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# WAEC GCE 2021 Further Mathematics Questions and Answers Update

Filed in Exam, WAEC News on February 23, 2021

WAEC GCE Further Mathematics Questions 2021 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:

11-20: ACBABDBDAB
21-30: BDDBCBACCD
31-40: ACBDACDBAB

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

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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.

If you need help with updated questions and answers at the right time about this examination, kindly provide us with your phone number and email address in the comment box below. Also, feel free to ask any questions pertaining to this guide.

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