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A EXAMHNO

Exetastik perodoc Ianouarou-Febrouarou 2012

Paraskeu 20/1, 18:30-21:00 (me zug A.M.)

Exetazmeno mjhma: Majhmatik

Eishghtc: Euggeloc Marinkhc

Jma 1o (2 mondec): Dnontai oi pnakec

A =

[2 10 1

]kai B =

[2 11 0

].

Na deiqje ti (BA AB)2 = 2I2.

Jma 2o (2 mondec): Dnetai to grammik ssthma

x1 x2 + x3 = 1x1 + c

2x2 + x3 = 2x1 + cx2 + 3x3 = 1

.

Qrhsimopointac ton kanna tou Cramer na breje gia poic timc thc paramtrou c toparapnw ssthma qei peirec lseic.

Jma 3o (2 mondec): Upologste ta paraktw arista oloklhrmata.

(a)

2x(x2 + 15)15dx, (b)

x+ 2

x2 9dx.

Jma 4o (2 mondec): H antoq enc xlinou dokario orjogniac diatomc enai sh me to

ginmeno thc mikrc distashc x ep to tetrgwno thc meglhc distashc y. Na brejonoi diastseic thc diatomc enc xlinou dokario mgisthc antoqc, an h diagnioc thc

diatomc enai 3.

y

x

3

Jma 5o (2 mondec): Dnetai h sunrthsh f(x) =3xex, x [2, 4]. Na upologisteo gkoc tou stereo pou pargetai ap thn peristrof thc grafikc parstashc thc

parapnw sunrthshc grw ap ton xona xx.

Prin fgete paradste to grapt sac KAI ta jmata. Kal epituqa.

Jma 1o

'Eqoume ti

BA =

[2 11 0

] [2 10 1

]=

[4 + 0 2 + 12 + 0 1 + 0

]=

[4 32 1

],

AB =

[2 10 1

] [2 11 0

]=

[4 + 1 2 + 00 + 1 0 + 0

]=

[5 21 0

],

= BA AB =[4 32 1

][5 21 0

]=

[ 1 11 1

].

Epomnwc

(BA AB)2 = 2 = =[ 1 1

1 1

] [ 1 11 1

]=

[1 + 1 1 + 1

1 + 1 1 + 1]

=

[2 00 2

]= 2

[1 00 1

]= 2I2.

Jma 2o

Upologzoume prta thn orzousa D tou sustmatoc.1oc trpoc: Anptuxh wc proc thn 1h gramm.

D =

1 1 11 c2 11 c 3

= c2 1c 3

+ 1 11 3+ 1 c21 c

= 3c2 c+ 3 1 + c c2 = 2c2 + 2 = 2(c2 + 1).2oc trpoc: Kannac tou Sarrus.

1 1 11 c2 11 c 3

1 11 c2

1 c.

D = 1 c2 3 + (1) 1 1 + 1 1 c 1 c2 1 c 1 1 3 1 (1)= 3c2 1 + c c2 c+ 3 = 2c2 + 2 = 2(c2 + 1).3oc trpoc: Idithtec orizousn.

D =

1 1 11 c2 11 c 3

221=1 1 10 c2 + 1 01 c 3

,opte me anptuxh wc proc th 2h gramm parnoume

D = (c2 + 1)

1 11 3 = (c2 + 1)(3 1) = 2(c2 + 1).Gia na qei to ssthma peirec lseic ja prpei opwsdpote D = 0. 'Omwc, gia kjepragmatik arijm c isqei ti

c2 + 1 > 0 c2 + 1 6= 0 2(c2 + 1) 6= 0 D 6= 0.'Ara to ssthma den qei peirec lseic gia kama tim thc paramtrou.

2

Jma 3o

(a) Jtoume

x2 + 15 = t d(x2 + 15) = dt (x2 + 15)dx = dt 2xdx = dt,opte

2x(x2 + 15)15dx =

(x2 + 15)15 2xdx =

t15dt =

t16

16+ c =

(x2 + 15)16

16+ c.

(b) 'Eqoume ti

x+ 2

x2 9 =x+ 2

(x+ 3)(x 3) =A

x+ 3+

B

x 3 =A(x 3) +B(x+ 3)

(x+ 3)(x 3)=A(x 3) +B(x+ 3)

x2 9 .Ap thn isthta tou prtou me to teleutao klsma parnoume

A(x 3) +B(x+ 3) = x+ 2.Gia x = 3 qoume

6A = 1 A = 16,

en gia x = 3 qoume

6B = 5 B = 56.

Epomnwcx+ 2

x2 9dx = (

1

6

1

x+ 3+5

6

1

x 3)dx =

1

6

1

x+ 3dx+

5

6

1

x 3dx

=1

6ln |x+ 3|+ 5

6ln |x 3|+ c.

Jma 4o

An S enai h antoq tou dokario, tte S = xy2. Epshc

x2 + y2 = 32 x2 + y2 = 9 y2 = 9 x2,dhlad

S = xy2 = x(9 x2) = 9x x3.'Ara h antoq tou dokario mpore na grafte wc sunrthsh mno tou x, opte

S(x) = 9x x3 S (x) = 9 3x2 S (x) = 6x.'Eqoume ti

S (x) = 0 9 3x2 = 0 3(3 x2) = 0 x2 = 3 x = 3.

H tim x = 3 aporrptetai giat prpei x > 0. Epshc S (3) < 0. 'Ara sto x = 3 hS(x) parousizei olik mgisto, dhlad h antoq gnetai mgisth. Gia x =

3,

y2 = 9 x2 = 9 3 = 6 y =6.

(H tim y = 6 epshc aporrptetai giat prpei y > 0).

3

Jma 5o

O zhtomenoc gkoc enai

V = pi

42

[f(x)]2dx = pi

42

3xexdx = 3pi

42

xexdx.

To antstoiqo aristo oloklrwma enaixexdx =

xd(ex) = xex

exdx = xex ex + c = (x 1)ex + c,

opte 42

xexdx = (x 1)ex42

= (4 1)e4 (2 1)e2 = 3e4 e2 = e2(3e2 1).

Epomnwc

V = 3pie2(3e2 1).

4