1
a r"" -0.832
b .As the number of tomatoes in a bag increases, the median
weight of tomatoes in the bag decreases.
j'(x)
Now
Y""
=
If x
d
Y""
13, then
-8.21(13)
=
ii If x
y""
20, then
-8.21(20)
P(X ~ 15)
P ( X ; f.£ ~
15;
f.£)
P ( Z ~ 15 ~ 13)
=
118 g.
60.2 g.
=
b f(l)
.'.
lim (7 - 2x - h)
=
7(1) _12
6
1'(1)
= 7 - 2(1) = 5
the tangent has equation
y-6=5
x-I
which is
0.613
a
y - 6
= 5x -
or
= 5x + 1
y
y=f(x)=-2x2+3
Let
=
dy
dx
lim f(x
+ h)
=
- f(x)
h
h~O
6.97
=
f.£) = 0.035
f.£
.
+ h? + 3J- [-2x2 + 3J
lirn [-2(x
h
lirn -2(x2
"" -1.812
=
=
h
.
~-4xh-2h2~
hrn ------------~--lirn -2J{(2x
= lirn
P (X < 33)
= 0.262
= 0.738
<
=
P(X ~ 33)
33;
f.£)
h~O
b When
0.738
.'.
C
(J'
+ h)
{as h
# O}
f.£ "" 0.6372(J' .... (2)
dy
-=-4(-1)=4
dx
x=-I,
the tangent has gradient 4.
When
.'.
x=-I,
y=-2(-I?+3=1
the point of contact is (-1,1).
Since the gradient of the tangent is 4, the gradient of the
- (2) gives
(24 - f.£) -
(33 -
normal is - ~ .
f.£) "" -1.812(J' - 0.6372(J'
So, the equation of the normal is
y -1
1
.'. -9 ""-2.449(J'
.'.
f.£ "" 24
-2(2x
= -4x
33 - f.£ "" 0.6372
33 -
h
+ h)
IJ{
h~O
24 - f.£ "" -1.812(J' .... (1)
P (Z
+ 2xh + h2):;Y1f + 2X2 A
h~O
(J'
Also,
5
h~O
24--
O}
the point of contact is (1, 6).
Now
P (X ~ 24) = 0.035
(z ~ 24;
#
h~O
h~O
P
{as h
IX
= 7-2x
2
-(J' "" 0.287
(J' ""
X(7 - 2x - h)
h~O
= 0.613
= 0.613
=
(J' ""
+ 1.812(3.67)
3.67
-4
x-(-I)
"" 30.7
. me mean f.£ "" 30.7 and the standard deviation
(J' ""
4y - 4
.'. 4y
3.67.
= -(x + 1)
= -x + 3
. -x+3
.'. y= -4I
SOLUTIONS TO TOPIC 6 (CALCULUS)
d The normal meets the curve where
-x+ 3
-4-=-2x
CALCULATORS
a
f(x
f(x) = 7x - x2
+ h) = 7(x + h) - (x + h)2
= 7x + 7h - (x2 + 2xh + h2)
= 7x
+ 7h -
X2 - 2xh - h2
-x + 3
8x2 - x - 9
(8x - 9)(x + 1)
.'. x
When
x=~,
=
2
_8X2
+3
+ 12
= 0
=0
=
-1 or
~
y=_2(~)2+3
=-2(*)+3
15
32
So, the normal meets the curve again at (~, ~).
Mathematics
+ X2
h
= lim
e The estimate in i is an interpolation whereas the estimate
in ii is an extrapolation. The estimate in i is therefore more
reliable.
3
X2 - 2xh - h2 - 7x
7h - 2xh - h2
= h~O
lim
+ 224
"" 60.2
So, the median weight is approximately
+ 7h -
h
+ 224
"" 118
So, the median weight is approximately
f(x)
h
7x
+ 224
-8.21x
+ h) -
f(x
lim
h~O
There is a moderate negative linear correlation between the
variables.
C
=
SL - Exam Preparation
& Pradice
Guide
(3rd edition)
3
a
3x + 1
1.
_1.
= --,fi = 3x + X 2
Y
c If
2
dy
3
-dx =
-x
_1.
=
dy
dx
0
then
V3-x-.
2~
1_1
2
-x
-
2
2
2
3
2,fi -
2+x
2.,;3=X
1
2x,fi
.'.
=
_(x4
+ 9)-2
y
=
x (1 - x )
2
2
X
+ 3)5
= 5(2x + 3)4
=
dy
dx
(4x3)
l(
2)_1.
2 1- X
2 x (-2x)
3
X
-;::==
.~--+---'-----I~
x
Sign diagram for f'(x):
~
~
(2x
There is a local maximum at (~,
X2
+ 3)4
= 10(2x
7
3x
X2 - 2x
a
g(x) = -x cos x
"
b g'
Ci)
= - cosen
+
=_1+2l:(v'3)
2
3
When
i sin( i)
.'.
2
1'(x)
=
=
x3 - 2X2
3x2 - 4x
1'(x)
=
3x2
i' (x)
= 0 when
f(x)
b
-
x = 1,
J3TI
Y =
y
=
(3x
+ 1)
!
dy
dx
,
=
dy
dx
when
x = 1,
y=
(5x_x2)1,
For
f(-I)
f(l)
.'.
4'
~ at their point of
intersection.
I
c The equation of the common tangent is
+/7'.- ~x
1
-1
0
\
y - 2
x-I
1
iJ.
which is
= (_1)3 - 2(-1)2 =-3
8
3x - 4y = -5
h(t)=100+32t-4t2
a
=
h'(t)
x = -1.
is defined when
4y - 8 = 3x - 3
4
x = O.
= 13 - 2(1)2 = -1
a y = (2 + x)V3 - x
+1
dy = ~(5X_X2)-!(5-2x)
dx
b h'(t)=0
32 - 8t
when
32=8t
.. t
6
(3)
3
Both curves have the same gradient
x = 0 or ~
the least value is -3 when
-!
dy
_1.
dx = ~ (4 2)3 = ~.
when x = 1,
4x
= 0, so the greatest value is 0 when
and
+ 1)
H3x
2V3x
or
Also,
1
3
3
= x(3x - 4)
f(O)
=
= 2.
"'1-
Sign diagram for f'(x)
on -1 ~ x ~ 1:
- X2
- X2
they meet at (1, 2).
b For
So, the tangent has gradient
a
+ 1 = 5x
+1= 0
.'. x
_ "v'3
1
--6--2
5
¥ vll).
(x - 1)2 == 0
.'.
g'(x)=-cosx+(-x)(-sinx)
= - cos x + x sin x
3
V3x + 1 = .j5x
a The curves meet where
.'.
4
2x
x=~
+x2 x
= 2xy rr=::
1 - X2 y
- x)
== 6 -
!
dy
2)!
dx = 2x(1 - x
d
+ x = 2(3
2
.'. 3x =4
dy
dx
c
=J3=X
.'. 2 + x
y=(x4+9)-1
b
=0
2 3-x
3 - x?
0
=4
Sign diagram for h'(t):
.'. x ~ 3
the domain is
b
{x
Ix
~ 3}.
So, the height is maximised when
i
y=(2+x)(3-x)2
dy = (1)(3-x)!
dx
Now
=
(3 - x)! - ~(2
= J3=X
+ x)(3
=
100
+ 32(4)
- 4(4)2
=
164
the maximum height is 164 m.
+(2+x)[~(-I)(3-x)-!J
{product rule}
h(4)
t = 4 s.
9
a
y=
- x)-!
..
_ ----:2 =+=x=
2J3=X
b
dy
-dx
.
Y
=
1- 2x
-~
1
--x
3
1- 2x
= --,-
=X
_1.
3
-2X3
1
X3
_1
3 -
= 2x(1 + 2x)
4-1.
-x
3
3
4
dy = 2(1 + 2X)4 + 2x(4)(1 + 2X)3(2)
dx
= 2(1 + 2X)4 + 16x(1 + 2X)3
III
Mathematics SL - Exam Preparation & ProcIice G::i::i. I:lF..:iiii:lli
a
-0.832
T ~
j'(x)
Now
b As the number of tomatoes in a bag increases, the median
weight of tomatoes in the bag decreases.
=
If x
d
13, then
y ~ -8.21(13)
+ 224
=
ii If x
20, then
y ~ -8.21(20)
+ 7h -
7x
lirn
lirn X(7 - 2x - h)
=
+ 224
b
60.2 g.
=
f(l)
lirn (7 - 2x - h)
h--->O
=
7(1) - 12
6
the point of contact is (1, 6).
.'.
e The estimate in i is an interpolation whereas the estimate
in ii is an extrapolation. The estimate in i is therefore more
1'(1)
Now
=
=
7 - 2(1)
5
the tangent has equation
reliable.
=
y-6
P(X ~ 15)
P ( X ; J.l ~
15;
J.l)
=
0.613
=
0.613
5
x-I
which is
y - 6 = 5x - 5
or
P ( Z ~ 15 ~ 13) = 0.613
2
=
y
5x
=
dy
dx
+ h) -
lirn f(x
=
f(x)
h
h--->O
17~ 6.97
=
= 0.035
P ( Z ~ 24;
J.l)
=
0.035
24 - J.l ~ -1.812
17
24 - J.l ~ -1.81217
P(X ~ 33)
P (X
P (Z
<
=
0.738
J.l)
=
.... (1)
h
lirn ~-4xh-2h2~
h
Urn -2J{(2x
h--->O
=
0.738
+ 2xh + h2) ;:Y'3'+ 2X20
_
h--->O
= 0.262
< 33) =
33;
=
+ 3]
h
lirn -2(x2
h--->O
[-2x2
+ h)2 + 3] -
lirn [-2(x
h--->O
P (X ~ 24)
+1
y=f(x)=-2x2+3
a Let
2
- ~ 0.287
17
Also,
{as hiO}
IX
= 7-2x
~ 60.2
So, the median weight is approximately
+ X2
h
h--->O
118 g.
X2 - 2xh - h2 - 7x
7h - 2xh - h2
lirn
=
3
h
h--->O
~ 118
So, the median weight is approximately
f(x)
h
h--->O
=
+ 224
+ h) -
f(x
lim
h--->O
=
There is a moderate negative linear correlation between the
variables.
c y ~ -8.21x
=
+ h)
IJ{
Urn -2(2x
h--->O
+ h)
{as hi
5
O}
= -4x
b When
dy
-=-4(-1)=4
dx
x=-I,
.' _ the tangent has gradient 4.
33 - J.l ~ 0.6372
17
(
33 - J.l ~ 0.637217 .... (2)
- ( ) gives
J.l ~
24
+
So, the equation of the normal is
y -1
1
-9 ~ -2.44917
SOumONS TO TOPIC 6 (CALCULUS)
y=-2(-I?+3=1
(-1,1).
i.
.. 17~ 3.67
1.812(3.67) ~ 30.7
the mean J.l ~ 30.7 and the standard deviation
x=-I,
thepointofcontactis
Since the gradient of the tangent is 4, the gradient of the
normal is -
(24 - J.l) - (33 - J.l) ~ -1.81217 - 0.637217
.'.
When
.'.
-4
x-(-I)
4y - 4 = -(x + 1)
:.4y=-x+3
-x+3
.'. y=-4-
17~ 3.67.
d
The normal meets the curve where
-x+3
-4-=-2x
1.ATORS
f(x)
f(x
-x + 3
8x2 - x - 9
(8x - 9)(x + 1)
= 7x - X2
+ h) = 7(x + h) - (x + h)2
= 7x + 7h - (x2 + 2xh + h2)
= 7x + 7h - X2 - 2xh _ h2
6
2
+3
= -8X2 + 12
=0
=0
.'. x = -1 or ~
When
x
=~,
y
=
_2(~)2
+3
=-2(*)+3
15
32
So, the normal meets the curve again at (~,
.iic;""""dics
SL - Exam Preparation & Pradice Guide (3,d edition)
H).
