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## WAEC 2017 Further Mathematics Answers

WAEC 2017 Further Mathematics Answers by : 12:00 am On March 24, 2017

FURTHERMATHS OBJ:
11-20: DCBDCCDBCC
21-30: ACDCAABBCB

(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

5)
USING THE BINOMIAL PROBABILITY FUNCTION
n(rp^rq^n-r
n=10
p=8%(0.08)
q=92%(0.94)
Prob(aprotimate – under >)
=10/2(0.08)^3(0.92)^7 10/4(0.08)^4(0.92)^6 10/5(0.08)^5(0.92)^5 10/6(0.08)^6(0.92)^4 10/7(0.08)^7(0.92)^3 10/8(0.08)^8(0.92)^2 10/9(0.08)^9(0.92)^1 10/10(0.08)^10(0.92)^0
=0.0343 0.052 0.0005 0.00003 0.00000195 …….Aproxtimately 0.0400

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 standtogether
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)

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

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)

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
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

14ai)
SKETCH THE DIAGRAM
14ii)
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