è¥0<Ïâ¦1ï¼åéa=(cosÏxï¼(â3)cosÏx)ï¼åéb=(sinÏxï¼cosÏx)ï¼è®°f(x)=a•b-(â3/2)ï¼
ä¸f(x+Ï)=f(x)ï¼(1)ãæ±f(x)ç解æå¼ï¼(2)ãè¥å
³äºxçæ¹ç¨3[f(x)]²+mf(x)-1=0å¨[-Ï/12ï¼5Ï/12]
ä¸æä¸ä¸ªä¸ç¸ççå®æ°æ ¹ï¼æ±mçåå¼èå´ã
解ï¼(1)ãf(x)=sinÏxcosÏx+(â3)cos²Ïx-(â3/2)=(1/2)sin2Ïx+(â3/2)(1+cos2Ïx)-(â3/2)
=(1/2)sin2Ïx+(â3/2)cos2Ïx=sin2Ïxcos(Ï/3)+cos2Ïxsin(Ï/3)=sin(2Ïx+Ï/3)ï¼
ç±f(x+Ï)=sin[2Ï(x+Ï)+Ï/3]=sin(2Ïx+2ÏÏ+Ï/3)=f(x)=sin(2Ïx+Ï/3)ï¼å0<Ïâ¦1å¯ç¥Ï=1ï¼
äºæ¯å¾è§£æå¼ä¸ºf(x)=sin(2x+Ï/3)ï¼
(2)ã设F(x)=3[f(x)]²+mf(x)-1=3sin²(2x+Ï/3)+msin(2x+Ï/3)-1
è¦æ±æ¹ç¨F(x)=0å¨[-Ï/12ï¼5Ï/12]ä¸æä¸ä¸ªä¸ç¸ççå®æ°æ ¹ï¼è®¾2x+Ï/3=tï¼
åF(t)=3sin²t+msint-1ï¼ä¸é¾æ¨å¾ï¼å½-Ï/12â¦xâ¦5Ï/12æ¶ï¼Ï/6â¦tâ¦Ï+Ï/6ï¼
注æå 个å
³é®çç¹ï¼F(Ï/6)=(m/2)-1/4ï¼F(Ï/2)=m+2ï¼F(Ï)=-1ï¼F(Ï+Ï/6)=-(m/2)-1/4ï¼
äºæ¯é®é¢æ¼å为è¦ä½¿F(t)=3sin²t+msint-1=0å¨Ï/6â¦tâ¦Ï+Ï/6æä¸ä¸ªä¸ççå®æ°æ ¹ï¼å¿
须使ï¼
F(Ï/6)=(m/2)-1/4â¦0ï¼å³mâ¦1/2...........â ï¼
F(Ï+Ï/6)=-(m/2)-1/4â§0,å³mâ¦-1/2.......â¡ï¼
F(Ï/2)=m+2>0ï¼å³m>-2ï¼....................â¢ï¼
â â©â¡â©â¢={mâ£-2<mâ¦-1/2}.è¿å°±æ¯mçåå¼èå´ã
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