Let f: R R be defined as f(x)=ae2x+bex+cx. If f(0)=-1, f' ( e2 )=21 and 0 e4(f(x)-cx) dx= 392, then the value of |a+b+c| equals
- 16
- 12
- 8
- 10
Answer
(3) 8
Solution
f(0)=-1 a+b=-1 …(i)
f'(x)=2ae2x+bex+c f'( e2)=8a+2b+c=21 …(ii)
0 e4(ae2x+bex) dx= [ a2e2x+bex ]0 e4= 15a2+3b= 392
15a+6b=39 …(iii)
(i) and (iii): 15a-6-6a=39 a=5, b=-6; (ii): c=21-40+12=-7
∴ |a+b+c|=|-8|=8
Answer: (3)