3
a
c If
=
dy
dx
0
then
V3-x-
2+x
2V3-x
=
0
2 +x
2v'3='X
=
v'3='X
2+x
..
= 2(3 -
x)
2+x=6-2x
b
.'.
=4
3x
x=~
c
x (-2x)
...•.•
~_+_--1-__
Sign diagram for J'(x):
d
~
+ 3)5
= 5(2x + 3)4
=
y
dy
dx
(2x
. a Ioca I·
There IS
maximum at
X 2
+ 3)4
= 10(2x
7
a The curves meet where
+ 1 = y'5x
+ 1 = 5x +1= 0
3x
X2 - 2x
4
a
g(x) = -x cos x
b g'
(i) =
- cos(
sinx)
.'.
x = 1,
When
i) + i sinG)
..
=_1+1!:(V3)
2 3 2
_ 7rV3 1
-6- -
b For
= x3 - 2X2
J'(x)
= 3x2 - 4x
2.
y=
(3x+1)!,
=
:~
~(3x+1)-!(3)
7rV3-3
=
x
dy
dx
1,
when
3
=
y=(5x-x2)!,
For
J'(x)
4:.
:~=~(5X_X2)-!(5-2X)
= 21(4-!)3
dy
dx
x = 1,
3
= 4:.
%
Both curves have the same gradient
intersection.
f' (x)
y - 2
3
-- 4 - x-1-
which is
4y - S
3x - 4y
f(O) = 0, so the greatest value is 0 when x = O.
f(-l)
and
.'.
f(l)
= (_1)3 - 2(_1)2 =-3
= 13 - 2(1)2 = -1
the least value is -3 when
x
=
8
a
het)
h'(t)
-1.
=
a y
=
(2
+ x)v'3=X
is defined when
.'.
the domain is
{x
Ix
X ~
0
when
32=8t
= (1)(3 -
~ 3}.
+ (2 + x)[~(
=v'3='X-
- ~(2
+ x)(3
h(4)
=
100
+ 32(4)
- 4(4)2
s.
= 164
the maximum height is 164 m.
-1)(3 - x)-~l
{product rule}
= (3 - x)!
=4
So, the height is maximised when t
Now
x)!
• t
Sign diagram for h'(t):
y=(2+x)(3-X)2
dy
dx
=4
3
1
b
3
= 32 - St
b h'(t)=0
3 - x?
= 3x = -5
100+32t-4t2
.'. t
6
at their point of
c The equation of the common tangent is
or
Also,
1
3
f(x)
Sign diagram for
on -l~x~1:
0
-6-
= 3x2 - 4x
= x(3x - 4)
J'(x) = 0 when x=Oor~
b
- X2
X2
they meet at (1, 2).
when
a
V3+I =
Y =
X
2
So, the tangent has gradient
5
=
=
.'. (x - 1)2
= - cos x + (-x)(= - cos x + xsinx
g'(x)
3
(4"3'"3
10 V"3
15).
V3x
.'.
1 x
9
a
Y
- x)-~
dy
..
2+x
2v'3=X
b
1 - 2x
= --
ijX
1 _i
-dx- - --x
3
Y
=
= 2x(1
1 - 2x
--1
-
X3
=
_ .1
X
3 -
2x
£
3
4-.1
3 -
-x
3
3
+ 2X)4
dy = 2(1 + 2X)4 + 2x(4)(1 + 2X)3(2)
dx
= 2(1 + 2X)4 + 16x(1 + 2X)3
IfiI
Mathematics
SL - Exam Preparation & Practice Guide (3rd edition)
s
10
3
a
3-2
dy
dx
2
dy
dX2
b
=
-6x-3
=
18
18
X4
-4
x
For x
>
When
x=l,
13
aCt)
d
x-4
= --2
is undefined when x = -2,
x+
is a vertical asymptote.
a f(x)
x-4
b
f'(x)
so x = -2
1,
+(x2) +-
(x - 4)1
2)2
{
quotient rule
}
14
Now
= 36 ms-1
Vmax
+ 2)2
2)2 =
when
_
x-3
.'.
.'. "6x - 25y
12
(x + ~)
a y =
equation is
x-(-l)
=
.'.
b
- 18
- 18
As x
-> 00,
y ->
As x
-> -00,
y
23
x=
0, so
x=
c Now
y = 5x + 3
0
2y
d
c Since
d
+ x-1)3
dX2
0
When
15
a
:2)
=
12t - 3t3
vet)
=
s'(t)
=
12 -
aCt)
= v'(t) =
-18t
b vet)
= 12 -
3x - 2 = 0
=
aCt)
when
or when
.'. x2
=
-1
or
3(4 - 3t2)
- tV3)
or
=
x=
& Pradice
Guide (3rd edition)
+
2
o
~t
v'3
-18t
which has sign diagram:
1
1- - =0
X2
X2
=
• t
o
1
Speed is decreasing when
opposite sign.
±1
.:.
SL - Exam Preparation
{using b}
0
ge
X
1
x
where
ge
which has sign diagram:
x =--
- 2
+1
(x + .!.)3 (1 _ ~)x2
.'.
+x
3x - 2
y = 23 + 2(2) + 1 = 13
set)
for all X:f= 0, the curve is concave
(x+~r=o
+5
+3
So, the tangent meets the curve again at (2, 13).
(1 _ x-2)
1-
x = 2,
= 3(2 + tV3)(2
~~ = 0
Mathematics
-
upwards either side of its asymptote.
dy = 4
dx
5x
.'. x=-10r2
r(
>
=
(x + 1)2(x - 2)=
00
(x + ~
= 4
-
-> 00
u= ( x+;; 1)4 = (x+x -1)4
4 (x
y
meets y = x3 + 2x + 1
3
x + 2x + 1 = 5x + 3
.'. x3
there are no horizontal asymptotes.
~~ =
= 5
(x + 1)2(x - 2) = (x2 + 2x + 1)(x - 2)
= x3 - 2X2 + 2x2 - 4x
= x3
is undefined when
4
has gradient 5, and its
.'. y+ 2 = 5x
is a vertical asymptote.
b
(-2)
y -
6
+ ~) = 6x
25y + 5 = 6x
t = 6 s.
the tangent at (-1, -2)
..
fg
-2"5
25(y
t
1'(X)~3X2+2
the tangent has equation
y-(-~)
I'
6
1'(-1)=3(-1)2+2=5
the point of contact is (3, - ~).
1'(3) = (3:
+1-
f(-I)=(-1)3+2(-1)+1=-2
Now
~1,
+~-,
I 72\"(if
Let f(x)=x3+2x+1
a
.'.
c Since f(3) =
>0
v(6) = 216 - 324 + 144 = 36 ms-1
6
(x
ms-2
t
v(2)=8-36+48=20ms-1
so Y = 1 is a horizontal
.'.
__ l(x
+ 24
where
= v'(t)
is the larger of v(2) or v(6)
Now
+=
ms-1
o
Vmax
->
24t
Sign diagram for v'(t):
x
f(x)
0 is at
=3(t-4)(t-2)
x
As x -> 00,
asymptote.
>
b v'(t)=3t2-18t+24=3(e-6t+8)
1-.1
-
(1++)4=16.
ge +
-
= 3t2 - 18t
dX2 = 2 sin 3x + 2x(3 cos 3x) + 6x cos 3x
+ 3x2(-3 sin 3x)
= 12x cos 3x + (2 - 9x2) sin 3x
-x+2-1+l
u=
vet) = t3
a
2y
-
0, there is a local minimum at x = 1.
{product rule}
= 2x sin 3x + 3x2 cos 3x
f(x)
is:
So, the minimum value of the function for x
(1, 16).
Y = X2 sin3x
~~ = 2x sin 3x + X2 (3 cos 3x)
11
dy
dx
Sign diagram of
Y = X2 = X
El
vet)
and aCt) have the
the speed is decreasing for 0:( t :( ~'
------------------------------~----=====Iiiiii
ii Velocity is decreasing when v'(t)::;; O.
Since v' (t) = a( t) ::;;0 for all t:;:, 0, the velocity
is decreasing for t:;:, O.
d2y
Sign diagram of dX2
x
When
16
a
g'(x)
So, there is a non-stationary
(2 - 2x - 2x
b
+ x2)e-(X+2)
g'(x)
=
(2x - x2)e-(x+2)
g' (x)
=
0
when
2x - X2
So, y
ii gl/(x)
= g(x).
=
.--~--r--~4---6~-~x
=
0
=
0 or 2
s" (x) = 0
..•••
_----'1
__
+
o
e The equation x3 - 12x2 + 45x - a = 0 has 3 real roots
if x(x2 - 12x + 45) = a has 3 real roots.
-'-1__
• X
.'.
X2 - 4x
+2 =
18
0
4
± -J16
x=
Sign diagram of gl/(x) is: •
= g(x)
x:;:' 2 +.J2.
17
f(2)
- 4(2)
2
Now
+ 1
2-v2
is concave up for x::;;
1'(x)
+ 45)
meets
54
= 8x-2
= 8 x T2 = 2
= -16x-3
1'(2) = -16
= -16
x3
= -2
8
1 + •
2+v2 x
2 -.J2 and
the gradient of the normal at (2, 2) is ~.
y - 2
1
-- 2 - x-2-
the equation of normal is
which is
2y - 4
or
x - 2y
=x= -2
2
y = x(x2 - 12x
a
= x3 - 12x2
dy
2
dx = 3x - 24x
dy
dx
+ 45)
+ 45x
19
a
+ 45
..
= 6x - 24
dx2
?
=
3x- - 24x
= 3(x2 - 8x
x
eX
y=
d2y
and
b
<a<
f(x)
x=2±V2
So, y
50
+ 2)e-(x+2)
when
y = x(x2 - 12x
This occurs provided
y = a in 3 places.
2
is increasing for 0::;; x ::;;2.
(x2 - 4x
y=a
(5,50)
-
Sign diagram of g' (x) is:
+ 45)
(3,54)
....... ---.-- -- .-
x(2-x)=0
x
y = x(x2 - 12x
y
- x2)e-(X+2) (-1)
+ 2)e -(x+2)
= (x2 _ 4x
inflection point at (4, 52).
The graph cuts the x and y-axes at O.
d
+ (2x
(2 - 2x )e-(x+2)
+ 180
= 52
= (2x - x2)e-(X+2)
s" (x) =
=
+ 45(4)
12(4)2
= 64 -192
_ x2e-(x+2)
= 2xe-(x+2)
.•.•
x
4
= 43 -
y
= x2e-(x+2)
= 2xe-(x+2) + x2e-(x+2)(_I)
g(x)
ii
= 4,
+
...••
------':-1
is:
ii
1(eX) - xex
e2x
2x + 1
e
dy
dx
y = e2x - 1
+ 45
+ 15)
..
(e2X)(2)(e2X - 1) - (e2X
(e2x - 1)2
dy
dx
= 3(x - 3)(x - 5)
b
I-x
J(x2
+ 1)(e2X)(2)
x3
+ e2X+1) dx = "3 + ~e2X+l + c
dy
S·19n diiagram fior dx:
20
When
x = 3,
y = (3)3 - 12(3)2
= 27 - 108
=
and when
x = 5,
+ 45(3)
y
=
+ 135
+ 45(5)
b
there is a local maximum at (3, 54) and a local
minimum at (5,50).
d2y
=6x-24
dX2
dX2
=0
when
=0
+ 2ax
+ 2a)
when x
=
-2a
0 or x = -3-
Thus p( x) has a stationary point at x = 0 for all a and b.
