F/maths Obj
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FURTHER MATHS THEORY
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)
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 = sqr 45 . 25 = 6 . 73
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
10a)
i ) ( x ^2-1 ) ( x +2 )= 0
( x -1 ) ( x +1 ) ( x +2 )
x =1, or – 1 or -2
ii ) 2 x -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
(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
11a)
Given
F(x)=§(4x-x^2)dx
F(x)=2x^2-x^3/3+k
F(3)=2(3)^2-(3)^3/3+k=21
18-9+k=21
K=21-9
K=12
Therefore
F(x)=x³/3+2x²+12
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
7)
m1=3
u1=8m/s
m2=?
u2=5m/s
v=6m/s
m1u1+m1u2=(m1+m2)v
3*8+m2*5=(3+m2)6
24+5m2=18+6m2
24-18=6m2-5m2
m2=6
(7b)
m2u2-m1u1=V(m1+m2)
6*5-3*8=V(3+6)
30-24=9v
9v=6
v=6/9
v=0.67m/
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
(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)
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