Let the function $$g:\left( { - \infty ,\infty } \right) \to \left( { - {\pi \over 2},{\pi \over 2}} \right)$$ be given by
$$g\left( u \right) = 2{\tan ^{ - 1}}\left( {{e^u}} \right) - {\pi \over 2}.$$ Then, $$g$$ is

The tangent to the curve $$y = {e^x}$$ drawn at the point $$\left( {c,{e^c}} \right)$$ intersects the line joining the points $$\left( {c - 1,{e^{c - 1}}} \right)$$ and $$\left( {c + 1,{e^{c + 1}}} \right)$$

If a continuous function $$f$$ defined on the real line $$R$$, assumes positive and negative values in $$R$$ then the equation $$f(x)=0$$ has a root in $$R$$. For example, if it is known that a continuous function $$f$$ on $$R$$ is positive at some point and its minimum value is negative then the equation $$f(x)=0$$ has a root in $$R$$.
Consider $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

The line $$y=x$$ meets $$y = k{e^x}$$ for $$k \le 0$$ at

If a continuous function $$f$$ defined on the real line $$R$$, assumes positive and negative values in $$R$$ then the equation $$f(x)=0$$ has a root in $$R$$. For example, if it is known that a continuous function $$f$$ on $$R$$ is positive at some point and its minimum value is negative then the equation $$f(x)=0$$ has a root in $$R$$.
Consider $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

The positive value of $$k$$ for which $$k{e^x} - x = 0$$ has only one root is