Objective — Question 5

Consider the statements: x: All wrestlers are strong. y: Some wrestlers are not weightlifters. Which of the following is a valid conclusion?

  • A. All strong wrestlers are weightlifters
  • B. Some strong wrestlers are not weightliftersCorrect
  • C. Some weak wrestlers are not weightlifters
  • D. All weak wrestlers are weightlifters

Explanation

Since some wrestlers (who are all strong) are not weightlifters, it is valid to conclude that some strong wrestlers are not weightlifters. This does not make an absolute claim about all wrestlers or all weightlifters, only acknowledging a possible overlap.

Further Mathematics 2022 Objective — Question 6

The functions f:x→5x²+7x−6 and g:x→2x²+3x−7 are defined on the set of real numbers, R. Find the values of x for which 3f(x)=g(x).

  • A. x=−3 or −5
  • B. x=3 or −5Correct
  • C. x=3 or −5
  • D. x=3 or 5

Explanation

3f(x)=g(x) gives 3(2x²+3x−7)... actually 3(5x²+7x−6)... solving 6x²+9x−21=5x²+7x−6 leads to x²+2x−15=0, i.e. (x−3)(x+5)=0, so x=3 or x=−5.

Further Mathematics 2022 Objective — Question 8

Given that (8x+m)/(x²−3x−4) = 5/(x+1) + 3/(x−4), find the value of m.

  • A. 23
  • B. 17
  • C. −17Correct
  • D. −23

Explanation

Combining the right side over a common denominator: [5(x−4)+3(x+1)]/[(x+1)(x−4)] = (8x−17)/[(x+1)(x−4)]. Comparing numerators, m = −17.

Further Mathematics 2022 Objective — Question 9

If x²+y²−2x−6y+5=0, evaluate dy/dx when x=3 and y=2.

  • A. −2
  • B. 2Correct
  • C. 4
  • D. −4

Explanation

Differentiating implicitly: 2x+2y(dy/dx)−2−6(dy/dx)=0, so dy/dx=(2−2x)/(2y−6). At x=3,y=2: dy/dx=(2−6)/(4−6)=(−4)/(−2)=2.

Further Mathematics 2022 Objective — Question 10

Evaluate ∫₀¹ x²(x³+2)³ dx.

  • A. 56/12
  • B. 65/12Correct
  • C. 12
  • D. 65

Explanation

Let u=x³+2, du=3x²dx. The integral becomes ∫u³/3 du = [u⁴/12], evaluated from x=0 to 1: [(1+2)⁴−(0+2)⁴]/12 = (81−16)/12 = 65/12.

Further Mathematics 2022 Objective — Question 11

If (2 −3; 1 4)(−6; k) = (3; −26), find the value of k.

  • A. −8
  • B. −5Correct
  • C. −4
  • D. −3

Explanation

From the first row: 2(−6)−3k=3, giving −12−3k=3, so k=−5 (this also satisfies the second row: −6+4(−5)=−26).

Further Mathematics 2022 Objective — Question 12

A linear transformation T is defined by T:(x,y)→(3x−y, x+4y). Find the image of (2,−1) under T.

  • A. (7,−2)Correct
  • B. (5,−2)
  • C. (−2,7)
  • D. (−7,2)

Explanation

T(2,−1) = (3(2)−(−1), 2+4(−1)) = (6+1, 2−4) = (7,−2).

Further Mathematics 2022 Objective — Question 14

Find the coefficient of x² in the binomial expansion of (x+2/x²)⁵.

  • A. 10Correct
  • B. 40
  • C. 32
  • D. 80

Explanation

The general term is ⁵C_r x^(5−3r)2^r. For the power of x to equal 2, 5−3r=2, so r=1. The coefficient is ⁵C₁×2¹ = 5×2 = 10.

Further Mathematics 2022 Objective — Question 15

Given that P={x: x is a multiple of 5}, Q={x: x is a multiple of 3} and R={x: x is an odd number} are subsets of μ={x: 20≤x<35}, find (P∪Q)∩R.

  • A. {20,21,25,30}
  • B. {21,25,27,33,35}Correct
  • C. {20,21,25,27,30,33,35}
  • D. {21,25,27}

Explanation

P={20,25,30}, Q={21,24,27,30,33}, R={21,23,25,27,29,31,33}. P∪Q={20,21,24,25,27,30,33}. Intersecting with R gives {21,25,27,33}.

Further Mathematics 2022 Objective — Question 17

If Uₙ=kn²+pn, U₁=−1, U₅=15, find the values of k and p.

  • A. k=−1, p=2
  • B. k=−1, p=−2
  • C. k=1, p=−2Correct
  • D. k=1, p=2

Explanation

U₁: k+p=−1. U₅: 25k+5p=15. Substituting p=−1−k into the second equation gives 20k=20, so k=1 and p=−2.

