Question

e range of the function f(x) =1/
|sin xl+1/|cos xl​

Answer 1

Answer:

1/|sinx|+|cosx| € [2√2 , ●●)hence range of f(x) is [2√2,●●) .

Answer 2

By $AM\geq GM$ inequality, we have,

$\longrightarrow\dfrac{\dfrac{1}{|\sin x|}+\dfrac{1}{|\cos x|}}{2}\geq\sqrt{\dfrac{1}{|\sin x\cos x|}}$

$\longrightarrow\dfrac{1}{|\sin x|}+\dfrac{1}{|\cos x|}\geq\dfrac{2}{\sqrt{|\sin x\cos x|}}$

Or we rewrite,

$\longrightarrow f(x)\geq\dfrac{2}{\sqrt{\dfrac{|\sin(2x)|}{2}}}$

$\longrightarrow f(x)\geq\dfrac{2\sqrt2}{\sqrt{|\sin(2x)|}}$

To get minimum value of $f(x),$ denominator of RHS should be maximum, so we need to take $\sin(2x)=1,$ i.e., its maximum value.

Then we get,

$\longrightarrow f(x)\geq2\sqrt2$

Hence the range is,

$\longrightarrow\underline{\underline{f(x)\in\left[2\sqrt2,\ \infty\right)}}$