The number of solutions of the equation $2^x = x^2$ is

The range of the function $f(x) = \left[ \frac{1}{\ln(x^2 + e)} \right] + \frac{1}{\sqrt{1 + x^2}}$ is,where $[*]$ denotes the greatest integer function and $e = \lim_{\alpha \to 0} (1 + \alpha)^{1/\alpha}$.

If the sets $A$ and $B$ are defined as $A = \{(x, y) : y = e^x, x \in R\}$ and $B = \{(x, y) : y = x, x \in R\}$,then:

If $f(x) = \frac{x - |x|}{|x|}$,then $f(-1) = $

If $e^{f(x)}=\frac{10+x}{10-x}, x \in(-10,10)$ and $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$, then $k$ is equal to

Let $f, g: R \rightarrow R$ be functions defined by $f(x) = \begin{cases} [x] & x < 0 \\ |1-x| & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} e^x - x & x < 0 \\ (x-1)^2 - 1 & x \geq 0 \end{cases}$ where $[x]$ denotes the greatest integer less than or equal to $x$. Then,the function $(f \circ g)(x)$ is discontinuous at exactly