Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```        if a+b+c=0 then the quadratic eqation
3x2+2bx+c has at least one root b/t (0,1)```
8 years ago

Ramesh V
70 Points
```										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

```
8 years ago
Vanya Saxena
18 Points
```										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) .
```
8 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details