+ 225
= 50
d2y
=
3x2
= x(3x
p' (x)
(5)3 - 12(5)2
+ ax2 + b
p(x) = x3
p'(x)
54
= 125 - 300
c
a
i If (-2, 6) is a second stationary point, then
-2a
-=-2
3
-2a =-6
.'. a
=3
Also, p( -2) = 6, so (_2)3
x
+ 3( _2)2 + b = 6
-8
=4
+ 12 + b = 6
.'.
El
~ Mathematics
b =2
SL - Exam Preparation & Practice Guide (3rd edition)
ii
p'(x)
2
+ 2ax
= 3x
f (x) =
a
p'(x)=3x(x+2)
= In(1 - 2x) -In(x2
Sign diagram for p'(x):
_2X2 - 4 - 2x(l2x)
(1- 2x)(x2 + 2)
minimum at (0, 2).
c
If (h, k) is the second stationary point,
-2a
-3
then
2X2 - 2x - 4
(1 - 2X)(X2
= h
f'(x)
=
+ ah2 + b = k
Now
X2 + 2
h) h2 + b = k
1 - 2x
3h
b
.'. a=-2
h3
Also, p(h) = k, so
h3
+ (-
3
3
+b =
3h
2
-2
0
always, so
f(x)
x
is defined when
< ~.
+
has sign diagram:
~
.~
I
-
-1
k
+b=
f(x)
k
24
b Area A
>
0 or in other words
f' (x)
So,
.'.
is decreasing on the interval where
f(x)
j' x
i'
is decreasing for -1:0:;; x
f'(x):o:;; O.
< ~.
a Total surface area
= 2(~x x x) + 2xy
P = 48 = 3.6x + 2y
y =
>
+ 1)
+ 2)
2
h3
.'.
2(x - 2)(x
(1 - 2x)(x2
+ 2)
2
h3 _
a Perimeter
+ 2)
,
-2
2x
f (x) = 1 _ 2x - X2 + 2
there is a local maximum at (- 2, 6) and a local
.'.
(1X2-+22X)
In
= X2
48 - 3.6x
2
Thus
{area of /::; = ~ base x height}
+ 2xy
X2 + 2xy = 27.
= xy + (0.6)x2
A = x (48 -23.6X)
+ 0.6x2
b V = area of end x length
= 24x - 1.8x2
+ 0.6X2
= ~x
= 24x - 1.2x2
dA
dx = 0 when
From a,
V=
dV
dx
.'.
dy
dx
>
G)
+ 1) (2x + 1)2
+ 2 - 2)
+ 1)2
e2x (2) (2x
(2x
dy
-d
x
)
+ 1)2'
2
y = 3 metres.
x ;» 0
= 2Inx+2,
(2x
4xe2x
..
~(g _ x
x =
So,
0
e2X[4x
d y = In (;2 ~ ~)
2
27 - 32
V is maximised when x = 3 and y = ~
= 3
= 2x In x as x
~; = (2)lnx+2x
z:
S·ign diiagram sor dV
= ~(3+x)(3-x)
x>-1
x
= In(x + 1) + -x+l
X2
4
3x2
4
27
--4
=
x>-1
i7
x3
4
c
:=(I)ln(x+l)+XC:l)'
~X2 C7~X2)
27
=-x--
The maximum area occurs when x = 10.
.'. the maximum area = 24(10) - 1.2(10)2 = 120 m2
y=xln(x+l),
27 - X2
.".y=~
z:
y = x In
X2 + 2xy = 27
x = 10
.
.
f
dA
rgn diagram or
a
m3
-- lx2y
2
dA .
dx = 24 - 2.4x = 2.4(10 - x)
x
x x x y
e2x (2)
f'(x)
f"(x)
A
0
+
B
0
0
C
+
0
+
-
D
I
x
=1-2
= In(x - 4) -In(x
+ 4),
1
2x
--4-~4'
xx +
x>4
x>
4
26
a y = f(x)
= x + 5 +.±
is undefined when
x
x = 0 is a vertical asymptote.
As x -;
00,
f(x)
-;
00
As x -;
-00,
f(x)
-;
-00
there are no horizontal asymptotes.
1"*'''''''''dicsSL -
Exam Preparation & Practice Guide (3'" edition)
x = 0, so
b
f(x)
.'. f (x)
b
+ 5 + -4x
= x
f(x)
j'(x)
4
+ 5 + -x
16)
3' 27
=0
inflection
(1,1) local max
1
X2
.-_-l------~~------lo~~x
(~, 0)
= 1- 4x-2
4x3 - 3x4 = k has exactly two distinct positive
solutions then the horizontal line y = k meets the
graph of y = 4x3 - 3x4
in two different points with
x-coordinate>
O.
Thus 0 < k < 1.
c If
=1-.i.
X2
X2 - 4
----;;2
(x
non-stationary
(2
+ 5x + 4 = 0
.'. (x + l)(x + 4) = 0
x = -1 or-4
1
= x + 5 + 4x.'.
c
x
= 0 when
y
+ 2)(x
- 2)
X2
j'(x)
0 when x
=
\
+/
·/~ W
ox
f(x)
=
0 when
=
f(2)
4x
at (-2, 1) and a local
So, there is a local maximum
=
0
=
X2 +4
~=1
X2 +4
+ 5 + ~= 9
2
1
1- ~
X2 +4
= -2 + 5 + !2 = 1
f( -2)
and
f(O)
= 1 - X2 + 4
= 1, so the y-intercept =
a f(x)
+/:\_
Sign diagram for j'(x):
Now
4x
±2
=
=0
=0
=2
X2 - 4x +4
minimum at (2, 9).
(x - 2)2
d
.'. x
So, the x-intercept is 2.
b
i' (x) =
2
0 _ (4(X
(-2, 1)
-.
+ 4) - 4X(2X))
(x2 + 4)2
8x2 - 4X2 - 16
-4
(x2
-1
+ 4)2
4x2 - 16
+ 4)2
4(x + 2)(x (x2 + 4)2
(x2
x=O
a
y
= 4x3
-
3
4x
-
3x4
4
3x
y
cuts the x-axis when
=
0
f' (x)
2)
has sign diagram:
=0
~3(4-3x)=0
.'. x
=0
the z-intercepts
f(-2)=1-(~8)=2
or ~
are 0 and ~, and the y-intercept
f(2)
is O.
ii ~;
=
12x2 - 12x3
=
=
1 - (~)
=
0
there is a local maximum at (-2, 2)
and a local minimum at (2, 0).
12x2(1 - x)
c
which has sign diagram:
f(x)=l--When
and when
.'.
x = 0,
x
=
y =0
1,
=4-
Y
3
=
1
.-----------~~--------.x
there is a stationary inflection at (0, 0)
2 "-local
and a local maximum at (1, 1).
iii
d2
dx;
=
24x - 36x2
=
• - 0I + 1-
which has sign diagram:
2
0
a
X
f(x)
=
3sin(x - 4)
j'(x)
=
3cos(x
- 4)
= 3cos(x
- 4)
3
x
= ~,
y
=4
(~)
3 _
3 (~)
b
4
=~-~
_
-27
min (2, 0)
12x(2 - 3x)
29
When
4x
X2 +4
16
there is a non-stationary
x
f(x)
=
12x - 2 cos (~)
j'(x)
=
12 - 2(-sin(~))
1
x ~
= 12 + ~sin (~)
inflection at (~, ~).
m
Mathematics Sl - Exam Preparation & Pradice Guide (3,d edition)
•
f(x)
c
+ 2x) (1 + 2x)2
=
2 cos 2x x (1
=
-- .fi
units
2
the minimum distance is
8( -2)
+ 1)
f(x)
.'.
32
Y4 =
8(0) =
.'.
= 48,
xy
so
=
~
I.....~.~..~~...l y m
+ x) + 30x
=
36 (~)
(~6+ x)
= 48
0, so x>
1.
b
+48x
+ x),
Since C = 48(36x-1
1
,
(1)(x-l)2
f (x) =
_1
-(x+2)(~)(x-l)
dollars
2(1)
= 48
(1- !~)
= 48
(X2 ~ 36)
x-I
+ 6)(x
48(x
9! -12(1)9-~
=
1'(10)
3 - 6(-31)
1
9
9
2
. 9
the gradient of the normal at x = 10 is
= ~ =
Now f(10)
or
=
dC
dx
= -9
has sign diagram:
= -9
y - 4
+y
9x
=
and y = 8.
+ 90
-9x
= 94
33
x2,
C(x, x2)
a RQ = vi X2 + 64 km
{Pythagoras}
so C is (x, x2).
~IX'-21
P(0,2)
82 = X2
Ixl
2
= x2
2
4
8
8
.'.
=
8 =
b Since 8
Now
X
82 =
2
d[8 ]
dx
-
J
>
+ (x2 - 2)2
+ X 4 - 4x2 + 4
X4
-
{Pythagoras}
= 5JX2
+ 64 + 3(11
+ 64 + 33-
+4
3x2 + 4
{as
>
8
b
O}
C'(x)
=
_
6x
= 2x(2x2
-
= 2x(v2x
C'(x)
n"
+ Vs)(v2x - Vs)
2
Sign diagram for
+ -
d[8
1\
dx.tif~~1
].
-2
2
+ 64)
5x
3)
\Y.
= 0
+ 64
when
2
(2x) - 3
_ 3
ylx2
5x
+ 64
.'.
5x
25x2
25x2
x
2
3x
_1
+
,\v
16x2
.'.
X2
.', x
Mothematics SL - Exam Preparation & Pradice Guide (3n1edition)
Q
S
+ 64) + 33 -
= ~(X2
ylx2
= 4x3
(ll-Xlkm!
3x
.1
5(x2
+4
3x2
-
C(x)
P
on land
- x)
millions of dollars
0, 8 is minimised when 82 is minimised.
X4
T1
xkm
+ cost
= cost under sea
2
3x
.::...8_km
R -.:::::- __
the total cost C(x)
= 5JX2
a
x = 6
C is a minimum when
y-4
-x -10
which is
C follows the curve y
- 6)
X2
4 so the point of contact is (10, 4).
the equation of the normal is
31
~9
+ 1)
dC = 48( -36x-2
dx
x + 21
(x - 1)2
=
f(x)
b
x m
= 36y + 48x
is defined if x-I>
-1
yX
when x = ±2.
y = ~
the total cost C = 18(2y
3
cos2(3x - 4)
f(x)
v's
=
v's units
the maximum distance is
Now
± ~.
x = 0 or at
a Let sides be x m and y m long, as shown.
d
-(3x
- 4)
1'()x = ---'Cdx
O=S'::-2 (;-:"3-x-_-4"7)
a
+4
=
x
2
x
- 4)
= tan(3x
= 8(2) = yl16 - 12
and
e2sinx
= 2e2sinx cos
units when
"i
The maximum distance occurs either when
one of the endpoints of the domain.
+ 1)
x 2cos(2x
= e2sinx x 2cosx
1'(x)
-3(~)+4
=
= V/1!_1!+4
4
2
+ 1))2
[sin(2x
Jsin(2x
f(x)
=8(~)
cos 2x - 2sin 2x
(1 + 2X)2
= ~ [sin(2x + 1)]-t
cos(2x + 1)
1'(x)
8(-~)
x 2
1
=
f(x)
sin2x
+ 2x)
2(1
30
J~
1 + 2x
1'(x)
d
Now by symmetry,
= sin2x
= 3
= 3 J X2 + 64
= 9(x2 + 64)
= 9x2 + 9 x 64
= 9 x 64
=
=
9 x 4
6
=
36
{as x> O}
1
Sign diagram for G'(x):
1
-
\
W
o
1+
~x
c
G(6) = 5J36 + 64 + 33 - 3 x 6
Now
= 5Ji06 + 33 -
j(X+~rdX
= j[x2+2X(~)+Gr]
dx
18
= j(X2+2+x-2)dX
50 + 15 = 65
the minimum cost is $65 million.
