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WAEC GCE Further Mathematics Questions 2019 | Check OBJ and Theory Answers Here

Filed in Exam, WAEC News by on March 29, 2019

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WAEC GCE Further Mathematics Questions 2019 | Check OBJ and Theory Answers Here.

Further Mathematics, Further Mathematics WAEC Questions 2019 | In this article, I will be showing you past Further Mathematics objective and theory random repeated questions for free. You will also understand how WAEC Further Mathematics questions are set and many other examination guides. Stay focus and read through.

WAEC GCE Further Mathematics Questions

WAEC GCE Further Mathematics Questions

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

WAEC GCE Further Mathematics Questions 2019

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FURTHER MATHS OBJ:

1-10: BBADDBABDA
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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