math ib sl answer sheet
TRANSCRIPT
-
8/10/2019 Math IB SL Answer Sheet
1/68
AnswersChapter 1Skills check
1 a yA
,.F,..,
v
,.,••
-s - 11.0-..
-a4
A c
• X
E 8
b A O, 2), B l , 0), C - 1, 0),D O, 0), E 2 , 1), F -2, -2) ,G(3, -1 , H -1, 1)
2 a 34 b 82
c 16 d _ _ 2
3 a 4
4 a yA
........-
A- J: ·
60
b -2
11 j)
J
c 10
.
v ,.R - b r.O 'I
X
I " vJ
b yA
.v
A
'"' ''
''I :
-
'02 1c y
6
A
I
' f,-,1 J\ 17
b h 1\. 1.. 0 ,/1'. v-
X
fL.
Answers
5 a x2 + 9x = 20b x2 - x + 3
c x2 + x 20
Investigation handshakes
a 6b
c
Numberof people
y
-
8/10/2019 Math IB SL Answer Sheet
2/68
xercise l C
1 Horizontal asymptote: y = 0
2 Horizontal asymptote: y =0 ,Vertical asymptote: x = 0
3 Horizontal asymptote: 0,Vertical asymptote: x = - 1
4 Horizontal asymptote: y = 2,Vertical asymptote: x = - 2
5 Horizo ntal asymptote: y = 2,Vertica l asymptote: x = 1
6 Horizo nt a l asymptote: y = 0,Vertica l asymptot e: x = - 3
xercise 1
1 Function , domain {2 3, 4, 5,6, 7, 8, 9, 1 0 } range {1 , 3, 6,10 , 15 , 21 , 28 , 36 , 45 }.
2 a domain {x: -4 < x < 4},range {y: 0 $ y $ 4}
b domain {x: - 1 < x < 5},range{y : 0 $ y $ 4}
C domain {x ; - oo X oo},range {y: 0 $ y < oo}
d domain{x:-oo < x < - 2< x < oo},range{y: - oo < y $ 3 4 $y < 8}
e domain{x:-5 < x < 5},range{y: -3 < y < 4}
f domain {x:-oo < x < oo}range {y: - 1 < y < I }
g domain {x: -2 < x < 2},range{y: -2 < y < 2}
h domain {x: - oo < x < oo},range {y: - oo < x < oo}
•domain xe R , x t 1, rangey e y;t: 0
3 a xe IR y e lRY.
8
4
by
16-
14-
•
- -·12- L
-10- _
8- -1\ 14-
- \ v... /-4 -2 0 2
cy
2
15
10
a-j -2 0
d X E JR, y Ey
5
-2 -1 0
- 10
-15
ey
10 - .. .-- .. .....v
8
6
4
2
--
.
t
v
·-
v/ v - -
v-t- - t
4
....
0 20 40 60 80 100 X
X
X
f X 4 , y > 0y
10
8
2
100 80 60 40 20 0 X
g xeJ Rx;t; O, y e lR y;t: Oy
,-1-
.
4
I 8 4 8-4
8
h X E JR, y > 0
y87-6 f-
543
-2 -1 0 1 2
i x e JR x ; t : - 2 , y e lR y:;t;O,y
0-r
l 8- 1-t- 6-
- 4- -\ -
-8 2 6 X-4 -t-6-
_Q
' 0
Answers
X
-
8/10/2019 Math IB SL Answer Sheet
3/68
j X E JR X: 2, y E JR y :# 1
y
1 B
_Q
'-- 0 ,.G. 10 20-..
4cv
0
)o-
X
k XE x:t: - 3 ,y E y:t: -6y
4/ v
/ vb 11 0 / vP'
'./ .....8
l X E JR, 0 < y $ 2y
;; \1'•• \
v •-f- ' • V ....
/
X
--
--5 -4 -3 -2 - 1 0 1 2 3 4 5 6 X
xercise l
• •• • •• 11 a I 5 II -5 Ill - 1 -2• - 2 a - 2V v
' 1b • •• •••I 21 II - 9 Ill 1 -2• 0 3aV v• 7 .. - 3 • •• 1c I II - Ill -
4 4 810V v - a4
d.
19•• -1 • •• 6II Ill
.5 2a + 5V v
2_ _51••
11• ••e I II Ill
4• 2 a2 + 2V v
2 a a2 - 4 b a 2 + l Oa + 21c a2 - 2a - 3 d a 4 - 4a 2
e 2 1 - lOa+ a2
3 a 2 b 11 c 2
4 a 1- -9
b x = 6, denominato r = 0and h (x) undefined .
Answers
5 a 125b The volume of a cube of
side 5.
6 • 1.. 5a I - - II
9 4... lIll - - iv 0.
2
b. - 4
..- 11 ... - 67II Ill
. -697 - 6 997V v•
- 6 9997I
c T he value of g(x) is gettingincreasingly smaller as xapproaches 2.
d 2e asymptote at x = 2.
yr , \
18r-....
0 A 0 \ 1 ) 1 &-;v
ir.
7 a - 9 m s - 1c 91 m s - 1
b 7 m s - 1
d 3s
8 a / ( 2 + 2h) - / ( 2 + h)h
b / (3 + 2h) - / (3 + h)h
xercise I F
' x
1 a 12 b 3 c - 15d 3x + 3 e 13 f 16
-17 h 3x + 1 • 18I•
38 k 3x2 + 6Jl 9x2 + 2 m 12 n 180 r+ 3 p x2 + x + 3
2 a 3 b 0 c - 12d -1 e - 5 f 48
g 3 - 4 x + x2
h - x
+ x2
3 a x 2 + 4x + 4 b 254 a 5x 2 + 5 b 5x 2 + 15 a x 2 - 8x + 19 b x 2 - 1
c 2.5
6 (ros) (x) = x 2 - 4 , x E JR, y > - 4
xerciselG
1 b , c
2 a y
l /0 // v
Jt....
4/'C
v , I..... ,(j -R ' X/ I IA
//
, A'
b y
l o/
/.....I /
//
'I // /-
/ /
R v . ........, v
l
4
X
v/
/
lo/ I , ..../
c y
i' /r-.... /........ \ /......... /........ \ //
/ \ .........../ .........-R kt ,(j \ .
........ X- . - \A/ \
/l > \
, i\
d y........ /
l > ......... /
//-
/ \, // \/
/ I"-./
/ /
-R - IL ,(jI - XI / I4 v - --//
// 8
/
e y....v
.A - /_-r/
/ .- ', ;
v ;;/ I,I l x4 -B - ,.:::[ 0 2 ' 4
... '--- A
-
8/10/2019 Math IB SL Answer Sheet
4/68
f y
-3 ; .2 1....-
Exercise l1 a i -2 an d 1
ii _ _ an d - 32
••• •Ill X IV X
b They are inverses of eachother.
2 a x+ 1 b3
c 4(x - 5) d (x + 3) 3
1ex+2
3xg l x
3 a 1 x
f Jx 3
hx+2Sx
b X C 1X
4 a 1 b -5 c . l20
S l +2xx-r6 a-c
y
8-
4- j .
/ r / -- I/ . ,.-- •- - I
,...-L
I _,J .- ' 0 2 4 X
-4- I I
d / x): X E IR y > 0/ - 1 x): x > 0, y E R
7 g- 1(x) = r . Th e range of g(x)is x 0 so the domain of
8
g - 1(x) is x 0. The domain off x ) is x E R so g - 1(x) # f x)
f x ) = mx + c thenf x)=l . .x -.£.m m
tn x , ,= not -1 so notperpendicular.
Invest igation functions Exercise
1 Changi ng the constant term 1 atranslates y = x along they-axis.
y
e x--1--1-J
2 Changing the x-coefficienta lt ers the gradient of the line.
y
X
V: l I I dJ3 y = Ix + hI is a translation of
- h a long the x-axisy
Y l x 3 7
X
-2
-3
b
c
d
-8 -6
I
y
8 ....._- ...
6
-4
y
6
4
-6
y6
- 4 t -4
4 2 4
-4-6-8
6 8
Answers
X
-
8/10/2019 Math IB SL Answer Sheet
5/68
g y
-15
2 g(x) = x) + 2h(x) = x ) - 4
1q(x) = 2 f x)
3 q(x) = f x + 4 ) - 2s x) = f x + 4)t(x) = f x - 2)
10 15 X
4 a D omain -1 < x < 7, range- 4 < y < 6
y86
b D omain - 3 ::; x ; 1, rangeO < y < 5
y
4
b
c
d
e
g
-4 f -
g
f -4 -
g
f -
y
64
g f 2
0-8 -4 -2
-4
y
4
2
-4
y
4
-4
2
2 4 X
4 X
4 X
4 6 8 10 X
-6 -4 -2 0 2 x 6 a Reflection in x-axis.-2
5 a y
4
f
g -4 2 4 X
-4
Answers
b Horizontal trans lation 3units .
c Ve rt ical st retch SF2,reflection x-axis, vertical
translation of 5 units.7 a b
yc...
A I....
/r )L _L2-
X)v 1 II
4 - i -2 - l1 0,
:2 3 X
Review exercise non-GDC
1 a 4a - 13 b 2 ; x
2 a 2x2 - l 5x + 28
b - 2x2 + 9
3 a 2x-173
b 2 3
4 f -1
(x) = - 5 x - 5y
' - 'c.
5 a
6 a
b
I
4
....(1 -
'l
L
A
x - 53
-- )
/
/
I
' n
6W
y
4 rr _ 1 )' '2 \
/
-- /
/
:2
2 \ \1
I + \
yA
•-3
1...
_ 1 /
-... l \o1
b . i l - 2
/
X
-
I/ ......
,..'_... . / r / f
X
,111 v X-I-,. I1
-
'>' '
7 a D omain x E IR, y > 0
b Domain x E JR. x * 3,Rangey E JR, y * 0
8 a f x )=2- l - 3x - 9 +2
1 ( x - 5 lb f x ) = - - 3- 3 1- 14 )
9 a Inverse function graph isthe reflection in y = x.
-
8/10/2019 Math IB SL Answer Sheet
6/68
b yA•
1 :2/
13 0 P_ '> In ,. .... X
lA v\0 -, -;: • J
I A
•
10 a - 2 b -13
c J - c x)=vx 2 3
l l a yI
10,J
A
1\
21
\ v-B -Q -1 0 , . J X
b P is ( 4, 1)
12 a (jog) x) = 3x + 6
b j -l x) = x g- 1(x) = x - 23
j l J2) = 1:= 4g - 1(12) = 12 - 2 = 10
j l (12) + g-l (12) = 4 + 101(12) + g 1 (12) = 14
13 a3(2x- l )
(hog)(x) = (2x -1 ) -2
_ 6 x - 3-
2x - 3
bl
x= -2
eview exercise GD
3 Domain x E JR, x :;t: - 2, rangey E JR, x:;t:. 0
4 a
y
1210
86
42
0-2 - t2-4-6-8
y
8
2
1 2 X
-3 -2 1 -20 1 2 X
-4
-6-8
b x-intercept -1.5,y-intercept 3.
5 a y
6
4
-6 -4 -2 0 2 4 6 X
1 Domain: x > -2 , range: y > 0 b 0
2 Domain: x E JR, range: y > -4
y
16
1412
10864
2
- 4 -3 -2 - 0 4 X
c Domain x E JR x :;t: lR,rangey > 0
6 a x = - 2 , y = 2
b y6
2
8
4f
-12 -8 -4 0 4 8 12 X-4
c (2.5, 0), (0, -2.5)
7 a y6
4
3 X
-4
-6
b X = +Ji
8 a 1x 3
b y...
j V
4, .I'
v11/ h 0
Jv
c 1.67
9 y·8
X -1
,v
j = - ' \-- - - - - 1- -p - f:l -B - 0
I f \V
l O a f - (x)= x 23
'' X
717
-- ,.X
b (g - 1o/ x) = 3 x - 2) + 3= 3x + 1
C ( / - 0 = x - + 2 = 31
x - 1 =3x+13
x - 1 = 3(3x + 1)x - 1 = 9 x + 3
8 x = - 41x = - -2
Answers
-
8/10/2019 Math IB SL Answer Sheet
7/68
d yI ' > '....
0
-- 4; _ - -.. - --
_
6 - 4 - \A
e x = 3, y = 3
Chapter 2Skills check
1 a a= 6
b X =+JSc n = -11
2 a 2k k - 5)
I
I1I
I \II
,_ .;. --:x - 3IIII
b 7a 2a 2 + a 7)
c 2x + 3) x+2y)
d Sa - b) a - 2)
e n + l)(n + 3)
f (2x - 3)(x + 1)
g m + 6) m- 6)
h Sx + 9y) 5x- 9y)
Exercise 2A1 a 1 2
b -8 , 7
c 5, 6
d -5 5'
e - 8 6'
f -3
2 4 1a3 2
b 4-2 - 5
c 5-1 - 2
d I9
- 2 2
e 2- 4 - - 3
f3 4
- 2 3
Answ rs
- - .
