Lets take f(x)=ax3 + bx2 + cx + d
so, f' (x) = 3ax2 + 2bx + c
Uding Rolles theorem, (Rolle's Theorem)   Let f(x) be continuous on [a, b], and differentiable on (a, b), and suppose that f (a) = f (b). Then there is some c with a < c < b such that f'(c) = 0
here (a , b) = (0 ,1)
f(0)= d and f(1) =a+b+c+d ; f(0) = f(1) which means a+b+c=0 which is true from given condition
hence there exists some c in btn (0, 1) where f '(c) =0
--
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and 

we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.
Regards,

Naga Ramesh

IIT Kgp - 2005 batch


										

										
This ques is solved using the ROLLE'S THEOREM
Let f '(x)= 3x2+2bx+c
=>  f(x)=x3+bx2+cx+k                where k is the integration constant
now f(0)=k    and f(1)=a+b+c+k =>          f(1)=0+k                   =>f(1)=k
So we get that f(0)=f(1)
  Using Rolle's Theorem we know that if f(a)=f(b)  then there exixsts at least one root c in (a,b)  such that f '(c)=0 
 
 Now we  can say that 3x2+ 2bx+c  has at least one root   b/t (0,1) .