Solution+V1 VG MVG MaD6NVC08+Trigonometry+and+Derivatives

8/3/2019 Solution+V1 VG MVG MaD6NVC08+Trigonometry+and+Derivatives

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Suggested solutions MaD6NVCO08 NV-College - Sjdalsgymnasiet

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

[email protected] to use for educational purposes. Not for sale. page 1 of 16


2/16



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]


( ) ( ) ( ) ( ) ( ) ( )( )

+=

+ 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


7/16



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

+=>=+=

=


8/16



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


9/16



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.


10/16



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


11/16


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]


+= babaxbxay ,,0,02sinsin

We may use Double Angle Identities: ( ) ( ) ( ) cossin22sin = ( ) 0cos2sin0cossin2sin02sinsin =+=+=+ xbaxxxbxaxbxa

===+

===+==

b

axaxbxba

xxxNnnxx

2coscos20cos2

360,180,018000sin 321 [0/1]

( )

( )

>

+===

+===

+

=

+

=

Solution+V1 VG MVG MaD6NVC08+Trigonometry+and+Derivatives

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