--
X
Exercise 28
1 a -5, 4
b - 23
3c2
d - 2 252
e - 9 , 41f 14
2 - 3 or 4
23 - or 3
5
Investigation perfectsquare trinomials
1 -5
2 -3
3 -7
4 4
5 9
6 10
Exercise 2C
1 - 4 ± M
2 5±J372
3 3± 2J2
4 - 7±- 652
5 l + J7
6 -l + J U
2
Exercise 2
1 - 3 + 2J3
2 1+ J 2
3
4- 3+-fi9
4
35 22
6 - 2+3J610
Exercise 2E
1 - 9 + J I938
2 4- 2 - 3
3 1- 15
4 3 ± J 55 no soluti on
6- 5+2.Jl0
3
7 3±. f 0
4
89 ±-Jl i3
4
9- 9 + J l 2 9
x
4
10 3+m4
Exercise 2F
1 18,32
2 24m, 11 tn
3 10
4 18 em, 21 em
5 2. 99 seconds
Investigation roots of
quadratic equations
ba 42
1c5
2 a -7 2 b'
3+ J89c10
4± F o3
3 a No solution
b No solution
c No solution
Exercise2G
1 a 3 7; two different real roots
b 8; two different real roots
c - 79; no real roots
d 0; two equal real roots
e - 23; no real roots
f - 800; no real roots
-
8/10/2019 Math IB SL Answer Sheet
8/68
2 a p < 4 b p < 3.125
c i I>4v 2 d2iP >-3
3 a k = 25 b k = 1.125
c k=± J lS d k = 0, - 0.754 a m > 9 b 2 < m < 2
c 16m> d m > 123
5 O < q < I
Investigation graphs o fquadratic functions
a Discriminant, 6. = 29
y
b 6 . = 12y
c 6. = 2 4
d 6. = 7 1y
0
X
X
y
0
X
X
e 6. = 0y
X
f 6. = 0
y
X
g 6. = 33
y
X
h 6. = 37y
X
f b2 - 4ac > 0, graph cuts
x-axis twice; if b2 - 4ac = 0,graph is tangential to x-axis; i f
b2 4ac < 0, graph do es not
intersect x-axis.
Exercise H
1 a x = - 4; 0, 5)
b x=3 ; (0 , -3 )
C X - 1; 0, 6)
d X S. (0 9)3
2 a (7,-2) ;(0,47)
b ( - 5, I) ; 0, 26)
c (1 ,6 ) ;(0,10)
d ( -2 , -7) ; (0 ,5)
3 a f x) = x + 5) 2 - 31y
0 0, -6) X
(-5,-31)
b f x) = x - 2.5) 2 - 4.25
y
,2)
0 X
2.5, -4.25)
C f x) = 3 x - 1)2 + 4y
0, 7)
1, 4)
0 X
d J x) = -2 x 2) 2 + 5
0
0, -3
2, 5)
X
Answers
-
8/10/2019 Math IB SL Answer Sheet
9/68
Exercise 2
1 a -3 ,0) ; 7 , 0); 0,-21)
2
b (4,0); (5,0); 0,40)
c -2 , 0); -1 , 0); 0 , -6)
d -6 ,0) ; 2 ,0) ; 0 , -60)
a y = x - 8) x + 1)Y
( 1 0) (8,0}
X
(-0, - 8}
b y = x - 3)(x - 5)
0, 15)
(3,0} (5,0}0 X
c y = 2 x + l ) x - 2.5)y
(0,5)
(-1, 0} (2.5,0)0 X
d y = 5 x + 2 x - ;
Y
( 2, 0) :.o)0 X
(0, -8
Answers
3 a y = x + 3) 2 - 25;
y = x + 8) x - 2)
Y
( 8 , 0) (2, 0)
0 X
0, -16)
(-3, -2 5)
b y = - x + 2) 2 + 25;
y = - x + 7) x- 3)
y(-2,25)
(0,21}
( 7 0) (3,0)
0 X
c y = - 0 .5 x - 3.5) 2 + 3.125;
y = - 0 x - 1) x - 6)
Y
3.5, 3.125)
0 X
(0, -3
d y = 4 x - 2.25) 2 - 12.25;y = 4 x - 0.5) x- 4)
Y
(0,8)
(0.5,0}0
4 a i 0
b x= 3
(4,0}
X
(2.25, - 12.25}
ii 6
c (3, - 1 8)
5 a f o g) x) = x 2) 2 + 3
b (2, 3)
c h x) = x 2 - 14x + 50
d 50
Exercise 2J
1 y = x 2 - 4x + 52 y = x 2 - 4 x 123 y = 3 x 2 - 6x + 5
y = 2.x 2 -32 2
5 y = 2x 2 + 7x + 46 y = -0 .4x 2 + 8x7 y = - x 2 + 4x + 218 y = 12x 2 - 12x + 3
Exercise K
1 a 14.5 metres
b 1.42 seconds
2 14 em, 18 em
3 a 1 0 x
c 50 cm 2
4 12.1 em
5 17m, 46 m
6 7, 9, 11
l + J572
8 28.125m 2
9 60 km, 70 h- 2
10 6 hours
Review exerci se non GDC
1 a -6, 2
b 8
7c - - 13
d 3, 4
e - 1±Ji3
f 7±J136
2 a 4
b - 4, 1c x =-1 .5
d -1 . 5
-
8/10/2019 Math IB SL Answer Sheet
10/68
3 a 5 1 1 11 92 ba -b - 2 5 32 32
4 -3 , - 6 3•
0.21••
0.33a I II
1 b 2525 a A = {3, 6, 9, 12, 15}
b -2 4 a 0.27 B = {1, 2, 3, 5, 6, 10, 15}
c 12 b No - the frequencies are4 7 8 1113 14
very different
5J3 A
c 4506 a f x ) = 2 x + 3 2 - 13 2 6 9 12 3 15 12 5 105 a b 0
b (1 - 5 11
y =I .x - x - 12c 5
7 1126 0.2 • 1 •• 2
Review exercise GD c I II -5 51 137 a b1 0 .907, 2.57 -a 2 40
b -4.35, 0.345 6 Ac - 2.58, 0.581 Exercise 8
d -1.82, 0.220 133 24
bl br
2 a 2 0 m
b 3 1 5 m 6 4 10
c 3.06 s 550
d 4.07 s 15 c3 21, 68 4
4 a= 0.4, b = 3, c = 2 7a 0.33 b 0.24
5 60 km h- 1 2 c 0.3Fr m
Chapter7
85 Exercise C
Skills check 151a
5250
1 4 b 1_i_a -7 35 8 b
53
c 2 d 22 25100
15 27 3 299A G c
19 3 500e f -27 7 27 11 2 a -
2 a 0.625 b 0.7 5
c 0.42 d 0.16 b 3-1 5e 15 f 4.84 Six have both activities. 1c -g 0.0096
6 112
a b -25 25 3 17-
Exercise A 4 20
1G p
1 b 1 4 9a - 4 a b2 4 13 2611 7 9
1 3c d - 2 14 4 c d -3 5
13 2e -
8 Five play neither. 5 a 0.5 b 0.5
nswer s
-
8/10/2019 Math IB SL Answer Sheet
11/68
660
7 a 14
8 a 0.6
c 0.9
Exercise 30
1
2
3
a Nc Ne Ng Nyes57
89
4 a 212
c 13
60Exercise 3E
b 34
b 0.4
b y
d y
f N
b 4760
1 HH H , HHT, HTH, HTT, THH,THT, TTH, TTT
a b 32 8
c . :.4
2BLUE
1 2 3 4
1 (1, 1) (1, 2) (1 , 3) (1, 4)RED 2 (2, 1) (2, 2) (2 3) (2, 4)
3 (3 , 1) (3, 2) (3 3) (3, 4)
4 (4, 1) (4, 2) (4, 3) (4, 4)
3 b 3a - -8 81 d 9c -4 16
3
Box 1
1 2 3
2 (2, 1) (2, 2) (2, 3)
Box 2 3 (3, 1) (3, 2) (3, 3)
4 (4 , 1) (4, 2) (4, 3)
5 (5, 1) (5, 2) (5, 3)
1 b1
a - -6 33 d 5c4 12
e 2-3
Answers
4 First draw
Second
draw
a 16
13c -
18
5e -9
5 a 16
c 29
0
1
2
3
4
5
Exercise 3F
1
2
125
2169
3 64125
4 0.6375
0 1
(0, 0) (0, 1)
(1, 0) (1, 1)
(2 0) (2, 1)
(3 0) (3, 1)
(4, 0) (4, 1)
(5 , 0) (5, 1)
b 2336
d 1336
b 19
2
(0, 2)
(1, 2)
(2 2)
(3 2)
(4, 2)
(5, 2)
5 a P(B)=0 .2 ;P(BnC)=0 .16b Not independent
6
7
512
1
59049
8 1256
9 a 0.4b P(E) x P(F) = P E F )
c P E F ) ;e 0
d 0.6410 _
27
2712 a 0.27
c 0.07
13 0.18, 0.28
14 a 11296
b 0.63
b 1216
3 4 5
(0, 3) (0, 4) (0, 5)
(1, 3) (1, 4) (1, 5)
(2 , 3) (2 4) (2 5)
(3, 3) (3 4) (3 5)
(4, 3) (4 , 4) (4, 5)
(5 3) (5 , 4) (5 5)
15 Rolling a six on four throwsof one dice
16 a 0.729 b 0.271
Exercise 3G
1 12 take both subjects
a8
27
4c5
2 a 0.2
c 215
3 3948
4 1a -3
c 3- >
56-95
6 1-6
7 a 0
c 0.63
8 67.3
9 3447
10 a 1-1
c13
11 0.3
123
b 2327
b3
b2-5
d1-2
b 0
b 43s
-
8/10/2019 Math IB SL Answer Sheet
12/68
-
8/10/2019 Math IB SL Answer Sheet
13/68
ChapterSkills check
1 a 1 b 81128 256
c 1 X 10- 9
2a 5 b 3 c 4
3 y
4
Y= x2y= x - 2) 2
-4 -2 0 2 4 6 X
Inve stigation - folding paper
Number Number Thickness As thick
of folds of layers km) as a
0 1 1 X 10 -7Piece of
paper
1 2 2 X 10 - 7
2 4 4 X 10· 7Cred itcard
3 8 8 X 10 - 7
4 16 1.6 X 10 6
5 32 3 .2 X 1Q-6
6 64 6.4 X 10 -s
7 128 1.28 X 10 · 5 Textbook
8 256 2.56 X 1 0-5
9 512 5. 12 X 1 0- 5
3 a 13 fol ds
b 15 fo lds
4 113 000 OOO km
Exercise 4A
1 a x
b 6p6q2
c 1 3 3x y3
d x4y6
2 a x3
b a4
ac
4
d 2x y3
3 a x l b 27 t 6
c 3x6y4 d y 6
Answers
Exercise 4 8
1 a 3
d 4
2 a 1-8
d 116
Exercise 4C
1 a
d
2 a
8a 3
d
3c
o .
a•
b
Exercise 4
1 a x= 5
b
e
b
e
b
e
c x = 3 -1
e x = 3
25
a x= -2
c - 3X - -5
3 x= -6
Exercise E
1 a x = 3 b
1
5 c 16
4-9
I 1- c4 3
25 - ) 916 16
2 c q3x l
- 4-p .3
4
b x= - 2
d3
x
2
b x = - 4
d4
x=
5
x = 2 1c x = -4
I 3-d X= 2 e X= 3 3 f x = -42 a x = 8 b x= 625
1c x= d x = 64
2561
X= 32 f x = -16
13
1 bx= x =3125 216
27X= 512 d x= -64
Investigation - graphs ofexponential functions 1
yy = wx y = sx
0, 1)
0 X
Investigation - graphs ofexponential functions 2
(1 )Xy= -3
(0, 1)
0
Investigation - compoundinterest
Half-2.25
yearly
Quarterly (1• 1 2.44 1 4062512
Mo nth ly
(1• 12.61 303529022 ..
Week ly r1-r2 2. 692 596 954 44 ..31)5
Daily (1•-1l 2. 714567 48202 .365Hourly
1 . , •1•- ]760 2. 718126 690 63 ..
Every[l- 516 ) 2.7182 79215 4 ..min ute
Eve ry . 3 1536000
ll•1 2.71 8282 4 7254 ..
second 31536000
Exercise 4F
1 Curves of
a y10
8
6g x) = 2x + 3
2 f X) = 2x
-3 -2 -1 0 1 2 3 X
b y5
4
g x) = 3 x 3 f x) = 3x
-3 -2 -1 0 1 2 3 X
-
8/10/2019 Math IB SL Answer Sheet
14/68
-
8/10/2019 Math IB SL Answer Sheet
15/68
3 a 2 b 3 c 2d 3 e 1
Exercise 4N
1 a p + q b 3p c q - p
2
dq
2e 2q - P
2
6 x - 3y - 6z3 a 1 log x b 2 - 2Iogx
1 1 1c - + - l o a x d - 1- l ogx2 2 b 2
4 y = 3a - 4 5 -3 - 2log 3x
Exercise 4
1 a 2.81 b -1.21 c 0 .325
d 0.514 e 12.4
2 2:2
3 a YX
d 2x+ y
4 a2
1
b Xy
e x+yy
logxY= log4
c 2yX
f y - xX
- 1 0 1 2 3 4 X- 1
-2
b
-4 -3 -2
5 a 2b
y2
1
bb -
2
c - 2b d
Exercise 4P
b
4
logx2
log5
3 4 X
1 a 2.32 b 3.56 c -1.76d 0.425 e 0.229 f -3.64g 1.79 h -11 . 0
2 a 6.78 b 2.36c -3 .88 d 0.263e 0.526 f 2.04
g - 99 9
Answers
Exercise4Q
1 a 1.16 b 1.41 c -0.314
d 0 .0570 e 11.1
ln5002 a
ln64
1n3
3 0 b Xln3a x=
Exercise4R
I1 a x = b x = 1
5
ln2
c x= d x = J 27
e x = 1.62
Exercise45
1 a x = 8 3 b x = 14
C X 9532
2 a x = 9 b x= 6c no solutions
3 A = x 2x + 7 = 2x2 + 7xX 0.5
4 x = 4
5X
16Exercise 4T
1 a 450 x 1.032
b 10 years
2 a i 121 ii 195
b 9.6 days (10 days)
3 49.4 hours
4 a v8
20v = 9 + 29e - 0.063t
-3 -2 - 1 0 1 2 3 t
b 38ms- 1
d 10. 7 m s - 1
5 a = b = 3
e 17
Review exercise GD
1 3.52
2 a 0.548 b 0 .954
c - 1 .183 a 5 b 2
c 3.60 d 1,4e I 00, 1
1004 a f x) > 0, range of g(x) is all
real numbers
b They are 1- 1 functions ;
f c ) I l n . I 2 xx =2 x ,g - x )=e C f o g ) x ) = x 3;
g o f ) x ) = 3 x
d x = J 3
5 a 218 393 insects
b 8.66 days
Review exercise non·GD
1 0
2log( )
loge:3 4.5
5 a x= 7
C X 1,4
n6 a m
c 2m
b x = 2
d x = 67
b m - n
d m + nn
7 Shift one unit to the right,1
stretch factor - parallel to3
x-axis, shift 2 units up.
