fbpx
EXAM NEWS WAEC NEWS

2020 WAEC FURTHER MATHEMATICS EXAMINATION QUESTIONS (Review)

2020 WAEC Further Mathematics past papers
2020 WAEC Further Mathematics past papers
Written by Nkedugist

2020 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 2020 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

2020 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.

2020 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
=================

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^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

WAEC Mathematics Questions and Answers 2020 Loading…

If you have any question about the 2020 WAEC Further Mathematics questions and answers, do well to let us know in the comment box.

Important Notice on WAEC Further Mathematics theory Answers

The above questions are not exactly 2020 WAEC Further Mathematics questions and answers but likely WAEC Further Maths repeated questions and answers.

These questions are strictly for practice. The 2020 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.

Facebooktwitterredditpinterestlinkedinmail

About the author

Nkedugist

I created Nkedugist.com.ng to serve and assist all Aspirants who are seeking school information or want to apply for admission into Nigerian Tertiary Institutions such as Universities, Polytechnics or Colleges of Education with information regarding the various education system.

2 Comments

Leave a Comment