Rolle's theorem is not applicable to the function $f(x) = |x|$ defined on $ [-1, 1] $ because

Verify Rolle's Theorem for the function $f(x)=x^{2}+2 x-8, x \in[-4,2]$

Functions $f(x)$ and $g(x)$ are such that $f(x) + \int\limits_0^x {g(t)dt = 2\,\sin \,x\, - \,\frac{\pi }{2}} $ and $f'(x).g (x) = cos^2\,x$ , then number of solution $(s)$ of equation $f(x) + g(x) = 0$ in $(0,3 \pi$) is-

If the function $f(x) = {x^3} - 6a{x^2} + 5x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1, 2] $ and the tangent to the curve $y = f(x) $ at $x = {7 \over 4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $x = 1$ and $x = 2$. Then the value of $a$ is

Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
  {{x^2}\ln x,\,x > 0} \\ 
  {0,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0} 
\end{array}} \right\}$, Rolle’s theorem is applicable to $ f $ for $x \in [0,1]$, if $\alpha = $

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 $[0, 1 ].$