Question

Rolle's theorem holds for the function f(x) = x3 + bx2 + cx, 1 x 2 at x = 4 3 , then the value of b + c is

Answer 1

Solution:

For a real valued function  f(x)  if it is continuous in the closed interval [p,q] and Differentiable in the open interval (p,q),such that , f(p)=f(q),then there must exist a point m such that ,a<m<b ,then f'(m)=0.

Now,coming to the Question

f(x)=x³+bx²+c x

f'(x)=3 x²+2 b x+c--------(1)

it is continuous in [1,2] and differentiable in (1,2).

f(1)=f(2)

→1³+b×1²+c×1=2³+b×2²+c×2

→1+b+c=8+4 b+2 c

→4 b-b+2 c-c=1-8

→3 b+c= -7------(2)

At, $m=\frac{4}{3}$that is,substituting in (1)

$f(\frac{4}{3})=0\\\\3\times (\frac{4}{3})^2+2 b\times \frac{4}{3}+c=0\\\\3\times \frac{16}{9}+\frac{8b}{3}+c=0\\\\48+24 b+9 c=0\\\\16+8 b+3 c=0$

→16+8 b+3(-7-3 b)=0-------using (2)

→16+8 b-21-9 b=0

→-5-b=0

→b= -5

Putting the value of b in equation (2)

→3 × (-5) +c= -7

→ -15+c= -7

→c= -7 +15

→c=8

So, b+c= -5+8

 b+c =3