The function $f(x) = x(x + 3)e^{-(1/2)x}$ satisfies all the conditions of Rolle's theorem in $[-3, 0]$. The value of $c$ is

Consider the quadratic equation $ax^2+bx+c=0$,where $2a+3b+6c=0$ and let $g(x)=\frac{ax^3}{3}+\frac{bx^2}{2}+cx$.
Statement-$I$ : The given quadratic equation $ax^2+bx+c=0$ has at least one root in $(0,1)$.
Statement-$II$ : Rolle's theorem is applicable to $g(x)$ on $[0,1]$.
Then

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f(0) = 2$,$g(1) = 2$,$g(0) = 0$,and $f(1) = 6$,then for some $c \in (0, 1)$:

From the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(x_1)$ where $a < x_1 < b$. If $f(x) = \frac{1}{x}$,then $x_1 = $

Let $f(1) = -2$ and $f'(x) \ge 4.2$ for $1 \le x \le 6$. The possible value of $f(6)$ lies in the interval

The value $C$ of the Lagrange's mean value theorem for the function $f(x)=x(x-1)(x-2)$ in the interval $[0, 1/2]$ is