Question

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

Answer: (A)

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