# WAEC Further Mathematics Questions And Answers 2024 | Theory & Objective

Have seen the WAEC Further Mathematics Questions and Answers for the 2024 May/June WAEC Examination? If you have not and you would like to have full access to it, all you have to do is read this article carefully to the end.

When it comes to WAEC Further Mathematics examination, candidates are faced with pressure owing to the fact that that the subject is usually seen as one of the difficult subjects. However, if candidates would granted access to the expected questions and answers for every year, there is going be great improvement when the 2024 WAEC Result will be released.

In this article, I am going be releasing some some EXPO on 2024 WAEC Further Mathematics Questions and Answers for both Essay and Objective. If you desire to get more detailed information, make sure you continue reading article.

See Also: Updated May/June WAEC timetable for 2024 | Nigeria, Ghana, Gambia, Sierra Leone & Liberia

## WAEC Further Maths Questions and Answers

(These answers are not for 2024 examination) move to the next section for correct answer to WAEC Further Mathematics 2024 questions.

Given:

f(x)={(4x-x^2)dx

f(x)=2x^2 – x^3/3 + K

f(3)=2(3)^2 – (3)^2/3 + K =21

18 – 9 + K=11

9+K=21

K=21-9

K=12

Therefore

f(x)= -x^3 + 2x^2 + 12

## 2024 Further Mathematics Objective Questions and Answers

The following are the exact WAEC Further Mathematics Objective Questions for 2024 examination:

1. A binary operation ∆ is defined on the set of real numbers R, by x∆y = √x+y−xy/4, where x, yER. Find the value of 4∆3

• A. 16
• B. 8
• C. 4
• D. 2

2. (3√6/√5+√54/3√) −1

• A. 5√3/6
• B. 3√15/6
• C. 5√6/12
• D. 5√3/123

3. If log10(3x+1)+log104 = log10(9x+2), find the value of x

• A. 1/3
• B. 1
• C. 2
• D. 3

4. Simplify 9∗3n+1−3n+2/3n+1−3n

• A. 3
• B. 9
• C. 27
• D. 815

5. Consider the following statement:

x: All wrestlers are strong

y: Some wresters are not weightlifters.

Which of the following is a valid conclusion?

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

6. The functions f:x → 2x2 + 3x -7 and g:x →5x2 + 7x – 6 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 5
• C. x = 3 or -5
• D. x = 3 or 5

7. Express 4π/2 radians in degrees.

• A. 288º
• B. 200º
• C. 144º
• D. 120º8

8. A straight line makes intercepts of -3 and 2 on the x and y axes respectively. Find the equation of the line.

• A. 2x + 3y + 6 = 0
• B. 3x – 2y – 6 = 0
• C. -3x 2y – 6 = 0
• D. -2x + 3y – 6 = 0

9. Which of the following is the semi-interquartile range of a distribution?

• A. Mode – Median
• B. Highest score – Lowest score
• C. 1/2 (Upper Quartile – Median)
• D. 1/2 (Upper Quartile – Lower Quartile)10

10. Evaluate ∫0−1 (x + 1)(x – 2) dx

• A. 7/6
• B. 5/6
• C. -5/6
• D. -7/6

Check: WAEC Literature Questions and Answers 2024 | Novels, Essay & Objective

11. If 36, p, 9/4 and q are consecutive terms of an exponential sequence (G.P), find the sum of p and q.

• A. 9/16
• B. 81/16
• C. 9
• D. 9⁹/16

12. Differentiate (5x3+x2)/x, x ≠ 0 with respect to x.

• A. 10x + 1
• B. 10x + 2
• C. x(15x + 1)
• D. x(15x + 2)13

13. Given that 8x+m/x2−3x−4 ≡ 5/x+1 + 3/x−4

• A. 23
• B. 17
• C. -17
• D. 17

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

• A. 2
• B. -2
• C. -4
• D. 415

15. Evaluate 10 x2(x3+2)3

• A. 56/12
• B. 65/12
• C. 12
• D. 65

16.

