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2017/2018 waec further maths questions and answers 2017/2018

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Topic: 2017/2018 waec further maths questions and answers 2017
imgMr.Net™ ¯ ¯↓
FURTHERMATHS OBJ:
1-10: CBADCBADBB
11-20: DCBDCCDBCC
21-30: ACDCAABBCB
31-40: CDCDDBCAAD
=========================

(1a)
g(x)=y
y=x+6
x=y-6
g^- f(x-6)
=4-5(x-6)/2=4-5x+30/2
=34-5x/2

(1b)
coodinate=(x1+x2/2 ,y1+y2/2)
=(7-2/2,7-5/2)
=(5/2,2/2)
=(5/2,1)

================================


(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


=======================================

(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

====================================



(9a)
1/1-cos tita + 1/1+cos tita

=1+cos tita + 1-cos tita//(1-cos tita) (1+cos tita)

= 2/1+cos tita - cos tita - cos^2 tita

= 2/1-cos^2 tita

Recall that :
Cos^2 tita + sin^2 tita = 1

.:. Cos^2 tita = 1-sin^2 tita

.:. 1/1-cos^2 tita + 1/1+cos tita

= 2/1-(1-sin^2 tita)

(9b)
At stationary points,
dy/dx=0.

y=x^0(x-3)

Let u=x^2,v=x-3.

du/dx=2x dv/dx=1.

dy/dx= Udv/dx + Vdu/dx

dy/dx=x^2(1)+(x-3)(2x)

.:. dy/dx=x^2+2x^2-6x

dy/dx=3x^2-6x

At stationary point,
dy/dx=0..

.:.3x^2-6x=0
Equation of line=> 3x^2-6x=0

===============================


(10ai)
(x^2-1) (x+2)=0
(x-1) (x+1) (x+2)
x=1, or -1 or -2

(10aii)
2x-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=6A, 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

(11b)
Given
AP: T2=a+d
T4=a+3d
T8=a+7d
Since they form the consecutive numbers of a G.P.
a+3d/a+d=a+7d/a+3d
a^2+6ad+9d^2=a^2+8ad+7d^2
6ad+9d^2=8ad+7d^2
6a+9d=8a+7d
9d-7d=8a-6a
2d=2a
a=d------Eqn(1)
Also:
S3+S5=20
3/2(2a+2d)+5/2(2a+4d)=20
Sub eqn(1) into eqn(2)
3/2(2a+2a)+5/2(2a+4a)=20
6a+15a=20
21a=20
a=20/21

===============================

(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

===============================



(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

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