Earboard Logo

Welcome, Guest: Join Earboard / LOGIN! / Advertise / Join Earboard Affiliate /Contact Us / News
Stats: 6557 members, 32061 topics. Date: Monday, 18th February 2019


HOME NEWSLETTER GENERAL NEWS TECHNOLOGY EDUCATION AFFILIATE
Click Here to Receive Our Next Post via Email ...
 

WAEC 2017 Further Mathematics Answers

Jamb to Postpone 2019 Mock Examination Day / 2019 UTME: Reprinting of JAMB Exam Slip Has Commenced / Junior WAEC (BECE) 2019 Timetable – Download PDF /

(0) (Reply) (Go Down)

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

FURTHERMATHS OBJ:
1-10: CBADCBADBB
11-20: DCBDCCDBCC
21-30: ACDCAABBCB
31-40: CDCDDBCAAD

(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
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

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

Click here to pay for our Runz

0

Copy the affiliate link below and Share with your Friends:

Publication author

offline 3 months

Editor

7
I am the brain behind Earboard.com
Comments: 41Publics: 581Registration: 13-02-2017

Leave a Reply

(0) (Reply)

Jamb to Postpone 2019 Mock Examination Day / 2019 UTME: Reprinting of JAMB Exam Slip Has Commenced / Junior WAEC (BECE) 2019 Timetable – Download PDF /  

(Go Up)


Sections: News (1) Technology Education (2)
Earboard - Copyright @ 2016 - 2019 George Gosmooth. All rights reserved. See How To Advertise. DMCA Content Removal.
Disclaimer: Every member is solely responsible for anything that he/she posts or uploads on Earbaord.