Μαθηματικά Κατεύθυνσης Απαντήσεις Θέματων...
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Transcript of Μαθηματικά Κατεύθυνσης Απαντήσεις Θέματων...
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2013
| www.lazaridi.info | tel. 6977.385.358 |
www.lazaridi.info Page 1 of 4
| | 2013 | |
|
1. 2. 3. 4. / / / /
|
1. 0=-2|2-z|+|2-z|2=|2-z|+|2-z|2=|2-z|+)2-z2)(-z(2=|2-z|+2)-z2)(-z( 22 ()
|z 2| = 1 (2,0) = 1
pi , pipi |z|
3
2. O z1, z2 w2+w+ = 0 Im(z1) = - Im(z2) pi 1=)zIm(1=|)zIm(|2=|)zIm(2|2=|)Im(z-)zIm(| 11121 Rmi,-m=z,i+m=z 21
z1 |z 2| =1 2=m1=1+)2-m(1=|i+2)-(m|1=|2-i+m|1=|2-z| 221 Rmi,-2=z,i+2=z 21 Vieta -=z+z=S 21 -4=
5==zz=P 21
,
2=|)Im(z-)zIm(| 21 z1, z2 ( pi ) pi (z1), B(z2) 2 = 2 (=1)
1)0,2(K
30
-
2013
| www.lazaridi.info | tel. 6977.385.358 |
www.lazaridi.info Page 2 of 4
3. o1
22
3 a-va-v-a=v
3+|v|3+|v|3|a|+|va|+|va||a-va-v-a|=|v| 2o122o1223 pi
1-3+|v|3+|v|3|v|3+|v|3+|v|3|v| 2323 -1 -1 1-)1+|v|+|v(|3|v| 23
-1
1+|v|+|v|1
-31+|v|+|v|
|v|22
3
)-1)(
1+|v|+|v|1
-31+|v|+|v|
1+|v|+|v||v(|22
2
-1|v1+|v|+|v|
1-3| 2
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2013
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www.lazaridi.info Page 3 of 4
H K(x) = (0, )+ () = ( )x(Klim
0x)x(Klim
+x) = (-2, + )
0 )+-2,( (x) = 0 = (0, )+
pi f(g(x)) = 1
3.
04pi
-x
dt)t(fx=)x(h [0, pi/4), pi (0, pi/4)
)4pi
-f(xx-dt)t(fx=)x(h 04pi
-x
h(0) = h(pi/4) = 0 pi .Rolle pi )
4pi
,0(x o 0=)x(h o
0=)4pi
-f(xx-dt)t(fx oo0
4pi
-x
o
o
)4pi-f(xx=dt)t(fx oo04pi
-x
o
o
)
4pi
-f(xx
x=dt)t(f o
o
o0
4pi
-xo
)4pi-f(xx=dt)t(f oo04pi
-xo
|
1.
0=h
h)-f(1-f(1)+f(1)-5h)+f(1lim0=
hh)-f(1-5h)+f(1
lim0h0h
0(1)f0(1)f60(1)f(-1)-(1)f5
===0=)h
f(1)-h)-f(1-
hf(1)-5h)+f(1(lim
0h
:
)x(f a=h
)f(x-)ah+x(flim o
oo
0h
)x(f
-
2013
| www.lazaridi.info | tel. 6977.385.358 |
www.lazaridi.info Page 4 of 4
2.
0>1-x
1-)x(f=)x(g * x > 1 g (1, + )
* .1.
1+xx
du)u(g=)x(h , x > 1 g(x)-)1+x(g=)x(h > 0 h (1,+ )
: < + 1 g
g() < g(+1)
g(+1) g() > 0
)5+x2(h>)5+x8(h 42 1>5+x2,1>5+x8 42 h
(1,+
) (0,2)(-2,0)x0>)x->x-x4>x2-x8x2>x85+x2>5+x8 242424242 (4x00 2
3.
f [1, x] pi )x,1(
1-x1-)x(f
=
1-xf(1)-)x(f
=)(f
pi 01)-(f(x)-1)-(x)(xf1)-(x)(xf1-f(x) >