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Further Maths OBJ:

1-10: BCADDDDDBB

11-20: DBABCCDDCC

21-30: BCABBAABAB

31-40: CBADBDDCAB

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(1)

X-3|2 2|+4|5 2 |+3|5 2|= -24

| -4 6-x| |2 6-x| |2 -4|

X-3(12-2x+8)+4(30-5x-4)+3(-20-4) = -24

X-3(20-2x)+4(26-5x)+3(-24)= -24

20x – 2x² – 60 + 6x + 104 – 20x – 72 = -24

-2x² + 6x – 60+104-72+24 = 0

-2x² + 6x – 4 = 0

Divide through by -2

X² – 3X + 2 = 0

X² – 2X – X + 2= 0

X(X-2)-1(X-2)=0

(X – 1)(X – 2)=0

X=1 OR X=2

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(2)

Log 3x – 3logx³+ 2 = 0

Log 3x – 3log3³/log3x + 2 = 0

Log3x – 3/log3x + 2 = 0

P – 3/p + 2 = 0 where P = log3x

P² – 3 + 2p = 0

P² + 2p – 3 = 0

(p + 3) ( p – 1) = 0

P = -3 OR p = 1

But log3x = p

when p = -3

Log3x = -3

X = 3-³ = 1/27

And when P = 1

Log3x = 1

X = 3¹ = 3

And X = 1/27 OR 3

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(3a)

U= x-2 ; hence X = u+2

Therefore :

X^3+5/(x-2)^4 = (u+2)^3+5/ (u+2-2)^4

=(u+2)^3+5/u^4

(3b)

(u+2)^3+5/u^4 = (u+2)^3/u^4 + 5/u^4

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(4)

Sn = n/2[2a+(n-1)d]

S12 = 12/6[2a + 11d]

S12= 6[2a + 11d] = 168

2a + 11d = 28——-(1)

Also

T3 = a + 2d = 7———(2)

Multiplying 1 by 2

2a + 4d = 14

Eqn 1 minus Eqn 3 gives

7d = 14

d = 2

Putting this in eqn 2

a + 2(2) = 7

a + (4) = 7

a = 7 – 4

a = 3

Common difference = 2

First term = 3

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(5)

TABULATE

cooking oil type|a|b|c|d|e|f|g|h

x|8|5|1|7|2|6|3|4

y|6|3|4|8|5|7|1|8

d|2|2|-3|-1|-3|-1|2|2

d²|4|4|9|1|9|1|4|4

£d² = 4+4+9+1+9+1+4+4 = 36

r = 1- 6£d²/n(n²-1)

= 1 – 6×36/8(8²-1)

= 1 – 216/504

= 1 – 0.43

= 0.57

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(6)

It follow that:

P(n) = 1/3 and p(k)¹ = 1 – 1/3 = 2/3

P(T) = 1/5 and P(T)¹ = 1- 1/5 = 4/5

Hence,

probability that only one if the them will be solve the questions will be:

= (1/3 × 4/5) + (1/5 × 2/3)

= 4/15 + 2/15

= 6/15

= 2/5

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(7)

m = 3i – 2j ; n = 2i + 3j ; p = i + 6j

Therefore:

4(3i – 2j) +2(2i +3j) -3(-i + 6j)

12i – 8j + 4i + 6j + 3i – 18j

12i + 4i + 3i – 8j + 6j – 18j

19i = 20j

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(8)

m1 = 20kg

u1 = 8ms-1

m2 = 30kg

u2 = 50ms-1

(a)

In same direction

m1u1 + m2u2 = (m1+m2)v

20 × 80 + 30 × 50 = (20+30)v

1600+1500 = 50v

3100/50 = 50/50

V = 62ms-1

(b)

In opposite direction

m1u1 – m2u2 = (m1 + u2)V

20×30 – 30×50 = (20+30)V

1600 – 1500 = 50V

100/50 = 50V/50

V = 2m/s

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(9)

Draw the pie chart diagram

radius is the same hence

(y – 2)² + (X – 3)² = (y + 4)² + (X – 5)² = (y + 1)² + (X + 2)²

Taking the first pair

Y² – 4y + 4 + x² – 6x + 9 = y² + 8y + 16 + x² – 10x + 25

– 6x – 4y + 13 = -10x+8y+41

4x – 12y = 28

X = 3y = 7——–(1)

Taking the second pair

Y² + 3y + 16+x² – 10x+25 = y²+ 2y + 1 + x²+ 4x + 4

-10x + 3y + 41 = 4x + 2y + 5

14x – 6y = 36

7x – 3y = 18———(2)

Eqn 2 minus Eqn 1

6x = 11

X = 11/6

Put this into Eqn (1)

11/6 – 3y = 7

3y = 11/6 – 7

3y = 11 – 42/6

3y = -31/6

y = -31/18

(a) coordinates of centre is (11/6, -31/18)

(b) Radius r = √(y-2)²+(x-3)²

r = √-31/18 -2)² + 11/6 – 3)²

r= √13.85 + 1.36

r = 3.9

(c) (X – 11/6)² + (y + 31/18)² = 3.9²

(X – 11/16)² + (y + 31/18)² = 15.21

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(15)

(a)

DRAW THE DIAGRAM

(b)

(2.2) + 80(0.6) = 300(1.2) + 100 (3)

2.2B + 48 = 360 + 300

2.B + 48 = 660

2.2B = 660 – 48

2.2B = 612

Reation at B = 278.18N

Taking moment about B

80(2.1) + 300(1) = 2.2A + 100 (0.8)

224 + 300 = 2.3A + 80

524 – 80 = 2.2A

A = 444/2.2

Reaction a+ A = 201.82N

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