=
x3
3
1 3
=}x
1
i= 0
X-I
+2x+-+c
= -
34
x
provided
1
- - +c
x
+ 2x
1
2x+l
-1
1
d
Let I be the length of the straight section.
Let r be the radius of the semi-circular sections.
The area A =
71T2 +
A-
1=
2rl
0
dx=
[~lnI2x+ll]~
=~ln3-~lnl
=~ln3-0
which is a constant
= ~ In3
7rT2
.... (1)
2r
Let p be the cost per unit length of straight wall.
e J(eX
J(e2X - 2ex
=
1)2 dx
-
2x
1
="2e
the total cost is
G
= (21 + 27rr(£))p
= (~
dG
dr
_
5;r)
71T+
=
(~+ 3; r)
=
37r)
(A r -1 +""2r
= (_
r2
p
38
8
J2 f(x)
ofJ
But
b Js f(x)
=
dx
+ Js8
f(x)
=
dx
1sin 3x + c
~ Sine;)
= ~A
-4
c
g(x)
39
+ 2)2
=
+ 2)2 dx
4
= J(x + 4x2 + 4)
(x2
J(x2
f(x)
dx
= 4~
= ~ sin3x + ¥
=
f'(x)
6 + (-4)
xS
=2
4x3
+ -3
= -
5
36
Since
a
Jo f (x) dx = 4,
the shaded area = 4
...
2
17ra
4
a2
=
a=
t
so
1.2
c =-5
{a>O}
f(x)
37
a
J (2X2 + X - 3) dx
b i"'(5x +4)1 dx
=
2x3
+4x+c
f(l)
.'.
7r
J,r
dx
= ft,
+ ~ + 4 + c = ft
But
=4
dx
+c = 4
-~ + c = 4
so
8
6
s
dx = J2 f(x)
xS
4x3
= 5+ 3 + 4x
- 5
X2
= 3 + 2 - 3x +
[l 1')'1:
c
40
f'(x)
=
J4x
+ 5,
a f'(x)
= J4x
b f(x)
= J J4x
=
f(O)
+5
0
is defined when
4x
x (5x
fs [(5X+4)~J:
=fs(9~-4~)
= fs (27 - 8)
=
38
15
4
[(X ~ 4)!]
"2
0
= ~ sin3x + c
g(%) = 4,
I--I---...L.I -+-_~r
.'.
C
J2cos3x
= 2 x
V3;
=
=
37r
(2A
a J: f(x) dx
+4)-! dx =
= 2 cos 3x
g(x)
the area of the shaded portion = 7rr2 = 7r (~:)
35
(x
0
g' (x)
= 2A
So, the minimum cost is
when. r =
14
=
dx
+4
x
p
h as SIgn
. diragram:
dG
dr
Jx
e
= [2Jx+4]:
= 2Vs - 201
= 4)2 - 4
2
dG = 0 when
dr
_1_
2
p
+ 37r)
A
r2
4
o
1
{using (I)}
p
-
+ 1) dx
+x +C
+5 ~
x
0
~-£
+ 5 dx
1.
=J(4x+5)2
1
="4
(4x
dx
+ 5) ~
3
"2
= i(4x+5)~
Mathematics
+c
+c
SL - Exam Preparation
&
Practice Guide (3rd edition)
so
=
f(O)
But
-1
12
Hence
i(5)!+c=-1
.
..
8 - 3x d
-x
V4-x
o
=-
C
=
f(x)
5y15 _
6
yI5 6 -
+ 5)!
i(4x
-
_
=
212
5
8 - 3x
2V4-x
---;=~dx
0
=2[xV4-x]~
(;:5
VO
= 2(2V2 - 0v4)
Vs
= 4V2
=
x(3 - 2x)~
f'(X)
= (1)(3 -
f{x)
+ x(~)(3
2x)~
vI3=2X
x
V3 -
1
3 - 2x - x
V3 -
1
/
I-x
(3-2x)2
1
x-I
(3-2x)2
1
b
dx=x(3-2x)~
=
dx
-ix(3
+1
2x3
- 2x)~
f2
+C
a v
47
=
--2 - dx
J5
X
~ - 3x-2
)
dx
= ~
10
=
Ixl - ---1 + C
- In
- In [z]
(2x
+ -X + C
dv
dt
=
1 - X2
=
where 1 - X2
.'. 4
.'.
Area
x
=
=
+C
C
y
dx
=
.1
= V4-x-
ID
1L
= [12 sin2x]
cos 7r) - (- cos 0)
47r
-4
(7r) - 21·sin ( -"27r)
2 sm"2
=
= ~(1)- ~(-1)
.
53
49
u=
dx
.r--+-~""
d~ (X2 In x)
-3
..
..
-2
+ x(~)(4
J 2x In x dx
2
50
1
- x)-2(-1)
{product rule}
2x In x
..
+
+x
+ x) dx = X2 In x + Cl
+ J x dx = x2 In x + Cl
~X2 + C2 = x21nx + Cl
2Jxlnxdx=x2Inx-~x2+C3
= 2xe
d x=2x
1 21
nx-4x
_x2
",2
+x
2
2
(e-'" )(-2x)
Thus
2V4 - x
8- 3x
(1 - x )
J 2x(1 - x2)e-",2 dx = x2e-",2
. . 2 X (1 - x 2) e- ",2 dx = x 2 e- ",2
.'. J x(lx2)e-",2 dx = ~x2e-x2
J
2V4 - x
Guide
(3n1 edition)
+c
{product rule}
2
= 2x(1 _,x2)e _",2
2(4 - x) - x
& Practice
1 2
_x2
y=xe
dy
dx = (2x)e-
x
2V4 - x
+ X2 (~)
J (2x In x
2 J xlnxdx
x(4 - x)~
=
= 2xlnx
I
... J xnx
M::i/i......,;!ics SL - Exam Preparation
8
LIJ:1!. cos2x dx
b
=2
x2) dx
=
+4 -
4V5t
ms-2
cosxl~
.'.
1(4 - X)2
=
=-(-1)+1
¥
- x
s
sinxdx
= 1O~ units''
= xV4
.'.
1
= - ~ - (-8 +~)
y
0 = 4v4 + C
C = -8
=1
(4 .3
m
4
= X2
= ±2
[4x _ 1x3]2
+C
.'.
-3
= J~2
16-
7r
y
((1 - x2) - (-3))
=
Jo
a
= [= (-
-3
= J~2
=
+ 4)!
(5t
48
f2 sin 6x +
y
ms-l
10(-~)(5t+4)-2(5)
-25
- ~ ( i ) sin 6x
meets
dx
.3.
[x2 _ .:
cos 6x) dx
-
1
+ J12 f(x)
dx
+ 4)~ + C
+ x -2) dx
a- ~
= ~x
f(x)
x (5t -; 4)~)
0, s = 0, so
t =
When
= 24~
=J
= ~x
10 (i
= 4(5t
3
b a=
3x dx
f2
+ 4)-2
10(5t
s = J v dt =
.'.
=
dx
3x-l
=(52-i)-(12-1)
2
f(x)
v5t+4
1
=
I sin
+ 4X[2
4~2
s:
whereas
= 2x
1
[:4 _ ~3_
No, the shaded portion has area
2
=
f(x) dx - J1 f(x) dx
+Cl
3 dx = / (12 -
2X2 -x2X -
J5
=
32
= 2x
b
+ 4) dx
"3
(3-2x)2
3/
J
a
J~2(X3 - X2 - 4x
= (4 - ~ - 8 + 8) - (4 + ~- 8 - 8)
2x
-
:.
2x
=
3(1 - x)
_
Thus
a J~2 f(x) dx
46
- 2x)-~(-2)
+ Cl
+ Cl
+C
51
a y = x2 + 2x - 3
meets
y = x-I
where
+ 2x - 3 = x-I
X2 + X - 2 = 0
+ 2)(x - 1) = 0
x2
.'.
(x
••
=
J v dt
J
(2t - 3~2 - ~) dt
3t3
2t2
=-
=
x
=
c set)
y
-2 or 1 .--\---F+-E&-:J4_
When x = -2, y = -3.
When x = 1, y = O.
x
= t
It+d
2(3)
2
1
d
- 2t+
213
- 2t
But when t = 0, s = 3
.'.
-3)
the graphs meet at (-2,
- ---
2
m
.'. 'd. = 3
s(t)=e-~t3-~t+3
m
and (1, 0).
ylx2+2x-3
=
f2
54
- 1) - (x2 + 2x - 3» dx
b Area = f2((X
+ 3)
(x - 1 - X2 - 2x
a v=2t-3e
Now
m s "!
s=Jvdt=J(2t-3t2)dt
dx
2
(2-x-x2)dx
=f2
3
= e - t +c
3
= [2x _ IX2 _ Ix3]
2
3-2
1
s(O) = c
= 2 - ~ - ~ - (2(-2)
- ~(_2)2
_ ~(_2)3)
and
So, the change in displacement is 0 m.
units''
+
a y = 4 - X2
meets
-2x
.'. X2 - 2x .'.
y = -2x
=48= 0
- 4
- 4
.'.
=
=
t.r + c,
s(l)
-2 or 4
y = O.
When x = 4,
y = -12.
y=4
b Area=f2((4-x2)-(-2x-4»)
a (x - 1)2(x + 2) = (x2
dx
= x3
= c
t.r + t.r = !r
m.
c+t.r
2x + l)(x + 2)
-
where
.'.
.'.
16 + 32) - (~ + 4 - 16)
-
3x
+2
meets y = x3 - 2x
3
x - 2x = x - 2
b y = x - 2
= [-~x3+X2+8x]~2
¥+
s.
= x3 + 2X2 - 2X2 - 4x + x + 2
- x2
+ 2x + 8) dx
= J~2 (_X2
i
c
55
the graphs meet at
(-2,0)
and (4, -12).
=
s(i)
The total distance travelled =
When x = -2,
= (-
"3
a change in direction occurs at t =
s(O) = c,
(x+2)(x-4)=0
.'. x
2
o
where
X2
t
1--~---1
Sign diagram of v:
52
s(l) = c
b v = 2t - 3t2 = t(2 - 3t)
=2-~-~+4+2-~
= 4~
3t3
2t2
=---+c
x3 - 3x
(x - 1)2(x
+2 = 0
+ 2) = 0
Since the factor (x - 1) is squared,
tangent to the curve when x = 1.
36 units''
y = x - 2 is a
The line meets the curve again when x = -2.
When x = 1, y = -1.
53
a
dv
a(t)=-=2-3t
dt
vet)
=
J(2 - 3t) dt
=
When x = - 2, y = -4.
-2
ms
.'.
3t2
c Consider
When x = 0, y = 0, so the y-intercept is O.
Now when t = 1, v = 0
When y = 0,
0=2-~+c
x(x
-~
vet) = 2t -
""2 -
1
2 ms
-1
3t2
2t - -
b Now v = 0 when
+ v2)(x
-
-
2)
v2)
the x-intercepts are
2
3t
x3 - 2x = 0
.'. x(x2
O=~+c
c=
and Pis(-2,-4).
y = x3 - 2x.
""2 + c
2t -
Tis(l,-l)
d
=
0
= 0
±v'2
and O.
y
y = x3 - 2x
- I = 0
2
2
.'. 4t - 3e - 1 = 0
.'.
