Question

The minimum value of the function $f\left(x\right)=\frac{1}{\sin x+\cos x}$ in the interval $[0,\frac{\pi }{2}]$ is :

• A. $\frac{\sqrt{2}}{2}$
• B. $-\frac{\sqrt{2}}{2}$
• C. $\frac{2}{\sqrt{3}+1}$
• D. $-\frac{2}{\sqrt{3}+1}$
• E. $1$

Answer 1

The correct option is A $\frac{\sqrt{2}}{2}$

Given, $f\left(x\right)=\frac{1}{\sin x+\cos x}$
On differentiating w.r.t, $x,$ we get
$f^{\prime }\left(x\right)=\frac{-1}{(\sin x+\cos x{)}^2}[\cos x-\sin x]$
For minimum, put $f^{\prime }\left(x\right)=0$
Therefore, $\cos x-\sin x=0$
$\Rightarrow \tan x=1$\
$\Rightarrow x=\frac{\pi }{4}$
$\Rightarrow x=\frac{\pi }{4},f^{\prime }\left(x\right)>0$, minima
Thus minimum value is
$f\left(\frac{\pi }{4}\right)=\frac{1}{\sin \frac{\pi }{4}+\cos \frac{\pi }{4}}=\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}$
$=\frac{1}{2/\sqrt{2}}=\frac{\sqrt{2}}{2}$