Question

The number of solutions for the equation \[{{\tan }^{-1}}\left( {{e}^{-x}} \right)+{{\cot }^{-1}}\left( \left| \ln x \right| \right)=\dfrac{\pi }{2}\] is:
(a) 0
(b) 1
(c) 3
(d) 2

Answer 1

Hint: We solve this problem first by taking the trigonometric equations out by using the standard formulas that is
\[{{\tan }^{-1}}x={{\cot }^{-1}}\left( \dfrac{1}{x} \right)\]
Then we use the composite angle formula of inverse trigonometric equations as
\[{{\cot }^{-1}}x+{{\cot }^{-1}}y={{\cot }^{-1}}\left( \dfrac{xy-1}{x+y} \right)\]
By using the above formulas we find the relation between to functions without trigonometric equations so that we can find the number of solutions using the graphs that is if \[f\left( x \right)=g\left( x \right)\] then the number of solutions to above equations will be number of points of intersections of \[y=f\left( x \right)\] and \[y=g\left( x \right)\]

Complete step-by-step solution
We are given that the equation that is
\[\Rightarrow {{\tan }^{-1}}\left( {{e}^{-x}} \right)+{{\cot }^{-1}}\left( \left| \ln x \right| \right)=\dfrac{\pi }{2}\]
We know that the inverse trigonometric relation that is
\[{{\tan }^{-1}}x={{\cot }^{-1}}\left( \dfrac{1}{x} \right)\]
By using this relation to above equation we get
\[\begin{align}
  & \Rightarrow {{\cot }^{-1}}\left( \dfrac{1}{{{e}^{-x}}} \right)+{{\cot }^{-1}}\left( \left| \ln x \right| \right)=\dfrac{\pi }{2} \\
 & \Rightarrow {{\cot }^{-1}}\left( {{e}^{x}} \right)+{{\cot }^{-1}}\left( \left| \ln x \right| \right)=\dfrac{\pi }{2} \\
\end{align}\]
We know that the formula of the composite angle formula of inverse trigonometric equations as
\[{{\cot }^{-1}}x+{{\cot }^{-1}}y={{\cot }^{-1}}\left( \dfrac{xy-1}{x+y} \right)\]
By using this formula to above equation we get
\[\begin{align}
  & \Rightarrow {{\cot }^{-1}}\left( \dfrac{{{e}^{x}}\left| \ln x \right|-1}{{{e}^{x}}+\left| \ln x \right|} \right)=\dfrac{\pi }{2} \\
 & \Rightarrow \dfrac{{{e}^{x}}\left| \ln x \right|-1}{{{e}^{x}}+\left| \ln x \right|}=\cot \dfrac{\pi }{2} \\
\end{align}\]
We know that from the standard table of trigonometric ratios that is
\[\cot \dfrac{\pi }{2}=0\]
By substituting this value in above equation we get
\[\begin{align}
  & \Rightarrow \dfrac{{{e}^{x}}\left| \ln x \right|-1}{{{e}^{x}}+\left| \ln x \right|}=0 \\
 & \Rightarrow {{e}^{x}}\left| \ln x \right|-1=0 \\
 & \Rightarrow \left| \ln x \right|={{e}^{-x}} \\
\end{align}\]
We know that if \[f\left( x \right)=g\left( x \right)\] then the number of solutions to above equations will be number of points of intersections of \[y=f\left( x \right)\] and \[y=g\left( x \right)\]
Now, let us draw the graph of \[y=\left| \ln x \right|\] and \[y={{e}^{-x}}\] then we get
Here we can see that the blue line represents \[y={{e}^{-x}}\] and the green line represents \[y=\left| \ln x \right|\]
Here, we can see that both the graphs intersect at 2 points therefore the number of solutions of the given equation is 2.
So, option (d) is the correct answer.

Note: We have a shortcut for solving the problem.
We are given that the equation as
\[\Rightarrow {{\tan }^{-1}}\left( {{e}^{-x}} \right)+{{\cot }^{-1}}\left( \left| \ln x \right| \right)=\dfrac{\pi }{2}\]
We have a standard result of inverse trigonometric ratios that is if \[{{\tan }^{-1}}x+{{\cot }^{-1}}y=\dfrac{\pi }{2}\] happens if and only if \[x=y\]
By using the above result we get
\[\Rightarrow {{e}^{-x}}=\left| \ln x \right|\]