第二节One possible versionWhen Norah heard of her mothers plan, she was terhappy. She couldnt wait to see Dan at once. Hardly hadthey met Dan in his yard when Norah shouted excitedlyShe ran right up to him as if she had known him all of herlife and she was climbing all over him like a monkeyAnd of course every time I turned around. she d comeback over and hug me again, " Dan said. When it was timeto leave, Norah and her mom passed the rose bush by thedoor. Dan picked a big red rose and gave it to Norah, saying, "This is a symbol of our friendship.The mother and her daughter hane been visiting himceely, erer since. Every time they went there, the old andthe young would play games and tell each other secretsorah considers Dan as a family member and Dan has neverbeen alone again. "Norah helped me recover from my sad-ness and gave me something to live for, Dan said, with afresh shave and haircut

21.【必备知识】本题考查的知识是“能利用基本初等函数的导数公式和导数的四则运算法则求简单函数的导数,并了解复合函数求导法则,能求简单复合函数的导数”,“了解函数的单调性与导数的关系;能利用导数研究函数的单调性,会求函数的单调区间(其中多项式函数不超过三次)”【关键能力】本题考查逻輯思维能力、运算求解能力【解题思路】(1)m=1f(x)=hnx-c-1求号/(x)=1+hnx-构造函数g(x)=f'(x)→g(x)≤0→f(x)的单调性→结果0 xlnxe>1,xlnx≤0f(x)<0将原问题转化为证构造函数xlnx→即证-Inx >0得证求导、构造函数p(x)=2e2(x-1)-x→p(x)的单调性p(1)=-1<0,pP(2)=0→h(x)在(1,+∞)上的单调性当x>1时,h(x)>0→f(x)<0解:(1)函数f(x)的定义域为(0,+∞)(1分)当m=一时,f(x)=xnx-e,则f'(x)=1+lnx-e(2分)记g(x)=1+lmx-e-1,则g(x)=1-e-1显然g'(x)在(0,+∞)上单调递减,且g'(1)=0(3分)所以当x∈(0,1)时,g'(x)>0,函数g(x)单调递增;当x∈(1,+∞)时,g'(x)<0,函数g(x)单调递减所以g(x)≤g(1)=1+ln1-1=0,即f'(x)≤0恒成立,(4分)所以函数f(x)在(0,+∞)上单调递减所以函数f(x)的单调递减区间为(0,+∞),无单调递增区间.(5分)(2)要证f(x)<0,只需证me>xlnx①当0 1,xnx≤0,m≥,不等式显然成立.(6分)②当x>1时,xnx>0,e'>e由m≥可得,me'≥-e于是原问题可转化为求证·e>xlnx,即证nx>0.(7分)令h(x)≈2e-2-1x,则h'(x)=2e2(x-1)-x(8分)令p(x)=2e2(x-1)-x,则p(x)=2xe"2-1,易知p'(x)在(0,+∞)上单调递增(9分)又p(1)=2-1<0,p(2)=30所以存在x∈(1,2)使得p'(x0)=010分)所以p(x)在(1,x)上单调递减,在(x,+∞)上单调递增,又p(1)=-1<0,p(2)=0,故当x∈(1,2)时,h(x)<0,h(x)单调递减当x∈(2,+∞)时,h(x)>0,h(x)单调递增,所以当x>1时,h(x)≥h(2)=1-ln2>0,即f(x)<0综上,f(x)<0.(12分)解题关键》求解本题第(2)问的关键有:(1)想到分0 1两种情况进行证明;(2)当x>1时,想到利用放缩法将愿问题转化为求证·e>xhx

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