【答案】
分析:(1)根據(jù)題意求出f′(x)由已知可知f(x)在(-∞,0)上是減函數(shù),在(0,1)上是增函數(shù),則x=0時,f(x)取到極小值,即f'(0)=0.求出b即可;
(2)由(1)得到f(x)的解析式,因為1是函數(shù)f(x)的一個零點,即f(1)=0,推出c=1-a,又f'(x)=-3x
2+2ax=0的兩個根分別為x
1=0,
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.f(x)在(0,1)上是增函數(shù),且函數(shù)f(x)在R上有三個零點,求出a的取值范圍,求出f(2)的取值范圍即可.
(3)要求直線y=x-1與函數(shù)y=f(x)的圖象交點個數(shù),需要把兩個解析式聯(lián)立求公共解,公共解有幾個交點就有幾個,再討論a的取值范圍,分不同情況討論出交點個數(shù)即可.
解答:解:(1)解:∵f(x)=-x
3+ax
2+bx+c,
∴f'(x)=-3x
2+2ax+b.
∵f(x)在(-∞,0)上是減函數(shù),在(0,1)上是增函數(shù),
∴當x=0時,f(x)取到極小值,即f'(0)=0.∴b=0.
(2)解:由(1)知,f(x)=-x
3+ax
2+c,
∵1是函數(shù)f(x)的一個零點,即f(1)=0,∴c=1-a.
∵f'(x)=-3x
2+2ax=0的兩個根分別為x
1=0,
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.
∵f(x)在(0,1)上是增函數(shù),且函數(shù)f(x)在R上有三個零點,
∴
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,即
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.
∴
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.
(3)解:由(2)知f(x)=-x
3+ax
2+1-a,且
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.
要討論直線y=x-1與函數(shù)y=f(x)圖象的交點個數(shù)情況,
即求方程組
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解的個數(shù)情況:由-x
3+ax
2+1-a=x-1,得(x
3-1)-a(x
2-1)+(x-1)=0.
即(x-1)(x
2+x+1)-a(x-1)(x+1)+(x-1)=0.
即(x-1)[x
2+(1-a)x+(2-a)]=0.∴x=1或x
2+(1-a)x+(2-a)=0.
由方程x
2+(1-a)x+(2-a)=0,(*)
得△=(1-a)
2-4(2-a)=a
2+2a-7.∵
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,
若△<0,即a
2+2a-7<0,解得
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.此時方程(*)無實數(shù)解.
若△=0,即a
2+2a-7=0,解得
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.此時方程(*)有一個實數(shù)解
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.
若△>0,即a
2+2a-7>0,解得
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.
此時方程(*)有兩個實數(shù)解,分別為
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,
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.
且當a=2時,x
1=0,x
2=1.
綜上所述,當
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時,直線y=x-1與函數(shù)y=f(x)的圖象有一個交點.
當
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或a=2時,直線y=x-1與函數(shù)y=f(x)的圖象有二個交點.
當
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且a≠2時,直線y=x-1與函數(shù)y=f(x)的圖象有三個交點.
點評:此題考查利用導數(shù)求函數(shù)的最值的方法確定函數(shù)解析式,函數(shù)與方程的綜合應用.培養(yǎng)學生解數(shù)學決問題的能力.