• A. -8
• B. -5
• C. -4
• D. -317

17. 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)
• B. (5, -2)
• C. (-2, 7)
• D. (-7, 2)18

18. Evaluate 4p2+4C2−4p3

• A. 18
• B. 6
• C. -6
• D. -18

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

• A. 10
• B. 40
• C. 32
• D. 8020

20. 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}, (P⋃Q)∩R.

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

21. A particle moving with a velocity of 5m/s accelerates at 2m/s2. Find the distance it covers in 4 seconds.

• A. 16m
• B. 26m
• C. 36m
• D. 46m

22. 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 = -2
• D. k = 1, p = 2

23. In how many ways can six persons be paired?

• A. 5
• B. 10
• C. 15
• D. 20

24. Solve: 32x−2 − 28(3x−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 = 325

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

• A. (10 units, 053º)
• B. (9 units, 049º)
• C. (10 units, 037º)
• D. (9 units, 027º)

26. If →PQ = -2i + 5j and →RQ = -i – 7j, find →PR

• A. -3i + 12j
• B. -3i – 12j
• C. -i + 12j
• D. i – 12j27

27. The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.

If a hunter is selected at random, find the probability that the hunter covered at least 6km.

• A. 3/5
• B. 2/5
• C. 3/8
• D. 9/40

28. The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.

What is the mode of the distribution?

• A. 5
• B. 6
• C. 7
• D. 8

29. 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 ≤ 1
• C. -1 < x < 130

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

• A. -80
• B. 10
• C. 40
• D. 80

31. 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. −1/2,8
• B. 1/2,−8
• C. −1/2,−8
• D. 1/2,8

32. 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 the collision, find their common velocity.

• A. 1/2 m/s
• B. 1/3 m/s
• C. 2m/s
• D. 3m/s33

33. The mean heights of three groups of students consisting of 20, 16 and 14 students each are 1.67m, 1.50m and 1.40m respectively. Find the mean height of all the students.

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

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

• A. 76o
• B. 53
• C. 37
• D. 14

35. Solve: 4sin2θ + 1 = 2, where 0º < θ < 180º

• A. 60º 0r 120º
• B. 30º 0r 150º
• C. 30º 0r 120º
• D. 60º 0r 150º

36. 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 ≤3/2
• C. -5 ≤ x ≤ 3/5
• D. 3/5 ≤ x ≤ -537

37. The probability that a student will graduate from 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. 0.78
• D. 0.8038

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

• A. (−1/4, 3/4)
• B. (1/4, 3/4)
• C. (−1/2, 3/2)
• D. (−1/2, −3/2)

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

• A. x2 –  6x – 5
• B. x2 –  6x + 5
• C. x2 –  6x – 3
• D. x2 –  6x + 7

40. A particle of mass 3kg moving along a straight line under the action of a F N, 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. 6N

41. If α and β are roots of x2 + mx – n = 0, where m and n are constants, form the

• A. mnx2  – n2 x – m = 0
• B. mx2  – nx + 1 = 0
• C. nx2  – mx + 1 = 0
• D. nx2  – mx – 1 = 0

42. A particle is acted upon by forces F = (10N, 060º), P = (15N, 120º) and Q = (12N, 200º). 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. 17.55j – 13.78i
• C. -17.55i + 13.78j
• D. -17.55i – 13.78j43

43. Evaluate: limx→−2 x3+8/x+2

44. If f(x-1) = x3 + 3x2 + 4x – 5, find f(2)

• A. 61
• B. 25
• C. 20
• D. 13

45. 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√3

## WAEC Further Mathematics Theory Questions 2024

1.

A binary operation * is defined on the set T = {-2,-1,1,2} by p*q = p2 + 2pq – q2, where p,q ∊ T.
Copy and complete the table.

2. Solve 2(2y+1)−5(2y) + 2 = 0

3. Two functions f and g are defined on the set of real numbers, R, by

f:x → x2 + 2 and g:x → 1/x+2.Find the domain of (g∘f)−1

4. Solve 3cos2x – sinx = 0 for 0º≤x≤360º

5. The probability that  Abiola will be late to the office on a given day is 2/5. In a given working week of six days, find, correct to four significant figures, the probability that he will:

(a) only be late for 3 days.

