The function $f(x) = x(x + 3){e^{ - (1/2)x}}$ satisfies all the conditions of Rolle's theorem in $ [-3, 0]$. The value of $c$ is

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

Mean value theorem $f(b) -f(a) = (b -a) f '(x_1);$ from $a < x_1 < b,$ if $f(x) = 1/x$ then $x_1 = ?$

Let $f(x) = \sqrt {x - 1} + \sqrt {x + 24 - 10\sqrt {x - 1} ;} $ $1 < x < 26$ be real valued function. Then $f\,'(x)$ for $1 < x < 26$ is

Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq$ 0 , and $f(x)=g(x) \sin x$.

$STATEMENT$ $-1: \lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.and

$STATEMENT$ $-2: f^{\prime}(0)=g(0)$.

If $c = \frac {1}{2}$ and $f(x) = 2x -x^2$ , then interval of $x$ in which $LMVT$, is applicable, is