Cambridge Additional Mathematics

(1 -2),

(3 0),

(0 -3),

x

y y=x -3x 2

y=x-3

O

y=ex- 1

y=2- 2 e-x

(0 0),

(21)ln¡,

y

x

y=-1

y=2

O

y

x

y=x 2

y=4x¡¡ 2

(_ _ 1)Qw,¡

O

40

50

t(sec)

v(t) = 50 - 10e-0.5t

v(t) (ms-1)

O

10

2 4 6 8 10 12 14 16 18 20

20

30

40

velocity km h()-1

t(mins)

O

50

y

3 x

3

-3

-3

O

x+y=9 22

Answers 501

EXERCISE 16A

1a 30 units^2 b^92 units^2 c^272 units^2 d 2 units^2

2a^13 units^2 b 2 units^2 c 6334 units^2

d(e¡1)units^2 e 2056 units^2 f 18 units^2

g^12 units^2 h 412 units^2 i (2e¡

2

e

)units^2

3 23 units^2

EXERCISE 16B

1a 412 units^2 b (1 +e¡^2 )units^2 c 1275 units^2

d 2 units^2 e 214 units^2 f(¼ 2 ¡1)units^2

21023 units^2

3a b(1,¡2)and

(3,0)

c 113 units^2

4 13 units^2

5a

b(0,0)and(ln 2,1)

cenclosed area= 3 ln 2¡2(¼ 0 :0794)units^2

6 12 units^2

7

enclosed area= 121 units^2

8aRearranging

x^2 +y^2 =9 gives

y=§

p

9 ¡x^2.

The upper half has

y> 0 ,so

y=

p

9 ¡x^2.

b^94 ¼

9a 4012 units^2 b 8 units^2 c 8 units^2

10 aC 1 isy= sinx,C 2 is y= 3 sinx b 4 units^2

11 a

R 5

3 f(x)dx=¡(area betweenx=3andx=5)

b

R 3

1 f(x)dx¡

R 5

3 f(x)dx+

R 7

5 f(x)dx

12 aC 1 isy=^12 +^12 cos(2x),C 2 isy= cos(2x)

bA(0,1),B(¼ 4 ,0),C(¼ 2 ,0),D(^34 ¼,0),E(¼,1)

cArea=

R¼

0 (

1

2 +

1

2 cos(2x)¡cos(2x))dx

13 If h(x)> 0 ona 6 x 6 b, the area betweeny=h(x)and

thex-axis is

Rb

ah(x)dx.Ifh(x)<^0 on a^6 x^6 b, the

area betweeny=h(x)and thex-axis is

Rb

a¡h(x)dx.

) the area between y=h(x) and thex-axis on a 6 x 6 b

is

Rb

ajh(x)jdx.

Letting h(x)=f(x)¡g(x), the area between

y=f(x)¡g(x) and thex-axis y=0on a 6 x 6 bis

Rb

ajf(x)¡g(x)jdx.

Equivalently, the area between y=f(x) andy=g(x) on

a 6 x 6 bis

Rb

ajf(x)¡g(x)jdx.

14 b¼ 1 : 3104 15 a=

p

3

EXERCISE 16C.1

1110 m

2a itravelling forwards

ii travelling backwards (opposite direction)

b 16 km c 8 km from starting point (on positive side)

3a b 9 : 75 km

EXERCISE 16C.2

1as(t)=t¡t^2 +2cm b^12 cm c 0 cm

2as(t)=^13 t^3 ¡^12 t^2 ¡ 2 tcm b 516 cm

c 112 cm left of its starting point

3

p3+2

4 m

4as(t)=32t+2t^2 +16m

bno change of direction

so displacement=s(t 1 )¡s(0) =

Rt 1

0 (32 + 4t)dt

cacceleration=4ms¡^2

5a 41 units b 34 units 6b 2 m

7a 40 ms¡^1 b 47 : 8 ms¡^1 c 1 : 39 seconds

dast!1,v(t)! 50 from below

ea(t)=5e¡^0 :^5 t and asex> 0 for allx,

a(t)> 0 for allt.

fg¼ 134 : 5 m

8av(t)=¡

1

(t+1)^2

+1ms¡^1

bs(t)=^1

t+1

+t¡ 1 m

cyan magenta yellow black

(^05255075950525507595)
100 100
(^05255075950525507595)
100 100 IB HL OPT
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Y:\HAESE\CAM4037\CamAdd_AN\501CamAdd_AN.cdr Tuesday, 8 April 2014 8:40:43 AM BRIAN