Μαθηματικά Κατεύθυνσης Γ΄Λυκείου ΑΝΑΛΥΣΗ (129-λυμένα-θέματα-πανελλήνιων-εξετάσεων-1983-2003)

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  • 1. (1983-2003)1983 199019971984 199119981985 199219991986 199320001987 199420011988 199520021989 19962003

2. 1983 x +1 x1) f + f(x)= x( x 2 + 1 -x) . f (0, + ) + .x+1x +1ln x x xx +1x +1 x>0 , x =e lim ( lnx)= lim lim lnx=x +xx + x x +x+1 x +1 ln x xx 1(+ )= + . lim x= lim e= + .x + x + x2 +1 x2 1 lim ( x 2 + 1 -x) = lim = lim = x + x +x + x +1 + x2 1 x ( 1 + 2 + 1)x 1 1 1lim lim =0 =0. x + x x +1 2 1+ +1 x2 x+1 111+xxx(+ )0 . : x =x = xx , x>0 11xxlim f(x)= lim [ xx (x 2 + 1 -x)]= lim x lim [ x( x 2 + 1 -x)]= x + x +x + x +ln x ln x x2 +1 x2 x lim e x lim ( x )= lim e x lim =x +x + x +x + x2 +1 + x1x ( 1 + 2 + 1) x+ 1 1ln x +(ln x) 1e0 = , lim = lim = lim =0 . 2 2 x +xx + ( x) x + x1 3. 19832) f , [ , ] , ( , ) f()=f()=0 . : f ( x)) F(x)= , c [ , ] , co ( , ) xc F( co)=0 .) c [ , ] , co ( , ) (co , f(co)) y=f(x) (c , 0) .) c [ , ] F [ , ] . F [ , ] ( , ) f ( x )( x c ) f ( x )( x c ) f ( x )( x c ) f ( x ) F(x)== .( x c) 2 ( x c) 2 f ( )f ( ) F()= =0 , F()==0 . Rolle c c F co ( , ) F( co)=0 .) y=f(x) (co , f(co)) y- f(co)= f (co)(x- co) , co () . F( co)=0f (co )(co c) f (co ), =0 f(co)( co-c)= f(co) (1)(co c) 2 y=f(x) (c , 0) 0- f(co)= f (co)(c- co) f(co)( co-c)= f(co) , (1) . 2 4. 19833) ) x>0 : lnx x-1 .) f [0 , + ) x ln x 1 x ,0 < x 1 f(x)= 0 , x = 0. : 1, x = 1 i) f , ii) (0 , 1) 1iii) f (1)= - . 2) g(x)=lnx-x+1 , x>0 .1 1 x g (x)=(lnx)-(x)+(1)=-1=, x>0 . g(x)=0 x=1 xxg (x)>0 01. x=1 g g(1)=ln1-1+1=0. x>0 g(x) g(1) lnx-x+1 0 lnx x-1 .) i) f (0 , 1) (1 , + ) .1 x ln xln x + (ln x) x2 lim+ f(x)= lim+ = lim+=lim+= lim+ x = lim+ x0 x0 1 xx0 1 x01 x0 1x0 x 1 ( 1) 2 xx x = lim+ (-x)=0=f(0) , f 0 . x0011 ln x + x x ln x 0( x ln x)x = lim (-lnx-1) lim f(x)= limx 1x 1 1 x = limx 1= lim (1 x) x 1 1 x 1 = -ln1-1= -1= f(1) , f 1 . f [0 , + ) .ii) x (0 , 1) ( x ln x ) (1 x) x ln x (1 x) (ln x + 1)(1 x ) + x ln x ln x x + 1f (x)== = . (1 x ) 2(1 x ) 2 (1 x) 2 () x>0 : lnx x-1 lnx-x+1 0 =3 5. 