Further Mathematics 2022 Objective — Question 19

Solve 3^(2x−2) − 28(3^(x−2)) + 3 = 0.

  • A. x=−2 or x=1
  • B. x=0 or x=−3
  • C. x=2 or x=1
  • D. x=0 or x=3Correct

Explanation

Dividing through by 3⁻² and letting y=3ˣ gives y²−28y+27=0, i.e. (y−27)(y−1)=0, so 3ˣ=27 or 3ˣ=1, giving x=3 or x=0.

Further Mathematics 2022 Objective — Question 20

Given that P=(−4,−5) and Q=(2,3), express PQ in the form (k,θ), where k is the magnitude and θ is the bearing.

  • A. (10 units, 063°)
  • B. (9 units, 049°)
  • C. (10 units, 037°)Correct
  • D. (9 units, 027°)

Explanation

PQ=Q−P=(6,8), so |PQ|=√(36+64)=10 units. The angle from the y-axis (bearing) works out to 037° (found via 90°−tan⁻¹(8/6)).

Further Mathematics 2022 Objective — Question 22

The table shows the distribution of distance (in km) covered by 40 hunters while hunting: Distance(km) 3,4,5,6,7,8; Frequency 5,4,x,9,2x,1. If a hunter is selected at random, find the probability that the hunter covered at least 6km.

  • A. 3/5Correct
  • B. 2/5
  • C. 3/8
  • D. 9/40

Explanation

Since frequencies sum to 40: 5+4+x+9+2x+1=40, giving 3x=21, x=7. P(at least 6km) = (9+2(7)+1)/40 = 24/40 = 3/5.

Further Mathematics 2022 Objective — Question 23

Using the same distribution, find the mode of the distribution.

  • A. 5
  • B. 6
  • C. 7Correct
  • D. 8

Explanation

With x=7, the frequencies are 5,4,7,9,14,1 for distances 3–8. The highest frequency (14) corresponds to a distance of 7km, so the mode is 7.

Further Mathematics 2022 Objective — Question 24

If g(x)=√(1−x²), find the domain of g(x).

  • A. x<−1 or x>1
  • B. x≤−1 or x≥1
  • C. −1≤x≤1Correct
  • D. −1<x<1

Explanation

For g(x) to be defined, 1−x²≥0, i.e. (1−x)(1+x)≥0, which holds for −1≤x≤1.

Further Mathematics 2022 Objective — Question 25

Find the coefficient of x³y² in the binomial expansion of (x−2y)⁵.

  • A. −80
  • B. −10
  • C. 40Correct
  • D. 80

Explanation

The term in x³y² is ⁵C₂x³(−2y)² = 10x³(4y²) = 40x³y², so the coefficient is 40.

Further Mathematics 2022 Objective — Question 26

The first, second and third terms of an exponential sequence (G.P.) are (x−4), (x+2) and (3x+1) respectively. Find the values of x.

  • A. −1/2, 8Correct
  • B. 1/2, −1
  • C. −1/2, −8
  • D. 1/2, 8

Explanation

For a G.P., (x+2)/(x−4) = (3x+1)/(x+2). Cross-multiplying: (x+2)²=(x−4)(3x+1), which simplifies to 2x²−15x−8=0, i.e. (2x+1)(x−8)=0, so x=−1/2 or x=8.

Further Mathematics 2022 Objective — Question 27

A body of mass 18kg moving with velocity 4ms⁻¹ collides with another body of mass 6kg moving in the opposite direction with velocity 10ms⁻¹. If they stick together after collision, find their common velocity.

  • A. −1/2 ms⁻¹
  • B. 1/3 ms⁻¹
  • C. 2ms⁻¹
  • D. 3ms⁻¹

Explanation

By conservation of momentum: 18(4)+6(−10)=(18+6)v, so 72−60=24v, giving v=12/24=½ ms⁻¹. None of the listed options equal ½ ms⁻¹ exactly — the correct computed value is 0.5ms⁻¹.

Further Mathematics 2022 Objective — Question 28

The mean heights of three groups of students consisting of 20, 16 and 14 students each are 1.67m, 1.5m and 1.40m respectively. Find the mean height of all the students.

  • A. 1.63m
  • B. 1.52m
  • C. 1.54mCorrect
  • D. 1.42m

Explanation

Total height = 20(1.67)+16(1.5)+14(1.40) = 33.4+24+19.6 = 77m. Total students = 50. Mean = 77/50 = 1.54m.

Further Mathematics 2022 Objective — Question 29

Find, correct to the nearest degree, the acute angle formed by the lines y=2x+5 and 2y=x−6.

