If $f: \mathrm{R} \rightarrow \mathrm{R}$ is a twice differentiable function such that $f^{\prime \prime}(x)>0$ for all $x \in \mathrm{R}$, and $f\left(\frac{1}{2}\right)=\frac{1}{2}, f(1)=1$, then

From mean value theorem $f(b) - f(a) = $ $(b - a)f'({x_1});$   $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $

Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be a thrice differentiable function such that $f(0)=0, f(1)=1, f(2)=-1, f(3)=2$ and $f(4)=-2$. Then, the minimum number of zeros of $\left(3 f^{\prime} f^{\prime \prime}+f f^{\prime \prime \prime}\right)(x)$ is....................

Let $f (x)$ and $g (x)$ are two function which are defined and differentiable for all $x \ge x_0$. If $f (x_0) = g (x_0)$ and $f ' (x) > g ' (x)$ for all $x > x_0$ then

Let $f (1) = - 2$ and $f ' (x) \ge 4.2$ for $1 \le x \le 6$. The smallest possible value of $f (6)$, is

Let $f(x)$ be a function continuous on $[1,2]$ and differentiable on $(1,2)$ satisfying
$f(1) = 2, f(2) = 3$ and $f'(x) \geq 1 \forall x \in (1,2)$.Define $g(x)=\int\limits_1^x {f(t)\,dt\,\forall \,x\, \in [1,2]} $ then the greatest value of $g(x)$ on $[1,2]$ is-