If $f(x) = x^2 - 2x + 4$ and $\frac{f(5) - f(1)}{5 - 1} = f'(c)$,then the value of $c$ will be

Let $f$ be a twice differentiable function on $(1,6)$. If $f(2)=8$,$f'(2)=5$,$f'(x) \geq 1$ and $f''(x) \geq 4$ for all $x \in (1,6)$,then:

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 $[0, 1]$.

Let $f(1) = -2$ and $f'(x) \ge 4.2$ for $1 \le x \le 6$. The smallest possible value of $f(6)$ is:

If the function $f(x) = x(x + 3) e^{-x/2}$ satisfies Rolle's theorem in the interval $[-3, 0]$,then find the value of $c$.

Let $f(x) = (x-4)(x-5)(x-6)(x-7)$,then -