Question

Find the domain of the function ​

Answer 1

Given Question

Find the domain of the function

$\rm :\longmapsto\:f(x) = \dfrac{1}{ \sqrt{ |cosx| + cosx} }$

$\green{\large\underline{\sf{Solution-}}}$

Given function is

$\rm :\longmapsto\:f(x) = \dfrac{1}{ \sqrt{ |cosx| + cosx} }$

Now,

$\rm :\longmapsto\:f(x) \: is \: defined \: if \: |cosx| + cosx > 0$

We know,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{ - x \: \: when \: x < 0} \\ \\ &\sf{ \: \: x \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |cosx| + cosx = \begin{cases} &\sf{ \: \: 0 \: \: when \: cosx \leqslant 0} \\ \\ &\sf{ \: \: 2 \: cosx \: \: when \: cosx > 0} \end{cases}\end{gathered}\end{gathered}$

So,

$\rm\implies \:f(x) \: is \: defined \: when \: cosx > 0$

So, from graph we concluded that cosx > 0 in the following intervals.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf cosx & \bf \: x \: \in \: \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf + ve & \sf \bigg( - \dfrac{\pi}{2} ,\dfrac{\pi}{2} \bigg) \\ \\ \sf + ve & \sf \bigg(\dfrac{3\pi}{2} ,\dfrac{5\pi}{2} \bigg) \\ \\ \sf + ve & \sf \bigg(\dfrac{7\pi}{2} ,\dfrac{9\pi}{2} \bigg) \end{array}} \\ \end{gathered}$

So, if we generalized this we get

$\rm\implies \:cosx > 0 \: when \: x \: \in \: \bigg(\dfrac{(4n - 1)\pi}{2} ,\dfrac{(4n + 1)\pi}{2} \bigg) \: \forall \: n \in \: Z$

Hence,

Domain of the function is

$\red{\rm\implies \boxed{\tt{ \: x \in \: \bigg(\dfrac{(4n - 1)\pi}{2} ,\dfrac{(4n + 1)\pi}{2} \bigg) \: \forall \: n \in \: Z}}}$

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More to know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

$\tt \huge \colorbox{lime}{❥Solution :-}$

️‍️

$\tt \red{Given} : - \: f(x) = \frac{1}{ \sqrt{ | \cos(x) | } + \cos(x) }$

️‍️

$\tt \red{ To \: Find} \: :- \: Domain \: of \: f(x)$

️‍️

$\tt \red{Solution} \: :-f(x) = \frac{1}{ \sqrt{ | \cos(x) | } + \cos(x) }$

$\tt \: | \cos(x) | + \cos(x) > 0$

$\tt \: = > \cos(x) > 0$

$\tt \: = > 2n\pi - \frac{x}{2} < x < 2n\pi + \frac{\pi}{2}$

$\tt \: = > x∈( \frac{(4n - 1)\pi}{2} , \: \frac{(4n + 1)\pi}{2} ,n∈ \: I$

Hence, the Domain is

$\tt \pink{( \frac{(4n - 1)\pi}{2} < x < \frac{(4n + 1)\pi}{2} )}$

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