8 a
b- 1f 1 x)= - 1ogx
3
c
9 a = 2 b = 4
-
8/10/2019 Math IB SL Answer Sheet
16/68
Chapter 5Skills check1 a -8x+ 20 b 12x -1 8
c -.x3 7 x
2
d x 4 + 6x 3 + 9x 2
e x 3 + 5x 2 - 24xy
- - 2 6- x•O 3•)(
y = 4....
2-y•O
T ,.-3 -Q - 1 0 1 2 4 X
-2
y - - 3 -4
3 A is a h orizo ntal shift of 4units to the right. Function Ais y = x - 4) 3B is a vertical shift of 2 units
down . Function B is y = .x3- 2Investigation graphingproduct pairs
X 24 12 8 3 6 4 2 1y 1 2 3 8 4 6 2 24
y
24
18xy= 24
12
6
-2 O 2 4 6 X-6
- 8
-2
As y gets bigger, x gets smalJerand vice versa.The graph gets closer and closerto the axes as x- andy -values.mcrease.
Exercise SAl1 a2
d - 1
2g
3
2 a
d
2
13
I
3x
b . .3
e 32
h7
b 1
e
X
1
4y
c 13
f7
1c -
f
y
92x
g 5 h 3d • tIa 2 d
x lJ
x l
3 a 16 x - = 1 b 3 4- x - = 1
45
6 4 3
c3d 2c
a 4 b X•
0 .5..
a I II 0.05...
0.005•
Ill IV 0.0005b y gets sm a ller, nearer to
zero.
c24 .
y so 1t can never beX
zero.
d • 0.5..
I II 0.05
iii 0.005.
0.0005V
e x gets sma ller neare r to zero.f 24 .x = - so 1t can never be
y
zero.
Investigation graphs ofreciprocal functions1 a y
2
6
4
2
2 4 X
The nu m erator indicates thescale factor of the stretchparallel to they-axis.
y
-6 -4 -2 0-2
-4
-6
Changing the sign of thenumerator reflects the graphsof the origi na l functions inthe x-axis.
3 aX 0.25 0.4 0.5 1 2 4 8 10 16
f x} 16 10 8 4 2 1 0.5 0.4 0.25
I
b T he values of xandf(x) are the samenumbers but in reverseorder.
c d e
'-14 /- - ' 1- -
//
i7l/
... \.............
i 1 0 1 2 1 1i>Xf The function reflects onto
itself.
g The function is its own.mverse.
Exercise 8
1 y10-I
18- 5II 6- y • X4- \ I2' .........
10 :-:8 -I, 0 2 6 xp - 10__. -2-
II -.. 8\
:_j_ - 10
Y
8-y •- I· 6- X -1-
4- \ i-2- -.....
•• -2 0 4 lOX0 -:S- u 2 6-2 I.._ •A
T
( llCAnswers
-
8/10/2019 Math IB SL Answer Sheet
17/68
-
8/10/2019 Math IB SL Answer Sheet
18/68
d y=O x .=-1Domain x E IR x t - 1
Range y E IR y t 0
e y = 2 x = -1Domain x E R x t - 1
Range y E lR y t 2
f y= - 2 x = 1
Domain x E x = - 1
Range y E IR , y t - 2
g y = 2 X = 3Domain x E IR x t 3Range y E IR , y t 2
h y = -2 X = -3Domain x E R x t - 3
Range y E IR y t - 2
ay
86
4
2
4y= -
x
2 4 6 8 X
Domain x E IR x 0RangeyE R y t 0
b y8
36y = X - 3
4
2
-8 -6 -4 - 2 2 4 6 8 x-2- 4
-6
Domain x E IR x :t; 3
Range y E R y 0
c
- 10 -8 -
-4y = - 8x 5
y
4
-4 -2 0-4
-8
- 12
-14
Domain x E R x 5Rangey E IR y = - 8
2 X
d y
8 16 y - 7 3
4
2
-2 0 2 4 6 8 10 12 X-2
Domain x E R x :t; 7
Range y E R y 3
e y
-12 -8 -4 x-4
6 -8y= - 6x 2
Domain x E R x 2Range v E JR v :t: - 6
f y8
6
4
-6 -4 -2 0
5y= - 4X
2 4 6 X
Domain XE R x = 0
Range y E IR y:t:
4g 1y - - 2x 12
y
1
-5 - 4 - - 2 -1 Q 1 X- 1
- 3
- 4
Domain x E R x = -3Range y E IR y :t: - 2
h y6
43
2y
X
- 6 -4 - 2 4 6 X
-4
- 6
Domain x E R x 0Range y E IR y :t; 0
3
•I y
10
86
42
-8 -6 -4 -2 °2-4-6
4 6 8 X
= 4 5Y 3x - 6
Domain xE IR x t 2
Range y E IR y 5
t
2.52
1.51
0.5
-20 - 10 0 10 20 30 40 50 c
b 3.9°C
4 a y86
4
- 10 -8 -6 - -2 O 2 4 6 8 10 X-2-4
-6
The linear function is a line ofsymmetry for the rational
function. The l inear function
crosses the x-axis at the sameplace as the vertical asymptote.
1y =X 1
y
0 2 4 6 8 10 X
The linear function is a line ofsymmetry for the rational
function. The linear function
crosses the x-a xi s at the same
place as the vertical asymptote ofthe rational function.
n sw rs
-
8/10/2019 Math IB SL Answer Sheet
19/68
Investigation - graphingrational functions 2
a y432 Xy = x 31
-10 -8 -6 -4 - 2 4 6 8 10 X1-2-3
y4
321
x ly= x 3
2 4 6 8 10 X-10 -8 -6 -4 --1-2-3
y86
4
2
4
-6
y8
642
2 4 6 8 lOX
x - 1Y = x 3
-10 -8 -6 - 4 -2 2 4 6 8 10 X
b
Rational Verticalfunction asymptote
XX= -3Y = x 3
x 1X = - 3y = X 3
2 xY = x 3 X= - 3
2x-1Y = X= -3
x 3
-4
-6
Horizontalasymptote
y = 1
y = 1
y = 2
y = 2
Domain Range
X E JR ., y E I {,X - 3 y;;; 1
X E JR, y e R,X ;t; - 3 y :1
X EJR y E R,X t; - 3 y :2
x e lR, y E I(,X ;t; - 3 y;o2
c The horizontal asymptoteis the quotient of thex-coefficients.
Answers
d The domain excludes thex-value of the verticalasymptote.
Exercise 5
1 a y = 1, x = 3Domain x E IR x :t. 3
3
Range y E IR, y 12 1
b y = 3 x = 3
D . 1omam x e R x :t. -2 3
Range y e IR ,y :t. -3
3 5C y = - ,X = - -
4 4
Domain x E R ,x - 24
3
Range y e R , y -417 1
d y s x - 4
a
a
b
Domain x E IR x _ 4
17Range y e IR, y -
8••• b i • d iill c IV
y
6x + 2
4 y = x 3
2
- 10 -8 -6 -4 -2 0-2
2 4 X
-4-6
Domain x E R , x :t. - 3
Range y E lR, y 1
y
XY = 4x 3
-6 -4 -2 2 4 6 8 X
Domain x E IR, x4
1Range y E y =1 -
4
c
d
e
f
y4
321
-15- 10 -5 0- 1-2
-3
x - 7y = 3 x - 8
10 15 20 X
Domain x E x :;t:3
1Range y E R , y -
3
y
86
4
9x 1y=3 x - 2
-20-15-10 - 50 5 10 15 20 X-2-4
Domain x E IR,x ::;t:3
Range y E lR , y 3
y
4
2
-3x 10y= 4x- 12
-6
Domain x E R , x ::;t:3
3Range y E R , y ::;t:.- -. 4
y4
35x 2
y =4x
2
X
-8 -6 -4 -2 2 4 6 8 X-1
-2
-3
-4
Domain x E IR,x 0
5Range y E R , y ::;t: -
4
-
8/10/2019 Math IB SL Answer Sheet
20/68
-
8/10/2019 Math IB SL Answer Sheet
21/68
b
c
d
4 a
b
x = O , y = - 3
Domain x E R, x :t 0Range y E JR, y ::t -3X= - 6 y = - 2Domain x E JR, x ::t - 6
Ra n ge y E JR, y ::t - 2
x = l , y = 5
Domain x E R, x t 1Range y E JR, y :t 5
2
2
1
1
c =
c40 -
00 -
60 -
20 -
80 -
40 -
300
s
-300c - s
\ -
.
0 5 10 15 20 25 30 5
c The domain and rangeare limited to JR+ and thedomain to z
5 a. 2I y = - = 2
ljj X= - 2
iii ( -2 , 2)
b] 1
(0, - 2 ) (2, 0)
cy
1:)( )
I I
u
= k 1c X l LJ A
/ • I I.../
8 -6I
2 p 2I
64 - 4I
x- -2I I
Review exe rcise GD
1 a yf IX)
l 1 2· - i
8 - 2 0 \ 8 X- - - 2-
f 1 -4- rr--- t----- ·-- ·-- -- r:--1---r- - - d- - t - -6 6-1 --... f X = - 5f \ )(
-I u-io-
Domain x E JR., x t 0
Range y E JR.,y t 5
Answers
j
8'
b
c
d
X
e
y( )
t::..
A\ _ f x ,
- -r- --r--8 -6 -tl. '·2,., ) X-
A
a 2u 1==.
( ) l X I
Dom ain X E JR, X ::f. 0
Range y E JR, y :t 3
y0
af x
u r.t-
II
f xA I - 5I•,., ill...
/ :\4 0 2 4: Vl
IA
I
c
0 __l -
Domain x E R, x t 5
Range y E JR, y t 0
y....
f x) rI
110.
-
0I
p 1: 1.... II
A If x) = 8 1
X 7 Ia 1\.v
I 'II1 \ I... v \ I..... I......
Domain x E JR, x ::f. 7
Range y E JR., y t - 8
y.l2
j.6
f( )iII \ -l > .I - ......I
e.. -6 -g 0 21 2I • ft.
\ I f( ) ..u
I It+
:D I
Domain x E R, x t - 3Range y E JR, y ::f. 0
,12 X
X
f ycv
A
II I ,.,
/...
..v ·ru -8 -6 - 4 Q_.? 2. t::. ,...
f(x)- x 4- 2 IL G I
I0Domain x E JR, x :t - 4
Range y E JR., y t - 2
2 a Using the equation
S ddistance
pee = .orne
X
5600distance= 5600, s
t
b s km h-1)
1200 f
1000
800 5600s= -
600 t
400
200
0 4 8 12 16 20
t hours)
c 560kmh - 1
3 a
m minutes)
300
250
200
150
100
22.2s + 14.28m =s
50
0 20 40 60 80 100120 s
b • 165 min..
57.9 minI...
36.5 minll
c m = 22.2
d The number of minutesthat can be spent in directsunlight without skindamage on a day whens = 1.
-
8/10/2019 Math IB SL Answer Sheet
22/68
4 a cX 106
x10 6750 OO m100 -m
x1Q6 v-x106
/
2
1
0 20 40 60 80 100
b i 187 500 Thai bahtii 750 000 Thai baht
m ( )
iii 6750 000 Thai baht.
c No. When m = 100, thefunction is undefined.
