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

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

Further Mathematics, Further Mathematics WAECQuestions 2019 | In this article, I will be showing you past Further Mathematics objective and theory random repeatedquestionsfor free. You will also understand howWAEC FurtherMathematics 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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