solution+v1 vg mvg mad6nvc08+trigonometry+and+derivatives
TRANSCRIPT
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8/3/2019 Solution+V1 VG MVG MaD6NVC08+Trigonometry+and+Derivatives
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VG/MVG-Level Test: MaD6NVCO08DirectionsWarning: There are more than one version of the test.Test time 90 minutes.Resources Calculators and Formulas for the National Test in Mathematics D, and your
personalized green-booklet.
Test material:The test material should be handed in together with your solutions.
Write your name on all sheets of paper you have in front of you.
The test The test consists of a total of 7 problems, i.e.:p8-p14.Note that for most of problems short answers are not enough. They require that
you write down what you do, that you explain your train of thought, that you,
when necessary, draw figures. When you solve problems graphically or
numerically please indicate how you have used your resources.
Problems 13 and 14 are larger problem which are also graded heavily.Try all of the problems. It can be relatively easy, even towards the end of the
test, to receive some points for partial solutions. A positive evaluation can be
given even for unfinished solutions.
Score and The maximum score is 36 points.
mark levels The maximum number of points you can receive for each solution is indicated
after each problem. If a problem can give 2 Pass-points and 1 Pass with
distinction-point this is written (2/1). Some problems are marked with, whichmeans that they more than other problems offer opportunities to show
knowledge that can be related to the criteria for Pass with Special Distinction in
Assessment Criteria 2000.
Lower limit for the mark on the test
Pass (G): 15 points from the G-Level test. Pass (G): 10 points in this test. VG Pass in the G-level-test and 20 points in this test MVG Pass in the G-level-test and 25 points in this test, and MVG-
quality in most of problems specially in P13 and P14.
8a 8b 8c 9 10a 10b 11 12 13 14 Sum
G 1 2 3
VG 2 2 4 3 3 2 2 3 4 5 33
MVG M1
M5
M1-
M5
M1
M5
M1-
M5
M1-
M5
G
VG
MVG
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8. Differentiate the following functions w.r.t. x
a. . [0/2]( ) DaxCBAy n ++= cosb. )ln(cosx [0/2]y =
c. First show that the derivative of ( )x
ax isf=
( ) aaxfx
ln=
. Thendifferentiate ( )ax [0/4]ay x ln
2
=
Suggested solutions:
a. Answer: ( ) 1nax cossin += CBaxnaCAy
Use the chain ruledx
dz
dz
dy
dx
dy= and ( ) ( )xaAxf = sin ( ) ( xaaAxf )= cos
Change of variable axCBz cos+ axadx
dzcos= ;
( )
=+
=+=++=
axaCdx
dzaxCBz
znAdzdyDzADaxCBAy nnn
sincos
cos 1
( ) ( ) ( 11 cossinsin +==== nn axCBaxnaCAaxaCznAdx
dz
dz
dy
dx
dyy )
Answer: ( ) 1cossin += naxCBaxnaCAy b. Answer: xyxy tan)ln(cos ==
)ln(cosxy =
Change of variable: xz cos
( ) xx
xx
zdx
dz
dz
dy
dx
dy
xdx
dzxz
zdz
dyzy
tancos
sinsin
1
sincos
1)ln(
=
===
=
==
Using:dx
dz
dz
dy
dx
dy= and ( ) ( )xaAxf = cos ( ) ( )xaaAxf = sin ;
( ) ( )x
AxfxAxf == ln
c. Answer: ( )axayx
ln
2
= ( ) xaxaaxyx
lnln2
2
= ax 12
+
To show that the derivative of ( ) xaxf = is ( ) xaaxf = ln , we may rewriteas and then calculate :( ) xaxf = ( ) kxx eaxf = k
( ) ( ) axxkxxkxx eaxfkakxekxaxeaeaxf lnlnlnlnlnln ======= ( ) ( ) xaxaxx aaeaxfeaxf ==== lnln lnln QED
Change of variable: axz
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( ) ( )
( )x
aax
azdx
dg
dx
dz
dz
dg
dx
dg
adxdzaxz
zdz
dgzzg
axxg
1111ln
ln
====
=
==
=
( )
( )
2
2
ln22ln
ln
22 xu
uu
x
aaxxaadx
du
du
dh
dx
dh
aadu
dhauh
xdx
duxu
axh
===
==
=
=
( ) ( ) ( )( ) ( )
( ) ( ) ( )
( )x
aaxaaxghghy
xxgaxxg
aaxxhaxh
xgxhaxay
xxxx
x
1lnln2
1ln
ln2
ln2222
2
+=+=
==
==
==
Answer: ( )x
aaxaaxyxx 1lnln2
22
+=
9. Find the limit value of( ) ( )
+ h
hxhx
h 2
coscoslim
0. Interpret your results. You may use
( ) 1sinlim0
=
h
h
h. [0/3/M1-M5]
Suggested solutions:
( ) ( ) ( ) ( ) ( ) ( )( )
+=
+ h
hxhxhxhx
h
hxhx
hh 2
sinsincoscossinsincoscoslim
2
coscoslim
00
( ) ( ) ( ) ( )
=
h
hxhxhxhx
h 2
sinsincoscossinsincoscoslim
0
( ) ( )xx
h
hx
h
hx
hhsin1sin
sinlimsin
2
sinsin2lim
00==
=
/
/=
Answer:( ) ( )
xh
hxhx
hsin
2
coscoslim
0=
+
Due to the fact that( ) ( )
+ h
hxhx
h 2
coscos
0lim is the definition of the
derivative of the function xcos , we found that the derivative of xcos is
xsin . i.e.: Answer:( ) ( )
xh
hxhx
dx
xd
hsin
2
coscoslim
cos
0=
+=
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10. The function ( ) 52cos3 += xxf is givena) Find the largest possible slope of a tangent to the curve ( ) 52cos3 += xxf . [0/3]
b) Find the equation of such a tangent. [0/2/M1-M5]
Suggested solutions:a) The slope of a tangent to a curve is equal to the value of the derivative
of the function at the point, i.e. ( )afk = where k is the slope of thetangent, and a is the x-coordinate of the point where the tangent is
tangent to the curve. ( ) 52cos3 += xxf ( ) xxf 2sin23 = ( ) xxfk 2sin6 == .