5 ay f ) 2 + 1X =
2x - 5A
:.., Io.J I --...... I
I I 2'
J
-I
I X = 2.p2 -1 0 I
,
'I
X
5b •I x = 2•Y = 2
•• 2.25I•• • 1.8ll
Chapter 6Skills check
1 a - 6 b - 3, 5
2 a k =1 5 - 3m4
c 5
bp
3 a 108 b - 12.22
4 a 5 b 163
c - -32
Investiga tion saving money
a Week Weekly Totalnumber savings •savrngs
1 20 20
2 25 45
3 30 75
4 35 110
5 40 150
6 45 195
7 50 345
8 55 300
b Sa vings in 1Oth week: $65;Savings in 17th week: $100
c Total saved in 1st year (52weeks): $7670
d 1 000 saved afte r 17 weeks. Exercise 6C
e M=20+5 n - l ) o r M = 1 5 + 5 n 1 d =0 .9
f T = n (35 + Sn) or T = Sn (7 + n) 2 d = - 3 u = 64' 12 2 3 5.5
Exercise 6A
1 a 19,23,27 b 16,32, 64
c 18,24,31 d 80 , - 160,320e 9 11 13
14 j 17 20
f 6. 01234 , 6.012345,6.012 345 6
2 a 10, 30,90, 270b 3, 7, 15, 31
c3 1 1 2
3 a u = 2 and u = u + 2I n +l nb u = 1 and u + 1= 3u11 n
uc u1 = 64 and u ,, 1 = ;:
d u 1 = 7 a n d u n 1=un+54 a 3, 9,27, 81.
b -3 -9 - 15 -2 1' ' '
c 1,2 ,4 , 8
d 1,4,27, 2565 a u = 2n b u = 3n-l
n n
c u = 2 7- ll d u = 5n + 2n n
n
f u = nxu = n + 1 11
6 a 610b u 1 = 1, u2 = 1, and
u = u + un+ l n n-1
Exercise 68
1 a i u 5 = 45
b i u 15 =235
••11 u = 3n
ii u =15n + 10nc i u 15 = 106
ii u = 5n + 31nd i u = - 8215
ii u = 113 - 13nne i u 15 = 14
ii u = 0.6n + 5nf i u 5 = x + 14a
••11 u = x +an - a
2 a 51d 15
17
b 169
e 27
n
c 37f 10
4 8
Exercise 6
1 11 a r = u =2 7 4
b r = - 3 u = - 2 916' 7
c r = 10, u 7 = 1 000 000
d r= 0.4, u7 = 0 .1 024
e r= 3x u = 1458x 6' 7
f r = .. u = ab 7a ' 7
Exercise 6E
1 r = 0.4, u 1 = 125
2 r = 2, u = 4 .53 a n = 12 b n = 9c n = 7 d n = 33
4 r = +4, u2 = +36
5 p = +276 x = 8
Exercise 6F8
1 a L nn=l
6 6
c I , (29-2n) d I,240(0.5 '-1 )n= JI ' J l
I() I 8
e I , an f I , (3n +1)1=5
I I;
I , 3 L h I na u=-1 n=J
g
2 a 4 + 7 + 1 0 + 1 3 + 1 6 + 19+ 22 + 25
b 4 + 16 + 64 + 256 + 1024
c 40 + 80 + 160 + 320 + 640
d x5
+ x6
+ x7
+ x8
+ x9
+x 1o + x l l
3 a 315 b 363c 140 d 315
Exercise 6 6
1 234
2 108
3 594
4 40 x + 152
Answers
-
8/10/2019 Math IB SL Answer Sheet
23/68
5 a n = 246 2292
b 1776
Exercise 6H
1 3
2 a3 a
3n 2 - 2n b 17
1.7 5n 2 - 31. 75n b 21
4 a5 a
1600 b 12600n = 24 b S
24= 1776
6 d = 2.5, S 20 = 575
Exercise 6
1 a 132 86 0 b 1228.5c 42. 65 625d 4095x + 4095
2 a 435 848 050b = 11 81 9 .58
c - 1048 575d 1og(a 1048575
3 a ib i
•C I
d i
9
6g
11
Exercise 6J
1 a 6 b 5
II
II
II
c 19 d 659048
2 r = 3 S =3 r= 3
15
4 a 1 5 b 215 20596 3
76684
3685.5
1.626 5375
885.73
Investigation converging•senes
1 i 1a r -22b r =5
- 1
c r = 4.Inspect va l ues on GDC
2 a T he va lues are approachi ng4 as n 7 oo
Exercise K
1 I rl < 1-2 a S 4 = 213.3,S 7 z 2 15.9,
an d S = 216
b S4 = 1476, S7 = 1975.712,an d S = 2500
c S 4 = 88 .88, S 7 = 88.888 88,and Soo = 88 .8
-d S4 = 10.83,S 7 z l2 .71 ,
and S = 13.5-
3 13.44 192
5 16 or 48
6 150
7 4118
Exercise 6L
1 - 20
2 a 26.25 em3 a 3984.62
c 4035.364 425 18
6 2327 ::=:19 6 years8 a 1,8 ,21
c 6n - 59 a 4, 12, 28
c 4(2''- 1)10 z 86 months
11 About 16. 30
Exercise6M
1 10
2 283 354 84
5 156 120
b 119b 4025. 81
b 1, 7,13
b 4, 8, 16
Investigation patterns inpolynomials
1 a + b
2c T he va lues are approachi ng
19 2 as n 7 oo3
b T he va lues are approachi ng125 as n 7 oo
a2 + 2ab + b2
a 3 + 3a 2 b + 3ab 2 + b3
. ( 1 )so3 Results like 1-2
. are
beyond the limit of the display.
Answers
4
5
a4 + 4a 3 b + 6a 2b2 + 4ab 3 + b4
a 5 + 5a 4 b + 10a 3b2 + 10a 2b3 +5ab 4 + b5
6 a 6 + 6a 5b + 15a 4 b2 + 20a 3 b3 +15a 2 b4 + 6ab 5 + b6
The co efficients are from Pascal'striang le.
a+ b f = a7 + 7a 6 b + 21a 5 b2 +35a 4 b3 + 35a 3b4 + 21 a2 b5
+ 7ab 6 b7
Exercise 6N
1 y 5 + 15y 4 + 90y 3 + 270y 2
+ 40 5y + 243
2 16b 4 - 32b 3 + 24b 2 - 8b + 1
3 729a 6 2916a 5 + 48 6 0a 4
+432 0a 3 576a 64
5 X 8 + 8x 7 y + 28 x 6 y 2 + 56x 5 y 3
+ 70x4
Y4
+ 56x3
y5
+ 28 x2
y6
8x y 1+ ys
6 8Ja• - 2 16a 3 b 216a 2 b2
- 96ab 3 +16b 4
7 243 cs + 810c4
+ 1080c3
+ 720c2
d d 2 d 3
240c 32+ d d s
Exercise6
1 336x 5
2 - 1280y 4
3 4860a 2 b4
4 - 5125 26 +4
7 179208 48 6 0
9 810 7
Review exercise non GDC
1 a 42 a 1-
43 a 44 a 305 120
6 a 1-4
7 +4
b 283
b 1
b 5b 262
b 32003
c 25c 256
3
-
8/10/2019 Math IB SL Answer Sheet
24/68
8 720.x3
9 a 17 b 323
Review exercise GDC
1 a 3 b 522 a 96 b 323 a u = 7 d = 2
Lb 720
4 a 2 b 115 186 u = 5 r = -3
1
7 - 945x•16
81-4
9 a 5.47 million
Chapter 7Skills check
b
1 a 3x(3x 3 - 5x 2 + 1)b (2x - 3) 2x + 3)c (x - 3) x - 2)
d (2x + 1 )(x - 5)
2 a x 3 + 6x 2 + 12x + 8b 81x 4 - 108x 3 + 54 x 2
- l2x + 1
2056
c 8x 3 + 36 x 2y + 54xy 2 + 27 y 3
3 a x-6 b 4x - 3I 5- d -c 5x 2 X ;3--e 7x 2
Investigation - creating asequence
Portion of the paper
Round you have at the endnumber of the round Fraction
Decimal 3 sf)
1 1- 0.333
3
24- 0.4449
313
0.48127
440- 0.49481
5121
0.498243
6364
0.499729
1 The portion gets closer to .
2 The portion gets closer and
closer to I , yet never reaches . ..2 2
Exercise 7A
1 Divergent
2 Conve rgent; 3.5
3 C on vergent; 0
4 Convergent; 0. 75
5 Divergent
Exercise 78
1 10
2 1
3 1
4 Does not exist
5 4
6 Does not exist
Investigation - secant andtangent lines
1 y
2
X
Point LineGradient
p
A
8
cD
EF
3 04
Coordinates
0 , 1)
- 1.5, 3.25)
-1 , 2)
-0 .5 , 1 .25)
(0.5, 1.25)
1, 2)(1.5, 3.25)
y
or slope
- -
AP - 1.5
BP - 1
P - 0 . 5
DP 0.5
EP 1FP 1.5
X
Exercise 7C
1[3( x +h)+4]- (3x +4) =
3h
2[2(x +h) 2 - I ] - 2x 2 - 1
= 4 x + 2hh
3[ (x +h) 2 +2(x +h)+3
J(x 2 +2x + 3
h= 2 x + h + 2
Exercise 70
1 2; m = 2
2 6x + 2 · m = - 16
3 2x - J· m = 1
ln v .estigation - the derivativeo f f(x)
=1 f ( x ) = x 2f (x) =lim (x + h) z - x z
IH O h
= l im(2x+h)h -+ 0
=2x
f ( x ) = X 3
( ) 1. ( x + h) ' - x3
X = 1mh - . 0 h
= lim(3x 2 + 3xh + h 2 )h - t O
f ( x ) = X 4
f ' (x ) = lim x + h) 4 - x•,, .. o h
= l im(4x 3 + 6x 2 h+ 4xh 2 + h 3 )h -+ 0
2 To find the derivative o ff (x ) = x", multiply x by theexponent n and subtract onefrom the exponent to get the
new exponent. If f (x) = x",thenf ' x) = x n - l
3 Prediction: f x) = 5x 4
f ( x ) = x 5
f ( x ) = lim (x + h i - xsh-+0 h
= lim 5x 4 + 1Ox 3h + 1Ox 2h2h-+0
+5xh 3 + h 4 )
Answers
-
8/10/2019 Math IB SL Answer Sheet
25/68
Exercise 7E
1 5x 4
2 8x 7
43 ·x s
1 12 or fx33 23x 3 . x
4
5 1 1- or - ; : : =2Fxl
2x 2
6
Exercise F
161 - -x92 0
4 5n x 4
5 2x - 8
6l 4
7 3-2x 3
83
-8 i 3
5 310 - , + - ,- -
6x 0 4x 4
11 12 x 3 - 4x
12 4x + 32 2
13 . _ + .3x 3 3x 3
14 6x 2 - 12x
15 3x 2 + 4x - 3
Exercise 7G
1 y + 3 = 2(x - 3);I
y + 3 = - - x - 3)2
y1y+3= - -- x-3)2 1
Answers
f(x) = x2 4x
2 a y - 4 = -4 (x + 3)
b 6 = l ( x - 1)
1c y - 5 = - ( x - 3 )
315
d y - 9= - - (x -1 )4
13 a y - 3 = - - (x -2 )
71b y+5= - (x + I)6
c y - 25 = (x - 2)20
d y + 2=-2_ (x - I)26
4 x = l · x = - 1
5 5
Investigation the derivativesof ex and In x
1 Co nj ecture: f'(x) = e
2 Co nj ecture: f ' (x ) =. .
Exercise H
14
X
12 e + - 1
2x 2
X
4 8x + 315 2e + -X
6 5e-" + 4
7 y - 5 = 12(x- In 3)
8 y-9=. . . (x +3)6
19 y - 1= - (x-e)e
10 y-7=- . . . (x -2)9
11 2e 3 ; 40.25
12 - ; 0.20824
X
Investigation the derivativeof the product of twofunctions
11
f (x) = 11x 10
u'(x) = 4x 3 ; v' (x) = 7x 6
u' (x) · v'(x) = 28 x 9
5
6
7
8
No
f ' (x ) = x 4 · 7x 6 +x 7 · 4x 3 = l l x 10
f ' (x ) = u(x) · v'(x) +v(x)·u'(x)
f(x) = (3x + l)(x 2 - 1)
= 3x 3 + x 2 - 3x - 1
f (x) = 9x 2 + 2 x - 3
f (x ) = (3x + 1)(x 2 - 1)f x) = (3x+ 1)(2x) + (x 2 -1) 3)
= 6x 2 + 2x + 3x 2 - 3
= 9x 2 + 2x - 3
This supports the co nj ecture.
Exercise 7
1x - 4 2
2 l 0x 4 + 4x 3 +9x 2 + 2 x + I
31- 1nx
x
ex4 - e l n x
X
65 (x + 4) 2
ex
9 -1
Exercise 7J
1 2x 2 -5
3
2 4 x 3
3 4xe- + 2 x 2e
2xe - 4e4
5 32 16
X - 9
ex
2x7
x2 + Y
8 3 + 31nx19 1 - -
x 2
-
8/10/2019 Math IB SL Answer Sheet
26/68
10
11X+ I
(x - 1)1
12 10x 4 + 12x 3 - 3x 2 - 18x - 15
I13 y = - - (x -1)
e14 y X ]
15 - 9n + 3.5
16 4n r 2
17 718 4
Investigation findingthe derivative of acomposite function
1 a I (x) = 2 - x) 3= 8 - 12x + 6x 2 - x 3
l ( x ) = 1 2 + 1 2 x - 3x 2b l ( x ) = 3 ( 2 - x)l · ( -1 )
2 a I (x) = (2x + 1 2
= 4x 2 + 4x + 1
l ( x ) = 8x + 4b I (x) = 2(2xl)· 2
3 a l ( x ) = (3x 2 + 1)2= 9x 4 + 6x 2 + 1
I x) = 36 x 3 + 12xb I x) = 2(3x 2 + 1) · (6x)
4 The derivative o f a compositefunction is the derivative ofthe outside function with
respect to tbe inside functionmultiplied by the derivativeof the inside function.
