If $f(x)$ satisfies the conditions of Rolle's theorem in $[1, 2]$ and $f(x)$ is continuous in $[1, 2]$,then $\int_1^2 f'(x) dx$ is equal to

If $a, b, c \in \mathbb{R}$ and satisfy $3a + 5b + 15c = 0$,then the equation $ax^4 + bx^2 + c = 0$ has:

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

Let $f(x)$ be continuous on $[0,6]$ and differentiable on $(0,6)$. Let $f(0)=12$ and $f(6)=-4$. If $g(x)=\frac{f(x)}{x+1}$, then for some Lagrange's constant $c \in(0,6)$, $g^{\prime}(c)=$

In the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(c)$. If $a = 4$,$b = 9$ and $f(x) = \sqrt{x}$,then the value of $c$ is:

Let $f(x) = \log(1 + x^2)$ and $A$ be a constant such that $\frac{|f(x) - f(y)|}{|x - y|} \leq A$ for all real $x, y$ where $x \neq y$. Then,the least possible value of $A$ is