NIGERIA 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)
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2.
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3a)
If f(x+2)=6x^2+5x-8)
To find f(5)
Therefore f(x+2)=f(5)
where x+2=5
x=5-2
x=3
therefore f(5)=6(3)^2+5(3)-8
=6(9)+15-8
=54+7
=61
3b)
(7root2+3root3)/(4root2-2roo3)
(4root2+2root3)/(4root2+2root3)
(24 2+14root6+12root6+6 3)/(16 2+8root6-8root6-4*3)
(48+26root6+18)/(32-12)
=(66+26root6)/20
=66/20+(26root6/20)
=33/10+(13root6/10)
=3.3+1.3root6
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10a)
i) (x^2-1) (x+2)=0
(x-1) (x+1) (x+2)
x=1, or -1 or -2
ii) 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=CA, A=-1/6
Let x+2=0 x=-2
2(-2)-3=C(-2-1)(-2+1)
-7=3C, C=-7/3
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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
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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
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)
===============
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
KEEP REFRESHING FOR THE ANSWERS
MORE LOADING…
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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