f(x)=min (tan x, cot x), 0 ≤ x ≤ (π)/2, then ∫limits_0^π / 2 f(x) d x…

f(x) = min (tan x, cot x), $\in\left[0, \frac{\pi}{2}\right]$

$f(x)=\tan x, 0 \leq x \leq \frac{\pi}{4}$

$=\cot x, \frac{\pi}{4}<x \leq \frac{4}{2}$

Hence

$=\int\limits_0^{\pi / 2} f(x) d x=\int\limits_0^{\pi / 4} \tan x~dx + \int\limits_{\pi / 4}^{\pi / 2} \cot x ~dx$

$I=|\ln (\sec x)|_0^{\pi / 4}+|\ln (\sin x)|_{\pi / 4}^{\pi / 2}$

$=(\ln \sqrt{2}-0)+\left(0-\ln \frac{1}{\sqrt{2}}\right)$

$2 \ln \sqrt{2}=\ln 2$

Hence (1) is the correct answer.