If $a + b + c = 0$,then how many roots does the equation $3ax^2 + 2bx + c = 0$ have in the interval $(0, 1)$?

Let $f: R \rightarrow R$ be a differentiable function such that $f(a)=0=f(b)$ and $f^{\prime}(a) f^{\prime}(b) > 0$ for some $a < b$. Then,the minimum number of roots of $f^{\prime}(x)=0$ in the interval $(a, b)$ is

Let the function $f:[-7,0] \rightarrow R$ be continuous on $[-7,0]$ and differentiable on $(-7,0)$. If $f(-7)=-3$ and $f'(x) \leq 2$ for all $x \in (-7,0)$,then for all such functions $f$,$f(-1)+f(0)$ lies in the interval:

If the function $f(x) = 2x^2 + 3x + 5$ satisfies the Lagrange's Mean Value Theorem $(LMVT)$ at $x = 3$ on the closed interval $[1, a]$,then the value of $a$ is equal to:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice continuously differentiable function such that $f(0)=f(1)=f^{\prime}(0)=0$. Then:

$A$ value of $c$ for which the conclusion of the Mean Value Theorem holds for the function $f(x) = \log_{e}x$ on the interval $[1, 3]$ is