# 2021 WAEC Further Maths Question And Answer

2021 WAEC Mathematics Questions And Answers.

Further math Objective and essay question and answer

PAPER 1: This paper will contain forty multiple-choice objective questions that will cover the whole course.

Candidates will have one hour to answer all questions for a total of 40 points. The following areas of the curriculum will be used to generate the questions:

30 questions on pure mathematics

4 questions about statistics and probability

6 questions on vectors and mechanics

PAPER 2: There will be two portions, Sections A and B, that must be completed in two hours for a total of 100 points.

Section A will include eight obligatory questions of elementary kind worth 48 points. The following is how the questions will be distributed:

2021 WAEC further Maths Theory

1)

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

2)

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

3)

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

(4)

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)

(5)

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

6ai)

P:F=4:1 =4x+1x=100

5x=100

x=100/5

x=20

pass=20*4=80%

fail= 20*1=20%

p(pass)=80/100=0.8

p(fail)=20/100=0.2

n=7

12ai)

P(at least 3passed)

P=0.8

Q=0.2

P(x=r)=n(rP^rq^n-r

P(x>/3)=1-P(x<2) P(x<2)=P(x=0)+P(x=1)+P(x=2) P(x=0)=7dgree (0.8)degree (0.2)^7 P(x=0)=0.0000128 P(x=1)=^7( (0.8)^1 (0.2)^6 =0.0003584 P(x<2)=7^C2 (0.8)^2 (0.2)^5 =0.0043008 P(X<2)=0.0000128+0.0003584+0.004300 =0.004672 P(x>3)=1-0.004672

=0.995321

=0.10(2d.p)

6aii)

P(between 3 and 6 failed)

P=0.2

q=0.8

P(36)

P(x=3) + P (x=4)+p(x=5)+P(x=6)

p(x=3) 7^C3 (0.2)^3 (0.8)^4

=0.114688

p(x=4)=7^C4 (0.2)^4 (0.8)^3

0.028672

P(x=5)=7^C5 (0.2)^5 (0.8)^2

=0.0043008

P(x=6)=7^C6 (0.2)^6 (0.8)^1

=0.0003584

p(36)

=0.114688+0.028672+0.0043008

+0.0003584

=0.1480192

=0.15(2d.p)

Today’s WAEC Further Maths OBJ Answers

11-20. DCBDCBCDCC

21-30. ABDCDABBCB

41-50. ABDCBACDDC

2021 WAEC Further Mathematics Possible Questions and answers

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