高一数学 已知函数f(x)=(1/2)^|x|,x为实数，关于x的方程f^2(x)-(a+1)f(x)+a=0有三个不同实数解，求a范围 已知函数f(x)＝|x+1x|?|x?1x|，关于x的方程f...

\u9ad8\u4e00\u6570\u5b66 \u8bbe\u51fd\u6570f(x)= 2^ +a\2^x-1 (a\u4e3a\u5b9e\u6570)

\u697c\u4e0a\u7684\u5f88\u597d\u5440\uff0c\u4f60\u89c9\u5f97\u662f\u6b63\u89e3\uff0c\u6211\u600e\u4e48\u89c9\u5f97\u4e0d\u662f\u6b63\u89e3\u5462\uff1f
\u697c\u4e3b\uff0c\u9898\u76ee\u53d9\u8ff0\u6ca1\u9519\u4e86\u5427\uff1f---------------f(x)= 2^x+a/(2^x)-1
1
a=0\u65f6\uff0cf(x)=2^x-1\uff0cg(x)\u7684\u56fe\u50cf\u4e0ef(x)\u7684\u56fe\u50cf\u5173\u4e8ex=1\u5bf9\u79f0\uff0c\u5373\uff1af(1-x)=g(1+x)
(\u5f53\u7136\uff1af(1+x)=g(1-x)\u4e5f\u662f\u53ef\u4ee5\u7684)\uff0c\u5373\uff1ag(1+x)=2^(1-x)-1\uff0c\u4ee4\uff1at=1+x\uff0c\u5219\uff1ax=t-1
\u6240\u4ee5\uff1ag(t)=2^(2-t)-1\uff0c\u6545\uff1ag(x)=2^(2-x)-1
2
a0\uff0c\u5219\u65b9\u7a0b\u4e3a\uff1at+a/t-1=0\uff0c\u5373\uff1at^2-t+a=0\uff0c\u5224\u522b\u5f0fdelta=1-4a
\u5f53a0\uff0c\u6545\u65b9\u7a0b\u6709\u5b9e\u6839\uff0ct=(1+sqrt(1-4a))/2\u6216t=(1-sqrt(1-4a))/2
\u56e0\u4e3a\uff1aa0\uff0c\u6545\uff1a1-4a>1\uff0c\u6545\uff1a-sqrt(1-4a)<-1\uff0c\u6545\uff1a1-sqrt(1-4a)<0
\u6240\u4ee5\uff1a2^x\u4e0d\u53ef\u80fd\u7b49\u4e8e(1-sqrt(1-4a))/2
\u6545\uff1a2^x=(1+sqrt(1-4a))/2\uff0c\u5373\uff1a2^(x+1)=1+sqrt(1-4a)
\u6545\uff1ax+1=log2(1+sqrt(1-4a))\uff0c\u5373x=log2(1+sqrt(1-4a))-1

\u89e3\uff1a\u5148\u6839\u636e\u9898\u610f\u4f5c\u51faf\uff08x\uff09\u7684\u7b80\u56fe\uff1a\u5f97f\uff08x\uff09\uff1e0\uff0e\u2235\u9898\u4e2d\u539f\u65b9\u7a0bf2\uff08x\uff09+a|f\uff08x\uff09|+b=0\uff08a\uff0cb\u2208R\uff09\u6070\u67096\u4e2a\u4e0d\u540c\u5b9e\u6570\u89e3\uff0c\u5373\u65b9\u7a0bf2\uff08x\uff09+af\uff08x\uff09+b=0\uff08a\uff0cb\u2208R\uff09\u6070\u67096\u4e2a\u4e0d\u540c\u5b9e\u6570\u89e3\uff0c\u2234\u6545\u7531\u56fe\u53ef\u77e5\uff0c\u53ea\u6709\u5f53f\uff08x\uff09=2\u65f6\uff0c\u5b83\u6709\u4e8c\u4e2a\u6839\uff0e\u6545\u5173\u4e8ex\u7684\u65b9\u7a0bf2\uff08x\uff09+af\uff08x\uff09+b=0\u4e2d\uff0c\u6709\uff1a4+2a+b=0\uff0cb=-4-2a\uff0c\u4e14\u5f53f\uff08x\uff09=k\uff0c0\uff1ck\uff1c2\u65f6\uff0c\u5173\u4e8ex\u7684\u65b9\u7a0bf2\uff08x\uff09+af\uff08x\uff09+b=0\u67094\u4e2a\u4e0d\u540c\u5b9e\u6570\u89e3\uff0c\u2234k2+ak-4-2a=0\uff0ca=-2-k\uff0c\u22350\uff1ck\uff1c2\uff0c\u2234a\u2208\uff08-4\uff0c-2\uff09\uff0e\u6545\u7b54\u6848\u4e3a\uff1a\uff08-4\uff0c-2\uff09\uff0e

答：
f²(x)-(a+1)f(x)+a=0
[ f(x)-a]×[ f(x) -1 ]=0
所以：
f(x)=(1/2)^|x|=a或者f(x)=(1/2)^|x|=1
由后面一个解得：x=0
所以：f(x)=(1/2)^|x|=a存在两个不同的解
因为：0<f(x)=(1/2)^|x|<=1恒成立
所以：0<a<1

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