.'.
3t2
-
4t
+1= 0
(3t - l)(t - 1)
.'. t
=~
or
t
=
0
=
1
t = ~ s is the other time.
El
Mathematics
SL - Exam Preparation
& Pradice
Guide (3rd edition)
•
i', ((x
3 -
e Area =
f2 (x
2
=
3
+
43
=
[
x4 -
+ 2)
3x
-
59
2x) - (x - 2)) dx
a
dx
Any odd function
has rotational symmetry
about O.
y
1
y=f(x)
f(Xl)
.
]1
+2x
-2
..........
- f(Xl)
+ Jo1f(x)
dx
=(~-~+2)-(4-6-4)
=
6~ units''
b
J,
a+l
a
56
X2
dx
=
i:
f(x) dx
1
2
( a+1 )
.-t?
3
-a
6a
+ 3a + 6a -
~
But by symmetry,
.'. f1f(x)dx=0
=2"
c
= 0
1=0
-6 ± )36
a=
- 4(6)( -1)
a>
0, so
If
v'6O -
a=
dx
f(x)
6
i'. (e-
+x3cos2x)
=
J~l e-
dx +
2x
J~l
f(8)
a
= 2 -. cos 8
sm8
= (_~e-2)
j'(8) = sin8 x sin8 ~ (2 - cos 8) cos 8
2
= 1(e
2
sm28
2
sin 8 - 2cos8 + cos2 8
sin28
60
1- 2cos8
sin28
b
_
.' . cos t + cos 2t = 0
cos t + (2 cos2 t - 1) = 0
.'. 2 cos2 t + cos t - 1= 0
""2
8 =
1-.
(2 cos t - 1) (cos t + 1) = 0
+cos8
= -sin38+2sin8cos28
-I!
x 2[sin8]1 x cos 8
cost = ~ or-1
.'. t
= sin 8[2 cos'' 8 - sin2 8]
= sin8[2cos2 8 - (1- cos" 8)]
= sin8(3cos2
.'.
<
8 ~ ~,
1'(8) = 0
so
cos 8
when
Sign diagram for
= 1-, rr,
5;
8 - 1)
= 0 when sin 8 = 0 or cos2 8 = ~
But 0
= sin27r + ~ sin d-r = 0 cm
X
ii The point comes to rest when v = 0 cms-1
= cos8 x [sin8]2
I' (8)
t = 27r,
v = cos Zsr + cos d-r = 2 cms-1
V3
j'(8) = -sin8[sin8]2
b
x = sin 0 + ~ sin 0 = 0 cm
1-
8=
the minimum value is .J3 when
a f(8)
t = 0,
v = cos 0 + cos 0 = 2 cms-1
is a minimum when
V3
dx
v = dt = cos t + ~ x 2 cos 2t
= cos t + cos 2t cm s -1
When
l' (8):
""2
5
~)
e2
Velocity
2_1
l
f(2!:3) = __ 2 = _2 =.J3
.ow
_ (_~e2)
x = sin t + ~ sin 2t cm
When
f(8)
dx
{using b, since x3 cos 2x is odd}
8-2!:
- 3
1'(8) = 0 when cos 8 = ~
Sign diagram for
a
= cos 8 for all 8}
x3 cos 2x dx
= [!2 e -2x] ~1 + 0
57
for any odd function .
is an odd function.
2X
d
12
Al = A2
f(x) = x3 cos 2x then
f(-x) = (-x)3cos(-2x)
= _(x3 cos( -2x))
=-(x3cos2x)
{cos(-8)
= - f (x) for all x
12
-6 ± v'6O
a=
12
But
(-A2)+A1
3
+ 3a + 3a + 1-;? = ~
2
=
"2
2
3a2
J~l f(x)
= Al - A2
1
3
=
>
0
and
sin 8
1---'+---'-'
----'--L-+-'--1r
Sign diagram for v is:
>
0
o
11"
:3
511"
7r
"3
L
t
27r
cos 8 = ~
l' (8):
·
A cce Ieration
+ /.\
. t - 2·sin 2 t cm s" 2
a = dv
dt = - sin
.'. a(1-) = - V; - 2(V;)
= _3f"
cms-2,
a(7r) = 0 cms-2,
f(8)
is a maximum when
At this time,
.'.
cos 8 = ~
and
sin28 = 1- cos2 8 = 1- ~ = ~
the maximum value of f(8)
is
Malhematics SL - Exam Preparation & Practice Guide (3ro edition)
~
x ~ =
a( 571") = V3 + 2V3 = 3V3 cm S-2
3
2
2
2
3~
El
2
b
At the time
t = 0,
the point is moving with velocity
65
2 cm S-l in the positive direction.
It moves in this
direction for
seconds, before coming to rest. It then
if
The parabola has x-intercepts ±7r, so its equation is
y = k(x - 7r)(x + 7r) for some k.
k( -7r)7r = a
The y-intercept is a, so
moves back to its initial position, arriving there after
tt seconds. It momentarily stops, then continues to move
in the negative direction until
5;
=
t
seconds.
k=-~
At this
=
Y
time it changes direction and moves back to the initial
position, arriving there at t = 27r seconds.
61
_~(X2
7r2
= _~X22
is a translation of
y = sin x
Consider the case a
>
O.
Using symmetry, the shaded area is 217r
217r
y = sin(x +~)
= sin (x
.'.
+a
7r
(-oi).
through
b y
_ 7r2)
We need to consider two possible cases:
i)
a f (x) = sin (x +
7r2
i)
+
meets y =!
on the given interval,
(x
when sin
The total shaded area = 2
JoT
%) -
(sin(x +
%)
a
[ --x 2
37r
= !
2;, and 27r.
x = -27r, - 4;,0,
2~
+
( - .; X2 +
3
a) dx
+ax
]7r
( - .; x2 +
a) dx.
= 4
=2
0
- a7r + a7r + 0 - 0 = 2
3
!) dx
.'.
%a7r = 2
2~
%) - !x];
5; - if) - (- cos %) )
.'. a= ~
= 2 [- cos(x +
= 2 ( (- cos
=
2
(vs _
-- 2v3 Jo"f
62
So, a
66
27r
3
= ±~.
X2 - 2x
units''
=
(-t case;>
=
(0+5)
=
5t
The volume of the solid of
revolution
+ 5 sin x]
!
+ 5sin(~»
y
-
(-t cosO
+ 5sinO)
=
7r
=
tt
=
tt
'-~~------~--.x
a
1
tanx
=
1
x cos2 x
= (f
J1[..
1
1.
2
k=2
_2_
sin2x
+ 3)2
(2x
= 7r
[!
(2x ;
=
dx
67
= ![In(tanx)ll
a
f'(x)
dx
3)3] :
-
497r
3
_
33)
=~
x
6
!(In v3 - In( Js»
= !(In v3 + In v3)
=
=
!In] v3 x v3)
=
!In 3 units''
b V =
3
(e2X)2 dx
3
e4x dx
tt J1
= 7r J1
= 7r
= %(53
_
-
4x3
+ 4X2)
dx
4~4+ 4~3]:
=7r(¥-16+¥)
ii F'(x)
b From ii,
=
with domain
= (l)lnx+x
=
F(x)
Jo1
(x4
1Ji.7runits3
15
6
sinx cos x
2
2 sin x cos x
2
sin2x
= tt
(x2 - 2X)2 dx
5
_1_ dx
sin2x
(!f
J1[..
y2 dx
= 7r [x5
6
=
12
12
12
y = X2 - 2x
b Shaded area
y = In(tanx)
cos x
1
= sinx x cos2 x
a V
are 0 and 2.
(-t +0)
-
dy
dx
64
= x(x - 2)
the x-intercepts
(sin 3x + 5 cos x) dx
= [- ~ cos 3x
63
+ vs)
2!:
232
By symmetry, the other possible solution is a = - ~.
x
>
O.
G)-l
lnx
J In x
J f(x)
provided
x>O
provided
x>O
dx = x In x - x + c
dx
=
is the antiderivative
F(x)
+ c
of f (x).
C
[~e4X] ~
7r (12
4" e
. 3
- e 4) units
-2
units''
I.
Mathematics
SL - Exam Preparation
&
Practice Guide (3rd edition)
o
d
Let P be a point on the curve C with x-coordinate
1 ~ t ~ e.
The shaded area A
= area 60PQ
- fIt In x dx,
= ~tlnt
- fIt Inx dx
= ~tlnt
- [xlnx
1
<t
t,
+ OQ + PQ = 40
a OP
69
cm
.'. x+x+8=40
=
40 - 2x
b A = ~OX2 where
8 = Ox
.'.
~ e
8
= ~ (~)
A
- xl~
X2
= ~X8
= ~t In t - tIn t + t - 1
= ~x(40 - 2x)
=t-~tlnt-1
units''
(12"In t + 2"t1 (1))
t = 2"I( 1 -In
dA = 1 dt
t)
dA
dx
=
c dA
= 0 when
dt
:--+-'---I~
t = e
x - ty
y -Int
x -
t
=t
68
\
20
x = 10.
Using
8 = Ox,
20 = 0(10)
.'. 0 = 2 radians.
70
v(t) = e2t - 3et
a
ms-I
v(O) = eO -3eo
= 1 - 3 = -2 ms-I
the initial velocity is -2 ms-I.
b Now
t(l-Int)=O
Conclusion:
10
x = 10, 8 = 40 - 2x = 20
1
t
t-tlnt=O
.'. t
7+C"\- rx
the area is a maximum when
- t In t.
= et (et, - 3)
v(t)
Since et
{as t>l}
Int=l
I
o
When
The tangent passes through (0, 0) if
.',
20 - 2x
t
ty - tIn t = x - t
or
20x - X2 cm2
h as SIgn
. diiagram:
.'.
d
1
dy
dx
the tangent has equation
which is
dA
dx
e
1
the area is a maximum when t = e.
dy
1
ii y=Inx
dx = ~
at (t, In t),
=
A
>
0 for all t, v(t) = 0 when
et = 3
which is when
=e
.'.
The shaded region is maximised in area
when [~P] is a tangent to C.
f(e2t - 3et) dt
8(t) = ~e2t - 3et + c metres
=
c 8(t)
••
a,b
t = In 3
the particle is stationary at t = In 3 seconds.
But
=
f v(t) dt
8(0) = 1,
~eo - 3eo + c = 1
so
.'.
~ -3+c=
.'. c=
1
3~
8(t) = ~e2t - 3et + ~ metres
8(ln5) = ~e21n5 _ 3e1n5 + ~
Thus
= ~eln52 _ 3(5) + ~
=
c The graphs meet where
.'.
2sinx
sin 2x = sin x
sinx(2 cos x -1)
.'.
So, the particle is 1 m to the right of O.
= 0
cos x = ~
= 0, i, tt
.'. x
71
Area A = fo"f (sin 2x - sin x) dx
= [- ~ cos 2x + cos
= (-~
cosen
x]
a
!
+ cos(i))
f() x
3x
= 3xe
eX
= -
=
J'(x)
- (-~
f;
b
(sinx - sin2x)
,
0 ~ x ~ 4
3e-x
-
3xe-x
-
3e-x
f' (x)
= 3 (1 - x) = 0
eX
Sign diagram of f'(x):
dx
+ 3xe-x
3(x - 2)
eX
cos 0 + cos 0)
= ~ unir'
-x
= -3e-x
f"(x)
= (-~(-~) +~) - (-~ + 1)
Area B =
+~
=lm
cos x - sinx = 0
.:. sin x = 0 or
¥ -15
= 16 - 15
3
when
I
o
= [- cos x + ~ cos 2x] :
x = 1
+C"\-
7
1
\
3
there is a maximum turning point at x = 1.
