高一数学。求解。两道题
\u4e24\u9053\u9ad8\u4e00\u6570\u5b66\u989819.\u89e3\uff1a
(1)
\u4ee4-x>0\u4ee3\u5165-x^2+2x
=-x^2-2x
\u56e0\u4e3af(x)\u4e3a\u5947\u51fd\u6570\uff0c
=>-x^2-2x=-(x^2+mx)
=>m=2
(2)\u6839\u636e\u9898\u610f
=>f(x)\u5728[-1\uff0c1]\u4e0a\u5355\u8c03\u9012\u589e
=>a-2=1
=>a=3
20
f(x)=-(x-a/2)^2+a^2/4-a/4+1/2
\u2460\u5f530\u2264a\uff0f2\u22641\u65f6
=>0\u2264a\u22642\u65f6,x=a/2\u65f6\u53d6\u5230\u6700\u5927\u503c
=>g\uff08a\uff09\uff1da^2/4-a/4+1/2
g(a)|min=7/16
\u2461\u5f53a/2\u22640\u65f6,x=0\u65f6\u53d6\u5230\u6700\u5927\u503c
g(a)=1/2-a/4
g(a)|min=1/2
\u2462\u5f53a/2\u22651\u65f6,x=1\u65f6\u53d6\u5230\u6700\u5927\u503c
=>g(a)=3a/4-1/2
g(a)|min=1
1.
x\u2208\uff08e-1\uff0c1\uff09
-1<a=lnx<0
\u56e0\u4e3a inx<0
\u6240\u4ee5b=21nx<a
c=ln3x=(inx)^2*inx
\u56e0\u4e3a0 <(inx)^2<1
\u6240\u4ee5 c>a
\u6240\u4ee5 b<a<c
2.
f\uff08x\uff09=ax2+2x+1
\u8fc7\u5b9a\u70b9\uff080,1\uff09
\u5206\u60c5\u51b5\u8ba8\u8bba\uff08\u7ed3\u5408\u56fe\u50cf\u770b\uff09
1.a<0 \u51fd\u6570f\uff08x\uff09=ax2+2x+1\u5728\uff08-\u221e\uff0c0\uff09\u4e0a\u4e00\u5b9a\u6709\u4e00\u4e2a\u96f6\u70b9
2.a=0 \u51fd\u6570f\uff08x\uff09=2x+1\u5728\uff08-\u221e\uff0c0\uff09\u4e0a\u4e00\u5b9a\u6709\u4e00\u4e2a\u96f6\u70b9\uff08x=-1/2\uff09
3.a>0 \u53ea\u9700 \u5bf9\u79f0\u8f74\u5927\u4e8e\u96f6
\u5224\u522b\u5f0f\u5927\u4e8e\u7b49\u4e8e\u96f6\uff08\u4e0d\u660e\u767d\u53ef\u4ee5\u753b\u56fe\u770b\uff09
\u8fd9\u79cd\u60c5\u51b5\u65e0\u89e3
\u89e3\u5f97 \u7b54\u6848 a<=0
1、f(5)=a*5^5+b*5^3+5c+7=1
f(-5)=-a*5^5-b*5^3-5c+7
相加得f(-5)+f(5)=14，所以f(-5)=13
2、与1类似
f(a)=(a^2+a+1)/(a^2+1)
f(-a)=(a^2-a+1)/(a^2+1)
f(-a)+f(a)=2，所以f(-a)=-1.
第一题13
1.13
2.-1
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