2023 WAEC FURTHER MATHEMATICS EXAMINATION QUESTIONS

2023 WAEC Further Mathematics past papers and answer. If you are a candidate who is in search of how 2023 Waec further mathematics questions and answers will look like then you are at the right place.

Nkedugists have been providing solutions to candidates who are eager to learn and make good grades in oncoming/going Waec exams. We provide you with samples of 2023 further mathematics past papers answer, just kindly read through to see what you will be asked.  The number of questions required Will be mentioned in the exam questions paper

2023 WAEC FURTHER MATHEMATICS EXAMINATION QUESTIONS SCHEME/SAMPLE

Your question type is divided into two papers, Papers 1 and 2, both of which must be taken.

PAPER 1: Will consist of forty multiple-choice objective questions, covering the entire syllabus. Candidates will be required to answer all questions in 1 hour for 40 marks. The questions will be drawn from the sections of the syllabus as follows:

Pure Mathematics – 30 questions
Statistics and probability – 4 questions
Vectors and Mechanics – 6 questions

PAPER 2: will consist of two sections, Sections A and B, to be answered in 2 hours for 100 marks.

Section A will consist of eight compulsory questions that are elementary in type for 48 marks. The questions shall be distributed as follows:

Pure Mathematics – 4 questions
Statistics and Probability – 2 questions
Vectors and Mechanics – 2 questions

Section B will consist of seven questions of greater length and difficulty put into three parts: Parts I, II, and III as follows:
Part I: Pure Mathematics – 3 questions
Part II: Statistics and Probability – 2 questions
Part III: Vectors and Mechanics – 2 questions

Candidates will be required to answer four questions with at least one from each part for 52 marks.

2023 WAEC FURTHER MATHEMATICS PAST PAPERS AND ANSWER SAMPLE

1. Given the matrix M=

2 -4 -4
1 8 2
1 1 -2

find |M|

A. -24

B. -8

C. 8

D. 24

E. 48

ANSWER: A

2. The gradient of a curve is 8x+2 and it passes through (1,3). Find the equation of the curve

A. y = 4x^2 + 2x + 3

B. y = -4x^2 + 2x -3

C. y = 4x^2 – 2x + 3

D. y = 4x^2 + 2x + 3

E. y= 4x^2 – 2x – 3

ANSWER: A

3. Given that y = 3x^3 + 4x^2 + 7. Find dy/dx at x = 1

A. 14

B. 15

C. 17

D. 30

E. 35

ANSWER: C

4. Integrate 3x^2 + 4x – 8 with respect to x

A. x^3 + 2x^2 + 8x + k

B. 6x + 4 + k

C. x^3 – 2x^2 + 8x + k

D. x^3 + x^2 – 8x + k

E. x^3 + 2x^2 – 8x + k

ANSWER: A

WAEC Past Further Mathematics Answers

Below are Past Sample Answers

5a)
pr(age)=4/5
pr(fully)=3/4
pr(must)=2/3
pr(age not admitted)=1-4/5
=1/5
pr(fully not admitted)=1-3/4
=1/4
pr(must not admitted)=1-2/3
=1/3
Therefore pr(none admitted)=1/5*1/4*1/3
=1/60
5b)
pr(only age and fully gained admission)=4/5*3/4*1/3
=1/5
================

12a)
tabulate
Marks| 1-10, 11-20, 21-30, 31-40, 41-50,51-60, 61-70, 71-80, 81-90, 91-100
F| 3, 17, 41, 85, 97, 115, 101, 64, 21, 6
C.B| 0.5-105, 10.5-205, 20.5-305, 30.5-405,40.5-505, 50.5-605, 60.5-705, 70.5-805,80.5-905, 90.5-1005
C.F| 0+3=3, 3+17=20, 20+41=61, 61+85=146,146+77=243, 243+115=358, 358+101=459,459+64=523, 523+21=544, 544+6=550
=================

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11a)
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
11b)
i) Tn=a+(n-1)d
T2=a+(2-1)d
T2=a+d
T4=a+3d
T8=a+7d
GP
Tn=ar^n-1
T1=ar^1-1
T2=ar^2-1=ar
T3=ar^2
a+d=a …..equation (1)
a+3d=ar …..equation (2)
a+7d=ar^2 …..equation (3)
T3+T5=20
a+2d+a+4d=20
2a+6d=20
a+3d=10 …..equation (4)
…..equation (2)/…..equation (1)
ar/a=a+3d/a+d
r=a+3d/a+d
…..equation (3)/…..equation (2)
ar^2/ar=a+7d/a+3d
r=a+7d/a+3d
but r=r
a+3d/a+d=a+7d/a+3d
(a+3d)^2=(a+d)(a+7d)
a^2+6ad+ad^2
a^2+7ad+ad+7d^2
a^2+8ad+7d^2
a^2+6ad+9d^2=a^2
+8ad+7d^2
6ad+9d^2=8ad+7d^2
6ad-8ad=7d^2-9d^2
-2ad=2d^2
ad=dd
a=d
===================

(9a)
1/1-cos tita + 1/1+cos tita
=1+cos tita + 1-cos tita//(1-cos tita) (1+cos tita)
= 2/1+cos tita – cos tita – cos^2 tita
= 2/1-cos^2 tita
Recall that :
Cos^2 tita + sin^2 tita = 1
.:. Cos^2 tita = 1-sin^2 tita
.:. 1/1-cos^2 tita + 1/1+cos tita
= 2/1-(1-sin^2 tita)
(9b)
At stationary points,
dy/dx=0.
y=x^0(x-3)
Let u=x^2,v=x-3.
du/dx=2x dv/dx=1.
dy/dx= Udv/dx + Vdu/dx
dy/dx=x^2(1)+(x-3)(2x)
.:. dy/dx=x^2+2x^2-6x
dy/dx=3x^2-6x
At stationary point,
dy/dx=0..
.:.3x^2-6x=0
Equation of line=> 3x^2-6x=0

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