1983 x=1 . x (0 , 1) : lnx-x+10 x2 f (x)0 (0) x>0 7x4+6x+5>0 , 7x4+3x+3>0 f (0 , + ) : lim f(x)=x + lim [( 7 x 4 + 6 x + 5 7 x 4 + 3x + 3 ) 63x 2 5 x + 20 ]=x + [(7 x 4 + 6 x + 5) (7 x 4 + 3x + 3)] 63x 2 5 x + 20 lim =x +7 x 4 + 6 x + 5 + 7 x 4 + 3x + 325 20x 2 (3 + ) 63 + 2(3x + 2) 63 x 2 5 x + 20xx x lim= lim =x + 7 x + 6 x + 5 + 7 x + 3x + 34 4x +65 33x2 ( 7 + 3 + 4 + 7 + 3 + 4 )x xxx25 20(3 + ) 63 + 2xx x(3 + 0) 63 0 + 0 3 7 9 9 lim=== .x +65 33 7+0+0 + 7+0+0 2 72 7+ 3 + 4 + 7+ 3 + 4x xxx21 23. 19872) f f(x)=x4-14x2+24x . C f . , , C , C , , xx . yy . , , C , C , , xx , f (x)=0 . : Horner f (x)=4x3-28x+24=4(x3-7x+6) = 4(x-1)(x2+x-6) , x f (x)=0 4(x-1)(x2+x-6)=0 ( x-1=0 x2+x-6=0 ) x=1 x=2 x= -3 . G yy , xG 0 . (1 , f(1)) , B(2 , f(2)) , (-3 , f(-3)) . 3 + 2 f (3) + f (2)1 f (3) + f (2) (, ) (- , ) AG = 2 GM (*) . 22 2 21 (*) xG-xA=2(xM-xG) xG-1=2(--xG) xG=0 .222 24. 19873) C f f(x)=x3+x2+9x-12 . , (2 , -10) C C 3 . : A C f(2)= -10 23+22+92-12= -10 2+= -4 . (1) f (x)=3x2+2x+9 , x . 3 f (2)= -3 322+22+9= -3 3+= -3 . (2) (1) , (2) =1 = -6 .23 25. 1988 11) f f(x)=x+1+ . x +1) .) C f , x x=2 , x=5 .) f = { 1} ( x+1 0) 1 x 2 + 2x x 2 + 2xf (x)=1+0-= , x . f (x)=0 =0 ( x + 1) 2 ( x + 1) 2( x + 1) 2x2+2x=0 x=0 x= -2 . f (x) : x --2 -1 0 +f (x) +0 --0+ f(x) .... f (- ,-2) , (0 ,+ ) [-2 , -1) , (-1 , 0] -2 f(-2)= -2 0 f(0)=2. lim f(x)= + lim f(x)= - . x + x ) x [2 , 5] f(x)>0 f [2 , 5] , 55 1x227 = f ( x)dx = ( x + 1 + )dx = [ + x + ln( x + 1)]5 =2+ln2 .. 22 x +1 2224 26. 19883x 2 5 x + 6, x 12) f f(x)= 2 x 2 + 3 , x > 1 xo=1 . f ( x) f (1)3x 2 5 x + 6 43x 2 5 x + 2 Horner : limx 1 x 1 = lim x 1x 1 = lim x 1x 1=( x 1)(3x 2) lim= lim (3x-2)=1 x 1 x 1x 1f ( x) f (1)2 x2 + 3 42( x 2 + 3 2)( x 2 + 3 + 2) lim+= lim+= lim+ =x 1 x 1x 1x 1x 1 ( x 1)( x 2 + 3 + 2) 2( x 2 + 3 4 ) 2( x 1)( x + 1)2( x + 1) lim+ = lim+= lim+ =1.x 1( x 1)( x + 3 + 2) 2x 1 ( x 1)( x + 3 + 2) 2 x 1 x2 + 3 + 2 f ( x) f (1)f ( x) f (1) lim= lim+=1 , f x 1x 1x 1x 1 xo=1 f (1)=1 . 25 27. 19882x 3 5x 2 + 13) : 1x dx . x 3 5x 2 + 1 f(x)= [1 , 2] x 22 x 3 5x 2 + 1 1 x3x231 dx = ( x 2 5 x + )dx = [ 5 + ln x ]1 = - +ln2 . 2 1 x 1x 3 2 626 28. 