Some difficult questions

1 f(x)=3x^4 - 8x^3 - 6x^2 + 24x + a
f'(x)=12x^3-24x^2-12x+24=12(x^3-2x^2-x+2)=12(x-1)(x+1)(x-2)
So a cannot be equal to -f(1),-f(-1) and -f(2) which are -13,19 and -52

2 f(x) = (x - a1)(x - a2)...(x - an) + (x - b1)(x - b2)...(x - bn)
f(a_1)>0,f(b_1)<0,f(a_2)<0,f(b_2)>...f(a_n)=(-1)^(n-1),f(b_n)=(-1)^n
So, f(x) crosses the x-axis n-times and just has n distinct real roots.

3 f'(x)=3x^2-3a^2. Let f'(x)=0=>x=a or -a. f''(x)=6x
when x=a, f''(a)>0. So f(a)=-2a^3+2b is the minimum point
when x=-a, f''(a)<0. So f(-a)=2a^3+2b is the maximum point
Noticethat f(infinity)=infinity and f(-infinity)=-infinity. So if f(x) has three real roots, then f(a)<0 and f(-a)>0. That is a^3>|b| while if f(x) has one real root, then f(a) and f(-a) should have the same sign. Actually only f(a) and f(-a) > 0 is make sence and this implies that a^3<|b|

2010-01-02 13:44:28 補充：
第一條沒有指明那些根一定要是實數

yes
by considering the graph

2010-01-03 01:55:11 補充：
你岩~~~~~~~~~~~~~~~~~

1. We should have f(-1)<0 and f(1)>0 and f(2)<0.
Hence -13 < a < -8.