己知函数y=X-2x+3在[0,a)上的最大值是3,最小值是2，求实数a的取值范围
\u5df2\u77e5\u51fd\u6570f\uff08x\uff09=x2-2x+3\u5728[0\uff0ca]\uff08a\uff1e0\uff09\u4e0a\u7684\u6700\u5927\u503c\u662f3\uff0c\u6700\u5c0f\u503c\u662f2\uff0c\u5219\u5b9e\u6570a\u7684\u53d6\u503c\u8303\u56f4\u662f______\u89e3\uff1a\u2235\u51fd\u6570f\uff08x\uff09=x2-2x+3\u662f\u5f00\u53e3\u5411\u4e0a\u7684\u629b\u7269\u7ebf\uff0c\u5bf9\u79f0\u8f74 x=1\u5f53 x=1\u65f6\u51fd\u6570\u53d6\u5f97\u6700\u5c0f\u503c f\uff081\uff09=1-2+3=2\u2235y=x2-2x+3\u5728[0\uff0ca]\u4e0a\u6700\u5c0f\u503c\u4e3a2\u2234a\u22651\u5f53x=0\u65f6 y=3 \u51fd\u6570y=x2-2x+3\u5728\uff081\uff0c+\u221e\uff09\u4e0a\u662f\u589e\u51fd\u6570\uff0c\u5f53x=2\u65f6 y=4-4+3=3\uff0c\u5f53x\uff1e2\u65f6 y\uff1e3\u2235\u51fd\u6570y=x2-2x+3\u5728[0\uff0ca]\u4e0a\u6700\u5927\u503c\u4e3a3\u2234a\u22642 \u7efc\u4e0a\u6240\u8ff0 1\u2264a\u22642\uff0e\u6545\u7b54\u6848\u4e3a\uff1a1\u2264a\u22642
f(x)=x^2-2x+3,\u7684\u5bf9\u79f0\u8f74\u65b9\u7a0b\u4e3a:X=1,
\u8ba8\u8bba:
1)\u5f53a\u22641\u65f6,x=0\u65f6,f(x)\u6700\u5927,f(0)=3,
\u5f53x=a\u65f6,f(x)\u6700\u5c0f,f(a)=a^2-2a+3=2,
a=1.
\u5219\u67090<a\u22641.
2)a>1\u65f6,f(a)max=a^2-2a+3=3,
a1=0,a2=2.
x=1,f(1)=1-2+3=2.
\u5373\u6709:1<a\u22642.
\u6545,\u5b9e\u6570a\u7684\u53d6\u503c\u8303\u56f4\u662f:0<a\u22642.
因此a>=1
由f(x)=3,得x^2-2x+3=3,
x(x-2)=0
x=0或2
即f(0)=f(2)=3
故有1=<a<=2
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