(b) not be late in the week:

(c) be late throughout the six days.

6. The table shows the scores obtained by a group of artistes in Vocal (X) and the instrument (Y) musical competition.

Calculate the spearman’s rank correlation coefficient between the scores.

7. A body of mass of 18kg is suspended by an inextensible string from a rigid support and is pulled by a horizontal force F until the angle of inclination of the string to the vertical is 35º. If the system is in equilibrium, calculate the:

i. value of F

ii. tension in the string

8. Given that p = (8N,030º) and q = (9N, 150º), find, in component from, the unit vector along(p – q).

9. Given that nC4, nC5 and nC6 are the terms of a linear sequence (A.P), find the :

i. value of n

ii. common differences of the sequence.

10.

A solid rectangular block has a base that measures 3x cm by 2x cm. The height of the block is ycm and its volume is 72cm3.

i. Express y in terms of x.

ii. An expression for the total surface area of the block in terms of x only;

iii. the value of x for which the total surface area has a stationary value.

11. A basket contains 12 fruits: orange, apple and avocado pear, all of the same size. The number of oranges, apples and avocado pear forms three consecutive integers.

Two fruits are drawn one after the other without replacement. Calculate the probability that:

i. the first is an orange and the second is an avocado pear.

ii.both are of same fruit;

iii. at least one is an apple

12.

The table shows the corresponding values of two variables X and Y.

a. plot a scatter diagram to represent the data

b i. Calculate:x̄, the mean of X and ȳ, the mean of Y;

ii. Caculate:
x̄1, the mean of X values below x̄ and ȳ1, the mean of the corresponding Y values below x̄

c.  Draw the line of best fit through (x̄,ȳ) and (x̄1,ȳ1).

d. From the graph, determine the relationship between X and Y;

ii. From the graph, determine the value of Y when X is 20.

13.

A particle initially at rest moves in a straight line with an acceleration of (10t – 4t2)m/s2
Find the:
a. velocity of the particle after t seconds;

ii. average acceleration of the particle during the 4th second.

b. A load of mass 120kg is placed on a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards at a constant velocity. [Take g = 10m/s2]

ii. with an acceleration of 3m/s2. [Take g = 10m/s2]

14. The vectors 6i + 8j and 8i – 6j are parallel to →OP and →OQ respectively. If the magnitude of →OP and →OQ are 80 units and 120 units respectively, express: →OP and →OQ in terms of i and j;

ii. |→PQ|, in the form c√k, where c and k are constants.

I hope that you are excited with questions you have seen so far.

## WAEC Further Mathematics Theory Answers 2024

Now that you have seen the expected theory questions, it is time to show you the correct answers to all the questions. the 2024 Further Mathematics theory answers are as shown below:

1.

