已知函數(shù)f(x)=x2+(a+1)x+lg|a+2|(a∈R,且a≠-2).
(1)寫出一個(gè)奇函數(shù)g(x)和一個(gè)偶函數(shù)h(x),使f(x)=g(x)+h(x);
(2)對(duì)(1)中的g(x).命題P:函數(shù)f(x)在區(qū)間[(a+1)2,+∞)上是增函數(shù);命題Q:函數(shù)g(x)是減函數(shù);如果命題P、Q有且僅有一個(gè)是真命題,求a的取值范圍;
(3)在(2)的條件下,求f(2)的取值范圍.
【答案】
分析:(1)由題意可得 h(x)=x
2+lg|a+2|; g(x)=(a+1)x.
(2)由函數(shù)f(x)在區(qū)間[(a+1)
2,+∞)上是增函數(shù)得
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,求出a的范圍為集合A,由函數(shù)g(x)是減函數(shù)得a+1<0,求出a的范圍為集合B,則(A∩
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)∪(
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∩B)即為所求.
(3)求出f (2),由函數(shù)在
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上遞增,可得f (2)>f (-
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),從而得到所求.
解答:解:(1)由題意可得 h(x)=x
2+lg|a+2|; g(x)=(a+1)x.
(2)由二次函數(shù)f(x))=x
2+(a+1)x+lg|a+2|的圖象是開口向上的拋物線,且的對(duì)稱軸為 x=
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,
在區(qū)間[(a+1)
2,+∞)上是增函數(shù),故有
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,解得
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,因?yàn)閍≠-2.
由函數(shù)g(x)是減函數(shù)得a+1<0,解得a<-1,a≠-2.
當(dāng)命題P真且命題Q假時(shí),由
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,解得a≥-1.
當(dāng)命題P假且命題Q真時(shí),由
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,即得-
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<a<-1.
故當(dāng)命題P、Q有且僅有一個(gè)是真命題,得a的取值范圍是
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.
(3)f(2)=4+2a+2+lg|a+2|=6+2a+lg(a+2),因?yàn)樵?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181948940904835/SYS201310241819489409048023_DA/12.png">上遞增,
所以,
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,即:f(2)∈(3-lg2,+∞).
點(diǎn)評(píng):本題考查函數(shù)的奇偶性和單調(diào)性,不等式的解法,求兩個(gè)集合的交集、并集和補(bǔ)集,準(zhǔn)確運(yùn)算是解題的難點(diǎn).