21.
前一题你显然会,这里不再重复.
(II)
f(1) = 2 - a
2 ≤ f(1) ≤ 8,2 ≤ 2 - a ≤ 8,-6 ≤ a ≤ 0
f'(x) = 3x² + 2ax - (2a + 3) = (x - 1)(3a + 2a + 3) = 0
此为开口向上的抛物线,与x轴交于A(1,0),B(-(2a+3)/3,0)
f(x)在AB间为减函数,要使其在[m,m+2]上为减函数,则须
(1) -(2a+3)/3 - 1 ≥ 2
或
(2) 1 + (2a+3)/3 ≥ 2
(1) -(2a+3)/3 - 1 ≥ 2
a ≤ -6
结合前提-6 ≤ a ≤ 0,a = -6
A(1,0),B(3,0),即m = 1 (减区间为[1,3])
(2) 1 + (2a+3)/3 ≥ 2
a ≥ 0
结合前提-6 ≤ a ≤ 0,a = 0
B(-1,0),A(1,0),即m = -1 (减区间为[-1,1])
结合二者,m = ±1