The function $f(x) = \frac{x^2 - 2}{\sqrt{1 + x^2}}$:

Let a function $f(x) = \begin{cases} -\ln(3x - [3x]) & ; 3x \neq n, n \in N \\ \ln(\operatorname{sgn}(3x)) & ; 3x = n, n \in N \end{cases}$,where $[.]$ and $\operatorname{sgn}(x)$ denote the greatest integer function and signum function respectively. Then the number of points where $f(x)$ is minimum in $x \in (0, 5)$ is:

The minimum value of $2x^2+x-1$ is:

Let $f, g$ and $h$ be real-valued functions defined on the interval $[0,1]$ by $f(x)=e^{x^2}+e^{-x^2}$,$g(x)=x e^{x^2}+e^{-x^2}$ and $h(x)=x^2 e^{x^2}+e^{-x^2}$. If $a, b$ and $c$ denote,respectively,the absolute maximum of $f, g$ and $h$ on $[0,1]$,then

The condition that $f(x) = ax^3 + bx^2 + cx + d$ has no extreme value is

If $g(x) = 2f(2x^3 - 3x^2) + f(6x^2 - 4x^3 - 3)$,$\forall x \in R$ and $f''(x) > 0$,$\forall x \in R$,then $g'(x) > 0$ for $x$ belonging to