# 2022 WAEC FURTHER MATHEMATICS EXAMINATION QUESTIONS (Review)

2021 Waec mathematics question and answer
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2022 WAEC Further Mathematics past papers and Answer. If you are a candidate who is in search on how 2020 waec further mathematics questions and answer will look like then you are at the right place.

Nkedugists having been providing solutions to candidates who eager to learn and make a good grade in oncoming/going waec exam. We provide you with the samples of 2022 waec further mathematics past papers answer, just kindly read through to see what you will be asked.  The numbers of questions required Will be mentioned in the exam questions paper

### 2022 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 1hours 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.

Check out 2020 waec further mathematics objective.

### 2022 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

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

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

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

### WAEC Past Further Mathematics Answers

Below are Past Sample Answers

5a)
pr(age)=4/5
pr(fully)=3/4
pr(must)=2/3
=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
=================

#### Related Article:

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=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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#### Important Notice on WAEC Further Mathematics theory Answers

The above questions are not exactly 2022 WAEC Further Mathematics questions and answers but likely WAEC Further Maths repeated questions and answers.

These questions are strictly for practice. The 2022 WAEC Further Mathematics questions will be posted on this page during the WAEC Further Mathematics examination. Keep checking and reloading this page for the answers or better still subscribe with us.

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