Question

已知二次函数f（x）=ax²+bx+4，集合A=｛x|f（x）=x｝

Answer 1

解：（1）因为A是单元素集合，所以方程f(x)
=
x有且只有一个实数解1。代入可得a
+
b
+
4
=
1，a
+
b
=
-3①，f(x)
=
ax
2
+
bx
+
4
=
x
=>
ax
2
+
(b
–
1)x
+
4
=
0
=>Δ=
(b
–
1)
2
–
16a
=
0②，联立解得b
2
–
2b
+
1
–
16(-3
–
b)
=
0
=>
b
2
+
14b
+
49
=
0
=>
(b
+
7)
2
=
0
=>
b
=
-7
=>
a
=
4
=>
f(x)的解析式为f(x)
=
4x
2
–
7x
+
4；
（2）因为1∈A，所以x
=
1是方程f(x)
=
x的解。代入可得a
+
b
=
-3①，即b
=
-3
–
a，所以f(x)
=
ax
2
+
(-3
–
a)x
+
4
=
a[x
–
(3
+
a)/2a]
2
+
4
–
(a
+
3)
2
/4a，二次函数的对称轴l：x
=
(3
+
a)/2a
=
(1/2)(1
+
3/a)，因为a∈[1，2]
=>
1/a
∈[1/2，1]
=>
3/a
∈[3/2，3]
=>
1
+
3/a
∈[5/2，4]
=>
x
=
(1/2)(1
+
3/a)∈[5/4，2]，分类讨论：
1）当x∈[5/4，2]时，f(x)min
=
f[(3
+
a)/2a]
=
4
–
(a
+
3)
2
/4a
=
5/2
–
(1/4)(a
+
9/a)
=
5/2
–
a/4
–
9/4a∈[0，7/8]；
（当1≤a≤4/3）f(x)max
=
f(5/4)
=
5a/16
+
1/4∈[9/16，2/3]，
或者（当4/3≤a≤2）f(2)
=
2a
–
2∈[2/3，2]；
2）当x∈[1/2，5/4]时，f(x)在x∈[1/2，5/4)上单调递减，所以f(x)max
=
f(1/2)
=
-a/4
+
5/2∈[2，9/4]；f(x)min
=
f(5/4)
=
5a/16
+
1/4∈[9/16，7/8]；
综上所述，M
=
f(x)max
=
f(1/2)
=
-a/4
+
5/2，m
=
f(x)min
=
f[(3
+
a)/2a]
=
5/2
–
a/4
–
9/4a，所以g(a)
=
M
–
m
=
9/4a，当a
=
2时，g(a)min
=
9/8
。

Answer 2

f(x)=ax²+x-1+3a(a属于R)在区间
[-1,1]上有零点
即ax²+x-1+3a=0在[-1,1]上有实数解
即a(x²+3)=1-x
即
a=(1-x)/(x²+3)有实数解
令
g(x)=(1-x)/(x²+3)则
a的范围即是g(x)的值域
g(x)=[-x²-3-2x(1-x)]/(x²+3)²
=(x²-2x-3)/(x²+3)²
=(x+1)(x-3)/(x²+3)²
∵-1≤x≤1∴
(x+1)(x-3)
≤0
∴
g(x)≤0
∴g(x)是减函数
∴x=-1
g(x)max=1