Question

The minimum value of the function
$f\left(x\right)=\frac{1}{sinx+cosx}$ in the interval $\left[0,\frac{Ï€}{2}\right]$ 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
$f^{′}(x)=\frac{−1}{(\sin x+\cos x{)}^2}[\cos x−\sin x]$
For minimum, put $f^{′}(x)=0$
$∴\cos x−\sin x=0$
$⇒\tan x=1$
$⇒x=\frac{π}{4}$
$∴x=\frac{π}{4},f^{′}(x)>0$, minima
$∴$ Minimum value is
$f\left(\frac{Ï€}{4}\right)=\frac{1}{\sin \frac{Ï€}{4}+\cos \frac{Ï€}{4}}=\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}$
$=\frac{1}{2/\sqrt{2}}=\frac{\sqrt{2}}{2}$