5 f(x) = x4 + x2)3= x 12 + 3x O + 3xs + x6
f x)
=2x 11 + 30 x 9 + 24x 7 + 6x 5
l ( x ) = 3(x 4 + x 2) 2 · (4x 3 +2x)
=3( x 8 + 2x 6 + x 4) ( 4x 3 + 2x)= (4x + 10 x 9 + 8x 7 + 2x 5)= 12x 11 + 30 x 9 + 24 x 7 +6x 5
Exercise 7K
1 x 5 · 3x 4 + 2x·I5(3x 4 + 2x) 4 (12x 3 + 2)
2 4x 3• 2x 2 + 3x + I·' .12(2x 2 + 3x + 1 )2 (4x + 3)
3
4
5
6
7
8
9
10
lnx; 3x ' ;X
2x+3;2
,•-
3(2x +3)1
3(1nx)2
x 1 ; In x;X
. 63 • 9x + 2· - - - - - : -
' I I-(9x + 2 3
e · 4 x ' · 12x ,e' '
Exercise L1 8x 2 (2x - 3) 3 + 2x(2x- 3) 4
or 6x(2x - l )( 2x - 3) 3
2
3
4
5
6
7
8
9
10
e'
-8x(x 2 + 3)2
-x 1 x+l+ ' or -3- -(2x + I ) (2x + 1)2 (2x + 1 2
6x 2
2x l
I
xlnx
- 2(e' - e ) - 2e ' (eh - I )or ,1(e' +e ' )2 (e · +I)
-2x+3
I 1
x s (x 2 +3 ) 2 +4.x3(x2 + 3) 25x 5 + 12x.1
or - - - - - - ,1,..-
(x 2 +3 ) 2
'11 a (2x-2)e ' 2 'b 2
c y - I = 2(x - 2)I-12 e
13 h (x) =6
. Since( l - 2x)
6 > 0 and(l - 2x) 4 > 0 forall x where his defined,the gradient of h is alwayspositive.
14 a 6b 8
Exercise 7M
31
2
3
4
5
6
7
8
x
3e J , (6n + 5)
8
-3
1
equals 0
d y = e · - e •dxdl_::...Y = e• +e rdx 1
d)_::... = e· - e 'dx 3
d4_::...Y = e' +e · ·dx 4
When n is odd
dY = e ' - e ' and when
dx .
n 1s even
d_::... = e ' +e ' .dx
9 dy -1
10
-
= -
•y {- l )"n
dx
-1 8
- · ·
Answers
-
8/10/2019 Math IB SL Answer Sheet
27/68
Exercise 7N
1 ab
c
2 ab
c
d
1 .4m;2 1 m
9 .8ms -
9.8ms - ; Om s - ; - 9.8ms - 1;The ball is moving upwardat 1 s at rest at 2 s anddownward at 3 s.
4000 litres; 1778 litres
- l l l l i t res/min; D uringthe time interval 0 to20 minutes, water is beingpumped out of the tank at
an average rate of111 litres per minute .
- 89 l i tres/min; at20 minutes, water is beingpumped out of the tank atan average rate of 89 litresper minute.
V (t) is negative for0 < t < 40 minutes, whichmeans water is flowingout of the tank during thist ime interval. Thereforethe amount of water in the
tank is never increasing
from t = 0 minutes tot = 40 minutes .
3 a 112 bacteria/dayb P' (t) = 25e 0 ·251
c 305 bacteria/ day; on day10 the number of bacteriaare increasing at a rate of305 bacteria/ da y .
4 a 20.25 dollars/unit;20.05 dollars/unit
b C (n) = O. ln + 10
c 20 dollars/unit ; t costs20 dollars per un it to
produce units after thelOOth unit.
Exercise 70
1 a Ocm· 9cms - 1
b 1 sand 3 s
ct = 3
t t = 1t = 0• •0 4
Answers
s
2 a 4ft
b s(2) = - 16(2) 2 + 40(2)+ 4 = - 64 + 80 + 4 = 20ft•
- 16t 2 + 40t + 4 = 20I1
II t = - 2s2
d.I
ds- = - 3 2 t 4 0
dt..40fts- 1II
. 5Ill - s
4
iv 29ft
3 a v(t) = s (t)
- e ( l) - t(e )- (e )2
-e 1 - t)
-e 2 t
v(t) = - tet
b l second
Investigation velocityacceleration and speed
1 a Let acceleration be 2 m s- 2 •
Time Velocity Speeds) m s- 1 ) m s -1 )
0 10 10
1 12 12
2 14 143 16 16
4 18 18
d Let acceleration be 2 m s- 2 •
Time Velocity
(s) m s- 1 )
0 - 10
1 -8
2 -6
3 - 4
4 - 2
2 a Spee d ing upb Slowing downc Spee ding upd Slowing down
3 a Speeding upb Slowing down
Exercise 7P
Speedm s- 1 )
10
8
6
4
2
1 a v ( t ) = 8 t 3 - l2t , t 2 : .0
a (t) = 24 2 - 12, t > 0b 84cms - 2 ; Velocity is
increasing 84 em s- 1 at
t ime 2 seconds .
c v(t) = 0 when t = 0 and1.22 s; a(t) = 0 whent = 0. 707 s; speeding upfor 0 < t< 0.707 s a n dt > 1.22; slowing down for0.707 < t< 1.22
2 a v(t) = - 3 t 2 + 2 4 t - 36,0 < t< 8a(t) = - 6 t + 24 ,0 < t< 8
b s(O) = 2 0 m ;b Let acceleration be - 2 m s - 2 . v(O) = - 36m s- 1;
Time Velocity
s) m s -1 )
0 10
1 8
2 6
3 4
4 2
Speedm s -1 )
10
8
6
4
2
a(O) = 24ms - 1;
c t = 2, 6 s; moving left on0 ; t ; 2 and 6 ; t ; 8,moving right 2 ; t:::; 6
d t = 4 s; speeding up on2 < t < 4 and 6 < t < 8,slowing down on 0 < t < 2and 4 ; t :::; 6
c Let acceleration be - 2 m s - 2 . 3 a v(t) = - 9.8 t + 4.9
Time Velocity
s) m s- 1 )
0 -10
1 -12
2 -14
3 - 16
4 - 1 8
Speedm s- 1 )
10
12
14
16
18
a(t) = .8b 2.01 s
c O.Ss; 11.2m
d v(0.3) = 1.96 > 0 anda(0.3) = -9.8 < 0. Since thesigns of v(0.3) and a(0.3)are different the particle isslowing down at 0.3 seconds.
-
8/10/2019 Math IB SL Answer Sheet
28/68
•a••II
b i
..II
1 1v(t) t
2 t+11 second
1 1a t) = - + -----,_-
2 ( t+1) 2
Since > 0 and2
1 > 0t+IY
1 1a t) = 2 + ( t+1) z > 0
for t :: 0 and so velocityis never decreasing.
Exercise 7Q
Decreasing (-oo, oo
2 Increasing (-oo, 2); decreasing(2, 00
3 Increasing ( -1 , 1) ; decreasing
(-oo, -1 ) and (1, ooDecreasing (-oo, 0); increasing(0, 00
5 Increasing (- 1, 0) and (1, oo ;decreasing (-oo, -1) and (0, 1)
6 Decreasing (-oo, 3) and (3, oo
7 Decreasing (0, oo
8 Increasing (- 3, oo ; decreasing( -oo, -3)
9 Increasing - oo, - J 3 ) and
( 3 oo ; decreasing (-J3, - 1 ,( -1 , 1) and (1,J3)1 Increasing (-oo, -2) and
(4, oo ); decreasing (-2, 4)
Exercise 7R
relative minimum (1, -5)
2 relative minimum (2, -21);relative maximum ( -2 , 11)
3 no relative extrema
relative minimum ( - 1, - 1and (1, -1); relativemaximum (0, 0)
(3 2187)5 relative minimum -4
, -256
6 relative minimum (0, 0);
relative maximum 2, )7 no relative extrema
8 relative minimum (1, 0);relative maximum ( - 3, - 8)
Exercise 7
concave up ( -oo, oo
2 concave up (0, 2); concavedown (- oo, 0) and (2, oo ;inflexion points (0, 0)and (2, 16)
3 concave up (2, oo ); concave
down (-oo, 2); inflexionpoint (2, 8)
concave up ( -oo, oo
5 concave up (-2, oo ; concavedown (-oo, -2); inflexion
point (- -: )6 concave up -oo - J3 and
3
7 a
J3 oo ; concave down3
J3J3- - - ; inflexion points3 , 3
J33- and3 , 4 J333 4
- 48xf ( x ) = (x z + 12) z
f (x)
_ x 2 + 12) 2 ( - 48) - (- 48x)[2 x 2 + 12)(2x)]
(x 2 + 12 t
_ x 2 +12) 2 (- 48 ) + 192x 2 (x 2 +12)- x
2 +I2t
Exercise 7T
y
4
2
3
4
( 4, 0)
-3 -2 -1 O 1 2 3 4 X-4
-2
y
8
6
-8)
y
-10
x= 4IIII
0, -2) 4 I
_-6 -4 -2 6 8 10 12 X
-2
y
1086
4
2
-4-6-8
0 3 4 5 x
_ 48(x 2 + 12)[- (x 2 + 12) + 4x 2 ] 5 y-(x 2 + 12)
_ 48(x 2 + 12)(3x 2 - 12)(x 2 + 12) 4
_ 144(x 2 +12)(x 2 - 4 )
x 2 + 12) 4
_ 144(x 2 - 4)
(x 2 +12) 3
b i relative maximum(0, 2)
inflection points
8 concave up (-oo, -2) and(4, oo ; concave down ( -2 , 4);inflection points at x = - 2 , 4
6
10864
2(0,0)
-4 -3 -2-4-6-8
(-1, 0
2 3 4 X
- ;Answers
X
-
8/10/2019 Math IB SL Answer Sheet
29/68
Exercise U
1 y y = f (x)I
y = f'(x)
; f\I I J 1 I I X
-3 -2 1 11 1 2
- - 1 X
y = f (x)
3 vY= f x)
6 8 X
y = f (x)
Exercise V
1 relative minimum (3, -75)
2 relative min imum (1, 0) and( - 1, 0); relative maximum 0 , 1)
3 relative minimum (3, -27)
relative minimum (
5 relative minimum (1, 0)
6 relative maximum (0, 1)
Exercise W
1 A - neither; B - relative andabsolute minimum;C - absolute maximum
2 A -neither; B - r e lativeminimum; C - relative andabsolute maximum;D - absolute minimum
absolute maximum 8;abso lute minimum - 8
Answers
4 absol ute maximum 16;absol ute minimum - 9
5 absol ute maximum 2;
abso lute minimum _2
Exercise X
79 11 - a n d -4 4
2 100 and 50
3 X = 50 ft; y = 200 ft3
Exercise Y
1 40cm by 40cm by 20cm
2 3 items
3 224 a
3 - 3hr=
5
b V(h) =reo5 3hJ h) orV(h) =
9r ( lO Oh - 20h 2 + h 3 )
25
c dV = 9P(100-40h + 3h 2 ) ·dh 25 '
d2V = 9p - 40+6h)dh 2 25
10d r = 4cm; h= - cm3
5 a p(x) = 4 2 x 2
b d2
pdx I d 2 3- X -x2 x z
c 0.630 th ousand unitsor 630 units
Review exercise non GDC
1 a 12x 2 + 6x - 2
12c - -x5d 10x 4 - 4x 3 - 3x 2 + 2x - 1
e
h
•I
•J
11
(x + 7) 2
2x + 3
1- 2ln x
4 1x - -
3 3
k ex 3x 2 + 6x + 1)
l 6e
3m.J2x - 5
n 2xe 2x(x + 1)
01
X
2 a x 3 + 3x 2h + 3xh2 + h3
b
f x)
1. [2(x+h/ - 6(x+h)] - 2x 3 - 6x)=
h-O h
1. 2x 3 +6x 2h+6xh 2 +2h 3 - 6x - 6h - 2x 3 6x=
h
=lim 6x2h+6xh
2+2 h
3-6h)
h
1. h(6x2
+6 xh - 2h2
- 6)= 1m ___ : ___________ :h
=lim (6x 2 + 6xh - 2h 2 - 6)h-o
c p = - 1 ;q= 1d f (x) = 12xe (O,oo)
3 y - 4 = - __ :_(x- 1)12
2../3 9- 2../3) -2../3 9+2../33 ' 9 3 9
5 a / ( 2 ) > / 9 ) > / ' ( 2 ) ;b f (2)> 0; since the graph
of f is concave up,/(2) = 0a n d / ' (2) < 0 since thegraph of fis decreasing
6 a.
4x 3 - 12x 2I
c
••12x 2 -24xI
• (0,0), (4,0)• •
(3, - 27)I...
(0,0),(2, - 16)llY
20
15
0-4 -3 -2 -1 5 2 3
101520 2 ,-16)
25
4,0)
X
-
8/10/2019 Math IB SL Answer Sheet
30/68
-
8/10/2019 Math IB SL Answer Sheet
31/68
Exercise SF
1 a 9 5 em b 6 7. 5d 92.5 e 35
c 57.5
y
. f" r • .j ·;
'
0 20 40 60 80 100 120 X
2 a 14d 82y
M
b 79e 7
1* II
c 75
Q3[ 1 a t*
71 75 79 82 85 X
3 a 19d 27y
1
...l
b 21e 15
iI
J ...l10 20
4 a 5 b 8d 10 e 3
S a m b i
Exercise 86
1 a 75cm
c 12
'" X' 3JI
...l
c 7
c 11
b (77.5 - 72) em= 5.5 em
c The middle 50% of da tahas a spread of 5.5 em.
2 y
3
4035
> .
30Q):::l0 25
';; 20>·.;::;ro 15:::l
E 10:::l
-
8/10/2019 Math IB SL Answer Sheet
32/68
c 2.47d The standard deviation
remains the same. This
is because the standarddeviation only measures the
spread of the numbers , andthat remains const a nt if thesame number is added to eachitem in the list.
e The mean is doubled.f 4.94g T he variance wi ll be
multiplied by 4 b ec au se thevariance is the stan d ar ddeviation squared .
