Function $f(x)={\left( {1 + \frac{1}{x}} \right)^x}$ then Range of the function f (x) is
$(0, \infty )$
$(- \infty , e)$
$(1, \infty )$
$(1, e) \cup (e, \infty )$
If $f(x) = \cos (\log x)$, then the value of $f(x).f(4) - \frac{1}{2}\left[ {f\left( {\frac{x}{4}} \right) + f(4x)} \right]$
The range of the function $f(x) = \frac{{x + 2}}{{|x + 2|}}$ is
If $\,\,f(x) = \left\{ {\begin{array}{*{20}{c}}
{3 + x;\,\,\,\,\,x \geqslant 0} \\
{2 - 3x;\,\,\,\,\,x < 0}
\end{array}} \right.$ then $\mathop {\lim }\limits_{x \to 0} f(f(x))$ is equal to -
Domain of the function $f(x) = \frac{{{x^2} - 3x + 2}}{{{x^2} + x - 6}}$ is
If $f(a) = a^2 + a+ 1$ , then number of solutions of equation $f(a^2) = 3f(a)$ is