The domine of definition of the function
[tex]f(x) = \frac{1}{ \sqrt{ |\cos(x)| + \cos(x) }} \: is \\ [/tex]
​

[tex]\large\underline{\sf{Solution-}}[/tex]

Given function is

[tex]\rm \longmapsto\:f(x) = \dfrac{1}{ \sqrt{ |cosx| + cosx} } [/tex]

Now,

[tex]\rm \longmapsto\:f(x) \: is \: defined \: if \: |cosx| + cosx > 0[/tex]

We know,

[tex]\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}[/tex]

So,

[tex]\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}[/tex]

So,

[tex]\rm\implies \:f(x) \: is \: defined \: when \: cosx > 0[/tex]

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

[tex]\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}[/tex]

So, if we generalized this we get

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

Hence,

Domain of the function is

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

[tex]\textsf{More to know :-} \\[/tex]

[tex]\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}[/tex]

Solve the quadratic equation
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There are 24 students in a math class and 20 students passed the last quiz. What was the ratio of the students who passed the quiz to those who did not pass the quiz?

Eva
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1/5
(252)
-5
COMPLETE
27
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DONE

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