If $f(x)=\frac{\left(\tan 1^{\circ}\right) x+\log _{\varepsilon}(123)}{x \log _{\varepsilon}(1234)-\left(\tan 1^{\circ}\right)}, x > 0$, then the least value of $f(f(x))+f\left(f\left(\frac{4}{x}\right)\right)$ is $...........$.

If $f(x) = \frac{{\alpha x}}{{x + 1}},x \ne - 1$, for what value of $\alpha $ is $f(f(x)) = x$

Show that the function $f: R_* \rightarrow R_*$ defined by $f(x)=\frac{1}{x}$ is one-one and onto, where $R_*$ is the set of all non-zero real numbers. Is the result true, if the domain $R_*$ is replaced by $N$ with co-domain being same as $R _*$ ?

The graph of function $f$ contains the point $P (1, 2)$ and $Q(s, r)$. The equation of the secant line through $P$ and $Q$ is $y = \left( {\frac{{{s^2} + 2s - 3}}{{s - 1}}} \right)$ $x - 1 - s$. The value of $f ‘ (1)$, is

The largest interval lying in $\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$ for which the function, $f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {\frac{x}{2} - 1} \right) + \log \left( {\cos x} \right)$  is defined is

Given the function $f(x) = \frac{{{a^x} + {a^{ - x}}}}{2},\;(a > 2)$. Then $f(x + y) + f(x - y) = $