
The line $y = \sqrt 3 $ meets the graph $y = \tan x$, where $x \in \left( {0,\dfrac{\pi }{2}} \right)$, in $k$points. What is $k$ equal to?
A.One
B.Two
C.Three
D.Infinity
Answer
582.3k+ views
Hint: The solution set for the two graphs , say $y = f\left( x \right)$ and $y = g\left( x \right)$ comprises of the solution of the equation $f\left( x \right) = g\left( x \right)$. Therefore, for the given equation, equate $\sqrt 3 $ to the $\tan x$ to form the solution set. Use the condition given,$x \in \left( {0,\dfrac{\pi }{2}} \right)$ to find the value of $k$.
Complete step-by-step answer:
The solution set for the intersection of the two graphs, say $y = f\left( x \right)$ and $y = g\left( x \right)$ will contain the solution of the equation $f\left( x \right) = g\left( x \right)$.
We are given two equations $y = \sqrt 3 $ and $y = \tan x$. The intersection of these two graphs will be the solution set for the equation $\sqrt 3 = \tan x$.
Taking ${\tan ^{ - 1}}$ on both sides, we get
\[{\tan ^{ - 1}}\left( {\sqrt 3 } \right) = {\tan ^{ - 1}}\left( {\tan x} \right)\]
Also, it is known that \[{\tan ^{ - 1}}\left( {\tan x} \right)\] is equal to \[x\].
Therefore, the equation \[{\tan ^{ - 1}}\left( {\sqrt 3 } \right) = {\tan ^{ - 1}}\left( {\tan x} \right)\] becomes,
\[{\tan ^{ - 1}}\left( {\sqrt 3 } \right) = x\]
Since, the \[\dfrac{\pi }{3}\] is the principle value as $\tan \left( {\dfrac{\pi }{3}} \right) = \sqrt 3 $.The general solution for the above equation will be
\[x = n\pi + \dfrac{\pi }{3},n{\text{ is an integer}}\]
The given range on the $x$ is given as $x \in \left( {0,\dfrac{\pi }{2}} \right)$.
The solution set of the intersection of the graphs is \[..\dfrac{{ - 2\pi }}{3},\dfrac{\pi }{3},\dfrac{{4\pi }}{3}..\].
But only one value of the solution set lies in the range $\left( {0,\dfrac{\pi }{2}} \right)$.
Since $k$is the number of solution points for the intersection of the graphs $y = \sqrt 3 $ and $y = \tan x$ in the range $x \in \left( {0,\dfrac{\pi }{2}} \right)$, we can say $k$ equals 1.
$k = 1$
From the graph, there is only one intersection point.
Note: The general solution for the equation $x = {\tan ^{ - 1}}y$ is $n\pi + \theta $ ,where $\theta $ is the principle solution of the equation $x = {\tan ^{ - 1}}y$. Alternatively the above solution can be solved using a graphing utility, counting the intersection points in the range $x \in \left( {0,\dfrac{\pi }{2}} \right)$ to find the value of $k$.
Complete step-by-step answer:
The solution set for the intersection of the two graphs, say $y = f\left( x \right)$ and $y = g\left( x \right)$ will contain the solution of the equation $f\left( x \right) = g\left( x \right)$.
We are given two equations $y = \sqrt 3 $ and $y = \tan x$. The intersection of these two graphs will be the solution set for the equation $\sqrt 3 = \tan x$.
Taking ${\tan ^{ - 1}}$ on both sides, we get
\[{\tan ^{ - 1}}\left( {\sqrt 3 } \right) = {\tan ^{ - 1}}\left( {\tan x} \right)\]
Also, it is known that \[{\tan ^{ - 1}}\left( {\tan x} \right)\] is equal to \[x\].
Therefore, the equation \[{\tan ^{ - 1}}\left( {\sqrt 3 } \right) = {\tan ^{ - 1}}\left( {\tan x} \right)\] becomes,
\[{\tan ^{ - 1}}\left( {\sqrt 3 } \right) = x\]
Since, the \[\dfrac{\pi }{3}\] is the principle value as $\tan \left( {\dfrac{\pi }{3}} \right) = \sqrt 3 $.The general solution for the above equation will be
\[x = n\pi + \dfrac{\pi }{3},n{\text{ is an integer}}\]
The given range on the $x$ is given as $x \in \left( {0,\dfrac{\pi }{2}} \right)$.
The solution set of the intersection of the graphs is \[..\dfrac{{ - 2\pi }}{3},\dfrac{\pi }{3},\dfrac{{4\pi }}{3}..\].
But only one value of the solution set lies in the range $\left( {0,\dfrac{\pi }{2}} \right)$.
Since $k$is the number of solution points for the intersection of the graphs $y = \sqrt 3 $ and $y = \tan x$ in the range $x \in \left( {0,\dfrac{\pi }{2}} \right)$, we can say $k$ equals 1.
$k = 1$
From the graph, there is only one intersection point.
Note: The general solution for the equation $x = {\tan ^{ - 1}}y$ is $n\pi + \theta $ ,where $\theta $ is the principle solution of the equation $x = {\tan ^{ - 1}}y$. Alternatively the above solution can be solved using a graphing utility, counting the intersection points in the range $x \in \left( {0,\dfrac{\pi }{2}} \right)$ to find the value of $k$.
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