Review exercise non-GDC
1 a 3 b 5 c 5d 9
2 a 4.2 b 4 c 43 Mea n = 27.5 yrs, st a n da r d
deviation = 0.4 yrs.
Type A
4 a 52 b 14 c 8Type B
a 525 a 426
b 8 c 3b 72 c 62
6 ay
140,_
c::
1000
8060
-
8/10/2019 Math IB SL Answer Sheet
33/68
Exercise 9C
1 f x) = + 4 r + 83
1 42 y - x 5 + - x 4 + 9
5 53 s(t) = f - f 6
4 115ncm 3
5 a -5ms- 2
b2
Exercise 901
2
3
4
5
6
7
8
9
10
2lnx + C x > 0
3 e + c. .In t + C, t > 04
_ _x2 +C2
i x 3 + 6x 2 + 9x + C32
- x3
+ 3x2 + 51n x + C, x > 01
. .u 3 + C31 _A __1 3 2- .x -- x- + - x - x+ C4 21- e + x) + C2
2 2 _- x z + - x 2 +5 3
Exercise 9E
1 . .c2x + 5) 3 + c6
2 - - 1 - 3 x + 5) 4 + C12
3
4
5
6
7
8
9
I- x -3
e 2 +C
_ _ln(5x + 4) + C x > _i5 5
3 7ln(7 - 2x) + C x > -2 , 2e x +l + C3
- 4 x - 3) 8 + C162 _
- (7x+2) 2 +C21
e4x + 4 In(3x- 5) + C4 3 '
5x> -
3
101
+ c12(4x - 5) 2
11 a 12(4x + 5) 2
b - 1 (4x + 5) 4 + C16
12 s = _ _ e-3t + 3f2 + . 23 3
Answers
Exercise 9F
1 . .(2x2 + 5)3 + c3
2 ln(x-1 + 2x) + C, x3 + 2x > 0
3
4
5
6
7
8
e + c1
- - - C
x2
+ 3x + 1e Fx + C
I- 2x3 + 5) 5 + c304- x 2 + x) 4 +C3
9 . .(0 - x2)4 + c2
10 - ln(x3 - 4x) + C x3 - 4x > 0
11 / (x) = ln(4x 2 + 1) + 4
2 f (x) = e + 4e
Investiga tion area and thedefinite integral
1 a i 0.5 ii 1; 1.25; 2; 3. 25• • •
ll 3. 75
b i 0.5 ii 1.25; 2; 3_25; 5...ll 5. 75
c 4.67; 3.75 < 4.67 < 5.75; thearea of the shaded region
1 22 2 (3)(6) = 9; 2x + 2)dx = 9;
- I
they are equala
3 f x)dxb
4 a . .(2.5 + 1)(3) = 5.25;2
5
( -. . x+ 3) dx = 5.251 2
b _ _1t(42) z 25 .1 ;2
4.
-J16- x 2 dx :::::25.1-4
Exercise 9G
16
(. . x + 1) dx = 16 ·2 .- 2
_ _(8)(4) = 162
0
2 (x3 - 4x)dx = 4; no area-2
formula3
3 3dx = 12; (4)(3) = 12- I
4
3
-J9 - x 2 dx :::::7.07;0
1 24 n(3 ) ::::::7.07
53 1
- d x z 1.10; no areaI X
formula6
6 ( . .x+2)dx = 18;0 3
_ _(2 + 4)(6) = 182
Exercise9H
1 12
2 143 -4
4 - 8
5 12
6 07 11
8 -39 20
10 12
11 a 412 a 4
b 12
b i a= 3· b = 7 ii
9 4 Exercise 9
12
3
4
5
6
7
8
9
1
10
31
2
36
5
4(e3 - 1)
1
16
316
a 24
10 12
Exercise 9J
1
2
3
4
ln 31 1
0
b 323
k = 3
-
8/10/2019 Math IB SL Answer Sheet
34/68
-
8/10/2019 Math IB SL Answer Sheet
35/68
8
9
0.3841
-1.952
' 2.68
y
3
2 - x - x-2) - e-"dx
Y.
8
4
2
-4 - 2 2 4 6 8 10 X-2
- 4
9.275 1 ) X+ 2x+6 - dx
1.725 2 x - 1
;:::9.68
10 ay
43
21
-2
b i
2 3 4 5 6 x
0.Jx- x )dx
..11 2.67 or -
k 3
2-Fx - x)dxor•
C I
Exercise LJ
1 x3 - 2x 2) - 2x2 - 3x) )dx +0
3
2x 2 - 3x) - x 3 - 2x 2)) dxI
3.08
Answers
1 Exercise M2 x - 1) 3 - x - 1))dx +
0
5
12 0
x - 1) - x - 1) 3)dx= 0.5 V n- 4 2) 5) 2513
3
4
l
0
- I .J3J
1.131
0( xe-x)- x3 - x))dx
1.18- 0.707l
- .0 + IO.x-2 - 9) --3
0.7071
2
3
4
n- 6 - 2x) 2 dx
0 1V
3n 6 2) 3) 113
2
:rr .J4-x 2 Ydx 33 .5;-2
4V 3 n 2
3) 33.5
4
:rr .J I 6 x 2 Ydx 134;0
- 9x2))dx + x-4- 9x2)V= :rr 4
3) ) 1340.7071
- - x-4 + 1 Ox-2- 9 )) dx +3
((-.0 + 10x 2 - 9) -0.7071
x-4- 9x2))dx 11 0
5 a i 4, 4)
ii f (x) = _ _ x2
m = / (4) = 2y - 4 = 2 x - 4)y = 2x - 4
b i 1.236, - 1.528)
Exercise N2
1 n( x3)2 dx = 127 1rI 7
1 281r2 n x2 + 1)2 dx =
150
381.1r
••II 1.236 )
0
±x2 x2) dx +3 3x-x2fdx= -
0 10
4 4 :rr .. ..)2 dx = 31Z4
1.236 4
I X 4
2.55 5 aln4 . 2
:rr e 4 dx0
Investigation: Volume ofrevolution
1
Interval
O x 1
1 :s:x:s:2
2 < x < 3
3 < x < 4
4 x:s:5 I5 < x < 6 I
2 7 1. 5; greater6
adius
f(1) = 0.5
f 2) = 1
f 3) = 1.5
f(4) = 2
f 5) = 2.5
f 6) = 3
3 n 0.5x) 2dx 56.50
b 2
Height Volume
1 0 = 1 n(0.5) 2 1) 0 . 785 4
2 - 1 = 1 n( 1)2 1) 3. 142
3 2 = 1 n( 1.5) 2 1) 7. 06 9
4 - 3 = 1 n(2 )2 1) 1 2.5 75 4 = 1 l 7r(2.5) 2 1) 19.636 5 = 1 n 3) 2 1) 28.27
6 a , {J;)'dx4 Volume= n- 3) 2 6) 56.5 b e
3
-
8/10/2019 Math IB SL Answer Sheet
36/68
Exercise9
1 a v t) = 2t - 6
bt= 3 C t=4 • t = 0
.. I I I I I I I I I I .,. S(t)-1 0 1 2 3 4 5 6 7 8
4
c 2 t - 6 ) d t = 8 m ;
0 4
12t - 6 Idt = 10 m0
2 a v t) = ? - 6t + 8b
t=4 st = 0 t = 2•0
3 36
t - 6•
12s(t)
c ? - 6t+ 1 2 m ;0
6
IP - 6 t+ 14.7 m• 0
3 a v t) = 3 t - 2?b t 0 t = 2 t = 4
...-- . - - - - . - - - . . - + s(t)-8 0 84
c 3 t - 2?dt = 1 6m ;0
4
l3 t - 2? Idt = 16 m0
4 a12 1
v t)dt = - (6)(6)
c
b
0
2 2
- . _(4 + 2)(2) = 12 m212 1
lv(t) ld t = - (6)(6)2 2
1+
2(4 + 2)(2) = 24 m
51
v t)dt = - (2)(2)0 21
+ s2 (3)(6) = 11 m
0
1Iv t) Idt = - (2)(2)2
I+ - (3)(6) = 1 1 m
212 1 1
v t)dt = 2(2)(2) +2
(6)(6)
1- 2 (4 + 2)(2) = 14m
12 1Iv t) Idt = - (2)(2) + . _(6)(6)
0 2 2
+ . _(4 + 2)(2) = 2 6 m2
5 a 2ms- 21
b s t) = -P - 9t + 123
8
c I? - 9 Idt11
9 m26 a 2ms- 2
b 2 < t < 1c 28
Exercise 9P
1
2
10I
18.4e 20 dt 239 billions0
of barrels1.5
1375P - P)dt 15500
spectators3 36.5 +
8
,te < O .Or OI3r
0240 cm 3
20
4 4000 +0
-133 1 - ) t1780 gallons
Review exercise non GDC1 a .0 - 4.x2 + 6x + C
3 2.
b- xJ + C7
1c - - +Cx 3
d 5 3 11- x - - n x +C x>O18 2 '1e - e 4 " +C4
1f - x 3 + l) s + C151 3
g l n 2x +3)+C, x > - -2 2
h . _ ln X ? + C, X > 02 .•I
.J
1- 3x 2 +1) 2 + C2
21n e + 3) + C3
k 2x - 5)2 + C
l
2 a
1 ( }- e 2x·+ C2
4
b 16c 8
d e6 - e3
e - 20
f InS2
2
3 a xl - l)dx1
b 4-
3 2 1c x2 - l ) d x - x2 - l)dx
I - J2
d Jr xl - 1)2dxI
4 f x) = - 2x+42
5 a 5b 28
6 s t) = 2e 2' + 2t + 67 13
Review exercise GD
1 107
2 a a t) = 4t - 11
b a= 1.5, b = 4
c 7.83ma y = 3xb (2, 6)c
-2
2
Y
864
2
-4
d 3x - x3- 2))dx = 6. 75- I
Chapter 1
Skills check1 a 32 b 27 c 343
d 181
e128 256
f 0.000000001 or 1 x I0 -9
X
2 a n = 4 b n = 5 c n = 3
d n = 4 e n = 3 f n = 3
nsw r s
-
8/10/2019 Math IB SL Answer Sheet
37/68
Exercise lOA
1 a Positive, strongb Negative, weakc Negative, strongd Posi tive, weake No correlation
a i Positive •• Linear...
Strongb i Negative
••
LinearI• • •
Strong..
LinearI.
c 1 Po s itive...
Moderate
d i No association••
No n-linear...
Zeroe i Positive •• Linear
• ••Weak
f•I Negative
Non-linear...
Strong
3 a Increases b Decreases4 a
Y Rainfall inTennessee60
I
·I-
01999 2001 2003 2005 2007 2009 X
b Strong, negativec As the year increases the
rainfall decreases.5 a
y
10 0
80
25 60c::Q)
b5 40
2 0
0
Scores
•-••
,
20 40 60 80 100 XMathematics
b Strong, positive, linear
Answe rs
Investigation - leaning tow ro f Pisa continued)
a,}j Scatterplot of lean vs year
750 •725 •
c:: • •C O 700) • ••
675 ••650 •• •
•
3 a 4, 6.67)
b
Q)j)
C O
.)c::
14
12
10
8
6
42
/
/
/
//
.
v/ ban oi t
I
87 .5 X5.0 77.5 80.0 82.5 85.0v/
0 xyear
b Strong, positive
c The lean is increasing. Thedanger with extrapolation
is that it assumes that thecurrent trend will continueand this is not always the
case.
Exercise lOB
1 a 96.7, 44.1)
b
yRelationship between leaf length
70
60
EE 40
3020
10
II
v1/I'
and width
M
_ A 4
•• v
• // •
0 40 80 120 160 XLength mm)
a i 175 em
b
y
190185180
E175
+-
170Q)
:r: 165
160155
0
..II 66 kg
I
e_@
/
L v/ •
//
;>
60 65 70
Weight kg)
/
/
75X
2 4Hours
c Strong, positive
6 8
d An increase in the numberof hours spent studyingmathematics produces an
increase in the grade .
Exercise lOC
1 a (x ,y)=(75 ,7 .03)
Y
-o 12.3 •Q)
j)C OQ)
j)
-og),.sc::Q)
2.3 ----=-. ...70
Temperature
b y =-
0. 9 x +79
c 7
80x
a £220000
b 75.4
c and d Note the values
140
120100
80
60
40
20
of m and b in the equationy = mx +bareapproximate.Y
y -x 300
0160 200 240
e Approximately 70 houses
-
8/10/2019 Math IB SL Answer Sheet
38/68
Exercise 1
The slope is - 0.3. As astudent plays one moreday of sport they do 18minutes less homework.They-i ntercept is 40,which means thatthe average studentwho does no hours ofsport does 40 hours ofhomework.
The slope is 6. For every timea person has been convictedof a crime they know 6 morecriminals.
They-intercept is 0.5,which means that peoplewho have not been
convicted of a crimeknow 0.5 criminals onaverage.
3 The slope is 2.4. For everypack of cigarettes smoked perweek there are 2.4 more sickdays per year.
They-intercept is 7, whichmeans that the average personthat does not smoke has 7 sickdays per year.
4 The slope is 100. 100 morecustomers come to his shopevery year .
They-intercept is - 5,which means that -5people visited his shop inyear zero; they-interceptis not suitable forinterpretation.
5 The slope is 0.8. Every 1mark increase in mathematic sresults in a 0.8 increase in
.science.
They-intercept is - 10 which isnot suitable for interpreta tionas a zero in mathematicswould mean a - 10 in
.science.
