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

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

Consider the function $f(x) = 8x^2 - 7x + 5$ on the interval $[-6, 6]$. The value of $c$ that satisfies the conclusion of the Mean Value Theorem is:

Let $f: D \rightarrow R$ where $D=[0,1] \cup [2,4]$ be defined by $f(x)=\begin{cases} x, & \text{if } x \in [0,1] \\ 4-x, & \text{if } x \in [2,4] \end{cases}$. Then,

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:

If $f$ is defined in $[1,3]$ by $f(x)=x^3+b x^2+a x$,such that $f(1)-f(3)=0$ and $f^{\prime}(c)=0$,where $c=2+\frac{1}{\sqrt{3}}$,then $(a, b)$ is equal to