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    • 利用自然对数的底数e（e=2.71828…）构建三个基本初等函数y＝ex，y＝lnx，y＝ex(x＞0)．探究发现，它们具 已知函数f（x）= lnx+k e x ...

    \u5df2\u77e5\u51fd\u6570f\uff08x\uff09=ex\uff08e=2.71828\u2026\u662f\u81ea\u7136\u5bf9\u6570\u7684\u5e95\u6570\uff09\uff0cx\u2208R\uff0e\uff08\u2160\uff09\u6c42\u51fd\u6570y=f\uff08x\uff09\u7684\u56fe\u8c61\u8fc7\u539f\u70b9\u7684\u5207\u7ebf\u65b9\u7a0b

    \uff08\u2160\uff09\u89e3\uff1a\u8bbe\u5207\u7ebf\u65b9\u7a0b\u4e3ay=kx\uff0c\u5207\u70b9\u4e3a\uff08x0\uff0cy0\uff09\uff0c\u5219kx0\uff1dex0k\uff1dex0\u2234x0=1\uff0ck=e\uff0c\u2234\u51fd\u6570y=f\uff08x\uff09\u7684\u56fe\u8c61\u8fc7\u539f\u70b9\u7684\u5207\u7ebf\u65b9\u7a0b\u4e3ay=ex\uff1b\uff08\u2161\uff09\u89e3\uff1a\u5f53x\uff1e0\uff0cm\uff1e0\u65f6\uff0c\u4ee4f\uff08x\uff09=mx2\uff0c\u5316\u4e3am=exx2\uff0c\u4ee4h\uff08x\uff09=exx2\uff08x\uff1e0\uff09\uff0c\u5219h\u2032\uff08x\uff09=ex(x?2)x3\uff0c\u5219x\u2208\uff080\uff0c2\uff09\u65f6\uff0ch\u2032\uff08x\uff09\uff1c0\uff0ch\uff08x\uff09\u5355\u8c03\u9012\u51cf\uff1bx\u2208\uff082\uff0c+\u221e\uff09\u65f6\uff0ch\u2032\uff08x\uff09\uff1e0\uff0ch\uff08x\uff09\u5355\u8c03\u9012\u589e\uff0e\u2234\u5f53x=2\u65f6\uff0ch\uff08x\uff09\u53d6\u5f97\u6781\u5c0f\u503c\u5373\u6700\u5c0f\u503c\uff0ch\uff082\uff09=e24\uff0e\u2234\u5f53m\u2208\uff080\uff0ce24\uff09\u65f6\uff0c\u66f2\u7ebfy=f \uff08x\uff09 \u4e0e\u66f2\u7ebfy=mx2\uff08m\uff1e0\uff09\u516c\u5171\u70b9\u7684\u4e2a\u6570\u4e3a0\uff1b\u5f53m=e24\u65f6\uff0c\u66f2\u7ebfy=f \uff08x\uff09 \u4e0e\u66f2\u7ebfy=mx2\uff08m\uff1e0\uff09\u516c\u5171\u70b9\u7684\u4e2a\u6570\u4e3a1\uff1b\u5f53m\uff1ee24\u65f6\uff0c\u66f2\u7ebfy=f \uff08x\uff09 \u4e0e\u66f2\u7ebfy=mx2\uff08m\uff1e0\uff09\u516c\u5171\u70b9\u4e2a\u6570\u4e3a2\uff0e\uff08\u2162\uff09\u8bc1\u660e\uff1af(a)+f(b)2\uff1ef(b)?f(a)b?a=(b?a+2)+(b?a?2)eb?a2(b?a)ea\uff0c\u4ee4g\uff08x\uff09=x+2+\uff08x-2\uff09ex\uff08x\uff1e0\uff09\uff0c\u5219g\u2032\uff08x\uff09=1+\uff08x-1\uff09ex\uff0eg\u2032\u2032\uff08x\uff09=xex\uff1e0\uff0c\u2234g\u2032\uff08x\uff09\u5728\uff080\uff0c+\u221e\uff09\u4e0a\u5355\u8c03\u9012\u589e\uff0c\u4e14g\u2032\uff080\uff09=0\uff0c\u2234g\u2032\uff08x\uff09\uff1e0\uff0c\u2234g\uff08x\uff09\u5728\uff080\uff0c+\u221e\uff09\u4e0a\u5355\u8c03\u9012\u589e\uff0c\u800cg\uff080\uff09=0\uff0c\u2234\u5728\uff080\uff0c+\u221e\uff09\u4e0a\uff0c\u6709g\uff08x\uff09\uff1eg\uff080\uff09=0\uff0e\u2235\u5f53x\uff1e0\u65f6\uff0cg\uff08x\uff09=x+2+\uff08x-2\uff09?ex\uff1e0\uff0c\u4e14a\uff1cb\uff0c\u2234(b?a+2)+(b?a?2)eb?a</