Exercise lOE
a
c::0
·_.:;
-::( )uc::0
u
14
12
10
8
6
4 /v
. /
/v
/ 'v
/
0 1 2 3 4 5 6Time hours)
b y = 1.84x 1.99c 8.43 3 sf)a
Y
30
8 200
15
-
8/10/2019 Math IB SL Answer Sheet
39/68
6 0.994. Strong, positivecorrelation.
Review exercise non-GDC
1 .. b11 v
c d.
111 1
a and bY.
60
50-40
•; ::::
03 30 •::::l
u
20
10
0 200 400 600 800Di stance km)
c 321itres3 a and c
}
13.6
13.2
- 12.8C/)
0g 12.4_)
Q)
. .. 12.0Q)
•
11.6 1 1.
11.2
••
• •
•
•0.85020 30 40
Age yea rs)
b Mean age = 34 years,mean time = 12 seconds
c Approximately 11.6 s
Review exercise GDC
1 a
•
•0 X
b As the time increases,the numbe r of push-upsdecreases.
An swers
3
4
N
c y = 1 29x + 9
d r = 0 .929. Thereis a strong, negativecorrel ation.
a w = -22.4 + 55.5h
b 66 .4kg
a r=0 785b y = 30.7 + 0.688xc 99 .5
This should be reasonablyaccurate since the product-moment correlation
coefficient shows fairly strongcorrelation.
aY.
50
40 •• ••t ) 30 ••
20 ••
10
0 20 40 60Test 1
b Positive, strong
c high
d y =0 50x
+ 0.48e 20.48
5 a c and fy
333
3E
c
c::
2
22
8642
086
42 v
I
L/vv
[7I
. //
80
/
0 1 2 3 4 5 6 7 8 XLoad kg)
b 4,30) d•I r= 0.986
ii (very) strong positivecorrelation
e y = 1 83x + 22.7g 30 .9cm
6
h Not possible to find ananswer as the value liestoo far outside the givendata set.
ay
40
35C/)
E 30Q)
. 0e 25Cl.
.o 20>ro
15co
10
5
0 1Agreeab leness
b Behavior problemsdecrease .
c -0.797
d Strong, negativecorrelation
e Fewerf y = - 10. 2x + 51.0
g 5.1
7 a y=10.7x+121
6 X
b i Ev e ry coat on averagecosts 10.65 toproduce.
••When the factory
does not produce any
clothes it has to pay
costs of 121.
c 870
d 4
ChapterSkills check
1 a x = 90b x= 50C X 68
70d x=
3
e x = 6.09 (3sf)
f x = 14.7 (3sf)
-
8/10/2019 Math IB SL Answer Sheet
40/68
Exercise 11A
1 b = 16, A 36.9°, B= 53. F2
3
A
B = 50°, a = 31.0, c= 48.3
A 35°, a = 2.58, b = 3.69A A
4 a = 36, A = 36.9°, B = 53.1°A
5 B = 55°, b = 15.7, c= 19 .2
6 c = 12. 9,A
41.2°,i
=48.8°7 X= 5, A= 22 .6°, B = 67.4°Exercise 118
1 a b= l2J3 A 30°,A
B = 60°
b B= 45° a = 9 c = 9 2\fL.C A = 3 0 ° a = 2 . 2 5
b = 9 J34
d a = 2J3 A = 30°,A
B = 60°
e b = sJ2 A= 45°,
2 x = 8J2 , y = 81 3 - 8, z = 16
3 X= 2 J3 + 2 AC = 4.J3 + 2J3 3
4 x = 1 AB = 3 2 or x = 3\j L.AB = 11 . J2
5 w = 9.8em, x = 13.9em,y = 6.5em, z = 15.4em
Exercise 11C
1 a 1 0 J 2 e m
b BAC = 70.5°A B C = 38.9°
2 a AE = 29.1, B E = 34.4
b AED = 74 .1 °,A
EBA = 54.5° ,AEB = 51.5°
3 7 5 8 m
4 71.5° and 108.5°
5 4.78 km, N2l . l 0 W
6 70 .7 m
7 44 .8 km, 243.5°
8 135. 7m, 202.2cm
9 91.2 m
10 40 .7 m
11 4.01 s
12 a 20.6°
c 35.1°
Exercise 11
d 50 .0°
1 a (0.940, 0.342)
b (0 . 956, 0.292)c (0.5, 0.866)
d (0.276, 0.961)
e (0, I)2 a 66°
c 45°
3 a 0.470
c 0.203
b 81°
d 14°
b 0.308
d 0.25
Investigation Obtuse angles
1y
-0 .766,0.643) (0. 766, 0.643)
0 X
2y
( 0.906, 0.423)155°
(0.906, 0.423)
X
3y
(-0.375, 0.927) (0.375, 0.927)
Exercise 11E
1 a B (0.866, 0.5),c ( -0 .866, 0.5)
b B (0.545, 0.839),c ( -0 .545, 0.839)
c 8(0 .7 07 ,0 .707),c ( -0 .707 0. 707)
d B (0.974, 0.225),c (- 0. 974, 0.225)
e B (0.087, 0.996),c (- 0. 087, 0.996)
2 a 70 .6°
b 17.3°
c 25.4°d 39.7°
3 a 0.2588, 165°b 0.5878, 144°c 0.9877, 99°
d 0.8988, 116°4 a 60.6°, 1 I 9.4°
b 25.8°, 154.2°
c 30.3°, 149.7°d 30° 150°
Exercise 11F
1 a 1.50c -0.910
b -1.92
d 1
2 a y = 1.09x, e= 48°b y = 1.87x,
e= 62°
c y = -2 .80x, e= 11 0°d y = - 1. 21x e= 129°e y = - 0 .75x e = 143°f y = 2.36x e= 113°
Exercise 11GA
1 a C = 5 0 ° a = 17.7em,c = 18.5A
b B = 68°, a = 1.69em,b = 2.44emA A
c B = 40.9°, C =84.1°,
c = 5.46emt
d A = 40° a = 149)
c = 190A
e c = 110°, a= 2.80, b = 4.212 26 .9 em
3 3.37 km, 2.24 km
4 15.8 m
Investigation Ambiguoustriangles
A A
1 C = 62°, C2
= 118°. Theangles are supplementary.
A A
2 B = 86° B = 30°1 2
b = 5.65 em, b2 = 2.83 em
Exercise 11HA A
1 a C = 61.0°, B = 89.0°,
h1
= 8.0emA A
C 2 = 119.0°, B 2 = 31.0°,
b2 = 4.1 em
An swe rs
-
8/10/2019 Math IB SL Answer Sheet
41/68
8 gob C, = 1.1°, A 1 = 5 . ,a = 19.0em
A - 0C2 =108 .9° A 2 =21 .1
a 2 = 8.0em
A 9 soC B 1 =68 .5° A 1 = 1. ,
a = 7.3emA
B2 = 111 .5°, A2 = 48.5°,a
2 = 5.5emA A
d C = 30.5°, B = 107.5°,
b = 47.0em
e Triangle does not existf B = 77.8°, c, = 32.2°,
c = 14.2emA A
B2 = 102.2°, C 2 = 7.8°,
c2 = 3.6 em
g i = 26 . 7°, c = 108.3°,c = 29.5 etn
h c, = 67. 1° A = 56.9°,a
= 45.5 em
C1
= l l .9° A 2 = 11.1°,
a2 = 10.4em
a BE= 8m CE= 6mDE= 15m
b EAB=53 .1°A
BCE= 53.1°,BCD= 126.9°,
A
ABD = 98.8°,CBD = 25.1°
c Given side BD = 1 7 min :::ABD and angle
A •
D = 28 .1 o and sideAB = 10, th en there are 2possible triangles, fittingth is data namely DBA andDBC.
3 b 5.80 km c 24 .9 km
d 143.5°
Exercise
1 a a= 65 . 7m B = 36.0°,
c = 80.0°A
b A = 28.9°, B = 52.8°,{; = 98.4°
Answ rs
A
A = 44.4°, B = 107.8°,A
c = 27.8°d b = 7.48m A= 43.5°,
A
c = 105 .5°e c = 92.8m A= 49.4°,
s = 60.6°f
A
A= 48.6°, B = 56.4°,
c = 75.0°12. 1 km
3 4.07 em 6.48 em
4 18.8 km
5 043.5° or 136. 5°
6 a 45°
b 71.8°c 63.8°
Exercise 11J
1 a 26.7 em 2
b 40. 8em 2
c 152cm 2
d 34 .1 cm 2
e 901 cm 2
f 435 cm 2
47 .8°
3 22.7 em
4 a 76.7°
b 81.4em 2
5 x=2 5 em
6 5.31 mm 18.5 mm
Exercise 11K
1 9.52 em
39 em
3 5 radians
43000 em
2, 220 em
5 22.95 em 2 , 21.3 em
6 8 = 1 .7 r=16
7 7.96 cm 2
Exercise
1 a 57r12
4 7rb
3
4 7rc
9
dl1.7r
6
a 0.977 rad
b I .87 rad
c 5.65 radd 4.01 rad
3 a 150°
b 300°c 270°
d 225°4 a 85.9°
b 20.6°c 136°
d 206°
Exercise 11M
J1 a -
2
Ib - -
2
cJ33
d J32
a 0.892b 0.949
c - 1.12d 0.667
3 a 9.76 em 2
b 5.45 emc 50.5 em 2
4 10.9 m 2
5 a 17.1 em 2 b 12.1 em 2
c 2.63 rad d 15 .8 em
Review exer cise non GDC
1 7J2 ema 30°
3 25
4 10 em 2
5 a 25 em
b 8·f3 em
b 125 em 2
-
8/10/2019 Math IB SL Answer Sheet
42/68
Review exercise GD
1 72.7 m
2 a (0.848, 0. 530)
b 72.9°
c ( 0 .600, 0.800)3 a 54 .7° b 10. 9cm
4 a 18.0 m b 34.3°
5 a 121 b 8.60 em
6 54.1km
7 a 3 1.9 b 1 3. 9 em
c 119 d 2 7.4 em
8 a 2 1. 6 em b 14.5 em
c 11 .16 em d 4 7. 3 em
hapter 12
Skills check
1 a (3, 0,0)b 3,4,0)
c 3,0,2)
d 3,4,2)
e 1.5,4,2)2 6. 71
3 a 20cm
b 101°
Exercise 12A
1 a x = - 2i + 3j
b y = 7j
c z = i + j - k__.. .. 2
2 a A B =3
- 1...b C D = 6
- 1
0...
c EF = 0
3 a =
b =
1
- 3= 3 i - 5j
- 5
- 24 = - 2i + 4j
3c = 8 = 3i + 8j
0d =
6= 6j
-3e =
6= - 3 i 6j
4 a 5b J1 = 3.16
c J 9 = 5.39d 5.3
e J 9 = 5.39
5 a J38 =6.16
b .J26=5.1 0
c 3
d 7
e F = 1.41
Exercise 128
1 a c = 3b1
d = a2
e = -5 b
f = -2 a
b They are per pendicular .
2 a , b, e
3 a - 247
bs 8
4 t = - 25, s = -55 a OG = j + k
....b BD = - i - j + k
....c A D = - i + k
_____ ... ... 1d OM
2i + j + k
,..6 a OG = 4j +3 k...
b BD = - 5i - 4j + 3k...c AD = - 5i + 3k
_____,...,..
d OM = i + 4j + 3k2
Exercise 12C_ ... -5 ... 5
1 PQ = QP =1 I
__,...,... - 42 a AB =
4
____.,.. 4b BA =
4
_.,.. -7c AC =
3
____.,.. 3d CB =
- 7
3 a 2i - 3j + 5kb - i + 5 j 6k
c - i + 5j - 6kd i - 5j + 6k
5...
4 L M - 4-3
5 us = 2i + 8j - 3k6 x O , y = 7 , z = 9
Exercise 1 2
- 3 3
5 AC = -5AB =-4 4
6...BC = - 10 . Any two of
8
these are scalar mult iples ofeach other
3
16
- 3... ... ....orBC= 2 soAB = B C
- 8
-3 - 6... ...
3 P P 2 = - 1 p lp3 = - 2
0 0
- 3
; 4 2_ 43
54 X = 3; AB : BC = 1 : 2
swers
-
8/10/2019 Math IB SL Answer Sheet
43/68
Exercise 2E
5...
1 AB = 0 ; J 9 = 5.39-2
____..,.. ...
2 IAB I = Jl29 IA C I = J42lo r;;::;;
IBC I = -v129. Two sidesequal length therefore
isosceles. Angle C A B = 46 .8°
3 t = +6
4 x= ±JS
5 a= +2
6 a 15b 10
c 13
Exercise 2F
1
2
3
4
5
6
7
8
(3.2
(4)2
5 + 5 = 1
J = 1. _(4i - 3j5
- 11
= = I -5J42
4
I.( 2i + 2 k )3
1
J5
_2_(2i j )J5
1
7---= -3J14
2
9 acosB
sinB
bcos a
0
sm a
Exercise 2G
1 a Si + jb 2i + 3j
c 2i + 4j
d 8i + 4j
e - i - 3jf 2i
Answers
2 a
b
c
d
e
- 22
l
8
- 1.5
- 3
-5
15
3- 34
3 a 8i j 3k
b - i + 2j + 3k
c i - 2j - 3kd 8i - 6j - 1 0k
4 19
4 X= -5.5 y = 3
z =-610
- 16
5 X - 4 .5 , y = 10.5
6 s = 4 . 5 t = 9 u = 9
Exercise 2H
4 a i b - a..11 b - a
iii 2b - 2a•IV b - 2a
v 2b - 3ab A B is parallel to and ha lf
the length of FC
c FD and A C are parallel__. .. ...
