If $f(x)=|x-2|, x \in[0,4]$ then the Rolle's theorem cannot be applied to the function because

Consider $f(x) = \int\limits_0^x {\left( {t + \frac{1}{t}} \right)\,dt}$ and $g(x) = f'(x)$ for $x \in \left[ {\frac{1}{2}, 3} \right]$. If $P$ is a point on the curve $y = g(x)$ such that the tangent to this curve at $P$ is parallel to a chord joining the points $\left( {\frac{1}{2}, g\left( {\frac{1}{2}} \right)} \right)$ and $(3, g(3))$ of the curve,then the coordinates of the point $P$ are:

Consider a 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 $2:$ The Rolle's theorem is applicable to function $g(x)$ on the interval $[0, 1].$

In which of the following functions is Rolle's theorem applicable?

The value of $c$ in the Lagrange's mean value theorem for $f(x)=\sqrt{x-2}$ in the interval $[2,6]$ is

Consider the function $f(x) = \begin{cases} x \sin \frac{\pi}{x} & \text{for } x > 0 \\ 0 & \text{for } x = 0 \end{cases}$. Then the number of points in $(0, 1)$ where the derivative $f'(x)$ vanishes is: