FURTHER MATHS OBJ
1-10. CBADCBADBB
11-20.DCBDCCDBCC
21-30. ACDCAABBCB
31-40. CDCDDBCAAD

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)

VERIFIED ANSWERS
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)
==================.

FURTHER MATH 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)

================

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(check whatsapp inbox for diagram)

14ii)
Using lami’s theory
T1/sin60=T2/sin30
48N/sin60=T2/sin30
48N/0.8660=T2/0.5
0.5(4 /0.8660=T2(0.8660)/0.86
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

======================

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

=================

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

=================
.
9a)X+6/(x+1)^2 = A/x+1 + B/
(x+1)² + C/(x+1)³
X+6/ (x+1)³ =A(x+1)² + B(x+1)
+ C / (x+1)³
X+6 =A (x+1)² + B(x+1) + C
Let X+1=0 , X=-1
-1+6= A (-1+1)² + B(-1+1) +C
5= C
Therefore:- C=5
X+6= A(X2+2x+1) + Bx + B +
C
X+6= Ax²+2Ax+A+Bx+B+C
comparing the coefficient of X²
A=0
Comparing the coefficient of X
1=2A+B
1=2(0)+B
B=1
x+6/(x+3)² = o/x+1 + 1/(x+1)²
+ 5(x+1)³
1/(x+1)² + 5/(x+1)³

.
9b)
S²1 1/(x+1)² + 5/(x+1)³ DX
1n (x+1)³ + 5 1n (X+1)⁴ |²1
1n 3(3) + 5 1n(3)⁴ – 1n +2³ – 5
1n +2⁴
3.2958+21.9722-2.0794-13.862
=9.3297

.
10a)
3x^2+x-2 <= 0
3x^2+3x-2x-2 <= 0
3x(x+1) -2 (x+1) <= 0
(3x-2)(x+1) <= 0
3x-2 <= 0 or 8+1 <= 0
3x <= 2 or x <= -1
X<= 2/3
Therefore range of value is
-1 >= x <= 2/3

.
15a)
at max height
V=0m/s
g=10m/s^2
V^2=u^2-2gh
0^2=30^2-2 10H
0=900-20H
20H=900
H=900/20
H=45m
.

15b)
time taken to get to max
height
V=u-gt
0=30-10t
10t=30
t=30/10
t=3secs
Timetaken to return=2t
=2 3=6secs
.

15c)
H=40m
H=ut-1/2gt^2=40
30t-1/210t^2=40
30t-5t^2=40
5t^2-30t+40=0
t^2-2t-4t+8=0
(t^2-2t)-(4t+ =0
t(t-2)-4(t-2)=0
(t-2)(t-4)=0
t-2=0 or t-4=0
t=2secs or t=4secs
.

11a)
Kp2=72
K!/(k-2)!=72
K(k-1)(K-2)!/(K-2)!=72
K^2-K=72
K^2-K-72=0
K^2-9k+8k-72=0
K(K-9)+8(k-9)=0
(K+ (K-9)=0
k=-8,K=9
We consider positive value of
K=9
.

11b)
The equation
2cos^2tita-5costita=3
Let cos tita=x
2x^2-5x=3
using quadratic formular
a=2,b=-5,c=-3
5+_root(25+24)/4
=5+_root(49)/4
=(5+_7)/4
=(5+7)4=3 or
(5-7)/4=-2/4=-1/2
since x cos tita
cos tita=-0.5
tita=cos^-1(-0.5)
tita=120degrees
——————
.

10a)
(1+x)^7
7Co(1)^7(x)^0 + 7C1(1)^6(x)
+
7C2(1)^5(x)^2 +
7C3(1)^4(x)^3 +
7C4(1)^3(x)^4 +
7C5(1)^2(x)^5 + 7C6(1)
(x)^6 + 7C7(1)^0(x)^7
= 1+7x + 21x^2+35x^3 +
35x^4+21x^5 +
7x^6+x^7
.
(10b)
35 21 7
a=35
d=T2-T1
=21-35
.
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