5 d M X = MP
Investigation cosine rule
Exercise 2
1 a - 18
c 20e - 13
b 5
d -13
2 a -9 b 20
3
c 20 d -5 8
e 13a Pe r p endicul ar
b Neither
c Parallel
d Neither
4
5
67
e Pe r pe nd i cu la r
f Parallelg Parallel
- 152
d = 1
3
45°
a 94.8 °
b 161.6°
c 136.4°- 1 1
8 a AB = A C =5 -2b - I I
- 11c J26JS
9 a 79.0°
b 90°
c l l 8 .1°
10 a A B = JU; A C = J 61
b cosBA C = 0 2617 26
c 10 .5
11 54.7°____..,.. .
12 a OA ·O = 0 thereforeperpen d ic u lar
b J6i13 A = 2.5
14 A = +915 p = ±3
Exercise 12J
1- 1
a r =2
b- 1
r=0
3
c r= 1- 2
3+ t
2
5+ t
- 2
3
+ t - 28
d r = 2j - k + t(3i - j + k)
24 - 1
a E.g. r = + t5 - 7
4b E .g. r =
- 2
-
8/10/2019 Math IB SL Answer Sheet
44/68
-
8/10/2019 Math IB SL Answer Sheet
45/68
Chapter 3Skills check
1 a f i2
b J3
f id -
f j2 2
2 af j b - 12
c 1
3 a - 1.48d - 0 .5b +2
4 a -0.182, 2.40 b +1.14
Investigation - Sine cosineand tangent o the unit circle
1 sin90° = 1 cos90° = 0 tan90°does not exist
2 sin180° = 0 cos180° = 1tan180° = 0
3 sin270° = -1 cos270° = 0tan270° does not exist
4 sin360° = 0 cos360° = 1tan360° = 0
5 sin -90° ) = -1, cos(- 90°) = 0,tan( -90°) does not exist
6 sin( -180°) = 0, cos( -1 8 0°) = -1 ,tan -180°) = 0
7 sinO = 0, cosO = I , tanO = 0
8 •1C 1C 0 1C
sm2 = 1,
cos2 = , ta n 2
does not exist9 sin1t = 0, cos1t = - 1, tan1t = 0
. 3rc 3rc 3rc10 s m - = -1, c o s - = 0, tan-
2 2 2does not exist
. 3rc 37r11 sm = 1 cos- - = 02 231Z d .
ta n - 2 oes not exist12 sin41t = 0, cos41t = I,
tan41t = 0
Exercise 3A1 a
b
nswers
c
d
e
f
g
h
2 a
b
c
d
-270°
5rc...--r-... 3
1l
2
e
f
g
h
1l
3
21l
For questions 3 to 8 , there are manyother possible correct answers.
3 a 120°, - 240°, - 300°
b 340° - 20° -160°
c 255° 285°, - 105°
d 65°, -245°, -295°
4 a - 35°, ±325°
b - 130°, +230°
c -295°, +65°
d 240°, ±120°5 a 230°, - 130°, - 310°
b 280°, - 80°, - 260°
c 40°, - 140°, - 320°
d 155°, 335°, -2 05°
6 arc 4rc 57r
3 3 3
b 7rc _ rc _ 37r4 , 4 4
c 31t - 4.1, 4 I - 2 1t , 1t - 4 .1
d 1t + 3, 21t - 3, 3 - 1t
7 a - 1Z + 111Z6 - 6
b -1, ± 1 - 21t )
c -2.5, ±(2.5 - 21t )
d 31Z + 7rc5 , - 5
-
8/10/2019 Math IB SL Answer Sheet
46/68
8 a51l 31l 71l- - - - -4 4 4
b 1.3 + 7r, 1.3 - 7r, 1.3 - 27r
c121l 21l 9rc
- - - -7 7 7
d 27r- 5 7 r 5 - 5 - 7r
Exercise 138
1 a 0.940c - 0 .342
2 a
c
12
1
b 0.342d - 0 .940
J3b - -2
d J32
3 a 0.8 b 0.6 c 0.6
d - 0.8
g - 0 .8
4 a .b
db
4e -3
h 43
b a
e a
g a h b
Exercise 13C
f
c b
f b
1 a -300°, -240°, 60°, 120°b ±120°, ±240°
c -315°, -135°, 45°, 22SO
4
3
d - 360°, - 180°,0°,180°,360°
e +45°, + 135°,+225°,+3 15°f +30°,+150°,±210°,±330°
2 a _11Jr rc _c _ rc6 6 6 6
b 0, n +2n
+ 1l 11 rcc _- , +-6 6
d - .1C 31l2 2.rc 21l 4rc 5.rc
e +- + - + - + -,_ , - , _3 3 3 3
71l 31l 1l 51Zf - - - 4 4 43 a 0°, 360°, 720°
b -135°, -45°,225°, 315°,585°, 675°
c -225° -45° 135° 315°,495° 675°
d ±60°, ± 120°, 240°, 300°,420° 480° 600° 660°
4 a2
5.rc .rcb - - ·6 ) 6
.1C 3.rcC +- .+-- -4 4
.rc 5.rcd + - , + -
6 6
Exercise 13
1 a +15°,+ 165°b -165° -105° 15° 75°c 90°d +180°
2 a 5 1C .1C 7 Z' ll.rc- - - 2 12 12,.1211.rc 7 rc .rc .rc 5 rc 3rcb - -
2 ) 12 4 ) 12 12 4.1Cc +-2
2.rc 4 rc3 a 3
3.rc 7rcc
4
Exercise 13E1 a sJU
2 a
3 a
c
4 a
c
5 a
c
6 a
c
7 a
184- 5
9Jl1
5
7-18j63
32
j6331
3-52425
7- -25
336-625
a
2ab
Exercise 13F
d 3.rc+-4
b 7 1C 3.rc 11.rc6 ) 2 6
d ..2
7b c - sJil7
b
18
1
9c 4 J5
b sJU18
d 5Ji.T.7
b
d
b
d
31
32
31163
512
4
57
25
b7
527d -
625
b
db 2 2- a
1 a 30° 90° 150°b 22.5°, 112.5°c 135°d 45° 135°
2 a -150° , -120° ,30° ,60°
b 90°
c + 150°, +30°
d -90°, 30°, 150°3 a 0 1r
b .1C 7.rc8 8
02.rcc -
3
4 a.1C 5.rc
8 8
c 0, 7r
6 k 6
7 b = 8
Exercise 13G
b .1C2rc 3rc
d 4 4
1 - 346° - 194 14° 166°
2
3
+27°, 333°
244° 296°
4 55°, 235°, 415°5 - 5.33, - 4.10, 0.955, 2.19
6 +1.71, 4.58
7 - 0 .739
8 - 0 .637, 1.41
Investigation: graphing tan x
1
Angle Tangent
measure x) valuedegrees) tan x)
0 0
- 30,+30 11
-J3'J3- 45,+45 - 1 , 1
-60, +60 - ../3 ) J3120 - /3135 - 1
1150 -
J3180 0
2101
-J3
225 1
240 .J3300 - ../3315 - 1
3301
- J3360 0
An swe rs
-
8/10/2019 Math IB SL Answer Sheet
47/68
3 ta n + 90° and tan + 270°are undefined. The limitof the tangent as the ang leapp roaches ± 90° or ± 270° isinfini te . Asymptotes are oftenshown on graphs for valuesth at do not exist.
Exercise3H
- 297° - 117° 63° 243°' ' '
-107° 73° 253°' '
3 124° 304°'
4 38°, 142°, 398°, 502°5 - 5.88, - 2.74, 0.405, 3.55
6 -1.88, 1.26
7 4.558 - 4.66, 1.20, 2.28, 4.77
Investigation: transformationso f sin x and cos x
Y
X
y
X
0 X
3
0 X
4 Y
0 X
Answe rs
5
Exercise 3
' I'
v"-,
' \ '..I•-2n -n
3 y
r-
J J'1.17' 0
".4
4
\
\11T [\.
\
5
v \
y
//
J
I/
I I\n.," . I - .rr \./
6 y' 11T 0-
\. ,..;\ 7
A
7
11T - r;r\I Iv
1.. /
X
y
0 TI"i -
4- 7 '/ -
y
A........
r\. /i' ' ./ v
0
J) TT X
y
1\v \0 \ I
I• \ v
y
11v \
I r \0 v \
[\, /
1\ J\ v
y
0 .T/ \
/ I[ .., I
/l l"'
rr x?_
/
.
II
2-
\..
2
\\
8 y
0
I i J vI I. rr - 0 ) 'TT Xl 4
-2
9 y =cos x -
2
; ) or
y =sin
1 y = sinx +l
y = tan ( x - : )
12 y = cos ( x - : ) - 1.5
Exercise 3J
I \.I \
' ftr- ' \
j \1 '
' ' -
J
3
-I
v- rr
4
1 1
1 \n- -Jfrr ir
y
1 \ 'I \.
I \() "\d \.y
A
I i\2
I
0 . 7-
"'-1 J
yA
,.,..
/15 I
I
A
y
I \ 1 1
0
A/
)
'
I/
)
1
71,.
\ 1 : - \ ;-. \ f r\ IJ: r•
5 y·"1 \ / 1
1 T T 0 T -21 -
J \ i/ \ J \
frr X
'rr X
'rr x
\
') rr X
prX
-
8/10/2019 Math IB SL Answer Sheet
48/68
6 y
,) ' I I '1\ /\--3-'-
- - - c- 1-2 1 - 1 .o
-
.
7 yr - r
v - \2 T 0 \..;. -
- I ',_
8
9 y=7.5sinx
10 y =cos(0.25x)
11 y=tan(0.2Sx)
-....:
12 y = - 3cos 0.5x) or
y = sin(x J-1 5
Exercise 13K
'
' i- ?
I
1 For questions 1 to 4, answers may vary.
1 y= 3.Ssin(x- 2; -I s ,y = 3.5cos( x +
5: )-1.5
2 ))- 2,
3 y =2 in(2x) +1,
y = 2cos( ))+1
4 ;)).
y x+: ))
X
5 y
6
X
72-
1\ \
1•
I c X' Iv v v v
- ,_2
8 y76-
Vi\ 5-4-
. I- - -· -- 1- - - -
-: 7l -1 -lQ )r 2f- 3rx
Exercise 13L
1 a ,b
f'lot2 f'lot),y1a4.8cos
-
8/10/2019 Math IB SL Answer Sheet
49/68
3 a, b
F"l.;.t2 F l.;.n,y1a0.8cos
-
8/10/2019 Math IB SL Answer Sheet
50/68
8
9
10
11
2sin (2x)
cos 2 (2x)
- 8.1Z COS .7rX)(1r x )
[cos( sin x)] cos x
3x 1a
b - 4cos 3 x sin x
12 a 3cos 3x - 4)
b - 9sin (3x - 4)
Exercise 148
1y - 1 = t(x-;-} -1 = -l(x-:)
2y - 2 = 4(x- ;} y - 2 = - ;)
3 - 2I4 a - - b - 2sin (2x)2
C y + i --13:.. 51Z3 ' 3
Exercise 4C
1
2
3
4
5
6
7
8
9
10
11
- 12sin (
1( l+cosx)
e sin2l cos 2t
2ex sin xt
- + tantcos t
3e 3... cos 4x - 4e 3... sin 4xI
cos2 2xJ tan 2x
cosx .lnxsmxX
.SLO X
or - tanxcosx
2 I Xa b - cos -
c
X 2 2
I 2 X 2 . X- In 3x c o s - + - s m -2 2 X 2
12 a= 1 b = 2'
Exercise 14
1 Relative minimum: -2)·3 ' '
relative maximum: ( f 2)2 relative minimums :
( I).e;, - 3 ; relativemaximums : ( :..).).(5 6 2 6 ' 2
3 d .creasmg: - < x < JZ;2
increasing 0 < x < :..; concave2
down: 0 < x < n; relative
maximum: (f·1)f(x}
1
1l 1l 37l- - -4 2 4
4 decreasing:1Z 1Z J1Z .
O 0relative minimum atx = 4.91
7 a j ' (x) = - r Sin X 2x COS Xb minimum : - 11.6;
maximum : 7.09
8 ad 1 fJ) 2 . 0 4sinBcosB=- Sill - r = = = = =
.J25 - 4sin 2 B
2. O 2sin2B
o r - sm r = = = = =
. J z s -4s in2
)
b
1r 3Jr 5Jr 7 f(x}0 < X < - < X < - - < X < ; r · (5.05 , 2.16}
f X)
1
8 8 8 8 2
relative maximum: ( ; 1}
relative minimums:
1
0-1
1l
451T M . 2n 64 2 4
n o) o ; -2 d'(O) - -2sin8 - -4sm6cos6(1.23, -2.16} h5- 4Sif1 28I I 4 )x points:
( T l ) (37T 1) (57T I) (71Z 1)8 1 2 I 8 2 J 8 1 2 J 8 l2
IT IT 37l 1l 57l 37l 71l IT X4 8 2 8 4 8
-3
•c 1 The blade is closestthe center of thewheel when d(8) hasa relative minimum orat an endpoint. Thereis a relative minimumwhen d (e) changes fromnegative to positive at8 =n. Testing theendpoints and critical
Answe rs
-
8/10/2019 Math IB SL Answer Sheet
51/68
numbers we findd O) = 7, d 2rr) = 7and d rr) = 3. So theclosest distance is 3metres and it occurswhen the angle ofrotation is rr.
ii The distance isch a n ging fastestwhen d ((J) has arelative minimum ormaximum. This occurswhen e s 1.23 radiansor 5.05 radians.
Exercise 4E
1 2sin x 3cos x + C
2 x 3 + 3s inC