Author Topic: 2017/2018 WAEC WASSCE Further Maths OBJ Essay Theory Questions And Answer Expo/Runs free (Read 2965 times)

JEFFREY96 · « **on:** March 24, 2017, 09:36:35 am »

F-MAThs
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Ghana members
FURTHER MATHS
1
a) 8m + 2³m
(2³)m + 2³m = 1/4
2³m + 2³m = 1/4
2(2³m) = 1/4
2³m = 1/8
2³m = 8 - ¹
2³m = 2-³
Since bases are the same, powers are same
3m = - 3
m = - 3/3
m = -1
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
(a)
Given
h=15.4t - 4.9t²
Velocity v = dh/dt = 15.4 - 9.8t
At maximum height, V =0
15.4 - 9.8t = 0
9.8t = 15.4
t = 15.4/9.8
t = 1.6secs
Time to reach maximum height = 1.6secs
b) maximum height = 15.4 t - 4.9t²
= 15.4(1.6) - 4.9(1.6)²
=24.64 - 12.544
= 12.096
Maximum 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
9b)
S²1 1/(x+1)² + 5/(x+1)³ DX
1n (x+1)³ + 5 1n (X+1)⁴ |²1
1n 3(3) + 5 1n(3)⁴ - 1n +2³ - 5 1n +2⁴
3.2958+21.9722-2.0794-13.8629
=9.3297
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
3x <= 2 or x <= -1
X<= 2/3
Therefore range of value is
-1 >= x <= 2/3
Good luck

NAIJA MEMBER

FOR NIGERIANS
1 a)
g ( x )= y
y = x + 6
x = y - 6
g ^- f( x - 6 )
= 4- 5 ( x - 6) / 2 = 4 - 5x +30/ 2
= 34-5 x / 2
1 b)
coodinate = (x 1+ x 2 / 2 , y 1+ y 2/ 2 )
= (7 - 2 / 2 , 7 - 5 / 2 )= ( 5 / 2 , 2 / 2 )
= (5 / 2 , 1)

=========================7)
m1=3
u1=8m/s
m2=?
u2=5m/s
v=6m/s
m1u1+m1u2=(m1+m2)v
3*8+m2*5=(3+m2)6
24+5m2=18+6m2
24-18=6m2-5m2
m2=6
(7b)
m2u2-m1u1=V(m1+m2)
6*5-3*8=V(3+6)
30-24=9v
9v=6
v=6/9
v=0.67m/
========
NAIJA

10a)
i ) ( x ^2-1 ) ( x +2 )= 0
( x -1 ) ( x +1 ) ( x +2 )
x =1, or - 1 or -2
ii ) 2 x -3/( x -1) ( x + 1) ( +2)
=A /x -1+B /x + 1+C /x +2
2x - 3=A ( x +1) ( x +2 ) +B ( x -1) ( x +2)
+C ( x -1) ( x + 1)
let x + 1=0, x =- 1
2( -1 ) -3=B ( - 1-1) ( - 1+2)
-5/2 =-2B /- 2 B =5/2
let x - 1 = 0 x = 1
2( 1) - 3=A ( 1 +1) ( 1+2 )
-1= CA , A = -1/6
Let x +2=0 x =- 2
2( -2 ) -3=C ( -2- 1) ( -2 +1)
-7= 3C , C =-7 /3
=================
(11a)
Given:
f(x)={(4x-x^2)dx
f(x)=2x^2 – x^3/3 + K
f(3)=2(3)^2 – (3)^2/3 + K =21
18 – 9 + K=11
9+K=21
K=21-9
K=12
Therefore
f(x)= -x^3 + 2x^2 + 12

——————————————-

(14ai)
SKETCH THE DIAGRAM(14aii)
Using lami’s theory
T1/sin60=T2/sin30
48N/sin60=T2/sin30
48N/0.8660=T2/0.5
0.5(48)/0.8660=T2(0.8660)/0.8660
T2=24/0.8660
T2=27.7N
(14b)
Using the equation of motion
H=U^2/2g
H=(20)^2/2*10
=20*20/20
H=20m
Timetaken to reach the maximum height
S=Ut+1/2at^2
20=0+1/2(100)t^2
20/5=5t^2/5
t^2=4
t=sqroot4
t=2S

5a)
pr(age)=4/5
pr(fully)=3/4
pr(must)=2/3
pr(age not admitted)=1-4/5
=1/5
pr(fully not admitted)=1-3/4
=1/4
pr(must not admitted)=1-2/3
=1/3
Therefore pr(none admitted)=1/5*1/4*1/3
=1/60

5b)
pr(only age and fully gained
admission)=4/5*3/4*1/3
=1/5

4)
(x^2+5x+1)sqroot(2x^3+mx^2+nx+11)=(2x-5)
remainder:30x+16
(x^2+5x+1)(2x-5)
=2x^3+10x^2+2x-5x^2-25x-5
=2x^3+10x^2-5x^2-25x-5
=2x^3+5x^2-23x+30x+16-5
=2x^3+5x^2+7x+11
Therefore m=5, n=7

12a)
tabulate
Marks| 1-10, 11-20, 21-30, 31-40, 41-50,
51-60, 61-70, 71-80, 81-90, 91-100
F| 3, 17, 41, 85, 97, 115, 101, 64, 21, 6
C.B| 0.5-105, 10.5-205, 20.5-305, 30.5-405,
40.5-505, 50.5-605, 60.5-705, 70.5-805,
80.5-905, 90.5-1005
C.F| 0+3=3, 3+17=20, 20+41=61, 61+85=146,
146+77=243, 243+115=358, 358+101=459,
459+64=523, 523+21=544, 544+6=550

13ai)
M=2
P=5
C=3
total=10
If the books of the same subject are to stand
together
No of arrangements=2!*5!*3!*3!
=2*120*6*6
=8640arrangements
(13aii)
Only the physics textbook must stand together
No of arrangements=5!*6!
=120*720
=86400arrangements

(13b)
P=13/20
q=1-13/20=7/20
pr(atleast 3 speak E)=1-Pr(2 speak E)
=(1-8C1p^1q^7+8C2p^2q^6)
=1-(8*(13/20)*(7/20)^7+28(13/20)^2*(7/20)^6
=1-(0.003346+0.0217467)
=1-0.0251
=0.9749
=0.975(3s.f)

JEFFREY96 · « **Reply #1 on:** March 24, 2017, 04:03:44 pm »

GOOD LUCK AND KEEP INVITING YOUR FRIENDS EXCLUDING YOUR ENEMIES

Earlyface

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Author Topic: 2017/2018 WAEC WASSCE Further Maths OBJ Essay Theory Questions And Answer Expo/Runs free (Read 2965 times)

JEFFREY96

2017/2018 WAEC WASSCE Further Maths OBJ Essay Theory Questions And Answer Expo/Runs free

JEFFREY96

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