19884) f f(x)=3x3-x2+x-3 , , . f 5 x1=1 x2= - , , .9 f f (x)=9x2-2x+ . 5 f x1=1 x2= - , 95 Fermat f (1)=0 f (- )=0 . :9f (1)=0 912-21+=0 -2+= -9 . (1) 5 255f (- )=0 9-2(- )+=0 10+9= -25 . (2) 9 819 (1) (2) =2 = -5 .27 29. 19891) f , g , : i) ,ii) f=g iii) 0 f(0)=g(0) . :) x , f(x)-g(x)=cx c .) f(x)=0 1 , 2 , g(x)=0 [1 , 2] .) x : f (x)=g(x) f (x)=g(x) +c , c f (x)=(g(x)+cx) f(x)= g(x)+cx+c1 , c1 . x=0 : f(0)= g(0)+c0+c1 c1=0 , (iii) . x , f(x)= g(x)+cx f(x)-g(x)=cx c .) () g(x)=f(x)-cx . g [1 , 2] . f(1)= f(2)=0 , :g(1)g(2)=( f(1)-c 1)( f(2)-c 2)= (-c 1)( -c 2)=c2 12 0 , 1 , 2 . g(1)g(2)0 x3-8>0 x>2 , f (x)0 , f .2 x +1) x [ , +1] x x x - 0 (x )dx 0 +1 +1 +1 +1 +1 x dx - dx 0 x dx 1dx x dx (+1-) +1 x dx . x [-1 , ] x x - x 0( 1 x )dx 0 1 dx - 1x dx 0 1dx - 1 1 x dx 0 [-(-1)] 1 x dx 1x dx .33 35. 1990x 3 1) f f(x)=+(+)x2+(-)x+ , , , , 3 2 + +=0 . 3 2 (0 , 1) f ( , f()) xx . f [0 , 1] (0 , 1) . f(0)= f(1)=+ ++ -+= ( + +)+=0 , 3 23 2 + +=0 . f(0)= f(1) . Rolle 3 2 (0 , 1) f ()=0 , f ( , f()) xx .34 36. 199012) f f(x)=3x+. 2x 2) f .) () f , y=3x x=1 x= >1 .) () + .) f = {0} . 1 lim f(x)= lim (3x)+ lim = + Cf x 0x 0x 0 2x 2 x=0 ( yy) . , + . f ( x) 1 = lim= lim (3+ 3 )=3+0=3 x + x x +2x 1 1lim (f(x)-x)= lim (3x+2 -3x)= lim=0.x +x +2x x + 2 x 2 y=3x+0 y=3x Cf + . y=3x Cf - . 1) f [1 , ] f(x)-3x= >0 x [1 , ] 2x 2 11 1 1 ()= ( f ( x) 3x)dx = 2xdx =[-]1 = - + =1 12 2x2 211= (1- ) .2 11 11) lim ()= lim [ (1- )]= (1-0)= . + +2 22 35 37. 19903) g , , g(-1)=7. f f(x)=3(x-2)2g(2x-5) , f f (2) . g(2x-5) , f(x) f (x)=[ 3(x-2)2]g(2x-5)+ 3(x-2)2[g(2x-5)]=6(x-2)(x-2)g(2x-5)+ 3(x-2)2g(2x-5)(2x-5)=6(x-2) g(2x-5)+6(x-2)2g(2x-5) . g(2x-5) , f (x) f (x)=[ 6(x-2) g(2x-5)]+[ 6(x-2)2g(2x-5)]== 6g(2x-5)+6(x-2)g(2x-5)(2x-5)+12(x-2)g(2x-5)+ 6(x-2)2g(2x-5)(2x-5)== 6g(2x-5)+12(x-2)g(2x-5) +12(x-2)g(2x-5)+12(x-2)2g(2x-5) . f (2)= 6g(-1)+12(2-2)g(-1) +12(2-2)g(-1)+12(2-2)2g(-1)= 6g(-1)+0+0+0= 42. 36 38. 19904) f x 4 2x 3 5f(x)=++(2-2+ )x2+(3+7)x-52 . 3 3 2 f . f 2 f , f (x)=0 . : x3 5 f (x)= 4+2x2+2(2-2+ )x+(3+7) , f (x)=4x2+4x+(22-4+5) . 