    \uff08\u2160\uff09 f\u2032(x)= 1 x -lnx-k e x \uff0c\u4f9d\u9898\u610f\uff0c\u2235\u66f2\u7ebfy=f\uff08x\uff09 \u5728\u70b9\uff081\uff0cf\uff081\uff09\uff09\u5904\u7684\u5207\u7ebf\u4e0ex\u8f74\u5e73\u884c\uff0c\u2234 f\u2032(1)= 1-k e =0\uff0c\u2234k=1\u4e3a\u6240\u6c42\uff0e\uff08\u2161\uff09k=1\u65f6\uff0c f\u2032(x)= 1 x -lnx-1 e x \uff08x\uff1e0\uff09\u8bb0h\uff08x\uff09= 1 x -lnx-1\uff0c\u51fd\u6570\u53ea\u6709\u4e00\u4e2a\u96f6\u70b91\uff0c\u4e14\u5f53x\uff1e1\u65f6\uff0ch\uff08x\uff09\uff1c0\uff0c\u5f530\uff1cx\uff1c1\u65f6\uff0ch\uff08x\uff09\uff1e0\uff0c\u2234\u5f53x\uff1e1\u65f6\uff0cf\u2032\uff08x\uff09\uff1c0\uff0c\u2234\u539f\u51fd\u6570\u5728\uff081\uff0c+\u221e\uff09\u4e0a\u4e3a\u51cf\u51fd\u6570\uff1b\u5f530\uff1cx\uff1c1\u65f6\uff0cf\u2032\uff08x\uff09\uff1e0\uff0c\u2234\u539f\u51fd\u6570\u5728\uff080\uff0c1\uff09\u4e0a\u4e3a\u589e\u51fd\u6570\uff0e\u2234\u51fd\u6570f\uff08x\uff09\u7684\u589e\u533a\u95f4\u4e3a\uff080\uff0c1\uff09\uff0c\u51cf\u533a\u95f4\u4e3a\uff081\uff0c+\u221e\uff09\uff0e\uff08\u2162\uff09\u8bc1\u660e\uff1ag\uff08x\uff09=\uff08x 2 +x\uff09f\u2032\uff08x\uff09= 1+x e x \uff081-xlnx-x\uff09\uff0c\u5148\u7814\u7a761-xlnx-x\uff0c\u518d\u7814\u7a76 1+x e x \uff0e\u2460\u8bb0r\uff08x\uff09=1-xlnx-x\uff0cx\uff1e0\uff0c\u2234r\u2032\uff08x\uff09=-lnx-2\uff0c\u4ee4r\u2032\uff08x\uff09=0\uff0c\u5f97x=e -2 \uff0c\u5f53x\u2208\uff080\uff0ce -2 \uff09\u65f6\uff0cr\u2032\uff08x\uff09\uff1e0\uff0cr\uff08x\uff09\u5355\u589e\uff1b\u5f53x\u2208\uff08e -2 \uff0c+\u221e\uff09\u65f6\uff0cr\u2032\uff08x\uff09\uff1c0\uff0cr\uff08x\uff09\u5355\u51cf\uff0e\u2234r\uff08x\uff09 max =r\uff08e -2 \uff09=1+e -2 \uff0c\u53731-xlnx-x\u22641+e -2 \uff0e\u2461\u8bb0s\uff08x\uff09= 1+x e x \uff0cx\uff1e0\uff0c\u2234 s\u2032(x)=- x e x \uff1c0\uff0c\u2234s\uff08x\uff09\u5728\uff080\uff0c+\u221e\uff09\u5355\u51cf\uff0c\u2234s\uff08x\uff09\uff1cs\uff080\uff09=1\uff0c\u5373 1+x e x \uff1c1\uff0e\u7efc\u2460\u3001\u2461\u77e5\uff0cg\uff08x\uff09\uff09= 1+x e x \uff081-xlnx-x\uff09\u2264\uff08 1+x e x \uff09\uff081+e -2 \uff09\uff1c1+e -2 \uff0e

    （Ⅰ）∵y＝
    e
    x
    （x＞0）的图象是反比例函数y＝
    e
    x
    （x≠0）的图象位于第一象限内的一支，
    ∴y＝
    e
    x
    （x＞0）的图象关于直线y=x对称．
    又y=ex，y=lnx=logex互为反函数，它们的图象关于直线y=x互相对称，从而可知：
    ①三个函数的图象形成的图形的一条对称轴方程为y=x．
    ②阴影区A、B关于直线y=x对称，故阴影区B的面积为1．
    ③M（1，e），N（e，1）．（6分）
    （Ⅱ）由于
    f(x1)+f(x2)
    2
    ＝
    ex1+
    e
    x1
    ?lnx1+ex2+
    e
    x2
    ?lnx2
    2
    ＝
    ex1+ex2+
    e
    x1
    +
    e
    x2
    ?ln(x1x2)
    2
    ，f(
    x1+x2
    2
    )＝e
    x1+x2
    2
    +
    e
    x1+x2
    2
    ?ln
    x1+x2
    2
    ＝e
    x1+x2
    2
    +
    2e
    x1+x2
    ?ln
    x1+x2
    2
    ，
    f(x1)+f(x2)
    2
    ?f(
    x1+x2
    2
    )＝
    ex1+ex2+
    e
    x1
    +
    e
    x2
    ?ln(x1x2)
    2
    ?e
    x1+x2
    2
    ?
    2e
    x1+x2
    +ln
    x1+x2
    2
    =
    ex1+ex2
    2
    ?e
    x1+x2
    2
    +
    e
    x1
    +
    e
    x2
    2
    ?
    2e
    x1+x2
    +ln
    x1+x2
    2
    ?
    ln(x1x2)
    2
     
    =
    ex1+ex2?2


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