For the Mean Value Theorem $f(b) - f(a) = (b - a) f'(x_1)$ where $a < x_1 < b$,if $f(x) = 1/x$,then $x_1 = ?$

According to the Mean Value Theorem,which of the following functions does not satisfy the conditions on the interval $[0, 1]$?

The abscissa of the points of the curve $y = x^3$ in the interval $[-2, 2]$,where the slope of the tangents is equal to the slope of the secant line as per the Mean Value Theorem for the interval $[-2, 2]$,are:

For all $x \in [0, 2024]$,assume that $f(x)$ is differentiable,$f(0) = -2$,and $f^{\prime}(x) \geq 5$. Then the least possible value of $f(2024)$ is:

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

If $f(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ 2-x, & 1 < x \leq 2 \end{cases}$, then Rolle's theorem is not applicable to $f(x)$ because