If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in the interval:

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then

Let $f:[1,3] \rightarrow R$ be continuous and differentiable in $(1,3)$ such that $f^{\prime}(x)=[f(x)]^2+4$ for all $x \in (1,3)$. Then:

If the function $f(x) = x^3 - 6x^2 + ax + b$ satisfies Rolle's theorem in the interval $[1, 3]$ and $f'\left( \frac{2\sqrt{3} + 1}{\sqrt{3}} \right) = 0$,then $a = $ ..............

The function $f(x) = x^3 - 6x^2 + ax + b$ satisfies the conditions of Rolle's theorem in $[1, 3]$. The values of $a$ and $b$ are:

Let $f:(a, b) \rightarrow R$ be a twice differentiable function such that $f(x) = \int_{a}^{x} g(t) \, dt$ for a differentiable function $g(x)$. If $f(x) = 0$ has exactly five distinct roots in $(a, b)$,then $g(x) g'(x) = 0$ has at least: