题目内容
(本小题满分12分)
已知函数f (x
)=ln(1+x)+a (x+1)2(a为常数).
(Ⅰ)若函数f (x)在x=1处有极值,判断该极值是极大值还是极小值;
(Ⅱ)对满足条件a≤
的任意一个a,方程f (x)=0在区间(0,3)内实数根的个数是多少?
已知函数f (x
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(Ⅰ)若函数f (x)在x=1处有极值,判断该极值是极大值还是极小值;
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(Ⅱ)对满足条件a≤
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(1)极大值;(2)2
(Ⅰ)f'(x)=
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…………2分
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—1<x<1时,f'(x)>0;x>1时,f'(x)<0,
∴f(x)在(—1,1)上为增函数,在(1,+∞)上为减函数
所以f(1)为函数f(x)的极大值 …………4分
(Ⅱ)
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…………5分
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+∞)是为减函数
因此f(x)在x=—1+
处取得区间(—1,+∞)上的最大值 ——6分
由f(—1+
)=0得a=—
…………7分
(1)当a<
时,f(—1+
所以方程f(x)=0在区间(0,3)内无实数根 …………8分
(2)当a=
时,f(—1+
所以方程f(x)=0在区间(0,3)内有且仅有1个实数根
—1+
…………9分
(3)当
a≤
时,
≤1,
又f(0)a<0,f(—1+
f(3)=ln4+16a≤ln4-2<0,
所以方程f(x)=0在区间(0,3)内有2个实数根. …………11分
综上所述,
当a<
时,方程f(x)=0在区间(0,3)内无实数根;
当a
时,方程f(x)=0在区间(0,3)内有1个实数根;
当
≤
时,方程f(x)=0 在区间(0,3),内有2个实数根.
…………12分
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—1<x<1时,f'(x)>0;x>1时,f'(x)<0,
∴f(x)在(—1,1)上为增函数,在(1,+∞)上为减函数
所以f(1)为函数f(x)的极大值 …………4分
(Ⅱ)
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因此f(x)在x=—1+
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由f(—1+
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(1)当a<
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
所以方程f(x)=0在区间(0,3)内无实数根 …………8分
(2)当a=

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所以方程f(x)=0在区间(0,3)内有且仅有1个实数根
—1+
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(3)当
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又f(0)a<0,f(—1+

f(3)=ln4+16a≤ln4-2<0,
所以方程f(x)=0在区间(0,3)内有2个实数根. …………11分
综上所述,
当a<
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当a
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当
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…………12分
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