32 =( 4)2-44(22-4+5) . =162-16(22-4+5)= -16(2-4+5)= -16(2-4+4+1)= -16[(-2)2+1]0 3 3 3 x [1,3] . = f ( x)dx = x e dx = x 2 (e x )dx =2 x1 1 13 3 3= [x2ex] 1 - ( x 2 )e x dx = [x2ex] 1 - 2 xe x dx =[x2ex] 1 - 2 x(e x )dx = 3 3 31 1 1 3= [x2ex] 1 -[2xex] 1 + 2e x dx =[x2ex] 1 -[2xex] 1 +[2ex] 1 =(9e3-e)-(6e3-2e)+(2e3-2e)= 3 3 3 33 1= 5e3-e ..38 40. 1991 41) = xdx , *, 01) >2 , = --2 1) 5 . 4 41) = 2 x 2 xdx = 2 x ( 1)dx = 0 0 2 x 4 1 44 1 x 4 = 2x dx - 2xdx = 2x (x)dx - -2=[] 0 - -2 = 0 2 x 00 1 1 1 =-0- -2 =--2 , >2 . 1 1 4111 1 1 1) () 5= 5 xdx = -5-2= -3= -(-3-2)= - +1= 05 1 44 3 14 2 1 4 1 4 x1 4 (x) 1 = - + xdx = - + dx = - - dx = - -[ln(x)] 04 =4 0 4 0 x 4 0 x4 1 2 12=- -(ln-ln1)= - - ln. 424 239 41. 1991 ln x2) f f(x)= x -, x>0 . 2 x) f .) f , x x=1 x=4 . 1 1 2 x ln x 2) x>0 f (x)=1 - x2 x = 2 x 2 + ln x . 2 2 x(2 x )4x x 1 g(x)=2x-2+lnx , x>0 . g(1)=0 . g(x)=2+>0 , x x>0 , g g(x)=0 x=1 . f (x)=0 g(x)=0 x=1 . g (0 , 1) (1 , + ) . 1 1g(2)=2+ln2>0 g()= -1+ln = -1-ln20 g(x)>0 x (1 , + ) f (x)0 , x . F F(x)= [f(x)]x , x . x 2 +1) >0 . g g(x)= , x . x ln f ( x )) F(x)=[f(x)]x= e , x . F(x)= e x ln f ( x ) (xlnf(x))=f ( x )f ( x )=e x ln f ( x ) (1lnf(x)+x )=[f(x)]x( lnf(x)+x) , x .f ( x)f ( x) x 2 +1 x 2 +11) g(x)=ln( x 2 + 1 )=ln (x2+1)= 2 x2 +1x 2 +1x = ln, x .x2 +144 46. 19916) f : [0 , ] f(x)=2x- 2 x+2 2 . 2 x [0 ,] f (x)=2xx- 2 x= x(2x- 2 ) 2 f (x)=0 ( x=0 2x- 2 =0 ) x= x= . f 2 4 [0 , ),(, ) . 44 2 3 1 3 : f ()= (2 - 2 )= (2 - 2 )=(1- 2 )0 . f 333222 :x 04 2f (x) - 0+ 0 f(x).... f [0 , ) 4 [, ] : 4 2 2 22 1 f( )=( )- 2+2 2 = - +2 2 ,44 222 f( )=12- 2 1+2 2 =1+ 2 ,22 0 f(0)=02- 2 0+2 2 =2 2 >1+ 2 = f(). 245 47. 1991 e x e, x < 1 7) f f(x)= ln x. ,x 1 x f , f , xx x=0 x=e . f (- ,1) , (1 , + ) . : ln x ln 1ln 1lim f(x) = lim (ex-e)=e-e=0 , lim+ f(x) = lim+==0 , f(1)==0 ,x 1x 1 x 1 x 1x 1 1 lim f(x) = lim+ f(x) = f(1) . f . x 1x 1 x [0 , 1) ex0 , x . () : g(x) 0 f(x)f (x) [f (x)]2 (x2+2)2 (2x)2 2x2+4-4x2 0 x2 2 x 2 - 2 x 2 . [- 2 , 2 ] . 47 49. 19922) ) f f(x)=x-x , x 00 : 0 0 x x +1x +1xf (t ) dt -4(x+1-x)>0 f (t )dt >4 . (1)x x +1 1 11 f(t) < 2 +4 2 +4 - f(t) >0 t t (tx2+ 4 f (t ))dt >0 x +1 x +1x +1 x