Let f be a real polynomial of degree n such that f(x)=f'(x)f''(x), for all x R. If f(0)=0, then 36 (f'(2)+f''(2)+ 02 f(x) dx ) is equal to:
- 42
- 46
- 56
- 66
Answer
(3) 56
Solution
Degrees: n=(n-1)+(n-2) n=3
Let f(x)=ax3+bx2+cx ( f(0)=0)
f'(x)=3ax2+2bx+c, f''(x)=6ax+2b
f'f''=18a2x3+18abx2+(4b2+6ac)x+2bc
Comparing with f: 18a2=a a= 118; 18ab=b (true); 4b2+6ac=c c=6b2; 2bc=0
2bc=0 with c=6b2 b=0, c=0
f(x)= x318, f'(x)= x26, f''(x)= x3
f'(2)= 23, f''(2)= 23, 02 x318 dx= 1672= 29
36 ( 23+ 23+ 29 )=36 149=56
Answer: (3)