1.å·²ç¥äºæ¬¡å½æ° f (x) = ax² + bx + c (aâ 0) 满足æ¡ä»¶ f (-x + 5) =f (x - 3)ï¼f(2)=0ï¼ä¸æ¹ç¨f(x)=xæ两个ç¸ççå®æ ¹ï¼é®ï¼æ¯å¦åå¨å®æ°mï¼n (mï¼nï¼ï¼ä½¿å¾f(x)çå®ä¹å为[mï¼n]æ¶ï¼å¼å为[3mï¼3n]ï¼å¦æåå¨ï¼æ±åºmï¼nçå¼ï¼å¦æä¸åå¨ï¼è¯´æçç±ã
解æï¼âµf (x) = ax^2+bx+c (aâ 0) 满足æ¡ä»¶ f (-x+5)=f(x-3)ï¼f(2)=0
f (5-x) = ax^2-(2a+1)5x+25a+5b+c=f (x-3) = ax^2+(b-6a)x+9a-3b+c
å¾a=-5/2, b=5, c=0
â´f(x)=-5/2x^2+5x
设åå¨å®æ°mï¼n (mï¼nï¼
f(x)=-5/2x^2+5x=3x==>x1=0,x2=4/5
â´m=0,n=4/5
2.å·²ç¥å½æ°f(x) = x åä¹ x²+2x+aï¼xâ[1ï¼+â]
ï¼1ï¼å½a=äºåä¹ä¸æ¶ï¼å¤æ并è¯æf(x)çåè°æ§
ï¼2ï¼å½a=-1æ¶ï¼æ±å½æ°f(x)çæå°å¼
(1)解æï¼âµf(x)=(x^2+2x+a)/xï¼xâ[1ï¼+â), a=1/2
令fâ(x)=(4x^2-2)/(2x)^2=0==>x1=-â2/2, x2=â2/2
f(x)å¨x1å¤åæ大å¼ï¼å¨x2å¤åæå°å¼
â´å½x<-â2/2æx>â2/2æ¶ï¼åè°å¢ï¼å½-â2/2<=x<0æ0<x<=â2/2æ¶ï¼åè°å¢åï¼
âµxâ[1,+â)
â´f(x) åè°å¢
(2)解æï¼âµa=-1
f(x)=(x^2+2x-1)/x==> fâ(x)=(x^2+1) /x^2>0
â´å½æ°f(x)åè°å¢ï¼å¨å®ä¹åå
æ æå°å¼ã
3.设å½æ°f(x)对任æxãyâRï¼é½æf( x + y)=f(x) + f(y)ï¼ä¸xï¼0æ¶ï¼f(x)ï¼0ï¼f(1)= -2
ï¼1ï¼è¯æï¼f(x)为å¥å½æ°
ï¼2ï¼è¯æï¼f(x)å¨Rä¸ä¸ºåå½æ°
ï¼3ï¼è¥ f( 2x + 5 ) + f( 6 - 7x ) ï¼ 4 ï¼æ±xçåå¼èå´
(1)è¯æï¼âµå½æ°f(x)对任æxãyâRï¼æf(x+y)=f(x)+f(y)
f(0+1)=f(0)+f(1)==>f(0)=0
f(x-x)=f(x)+f(-x)==>f(-x)=-f(x)
â´f(x)å¨Rä¸ä¸ºå¥å½æ°
(2)è¯æï¼âµxï¼0æ¶ï¼f(x)ï¼0ï¼â´å½x<0æ¶ï¼f(x)>0
设x1<x2ï¼x1-x2<0
f(x1-x2)=f(x1)+f(-x2) =f(x1)-f(x2)>0==>=f(x1)>f(x2)
â´å¨Rä¸ï¼f(x)åè°å
(3)解æï¼âµf( 2x + 5 ) + f( 6 - 7x ) ï¼ 4
f( 2x+5)+f(6-7x)=f(2x)+f(5)+f(6)+f(-7x)=f(11)+f(-5x) =f(11)-f(5x)>4
â´f(5x)<f(11)-4<0==>5x>0
â´xçåå¼èå´ï¼x>0
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