 # if a+b+c=0 then the quadratic eqation 3x2+2bx+c has at least one root b/t (0,1)

12 years ago

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

12 years ago

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) .