= (- cos 7r + ~ cos 27r)
-
(- cos(i)
= (1+~)
=
+ ~ cosen)
When
- (-~+~(-~))
,
3(1)
3
Y = -= eI
e
So, the maximum turning point is at
2~ units''
ics SL - Exam Preparation & Pradice Guide (3,d edition)
x = 1
~
~
(1, ~),
I~x
4
= 3(x -
ii j"(x)
=
2)
eX
0
when
Sign diagram of j"(x):
c When x = a, Y = _a2 + 4a
The equation of the normal is
=2
x
1f.-----'2--+--1
o
there is a non-stationary
at x = 2.
When
=
x
So, there
4
inflection point
3(2)
2,
d
= 0,
2a - 12a + 17a = 0
.'. a(2a2 -12a+ 17) = 0
e2
is a non-stationary
The normal passes through the origin.
x
When
6
=7 =
Y
x - (2a - 4)y = a - (2a - 4)( _a2 + 4a)
.'. x - (2a - 4)y = a - (-2a3 + 8a2 + 4a2 - 16a)
.'. x-(2a-4)Y=2a3-12a2+17a
x
=
0 and y
3
inflection
point
at
(2, ~).
2
Since a =1= 0,
c
(1, ~)
Y
± )(-12)2
-(-12)
=
a
- 4(2)(17)
2(2)
± yf144
12
- 136
4
1
± v'8
12
4
V;
=3±
1
2
a = 3 ± jz
3
e
d
If
F(x) =
3(x
+ 1)
eX
When
F(x)
e The area
x-
(2(3+ jz)-4)y=0
x
y
=
~
When
y
2:.J2
y=
y = -(x - 2)2
+4
=
.'. x -
+4
Area of shaded region
0
1
v'2
1
(_X2
+
(
2
x
+ y'2
4- 2
)
dx
+1y'2 x
)
)
dx
4
73
(2,4)
y
3+1- ( _X2 + 4x -
= 13+
0
=4
2 = ±2
x = 0 or
(x - 2)2
4
jz))
+ eO
= 0,
Y = -4 + 4 =
= 0, -(x - 2)2 + 4 =
y
+
3
15
. 2
3- units
e4
f(x) = -(x - 2)2
x
(3 + jz,f(3
J:
The vertex is at (2,4).
When
(2,4)
4
= [ 3(Xe~ 1)
= 0
y= 2+v'2
= J04 f(x) dx
= - e4
+ v'2)y
x - (2
of f(x).
is the antiderivative
15
the normal L has equation
X-(6+~-4)Y=0
= [F(x)l~
a
= 3 + jz,
X
F'(x)=-3e-x+3(x+l)e-X
= 3xe-x
= f(x)
then
72
a
+ l)e-
= -3(x
as required
= -(x -
2)2
a Suppose the line and the arc meet at point P where x
.'.
+4
P has coordinates
=
k.
(k, v3k).
y
r
~~~--------~--------\---.x
2
b
f(x) = -(x - 2? + 4
= _(x2 - 4x + 4)
= _x2 +4x
J'(x)
j'(a)
+4
tan
e = v3k
=
v'3
k
= -2x + 4
= -2a + 4
.'. e = i
and a
I
The arc length
So, the normal at P(a, f(a)) has
1
1
gradient =
-2a + 4
2a - 4
Area A
ID
= "i - e =
= er =
T6'"
units
= ~ar = ~("if )r =
2
Mathematics
2
"if
2
"'{2 units''
SL - Exam Preparation
& Pradice
Guide
(3rd edition)
•
b Using Pythagoras,
+ (v'3k)2 = r2
k2
k2(1+3)
.'.
> 0,
But k
.'.
=
0, y
since y>
=
8
<
n
- 2:
,+~-:
~r
: /l\
r2 - X2
=
y2
<
= g(8)
ii Sign diagram for 1'(8)
k2 = ~r2
so k
The equation of the arc is
If 1'(8) = 0 then g(8) = 2
.". from the graph, 8 ~ 1 {O
e
=r2
,8
o
~
f (8)
So, the maximum value of
vr2 - X2
is when
8 ~ 1.
CALCULATORS
a
=
=
f(x)
f'(x)
+ 4x
f(x + h)
_X2
lim
h
h~a
=
3
- f(x)
+ h)2 + 4(x + h)
lim -(x
=
A
fa'§: (vr2 - X2 - V3x)
=
dx
= fa'§: vr2 - X2 dx -
V3 fa'§:
= fa'§: vr2 - x2 dx -
V3 [~x2l!
= fa'§:
= fa'§:
vr2 - X2 dx -
V3 x ~ (~)2
vr2
1r2
-;? -
lim
2xh - h2 +M
h
h~a
x dx
= lim )(( -2x
.
c Usinz a and b
o
= -12 =
'§: rr==:
V r2 - X2 dx
a
Letting
r = 2 in
t'
la
1'§:
7rr2
1
.
..
d
A
,
X2 dx -
-
a
t' (x)
rz=:":
- X2 dx
V3r2
- -8
31
11
y
8
a
=
=
7r(1)2 -7r(x)2
=
tt
.'. S
Also,
sin8
=
'JL = y
1
'
so
S
+ 4k
y = _k2
=-
+ 4,
2k
+ 4k).
the tangent has equation
=-2k+4
- 4k
+ k2
= _ 2k
+4
- 4k = -2k
k2 - 4k
=
k2 - 8k
1
1- x2
.'.
= 7ry2
= 7rsin2
4
+
+ 9 = (- 2k + 4) (4 - k)
= -8k + 2k2 + 16 = 2k2 - 12k + 16
+7 = 0
.'. (k - 7) (k - 1)
k
=
0
=
1 or 7
4k
However, the tangent has positive gradient.
-2k+4>0
.'. 2k
8
The shaded area below [AB]
= area of sector
= ~T2 (28 - sin
= 8 - ~ sin 28
k,
4-k
+l =
.'. y2
+ k2
9
+ 3"f
X2
=
= f (x)
c The tangent passes through (4, 9)
7r(1 - x2)
But in the right angled 6.,
= f (x).
x-k
3"f
V36 - 9x2 dx
x
Since t' (k)
2
y-(-k +4k)
x-k
2
~dx=7r+
# O}
So, the point of contact is (k, _k2
V3r2
+ V3(4)
3
{as h
is the gradient function of y
b When
=2!:+,j3
1
+ 4)
represents the gradient of the tangent to y
at the point (a, f(a)).
c,
7r(4)
12
-M
t' (a)
= -12 + -- 8
V4-X2dx=
+4h+~
)(
= -2x + 4
V r2
7rr2
- h
h~a
as required
+ 4x)
h
h~a
ow
- (_x2
28
.'.
- area of triangle
So, k
=
k
<4
<2
4
1.
28)
2
y
a
1
c The shaded area between [AB] and [CD]
lM:::=ciics
=S -
=
=
_'ow
e-
+ sin28 - 28
g( 8) = tt sin 28 + 2 cos 28
i' (8) = 7r(2) sin 8 cos 8 + cos 28(2) = 27r sin 8 cos 8 + 2 cos 28 - 2
= (7rsin28 + 2cos28) - 2
= g(8) - 2
So, 1(8)
Let
2 (8 - ~ sin 28)
2
7rsin
2(8 - ~ sin 28)
2
7rsin 8 - 28 + sin 28
= 7rsin2
8
2
C has coordinates
.'.
(x, cos x) .
rectangle ABCD has area A
= 2x
cos x
SI..- Exam Preparation & Pradice Guide (3rd edition)
e
b
dA = 2 cos x + 2x( - sin x)
dx
dA = 0 when
dx
et
{product rule}
W
100
2xsinx
= 2cosx
80
xtanx
= 1
60
x R; 0.860
Wt = 100e
--10
40
{using technology}
20
dA
dx
Sign diagram for
!+~-:x
: 10.860
C has coordinates
3
\
o
10
~
"2
5
= xe~
= 1e~ x
l' (x) > 0
+xe~2x
(-4x)
for x
>0
= e~2x2 (1 + 2x)(1 - 2x)
where e~2x2 is always positive.
as each term is positive.
the function is increasing over its whole domain.
C
The function cuts the x-axis when
1
x--=O
VX
1
x=VX
=
:. x~
:.
f' (x)
y = O.
f(x)
This is at (~, ~e~~)
b
So, the x-intercept
= x - VX
is undefined when
1
VX
--> 00,
-->
0
and
f(x)
2
is l.
>
f(O) = Oeo = 0
.'.
O.
or (~, 0.303).
and
f(2)
= 2e~8 R; 0.000671
the maximum value R; 0.303
Volume of revolution = 7r
C
x = 0, so x = 0
= 7r
is a vertical asymptote.
As x
-,
There is a local maximum when x = ~.
The function does not cut the y-axis since x
d
'~]+ /;"\
L/l\Lx
o
has sign diagram:
2
1
x = 1
1
t
= e~2x2 (1 _ 4X2)
1
3
= 1+ "2X~2 = 1+ 2xVX
f (x)
40
2X2
222
1'(x)
- X~2
1
I
f(x)
a
1
=x
f(x)
30
~ mass decayed
(0.860, 0.652).
a The function is defined for x > O.
b
20
= tt
--> 00
J:
J:
J:
and the minimum is O.
y2 dx
(xe~2x2
r
(x2e~4X2)
dx
dx
ii Volume of revolution
there is no horizontal asymptote.
R; 0.174 units3
{using technology}
et
6
a
--r
The bin has capacity 500 litres
=
0.5 kL
r
7rr2h = 0.5
h=_l_
27rr2
Surface area A = 27rrh + 7rr2
2
= 27rr (27r~2)
J14(x
Required area =
- X ~ ~ ) dx
= 5 ~ units''
4
a
1
A(r) = - + 7rr
W(t) = 100e~20
dA = 0 when
dr
W(O) = 100eo = 100 g
27rr =
W = ~Wo = 50 g
then
r
t
20
:.
dW
-
dt
Since
d
=
dt
=-
1
27r
1
h R; 27rr2 R; 0.542
= 1nl = -ln2
2
So, the surface area of the bin is minimised when the bin
has a base radius and height of 54.2 cm.
t=20In2R;13.86
100~..L ( 1)
e 20 - 20
dW
-
12
r R; 0.542
2
it takes about 13.9 days for half of the mass to decay.
C
3
50 = 100e~20
e~!ri - 1
-~
2
r
Initially there is 100 g of radioactive substance present.
-
m
dA
1
b = -- 2 +27rr
dr
r
t
b If
2
r
{using technology}
+ 7rr
= - 5 e ~..L
20
7
f(x)
a
<
0
for all t, the weight of radioactive
b
=
3x3
+ 3x2
f(O) = -1,
l' (x)
3x - 1
-
so the y-intercept is -l.
2
= 9x
+ 6x + 2x
substance is always decreasing.
= 3(3x2
As t --> 00, W --> O. The amount of radioactive
substance decays to nothing.
= 3(3x - l)(x
1'(x)
m
=0
when
Mathematics
3
- 1)
+ 1)
x=-l
or ~
SL - Exam Preparation
&
Pradice Guide (3rd edition)
•
Sign diagram for f'(x):
f(-I)
= 2
Ne
=-11-
f(~)
and
b
there is a local maximum
at (-1,
t" (t)
2) and a local
f"(x) = 18x + 6
(1 + 2e-!
- 1)
=6(3x+I)
f"(x) = 0 when x = -~
+ •x
~
f"(t)
e-! --
when
t
y
inflection point at (-~,
--+--+---+--+.>o,-EIj--+-I--
x
~).
= 3x3 + 3x2
-
!]
