RJESENJA 2. MEDUISPITA IZ MATEMATIKE 1

1. (a) thx = shxchx =exexex+ex

(b) y = exexex+ex ye2x + y e2x = 1 x = 12 ln(1+y1y )

arthx = 12 ln(1+x1x), x 1, 1.

2. limn(nn2 3n+ 2) = () = limn(n

n2 3n+ 2) (n+

n23n+2)

(n+n23n+2) =

limn 3n2n+n23n+2 =32 .

3. limx0[ln(1+2x)

x arcsin(ex

2

)]= limx0

ln(1+2x)22x limx0 arcsin( e

x

2 ) =

= 2 arcsin(12) = pi3 .4. lim

x0+f(x) = f(0) = lim

x0f(x):

limx0

f(x) = limx0

(12+sin(ax)

x

)=

12+ lim

x0sin(ax)ax

a = 12+ a = f(0) = b

limx0+

f(x) = limx0+

shx

x= 1

= 1 = 12+ a = b = b = 1, a = 1

2.

5. (a)

f (x0) = limh0

f(x0 + h) f(x0)h

(b) f (5) = limh0

f(5 + h) f(5)h

= limh0

15+h 15

h= lim

h01

5h+ 25= 1

25.

6. t....y 0 = y(0)(x 0) y = y(0)x

yey + yeyy + 1 + 2x+ y = 0 y|T = 12xey+yey+1 |T = 12 t....y = 12x

7. y(x) = y(t)x(t) =3t2+2et+1 ;

y(t) =6t(et+1)(3t2+2)et

(et+1)2

et+1 =6tet+6t3t2et2et

(et+1)3|t=1 = e+6(e+1)3

8. Postoji tocka c na krivulji y = lnx takva da je tangenta u toj tockiparalelna sekanti kroz tocke (1e ,1) i (e2, 2).

f(e2) f(1e ) = f (c)(e2 1e ) 2 + 1 = f (c)( e31e )

f (c) = 3ee31 ;

1c =

3ee31 c = e

313e

9. (a) f(x) = Tn(x) +Rn(x) =n

k=0f (k)(c)k! (x c)k +Rn(x),

gdje je Rn(x) =f (n+1)(x1)(n+1)! (x c)n+1, a x1 c, x ili x1 x, c

(b) f(x) = 14 + 14(x 3) + 6(x 3)2 + (x 3)3

1

10. Izracunati:

(a) limx0

10x 10xx

= limx0

10xln10 + 10xln101

= 2ln10

(b) limxx th

(1x

)= lim

xth(1x

)1x

= limx

1

ch2(1x

) ( 1x2)

1x2

= 1.

PRIMJEDBA: U rjesenjima nedostaju slike.

2