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2018 WAEC SSCE Further Maths Essay/Essay Questions & Answers On Further maths 2018 –  Related to Essay Questions & Answers on Further Maths 2018 are WAEC Further Maths past papers 2017, WAEC questions on Further Maths, Further Maths past papers 2016, marking scheme for WAEC physics practical, and area of concentration for WAEC 2018. These are essential to your making  good grades in this 2018 May/June WAEC examinations to  qualify you for university admission this year.  2018 WAEC SSCE Further Maths Essay/Essay Questions & Answers on Further maths 2018.

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In this post are verified regularly repeated Further Maths Questions & Answers for 2018 May/June SSCE exam that holds on 9/4/2018.  Get them in your finger tips and you are sure of a high grade in this exams.

The West African Examinations Council (WAEC)  is the West African countries  Examination Body. It conducts the West African Senior School Certificate Examination for in-house school students and General Certificate Examination for private candidates in West African Countries. These are entry qualification examinations for higher education institutions in these countries.

As a matter of fact, no University, Polytechnic, or college of education in these countries would admit you into any of their courses without a good grade in any of these examinations. That is why over three million candidates register and take these subject exams annually. 2018 WAEC SSCE Further Maths Essay/Essay Questions & Answers On Further maths 2018.

Consequently, our goal here is to help you achieve this your life dream. Listed here are Further Maths  questions and answers relevant to 9/4/2018 Further Exam Questions.

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

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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

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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

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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

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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

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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

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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

2018 WAEC SSCE Further Maths Essay/Essay Questions & Answers On Further maths 2018

Requirements

2018 WAEC SSCE Further Maths Essay/Essay Questions & Answers On Further maths 2018