Question

The range of the function $f\left(x\right)=\frac{1}{|\sin x|}+\frac{1}{|\cos x|}$ is

• A. $[2\sqrt{2},\infty )$
• B. $(\sqrt{2},2\sqrt{2})$
• C. $(0,2\sqrt{2})$
• D. $(2\sqrt{2},4)$

Answer 1

Answer: (A)

$[2\sqrt{2},\infty )$

$f\left(x\right)=\frac{1}{|\sin x|}+\frac{1}{|\cos x|}$
Using $AM\ge GM$, we get
$\frac{\frac{1}{|\sin x|}+\frac{1}{|\cos x|}}{2}\ge {(\frac{1}{|\sin x|}\cdot \frac{1}{|\cos x|})}^{\frac{1}{2}}$
$\Rightarrow \frac{1}{|\sin x|}+\frac{1}{|\cos x|}\ge 2(2|\text{cosec}2x|{)}^{\frac{1}{2}}$ ......... [where $\left(\text{cosec}2x\right)\ge 1]$
$\therefore \frac{1}{|\sin x|}+\frac{1}{|\cos x|}\ge 2\sqrt{2}$
$\therefore$ Range of $f\left(x\right)\in [2\sqrt{2},\infty )$
Hence, (a) is the correct answer.