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### 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

**WAEC Mathematics Questions and Answers 2023 Loading…**

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