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### 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

=================

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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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