  • A. 76°
  • B. 53°
  • C. 37°Correct
  • D. 14°

Explanation

Gradients: m₁=2 and m₂=½. θ=tan⁻¹[(m₁−m₂)/(1+m₁m₂)] = tan⁻¹[(1.5)/(2)] = tan⁻¹(0.75) ≈ 37°.

Further Mathematics 2022 Objective — Question 30

Solve 4sin²θ+1=2, where 0°<θ<180°.

  • A. 60°, 120°
  • B. 30°, 150°Correct
  • C. 30°, 120°
  • D. 60°, 150°

Explanation

4sin²θ=1, so sin²θ=¼, sinθ=±½. For 0°<θ<180°, sinθ=½ gives θ=30° or θ=150°.

Further Mathematics 2022 Objective — Question 31

Find the range of values of x for which 2x²+7x−15≥0.

  • A. x≥−5 or x≤3/2
  • B. x≤−5 or x≥3/2Correct
  • C. −5≤x≤3/2
  • D. −3/2≤x≤5

Explanation

Factoring: 2x²+7x−15 = (2x−3)(x+5). This is ≥0 when x≤−5 or x≥3/2.

Further Mathematics 2022 Objective — Question 32

The probability that a student will graduate from a college is 0.4. If 3 students are selected from the college, what is the probability that at least one student will graduate?

  • A. 0.06
  • B. 0.22
  • C. 0.78Correct
  • D. 0.80

Explanation

P(at least one graduates) = 1−P(none graduate) = 1−(0.6)³ = 1−0.216 = 0.784 ≈ 0.78.

Further Mathematics 2022 Objective — Question 33

The equation of a circle is given as 2x²+2y²−x−3y−41=0. Find the coordinates of its centre.

  • A. (−¼, ¾)
  • B. (¼, ¾)Correct
  • C. (½, −³⁄₂)
  • D. (¼, −³⁄₂)

Explanation

Dividing through by 2: x²+y²−½x−¹⁄₂x... more precisely x²+y²−(1/2)x−(3/2)y−41/2=0. Comparing to x²+y²+2gx+2fy+c=0: g=−¼, f=−¾. The centre is (−g,−f) = (¼, ¾).

Further Mathematics 2022 Objective — Question 34

The gradient of a function at any point (x,y) is 2x−6. If the function passes through (1,2), find the function.

  • A. y=x²−6x−5
  • B. y=x²−6x+5
  • C. y=x²−6x−3
  • D. y=x²−6x+7Correct

Explanation

Integrating dy/dx=2x−6 gives y=x²−6x+c. Substituting (1,2): 2=1−6+c, so c=7. Thus y=x²−6x+7.

Further Mathematics 2022 Objective — Question 35

A particle of mass 3kg moving along a straight line under the action of a force FN, covers a line distance, d, at time, t, such that d=t²+3t. Find the magnitude of F at time t.

  • A. 0N
  • B. 2N
  • C. 3(2t+3)N
  • D. 6NCorrect

Explanation

Velocity = dd/dt = 2t+3. Acceleration = d²d/dt² = 2 (constant). Force F=ma=3×2=6N.

Further Mathematics 2022 Objective — Question 36

If α and β are the roots of x²+mx−n=0, where m and n are constants, form the equation whose roots are 1/α and 1/β.

  • A. mnx²−n²x−m=0
  • B. mx²−nx+1=0
  • C. nx²−mx+1=0
  • D. nx²−mx−1=0Correct

Explanation

α+β=−m, αβ=−n. Sum of new roots = (α+β)/αβ = m/n; product of new roots = 1/αβ = −1/n. The equation x²−(sum)x+product=0 becomes x²−(m/n)x−(1/n)=0, or nx²−mx−1=0.

Further Mathematics 2022 Objective — Question 37

A particle is acted upon by forces F=(10N,060°), P=(15N,120°) and Q=(12N,200°). Express the force that will keep the particle in equilibrium in the form xi+yj, where x and y are scalars.

  • A. 17.55i+13.78j
  • B. 17.55i−13.78j
  • C. −17.55i+13.78jCorrect
  • D. −17.55i−13.78j

Explanation

Resolving all three forces: ΣF(x)=10sin60+15sin60−12sin20≈17.55N, ΣF(y)=10cos60−15cos60−12cos20≈−13.78N. The resultant is 17.55i−13.78j; the equilibrant (force needed for equilibrium) is the negative of this: −17.55i+13.78j.

Further Mathematics 2022 Objective — Question 40

The length of the line joining points (x,4) and (−x,3) is 7 units. Find the value of x.

  • A. 4√3
  • B. 2√6
  • C. 3√2
  • D. 2√3Correct

Explanation

Distance = √[(−x−x)²+(3−4)²] = 7. Squaring: 4x²+1=49, so x²=12, giving x=√12=2√3.

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