If f(x) = (tan 1 ^ ° )x + _ e (123) x _ e (1234) - (tan 1 ^ ° ) , x...

Given,

$f\left(x\right) = \dfrac{\left(tan{1}^{°}\right)x + \log_{e}\left(123\right)}{x{\log}_{e}\left(1234\right) - \left(tan{1}^{°}\right)}, x > 0$

Now, let $tan{1}^{°} = a, \log_{e}\left(123\right) = b$ and $\log_{e}\left(1234\right) = c$

So, $f\left(x\right) = \dfrac{ax + b}{cx - a}$

$⇒ f\left(f\left(x\right)\right) = \dfrac{af\left(x\right) + b}{cf\left(x\right) - a}$

$⇒ f\left(f\left(x\right)\right) = \dfrac{a\left(\dfrac{ax + b}{cx - a}\right) + b}{c\left(\dfrac{ax + b}{cx - a}\right) - a}$

$⇒ f\left(f\left(x\right)\right) = \dfrac{{a}^{2}x + ab + b\left(cx - a\right)}{acx + bc - a\left(cx - a\right)} = \dfrac{{a}^{2}x + bcx}{bc + {a}^{2}} = x$

$⇒ f\left(f\left(x\right)\right) + f\left(f\left(\dfrac{4}{x}\right)\right) = x + \dfrac{4}{x}$

$∵ x > 0$, then least value $= \sqrt{x \cdot \dfrac{4}{x}} = 2$