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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
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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) .
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