2
2
+~~ 1Tn4\
i" (t):
Sign diagram for
the x-intercepts are
~ -1.515, -0.278,
and 0.793 .
f' (t)
is a maximum when
N
when f(t) =
(1)
1 + 2 "2
(~, -I~)
/----+--f(x)
-6
+ 4e -
t = In4
e Using technology,
(1, 2)
- 2e -!
1.
=
e2
.'.
d
t = In 4,
•t
and this occurs
N
2
So, the maximum rate of growth of the population occurs
3x - 1
when
(-2, -7)
a
=0
=~
there is a non-stationary
8
(2e-!
(-~)
(I+2e-!r
inflection point at x = - ~.
there is a non-stationary
f(-~)
~Ne-!
-! [-1
(~ ) Ne
( 1 + 2e -!)
Sign diagram for f"(x):
)
(
- Ne-h
(1 + 2e-!) (2e-!)
= --'-----'-----'-.,.-'------'-----'-(1 + 2e-!
minimum at (~, -I~).
c
r
r
r
-! (-~)(1 + 2e -!
t = In 4 .and the size of the population
is
~
at
this time.
f(x)=xlnx+I,0<x:(2
f'(x)
= llnx
+ x (~)
f'(x) = 0 when
+ 0 = ln z + 1
x = e-1
N
f(O) = 1 + 2 ="3
f(2) ~ 0.58N,
f(6) ~ 0.9IN,
lnx = -1,
which is when
N
c Now
and as t
f(t)
-> 00,
->
N
f(4) ~ 0.79N,
f(8) ~ 0.96N
f(t)
N····················································
.
Sign diagram for f'(x):
N
3
the minimum ability occurs when the child is
1
- ~ 0.3679 years ~ 4.41 months
e
The minimum memorising ability occurs in the 5th month.
b
Using technology
we graph
3
2
y
10
y = f(x)
2
4
.'. x
_~--~--~-.x
1
2
The maximum ability occurs at x = 2, or at the end of
the 2 year interval.
8
10
lnx
a g(x) = for 0 < x :( 5
x
If y = 0 then
In x = 0
1
9
6
=
1
So, the graph cuts the x-axis at (1, 0).
-,
G)x-Inx(I)
b 9 (x) =
X2
I-Inx
=
-x-2-
.:. g' (x) = 0 when In x = 1 : . x = e
Sign diagram for g'(x):
a
.'.
the stationary point is a local maximum at
gl/(x)
=
(_1)x2-(I-lnx)2x
-'--..::.X~_--:-
(e, ~).
_
x4
(1 + 2e-!
N> 0 is given.
all t.
Thus f'(t)
Mathematics
>
e-!
-x - 2x + 2xlnx
r
>0
x4
2lnx
and
(1 + 2e-!)
2
>0
for
& Practice Guide (3rd edition)
gl/(x) = 0
when
lnx
X
0 for all t.
SL - Exam Preparation
- 3
x3
=~
=
3
e2
~
4.482
El
s
Sign diagram for g" (x):
l~x
VJ
o
there is a non-stationary
b
As x
5
.". f(x)
At the point of inflection, the gradient of the graph is
,J.
9 (e2)
=
c As x
-+
d
,1
1-lne2
e3
0 (right),
1-2
= -- 3
e
y
= --
1
2e3
f(x)
!,(x)
-+
0
2
3
5
4
!,(x) =0
J g(x) = I:X
-0.4
x-1)e- (_1)
= x-2e-x + (2 (~
X
2
+ 2.)
X
2X2
+ x)
X2
= e:2" (1 1
1
1
0 and 2 - - -+ 2
x
---;
= e-x
0.2
-0.2
-+
X
=e-X(x-2_2+x-1)
(4.48, 0.335)
\ (5, 0.322)
(e, ~)
and e-
00
f(x) = (2 - x-1)e-x,
c Since
~ -0.0249.
-+ -00.
y
0.4
3
x
-+ -00
As x ---;00, e-X
inflection at (4.48, 0.335).
2. -+
0 (right),
-+
-2x2+x+1=0
when
.'.
2X2
(2x
+ l)(x
-
x - 1
=
0
- 1)
=
0
= -~
.'. x
e Since there is a local maximum at
g(x)(;~
Now
x>O
forall
lnx
1
-,,:::X
-c e
So
In x (;::
e
f'(~)
f' (x)
=4e-~
>0
f' (2)
and
=
or 1
2
e- ( -5) < 0
4
+10-
has sign diagram:
for x > 0
there is a local maximum at (1, ~).
11
a f(x) = In(xv'l
- 2x)
d
1
=lnx+ln(1-2x)'
{lnab=lna+lnb}
2x)
= lnx + ~ In(l-
f(x) exists provided
the domain is
{In u" = nlna}
x> 0
and
1 - 2x > 0
.'. x> 0
and
x < ~
I0 <x <
{x
n.
-2
-2
f(x)
= 2. + ~ (~)
b !,(x)
1- 2x
1
---x
1- 2x
~)e-X
= (2 -
x
-4
1
1- 2x - x
x(I - 2x)
1- 3x
x(I - 2x)
e Y= (2-;)e-X
( 2-;I)_xe
c For points where the gradient of the normal is
=
1- 3x
x(1-2x)
Area ~ 11.33
6
But 0
< x < ~,
f(io)
= In(
13
a
2.) e-
X
(x - 1)] dx
-
x
~ 0.373 units?
{using technology}
1
f(x) = 1 + 2x _ 1 = 1 + (2x - 1)-1
J'(x)
= -(2x - 1)-2(2) =
-2
(2x - 1)2
so x = ~
iov1-
~) = In(
f' (x)
ioA) ~ -1.66
b
e-x,
f(x) = (2-;)
1
f(x) = 0 when 2 - x
=
is never zero, and so
f (x)
has no turning
points.
So, the point is (0.3, -1.66).
a Consider
[(2 -
0.342
6 - 18x = 5x - 10x2
10x2 - 23x + 6 = 0
(x - 2)(lOx - 3) = 0
:.x=20r~
where
=x-1.
§.
6(1 - 3x) = 5x(1 - 2x)
.'.
y=x-I
meets
Using technology, the points of intersection are
(0.342, -0.658)
and (1.33, 0.330).
= ~.
!,(x)
12
x
4
g(x)
f(x)
x>O
0 {as e-x > 0 for all x}
= eX
= 1+
1
2x _ 1
'-~--4~-=-~2~~~---2~---4--'x
2. = 2
x
x-.!.
-
2
El
Mathematics
SL - Exam Preparation & Pradice
Guide (3rd edition)
#
b
X
If (2x - l)e
=
2x
-
0
then
2x = (2x - l)e
2x
--=e
2x - 1
.
..
16
X
1
4
x - 2 - 4x
-3x - 2
= ~ - x _ 2 = x(x - 2) = x(x - 2)
a f(x)
So, f(x)
x
= 0 when
=
=
=
f(x)
b
e"
1'(x)
1
l+--=e
x-1
-0.603
x:::::;
a v(t) = 2Vt - t = 2t! =
1
Vt
2d - t
1 ms
=
l' (x)
has sign diagram:
-2
t
,
f(-2)
= ~ and f(~)
0
>
t = 0
t;::: 0
t
=
Sign diagram of v:
or when
0
or
t
minimum at (~, 4~).
t2 = 2
(
=4
"
f
01
2x-3
=
f"(x)
~t
4
J:
= J:
( The total distance travelled =
-t)
So,
dt
(-3.40,0.45).
t
y~
e
2t~
-l~V-~ min (~, 4~)
max (-2~)
F
~~~~==~~~~-----.x
1.
\
non-stationary
inflection
= 0
( -3.40,
f(x)
4
x=2
3
1 2
= 3t"2 -:it
m
e Area =
15
a
1
4
= ~ - -x---2
x = 0
0.45)
.'. c = 0
.'. s(t)
{using technology}
:::::;0.4464_
f"(x)=O at
dt
=-3---+c
:i
2
4 a
1 2
= 3t2 -:it + c
The particle starts at the origin.
-3.4048
x:::::;
d
m
=9~
= J(2d
f( -3.4048)
tl
12Vt -
8
= 4x3
(x - 2)3
.'.
Iv(t)1 dt
2
x3 - (x _ 2)3
2
8
x3 = (x _ 2)3
\
.'.
=
8(x - 2)-3
-
.
(x) = 0 when
There is a reversal of direction at t = 4 s.
d s = Jvd(t)
= 4~
there is a local maximum at (- 2, ~) and a local
1
s(O)
- 2))(2x - (x - 2))
x2(x - 2)2
(3x - 2)(x + 2)
x2(x - 2)2
-1
(2- d),
t!
= 0 when
V
(x - 2)2
- 2)2
- 1
= -1 v =
(x - 2)2
+ (x
(2x
or 0.864.
t ms
-
x2(x
v'(t)
= t-"2
b
4X2
_ 2)-2
4
---+
X2
The solutions of the equation correspond to the points of
intersection of y = f (x) and y = 9 ( x ) . There are
hence two real solutions.
a(t)
+ 4(x
_x-
1
2x -1
14
4(x - 2)-1
2
X
Using technology,
- 2= 0
.'.
x=-~
(2x - 1) + 1
2x -1
(
-3x
y
J
9
1
Area =
X
-1
2
dX
r 1! (- - -)
}.!,;
1
4
x
x - 2
dx
"2
=
r~ (.!. + _4
}J,"2
x
2 -
) dx
X
3
=
[2Vxli
2
=6-2
=4
1 (]x)
=
=
7r19 .!.
tt
1
X
[In
xli
:::::;6.90 units3
SL - Exam Preparation & Pradice Guide
2
dx
+ 41n
=5In~-5In~
= 5(ln ~ - In ~)
dx
=7r(ln9-ln1)
Mathematics
units''
- (In 1
- 41n 1)
- (In 1
- 41n 1)
2
2
2
2
= In ~ - 41n ~ - In ~
9
b Volume of revolution = 7r
= [lnx-4In(2-x)]:
(3rd edition)
= 5(ln 3 - )n-"2 -)Hi +)n-"2)
= 51n 3 units''
~
17
b
dy = 2x - 3
dx
a
The gradient is 1 when
2a - 3
••
2a
.'. a
x = a.
t = 0, 8 = 0
.'. d = 0
2 + 27t cm
- 15t
v = 3(t2 - lOt + 9) = 3(t - 1)(t - 9)
But when
=1
=4
=2
••
=
8
c Now
ii When x = 2, Y = 22 - 3(2) = -2
So, the point of contact is (2, -2).
Since
8 = J v dt = J (3t2 - 30t + 27) dt
= t3 - 15t2 + 27t + d cm
dy = 1
dx
at (2, -2),
t3
So, v has sign diagram:
the normal has
the particle changes direction at t = 1 and t = 9 s.
8(0) = 0, 8(1) = 13,
8(9) = 729 - 1215 + 243 = -243
gradient -1.
the equation of the normal is
I
-243
x + y = 2 +-2
.'. y=-x
b y = X2 - 3x meets
21
=0
+ 2)2 dx =
a J(x3
X
J(x6+ 4x3
7
x
=7"+x
x(x - 4) = 0
.'.
0
= 0 or 4
b
Area = J0 (x - (x2 - 3x))
4
4
+2= 0
.'. x=-if:2
x
4
1
22
a v(t)
=
a(t)
=
E
3 units''
e-2t ms-1
v'(t) = e-2t(_2)
when
ms
-i ms-2,
a=
dx
2
-2
[-cosxl~=0.42
-cosk+cosO
+ 2)
e-2t(_2)
-2t
e
Jo sin x dx = 0.42
.'.
