verified 2017 waec Further mathematics Obj | theory answers | Runz

                                                    

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 NIGERIAN ANSWERS

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)

 (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

( 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
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(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)

( 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

 FURTHERMATHEMATICS NIGERIAN ANSWERS

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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)

9a)1/1-cos tita + 1/1+cos tita
=1+ cos tita+ 1- cos 0/1-cos^2 tita
= 2/1- cos^2 tita
= 2/sin^2 tita

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
i)T1=a=20/21
T2=a+d=40/21
T3=a+2d=20/7
T4=a+3d=80/21

ii)S10=10/2[2(20/21)+9(20/21)]
=(10/2)(20/21)(11)
=1100/21

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(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

 (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)

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(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

 10b)
X1 Y2
(3, 1)
r=sqr(x2-x1)^2+(y2-y1)^2
r=sqr(3+3)^2+(1-1)^2
r=sqr6^2+0=sqr36=6
the equatuon of a circle
(x-a)^2+(y-b)^2=r^2
(x-(-3))^2+(y-1)^2=6^2
(x+3)^2+(y-1)^2=36
x^2+6x*9+y^2-2y+1=36
x^2+y^2+6x-2y+9+1-36=0
x^2+y^2+6x-2y-26=0

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GHANA ANSWERS
2)

x+1/3x^2-x-2

3x^2-x-2/-6(-3 2)

3x^2-3x+2x-2

3x(x-1)+2(x-1)

(3x+2)(x-1)

x+1/(3x+2)(x-1)=A/3x+2+B/x-1

x+1/(3x+2)(x-1)=A(x-1)+B(3x+2)/(3x+2)(x-1)

x+1=A(x-1)+B(3x+2)

3x+2=0

x=-2/3

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8a)
60,56,70,63,50,72,65,60
mean=£x/n=60+56+70+63+50+72+65+60/8
mean=62
8b)
variance=£(x-x^-)^2/n
=(62-60)^2+(62-56)^2+(62-70)^2+(62-63)^2+(62-50)^2+(62-72)^2+(62-63)^2+(62-60)^2/8
=362/8=45.25
SD=sqr variane =sqr45.25=6.73
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