# WAEC GCE 2021 Further Mathematics Questions and Answers Update

**WAEC GCE 2021: Those who have written Further Mathematics in the WAEC GCE examination know just how tough the exam tends to be. This post is for those preparing to write Further mathematics in 2021 WAEC GCE. Read on.**

**Further Mathematics expo questions are now available on our website. In this article, you will get to see past WAEC GCE Further Mathematics random repeated questions for free, and also understand how WAEC GCE Further Mathematics questions are set. Stay focus and follow this guide.**

The *West African Examinations Council* (WAEC) is an examination board that conducts the West African Senior School Certificate Examination (WASSCE), for University and *Jamb entry examination* in West African countries. In a year, over three million candidates registered for the *exams coordinated by WAEC*.

**WAEC GCE Further Mathematics Questions and Answers**

**FURTHER MATHS OBJ:**

1-10: BBADDBABDA

11-20: ACBABDBDAB

21-30: BDDBCBACCD

31-40: ACBDACDBAB

**Essay Questions and Answers:**

1a)

8m + 2³m = 1/4

2²m + 2²m = 2-²

4³m = 2-²

Equate the base

6m = – 2

m = -2/4 = -1/3

1b)

log15 base 4 = x

x = log15 base/log4 base 10 = 1.1761/0.6020

= 1.9536*therefore x approximately* 1.954

========

2a)

(-2) = m(-2)² + n(-2)+2=0

4m-2n = – – – – – – – – (1)

F(1) = m(1)²+n(1)+2=3

m+n+2=3

m+n=3-2

m+n=1 – – – – – – – – – – – -(2)

X Equate (2) by 4

4m+4n=4 – – – – – – – – – – (3)

4m-2n= -2

-4m+4n=4/-6n = – 6

n=1

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

4)

Given y= 5/x² + 3

Y = 5(x² +3)-1

dy/dx = anxn-1

y = 5(x² + 3)-1

dy = 5(-1) (x2 +3)-1^1

dy/dx = 10x (x² + 3)-²

dy/dx = – 10/(x² +3)²

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

5a)

^nP5 ÷ ^nC4 = 24

n!/(n-5)! ÷ n!/(n-4)!4! =24

n!/(n-5)! * (n-4)!4!/n! =24

(n-4)!4!/(n-5)!=24

(n-4)(n-5)!4!/(n-5)! = 24

n-4=1

n=4+1

n=5

b)PR= 5C3 (1/6)³ (5/6)^5-3

5!/(5-3)!3! (1/6)³ (5/6)²

5!/2!3! (1/6)³ (5/6)²

10 (1/216) (25/36)

=0.03215

Pr = 1-0.03215

=0.9678

==========

6) make a table containing

MARKS,TALLY,F & FX

UNDER F – 2,9,4,2,2,1

UNDER FX – 2,18,12,8,10,6

€F= 20

€FX = 56

6b) Mean = €fx / €f

= 56 / 20

= 2.8

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

7)

Given

h=15.4t-4.9t

Velocity V=dh/dt =15.4-9.8t

At maximum height V=0

Therefore 15.4-9.8t=0

9.8t=15.4

t=15.4/9.8

t=1.6secs

Time to *reach maximum height* is 1.6secs

7b)

maximum h=15.4t-9.4t^2

15.4(1.6)-4.9(1.6)^2

22.64-12.544

=12.096

Max height = 12.1m

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

9a)X+6/(x+1)^2 = A/x+1 + B/(x+1)² + C/(x+1)³

X+6/ ~(x+1)³~ =A(x+1)² + B(x+1) + C / ~(x+1)³~

X+6 =A (x+1)² + B(x+1) + C

Let X+1=0 , X=-1

-1+6= A (-1+1)² + B(-1+1) +C

5= C

Therefore:- C=5

X+6= A(X2+2x+1) + Bx + B + C

X+6= Ax²+2Ax+A+Bx+B+C

comparing the coefficient of X²

A=0

Comparing the coefficient of X

1=2A+B

1=2(0)+B

B=1

x+6/(x+3)² = o/x+1 + 1/(x+1)² + 5(x+1)³

1/(x+1)² + 5/(x+1)³

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

10a)

3x^2+x-2 <= 0 3x^2+3x-2x-2 <= 0 3x(x+1) -2 (x+1) <= 0 (3x-2)(x+1) <= 0

3x-2 <= 0 or 8+1 <= 0

DISCLAIMER! These are not real WAEC GCE Further Mathematics questions but likely repeated questions over the years to help candidate understand the nature of their examinations. Ensure to take note of every questions provided on this page.

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