13ai)
M=2
P=5
C=3
total=10
If the books of the same subject are to standtogether
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)

10ai)
(x^2-1) (x 2)=0
(x-1) (x 1) (x 2)
x=1, or -1 or -2
10aii)
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=6A, A=-1/6
Let x 2=0 x=-2
2(-2)-3=C(-2-1)(-2 1)
-7=3C, C=-7/3

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

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

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