The largest value of ( ) xxfk 2sin6 == is 6maxmax == fk .
Answer: The largest possible slope of a tangent to the curve( ) 52cos3 += xxf is 6max =k .
b) The largest value of ( )xf xk 2sin6 == is 6maxmax == fk and it occurs at
a point where 12=
x .sin12sin =x ( )1sin2 1 = x
2.
2
32 nx += Nnnx +=
.
4
3
One such a tangent occurs at4
3=x , where the tangent and the
function share the point, i.e.
( )2
95
2
9
4
36
4
3
55054
32cos3
4
3
6
52cos3
=
+=+
=
=+=+
=
+=
+=m
mmy
f
mxy
xxf
The equation of a tangent with largest possible slope is 5.456 += xy .
The next such a tangent is made at radx4
7
4
3
=+= .
( )
5.1056
5.104
76
4
7
55054
72cos3
4
7
6
52cos3+=
+=+
=
=+=+
=
+=
+=xy
mmy
f
mxy
xxf
-2
0
2
4
6
8
10
-1,57 0,00 1,57 3,14 4,71 6,28
x rad
( )52cos3 += xxf
5.456 += xy 5.1056 += xy
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11. The graph of the function11. The graph of the function xy cos= intersects the lines ky = atfour points in the
interval radianx 20 . Find the sum 4321 xxxxs +++= . Give the answer in
radian. [0/2/M1-M5]
Suggested Solutions: 1x 2x 3x 4x
Answer: radianxxxxs 44321 =+++= Using the properties of the unit circle andthe fact that
( ) ( ) 14 2coscos2cos xxk ==== ,( ) 12coscos xx == ( ) 13coscos xx +==+
we may conclude that: 42 11114321 =++++=+++= xxxxxxxxs
-1,5
-0,5
0,5
1,5
0,00 0,52 1,05 1,57 2,09 2,62 3,14 3,67 4,19 4,71 5,24 5,76 6,28
x
4x
3x2x
1x
xy cos=
ky =
ky =
( )yxP ,( )yxQ ,
( )yxR , ( )yxS ,
x
y
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12. Find the equation of the tangent to the curve of the function ( ) xxxf 2sin = at the point( ) 2; . [1/3]
Suggested solutions: Answer: The equation of the tangent to the curve ofthe function at the point( ) xxxf 2sin = ( ) 2; is: += xy 3 .
The tangent mkxy += to the curve of the function ( )xf at the point( ) 2; , and the function ( ) xxxf 2sin = share two important properties:
( ) ( )( )
=
==
fk
fy 2
( ) ( ) ( ) 3212cos2cos2sin ====== fkxxfxxxf 3=k
( )
=+==+
=
+=mmm
y
mxy3223
2
3 += xy 3
Answer: The equation of the tangent to the curve of the function( ) xxxf 2sin = at the point ( ) 2; is: += xy 3 .
