Rolle's Mean Value Theorem

Question

Answer 1

Rolle’s Theorem states that, for a function $f$ if:

1. $f$ is continuous in range $[a,b]$
2. $f$ is differentiable in range $(a,b)$
3. $f(a) = f(b)$

then, $f’(x)=0, for some$  $a\leq x \leq b$

Here, $ax^2 + bx + c = 0$ means that we have been given $f’(x)$

$\therefore After\:integrating\: f’(x),\: we\: get\: f(x) $

$\therefore f(x)= \frac{ax^3}{3}+\frac{bx^2}{2} + cx + d$

$Now,$

$f(0) = d$

$f(1) = \frac{a}{3}+\frac{b}{2} + c + d = \frac{2a+3b+6c+6d}{6} = \frac{6d}{d} = d\;(\because 2a+3b+6c=0)$

$So,f(0)=f(1) \:means\:that\:there\:is\:atleast\:one\:root\:between\:0\:and\:1$

$So, option\: A\:is\: correct\: choice.$