Question

设f(x)=ax²+bx+ln(x+1) (a＞0) 1°当b=0时，讨论函数单调性

设f(x)=ax²+bx+ln(x+1) (a＞0) 1°当b=0时，讨论函数单调性 2° 问是否存在a,b使函数f(x)无极值且f'(x)存在零点，若存在，求得a,b若不存在说明理由

Answer 1

1.
f(-1)
=
a
-
b
+
1
=
0,
b
=
a+1
且函数f(x)的值域为[0,﹢无穷),
显然x
=
-1为该抛物线的对称轴,且抛物线开口向上,
a
>0
对称轴x
=
-1
=
-b/(2a)
=
-(a+1)/(2a)
a
=
1
f(x)=x²+2x+1
F(x)
=
x²+2x+1,
x
>
0
=
-(x²+2x+1),
x
<
0
2.
g(x)
=
x²+2x+1
-
kx
=
x²+(2
-
k)x+1
g(x)的对称轴为x
=
-(2-k)/(2*1)
=
(k-2)/2
当x∈[-2,2]时,
g(x)是单调函数,
对称轴x
=
(k-2)/2
≥2
或
x
=
(k-2)/2
≤
-2
即k
≥
6或k
≤
-2
3
不影响结果,不妨设m
>
0
mn
<
0,
n
<
0
m
+
n
>
0,
m
>
|n|,
m²
>
n²
f(x)为偶函数,
b
=
0,
f(x)
=
ax²
+
1
F(m)
=
f(m)
=
am²+
1
F(n)
=
-f(n)
=
-an²
-
1
F(m)+F(n)
=
a(m²
-
n²)
>
0