(X3
~ 10.2 units3
k
18
1["
-V2
=32-¥
--
+4x+c
0
c Volume of revolution =
~3J:
[4~2 _
+ 4) dx
3
dx
= J0 (4x - x2) dx
=
13
the total distance travelled = 13 + 256 = 269 cm
y = x where
X2 - 3x = X
X2 - 4x
f:::±::J
4
-2t
= 0.42
.'.
=-i
1
8
= -ln8
t = ~ In 8 ~ 1.04 s
cos k = 0.58
••
8(t) = J v(t) dt = J e-2t dt = _~e-2t
b
k ~ 0.952
When
19
a
Y
= -"2e1 0 + c
2 = -~ + c
20
c=
I
S0,
10
...---=.IEt;==:::::;:=--"::"'-F--'--+-_
2
3
1
b Area = J0 (3x2 + 1) dx
= 30
units"
k
Jo (3x2
then
.'.
23
k
(0
+ 0)
a
-10
.'.
=
v
=
J a dt
=
m
8 = _~e-2
+ 2.5 ~ 2.43 m
ms-1
-
30t + c cms-1
Now when t = 0, v = 27
.'. v=3t2-30t+27
cms-1
16.59 ms-1
~ 3.466
.'. t ~ 3.47 s
J (6t - 30) dt
= 3t2
~
20e -0.2t = 10
-0.2t
1
e
"2
-0.2t = -ln2
.'. t = 5ln2
0
cms-2
a=6t-30
+.2 5
20 = 30 - 20e-O.2t
If v = 20 then
10
for which the only real solution is k = 2
{using technology}
20
t = 1,
b v(2) = 30 - 20e-O.4
C
+ 1) dx =
+k
-"2e -2t
v(O) = 30 - 20eo = 30 - 20(1) = 10 ms-1
[x3+xl~=1O
3
=
v(t) =30_20e-O.2t
a
[x + xl~
+3-
2~
The particle is 2.43 m to the right of O.
3
= 27
8
ii When
x
3
c If
t = 0, 8 = 2
2
(3,28)
30
=
+ c
d As t
.'. v
--> 00,
e v'(t)
=
.'.
c = 27
-->
e-O.2t
30 m
S-l
-->
0
(below)
0 - 20(e-O.2t)(_0.2)
4
a(t) = eO.2t ms
= 4e-02t
-2
Both numerator and denominator
positive for all t? O.
El
Mathematics
are positive so a(t)
SL - Exam Preparation
& Practice
Guide
is
(3rd edition)
-------------------------------------------------------~
vet)
2
9
set)
=
4
6
8
J vet) dt = f[30
10
the pollutants are minimised when
+c
= 30t + 100e -0.2t + c
so 0 + 100 + c = 10
=
= 30t + 100e-0.2t -
set)
m
26
N=(8-t)et-6,
a
dN
---
dN
dt
_.
-- =
b
2a
1
r
X2
dx
+ (8 -te )
+8-
= et-6(_1
90 m
a
a
x
= ~.
0:(t:(8
_ ( - 1) e t-6
dt
-90
t-6
t)
t 6
-
t)e
= (7 -
1
r~t~x
Sign diagram for C/(x):
-0.2
.'. c
Hence
_x_=2
2-x
x = 4 - 2x
3x = 4
8,
- 20e-O.2t] dt
= 30t -
~----...
x=~
20e-0.2t
But s(O) = 10,
=
(_X_)3
2-x
c Now when
30 ----------------------------------------------d
25
20
15
10 a
5
..
=
t
0 when
7
dN
dt
= 2
SIgn diagram for --:
= 2
N(7) = (1)e1 = e
there is a local maximum at (7, e).
a
[x~!
~ ((2a)! -a!)
2
=2
3
d N _ ( - 1) e·t-6
----
ii
dt2
3
.'. (2a)2-a2=3
Using technology,
a
x
1.391,
b When
=
a
R:!
1.391
R:!
d2N
Y R:! -11.391
R:!
dt2
1.179
y=y'X=x~
dy
dx
-
At
x
I
R:!
=---
2
2
dy
dx
1.391,
-
.'.
R:!
~
1
0.424
R:!
.'.
2v 1.391
R:! -
.'.
Y
R:!
-2.36x
=
6
2
dN
+
11
--2-:
dt
-2.359
inflection point at (6, 2).
(8 -
t)et-6
=
=
N
=
0
0
8 as et-6
>
0 for all t.
When the bacteria are all dead, N = 0
.'. t = 8 hours
ii The maximum number of bacteria in the sample is
e million
-1.179
+ 4.46
- IJ t
I
6
8
=
= 2(1) = 2
there is a non-stationary
0.~24
the equation of the normal is
-2.359(1.391)
t
when
.'. t
c
R:!
R:!
0
iii The graph cuts the t-axis when
the normal at this point has gradient
-2.359x - y
=
2eo
N(6)
1
2y'X
_1
= -x
t-6
= et-6(_1
= et-6(6 - t)
..
Sign diagram for
So, the point of contact is (1.391, 1.179).
+ (7 -te )
+ 7 - t)
R:!
2.72 x 106 bacteria.
d2N
dt2
iii The rate of increase is a maximum when
=
O.
This is at t = 6 hours.
8
C(x)
=
X2
+
= 8x-2
C/(x)
d
1
+ (2
- x)-2
=
-16x-3
=
2
-- 16 + -,---...,...".
x3
(2 - x)3
- 2(2 - x)-3(-1)
2
b C'(x) = 0 when
29
N (million)
(2-x)2
(2-x)3
ci
_f1+-==-------------=--=-4.---.
t
16
x3
x3
(2_X)3=8
a
. (_X
)3 =8
2 - x
f(x)
=
acosbx,
-7T:(
X :( 7T
The amplitude of the cosine curve is 3, so a = 3.
-.
The period of the curve is
2 x 27T= 47T
b
_=-=dicsSL
- ExamPreparation
&
PracticeGuide (3rd edition)
(hours)
(8,0)
-
=
27T = .!
47T
2
b
When
x = c,
So, the point of contact is
Now
so
d
y = 3 cos (~)
(c,
y
3 cos (~) ) .
y= x
y=3cos(~)
~~
=
+ sinx
3 ( - sin ( ~ )) (~)
= -~ sin (~)
x = c,
When
(£)
dy = -~ sin
dx
2
. 2( )
3sm ~
the equation of the normal is
2x - 3 sin ( ~)y = 2c - (3 sin ( ~ )) (3 cos (~) )
2x - 3 sin (~)
y = 2c - 9 sin (~) cos (~)
2x - 3 sin ( ~)
y = 2c - ~ sin c
sin
a
(x
+ sinx
30
a
2 sin
{using technology}
y
0.5
-0.5
2e =
- x2) dx
::::::0.944 units''
the normal has gradient
{using
Jo
Area =
2
y = sin2t - sint
e cos e}
ii If the normal also passes through the origin, then
b The graph meets the t-axis when
2c - ~ sine = 0
sine = ~c
Using technology,
c::::::±2.0234,
c
- 1) = 0
sint(sint
0
y = 0
sin2t-sint=0
••
.:. sin t = 0 or 1
y
So, for
J02
Area =
=
0:(
(sin2
[0 -
Since
y =
Jo
k
{J (t) touches the t-axis at t = ~}
-,
[sin t - sirr' t] dt
2
J0
+1
X2
= 0 or ~.
t - sin t)] dt
::::::0.227 units '
31
t :( 2, t
{using technology}
is symmetric about the y-axis,
2
(x +1)dx=12
[x3 + x J: = 12
3
y
28
3
(k3 + k) x
..
3
k
(;
+ 3k
4
32
+ 3k = 36
- 36
=
0
a a=4costms-2
.. v = 4cost dt = 4sint
J
{using technology}
But when
a
12
( sin x) dx
::::::4.34 units"
29
=
Using technology, the only real solution is k = 3.
2
V = 7r ) 1<-
k3
(0)
y
f(x)
= x
ms-1
t = 0, v = 2
=
..
2
..
c=
v = 4sint
+2
+ sinx
+c
+c
4(0)
2
ms-1
1
v(4) ::::::-1.03 ms7r
b The total distance travelled =
x
=
b Area =
Jo~ (x + sinx)
::::::2.23 units''
-,c We need to solve
Using technology,
So, a:::::: 1.618
dx
+ sinx
J:
Iv(t)1 dt
14sint
+ 21 dt
::::::16.72 m
{using technology}
x
J05
= x2
x = 0 or x:::::: 1.618
33
a
f (x)
l' (x)
= 2 cos x - sin x
+ 2x sin x
+ 2 sin x + 2x cos x
= - 2 sin x - cos x
= 2x cos x - cos x
= (2x - 1) cosx
Mathematics
SL - Exam Preparation & Practice Guide (3rd edition)
•
J'(x) =0
b
2x-1=0
when
for 0 ~ x ~ 2,
:.
J'C~)
x = ~
3
7'
x-2!.
- 2
or
<0
= -~cos:t
1'(1) = cos
cosx=O
or
1> 0
<0
1'(1.9) = 2.8 x cos 1.9 ~ -0.9052
Sign diagram for f'(x):
x =
f(~)
and
f(2) ~ 1.896
~ 1.755
=
2
[- cosxl:+
and
z = ~.
(t) (~) + (i)(~)
2
x = ~.
:3
c
sin x dx = 0.3
2
y =:t,
=~+i
the smallest value is ~ 1.76 when
J:+
t,
= P(A n B) + P(A' n B)
b P(B)
ow
P(A' I B)
= 0.3
=
P(A' n B)
P(B)
_ (i) (~)
- cos(a + 2) + cos a = 0.3
2
:3
cos a - cos(a + 2) = 0.3
1
Y = cosx - cos (x
Y
4:
+ 2)
It
1
a
sin 2e = tan e,
If
.
2smecose
2sine
-1
then
sine
= --e
cos
cos2 e = sin e
2
e - sine = 0
sin e(2 cos'' e - 1) = 0
2sinecos
So, a ~ 1.96 or 5.46
{using technology}
SOWTIONS TO TRIAL EXAMINATION
1
cos e = ~
sine = 0 or
cose =
±~
±~, ±3;, ±7r
e = 0,
b
2
sine = 0 or
, • Ho calculators
A
a
x
6
24
X
=
X2
144
x = ±12
But
x>
0,
x
=
5
12
a The vertex is at (3, 7), so
f(x)
=24x
:.
(~t
24
1-~
24
-~
- 16
=~
T
"2
= a(x - 3? + 7
= 3, k = 7
= 2.5
a( _3)2 + 7
:. 9a
so
-1) cos x + e-X(
= -e-X(cosx
-
:. f(x) = 0
when
sinx)
6 f (x)
=-e-~(0+1)
=
0 at A
J (2X -
3x
-!) dx
2X2
3x!
=T---Y-+C
c h(~) = e-~ cos ~
2
=0
=
the point of contact is ( ~, 0).
SL - Exam Preparation & Pradice
>
But x
b h'(~) = -e-~ (cos~ +sin~)
Guide (3,dedition)
X2
But
-
:.
2
6v'x
f(4)
16 - 12
y - ~ = -e-~
x-2
. e~y=-x+75.
..
2.5
-4.5
c f(x)=-~(x-3?+7
+ sinx)
the tangent has equation
=
=
a= -~
h(x) = e-xcosx
h' (x) = e-X(
~cs
h
f(O)
b
=48
a
•
+C
= 3
+C = 3
C
=
-1
:. f(x)=x2-6v'x-1
~(x - 3)2 = 7
(x - 3)2 = 14
x - 3
= ±v'14
:. x
= 3 ± v'14
V14, 0).
A is (3 +