Question

The minimum value of the function f(x) = ${\int }_0^x\frac{d\theta }{cos\theta }+{\int }_x^{\pi /2}\frac{d\theta }{sin\theta }$ where $x\in [0,\frac{\pi }{2}],$ is

• A. $2ln(\sqrt{2}+1)$
• B. $ln(2\sqrt{2}+2)$
• C. $ln(\sqrt{3}+2)$
• D. $ln(\sqrt{2}+3)$

Answer 1

The correct option is A $2ln(\sqrt{2}+1)$

f(x) = ${\int }_0^x\frac{d\theta }{cos\theta }+{\int }_x^{\pi /2}\frac{d\theta }{sin\theta }$
$f^{{}^{\prime }}\left(x\right)=\frac{1}{cosx}-\frac{1}{sinx}$
= $\frac{sinx-cosx}{sinxcosx}$
$f^{{}^{\prime }}\left(x\right)$ change sign $-t_0+atx=\frac{\pi }{4}$
so f(x) takes minimum value at x = $\frac{\pi }{4}$
$f\left(x\right)={\int }_0^{\pi /4}sec\theta d\theta +{\int }_{\pi /4}^{\pi /2}cosec\theta d\theta$
$=\left(ln\right(sec\theta +tan\theta ){)}_0^{\pi /4}+(ln(cosec\theta -cot\theta ){)}_{\pi /4}^{\pi /2}$
$=ln(\sqrt{2}+1)-ln(1+0)+ln1-ln(\sqrt{2}-1)$
$=ln\left[\frac{\sqrt{2}+1}{\sqrt{2}-1}\right]=ln(\sqrt{2}+1{)}^2=2ln(\sqrt{2}+1)$