-20,0
-15,0
-10,0
-5,0
0,0
5,0
10,0
-3,14 -1,57 0,00 1,57 3,14 4,71 6,28
x rad
f(x),y(
x)
( ) xxxf 2sin =
+= xy 3
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13. You are going to study the function Bx13. You are going to study the function BxAy Ay += cos . Show that min is always
true if
max 5 yy =
BA =3
2, where B is a positive number. [2/4/M1, M2, M3, M5]
You may choose to solve the problem in general without going through the following
steps which is time-consuming. Otherwise you may follow the following steps (lengthy
but possible!):a. Show that the maximum value of the function is five times as large as functionsminimum value when 2=A and 3=B [1/0]
b. Let 8.1=B . Find A if minmax 5 yy = . [0/1]
c. Let 4=A . Find B if minmax 5 yy = . [1/1]
d. Show that min is always true ifmax 5 yy = BA =3
2. [0/2/M1, M2, M3, M5]
Suggested solutions: General method:
BxAy += cos
=
=
0
sin
y
xAy 0sin = xA Nnnx =
xAy sin= xAy cos= ( )( )( )
NnAnAy
AnAy
=+=
A
( )( )( ) ( )
NnnxyAnAy
nxyAnAy
+=>=+=
=
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Answer: For all positive values of BA3
2= , the maximum value of the
function is five times as large as its minimum value: min .max 5 yy =
QED
If :0=+=
=>==
12atmaximumahas012cos
2atminimumahas02cos
The function Bx has a minimum at NnAy += cos nx = 2 . The
minimum value of the function at this point is( ) BABnAy +=+= 2cosmin .
The function has a maximum at ( ) Nnnx += 12 . The maximumvalue of the function at this point is ( )( ) BABnAy +=++= 12cosmax .
If ( ) BBAABABABABA
BAy
BAyyy
=+=++=+
+=
+==
555555
min
max
minmax
BABABABA3
2
3
22346 ====
Answer: For all negative values of BA3
2= , the maximum value of
the function is five times as large as its minimum value: min5 y= .maxy
QED
Alternative General method (for full points):BxAy += cos ; BA
3
2=
If both A and B are positive: BA3
2= BA
3
2= .
Due to the fact that cosine is a bounded function 1cos1 x , itsmaximum value is one and its minimum value is minus one. Therefore,the local maximum and local minimum values of the function
are:BxAy += cos
minmaxminmaxmin
max
553
15
3
5
3
2
3
1
3
2
3
5
3
2
yyyBBy
BA
BBBBAy
BBBBAy
====
=
=+=+=
=+=+=
QED
Answer: For all positive values of BA3
2= , the maximum value of the
function is five times as large as its minimum value: min .max 5 yy =
QED
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IfA is negative but B is positive: BA3
2= BA
3
2= .
Due to the fact that cosine is a bounded function 1cos1 x , its
maximum value is one and its minimum value is minus one. Therefore,the local maximum and local minimum values of the function
are:BxAy += cos
minmaxminmaxmin
max
553
15
3
5
3
2
3
1
3
2
3
5
3
2
yyyBBy
BA
BBBBAy
BBBBAy
====
=
=+=+=
=+=+=
QED
Answer: For all negative values of BA3
2= , the maximum value of
the function is five times as large as its minimum value: min5 y= .maxy
QEDAlternative (step by step method) solutions:
a. min [1/0]maxmin
max5
132
5323sin2
3
2
sin
yyy
yxy
B
A
BxAy
=
=+=
=+=+=
=
=
+=
b.
=+=
+=
=
=
=
+=
minmax
min
max
minmax
2
8.1
8.1
?
8.1
5
sin
yy
Ay
Ay
A
B
yy
BxAy
( )
=
===+
+=++=+
2.1
6
2.72.768.195
958.18.158.1
A
AAAA
AAAA
Note that minmax 53
2
3
2
8.1
2.1
8.1
2.1yyBA
B
A
B
A====
=
= [0/1]
c.
( )
=
=
=+
+=+
+=+
+=
+=
=
=
=
+=
6
244
5204
5204
454
4
4
?
4
5
sin
min
max
minmax
B
B
BB
BB
By
By
B
A
yy
BxAy
BB
Note that minmax 53
2
3
2
6
4
6
4yyBA
B
A
B
A==
=
=
=
= [1/1]
The rest is as the General method presented at the beginning of thesolution.
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MVG-Quality In solving the problem number 13,
the student shows in general the
highest MVG quality by
M
Developing the problem in general
and showing: minmax 53
2yyBA == .
The student finds y , solves
and uses the required
(or/and ) table to find the
maximum and minimum value of
the function. The student studies
both and
0=y
yyx __
"y
0>A 0
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Suggested solutions MaD6NVCO08 NV-College - Sjdalsgymnasiet
14. Investigate how number of zeroes of the function14. Investigate how number of zeroes of the function
+= babaxbxay ,,0,02sinsin
varies depending on the chosen values of constants and b .a
[0/5/M1,M2,M3, M5]
Suggested solutions:
+= babaxbxay ,,0,02sinsin
We may use Double Angle Identities: ( ) ( ) ( ) cossin22sin = ( ) 0cos2sin0cossin2sin02sinsin =+=+=+ xbaxxxbxaxbxa
===+
===+==
b
axaxbxba
xxxNnnxx
2coscos20cos2
360,180,018000sin 321 [0/1]
( )
( )
>
+===
+===
+
=
+
=