p*q = p2 + 2pq – q2

when p = -2 and q = -2

p*q = -22 + 2(-2)(-2) – (-2)2
p*q = 4 + 8 – 4 = 8

when p = -2 and q = -1
p*q = -22 + 2(-2)(-1) – (-1)2
p*q = 4 + 4 – 1 = 7

when p = -2 and q = 1
p*q = -22 + 2(-2)(1) – (1)2
p*q = 4 – 4 -1 = -1

when p = -1 and q = -2
p*q = -12 + 2(-1)(-2) – (-2)2
p*q = 1 + 4 – 4 = 1

when p = -1 and q = -1
p*q = -12 + 2(-1)(-1) – (-1)2
p*q = 1 + 2 -1 = 2

when p = -1 and q = 2
p*q = -12 + 2(-1)(2) – (2)2
p*q = 1 – 4 – 4 = -7

when p = 1 and q = -2
p*q = 12 + 2(1)(-2) – (-2)2
p*q = 1 – 4 – 4 = -7

when p = 1 and q = -1
p*q = 12 + 2(1)(-1) – (-1)2
p*q = 1 – 2 – 1 = -2

when p = 1 and q = 1
p*q = 12 + 2(1)(1) – (1)2
p*q = 1 + 2 – 1 = 2

when p = 1 and q = 2
p*q = 12 + 2(1)(2) – (2)2
p*q = 1 + 4 – 4 = 1

when p = 2 and q = -2
p*q = 22 + 2(2)(-2) – (-2)2
p*q = 4 – 8 – 4 = -8

when p = 2 and q = -1
p*q = 222 + 2(2)(-1) – (-1)22
p*q = 4 – 4 – 1 = -1

when p = 2 and q = 1
p*q = 222+ 2(2)(1) – (1)2
p*q = 4 + 4 – 1 = 7
when p = 2 and q = 2
p*q = 22 + 2(2)(2) – (2)2
p*q = 4 + 8 – 4 = 8

2.

3.

4.

3cos2x – sinx = 0
cos 2A = 1 – 2 sin2A
3(1 – 2sin2x) – sinx = 0
3 – 6sin2x -sinx = 0
let sinx = p
3 – 6p2 -p = 0
-6p2 – p +3 = 0

factors of p in the eqn:

p = 0.7953 or 0.6287
when p = 0.7953
p = sinx
x = sin−1 p
x = sin−1(0.7953)
x = 52.68º

when p = 0.6287
p = sinx
x = sin−1 p
x = sin−1(0.6287)
x = 38.95º

5.

he will be late to office = 2/5
he will not be late to office is = 1 – 2/5 = 3/5

If he will be late for 3 days only, he will also not be late for 3 days

2/5 x 2/5 x 2/5 x 3/5 x 3/5 x 3/5

p = 0.0138

(b)   not be late to office is = 1 – 2/5 = 3/5

3/5 x 3/5 x 3/5 x 3/5 x 3×5 x 3/5

p = 0.0467

(c) be late to office = 2/5

2/5 x 2/5 x 2/5 x 2/5 x 2/5 x 2/5

p = 0.0041

6.

7.

If the system is in equilibrium, then the weight of the body = the force pulling it
and the force = the vertical component of force F
since Force F is inclined to vertical then the vertical component = Fcos35º
the weight of the body = mg
Fcos35º = 18(10)
0.8192F = 180

F = 180/0.8192

F = 219.7N

tension in the string

T = mg
T = 18kg × 10
T = 180N

8.

p = 8cos30i + 8sin30j
q = 9cos150i + 9sin150j
p = 6.9282i + 4j
q = -7.7942i + 4.5j
pq = (-7.7942 – 6.9282)i + (4.5 – 4)j
pq = -14.7224i + 0.5j
|pq| = √((-14.7224)2 + (0.5)2)
|pq| = √(216.7491 + 0.25)
|pq| = √216.9991
|pq| = 14.7309

the unit vector in the direction of p – q =

-0.9994i + 0.0339j ≌ -i + 0.0339j

9.

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)
6n – 54 = n2 -11n – 4n + 44
6n – 54 = n2 – 15n + 44
n2 – 21n + 98 = 0
n2 – 7n – 14n + 98 = 0
n(n – 7) -14(n – 7) = 0
(n-7)(n-14) = 0
n = 7 or 14

10.

12.

13.

15.

a. a = (10t – 4t2)m/s2, t = t seconds, u = 0
from first equation of motion,
v = u + at
v = 0 + (10t – 4t2)t
v = (10t2 – 4t3)m/s2

ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4
Δa4 = 10(t4-t3) – 4(t4-t3)2
Δa4 = 10(4-1) – 4(4-3)2
Δa4 = 10(1) – 4(12)
Δa4 = (10-4)m/s2
Δa4 = 6ms-2

b. R = m(g+a)
When the lift moves with a constant velocity, a = 0
R = mg
R = 120 × 10
R = 1200N

ii. when the lift moves upward with an acceleration of 3m/s2
R = m(g+a)
R = 120(10+3)
R = 120 × 13
R = 1560N

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