Question

The number of solutions for the equation $${\tan }^{-1}\left(e^{-x}\right)+{\cot }^{-1}\left(|\ln x|\right)={\displaystyle \frac{\pi }{2}}$$ 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
$${\tan }^{-1}x={\cot }^{-1}\left({\displaystyle \frac{1}{x}}\right)$$
Then we use the composite angle formula of inverse trigonometric equations as
$${\cot }^{-1}x+{\cot }^{-1}y={\cot }^{-1}\left({\displaystyle \frac{xy-1}{x+y}}\right)$$
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
$$\Rightarrow {\tan }^{-1}\left(e^{-x}\right)+{\cot }^{-1}\left(|\ln x|\right)={\displaystyle \frac{\pi }{2}}$$
We know that the inverse trigonometric relation that is
$${\tan }^{-1}x={\cot }^{-1}\left({\displaystyle \frac{1}{x}}\right)$$
By using this relation to above equation we get
$$\begin{array}{rl}& \Rightarrow {\cot }^{-1}\left({\displaystyle \frac{1}{e^{-x}}}\right)+{\cot }^{-1}\left(|\ln x|\right)={\displaystyle \frac{\pi }{2}}\\ & \Rightarrow {\cot }^{-1}\left(e^x\right)+{\cot }^{-1}\left(|\ln x|\right)={\displaystyle \frac{\pi }{2}}\end{array}$$
We know that the formula of the composite angle formula of inverse trigonometric equations as
$${\cot }^{-1}x+{\cot }^{-1}y={\cot }^{-1}\left({\displaystyle \frac{xy-1}{x+y}}\right)$$
By using this formula to above equation we get
$$\begin{array}{rl}& \Rightarrow {\cot }^{-1}\left({\displaystyle \frac{e^x|\ln x|-1}{e^x+|\ln x|}}\right)={\displaystyle \frac{\pi }{2}}\\ & \Rightarrow {\displaystyle \frac{e^x|\ln x|-1}{e^x+|\ln x|}}=\cot {\displaystyle \frac{\pi }{2}}\end{array}$$
We know that from the standard table of trigonometric ratios that is
$$\cot {\displaystyle \frac{\pi }{2}}=0$$
By substituting this value in above equation we get
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{e^x|\ln x|-1}{e^x+|\ln x|}}=0\\ & \Rightarrow e^x|\ln x|-1=0\\ & \Rightarrow |\ln x|=e^{-x}\end{array}$$
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 $$y=|\ln x|$$ and $$y=e^{-x}$$ then we get
Here we can see that the blue line represents $$y=e^{-x}$$ and the green line represents $$y=|\ln x|$$
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
$$\Rightarrow {\tan }^{-1}\left(e^{-x}\right)+{\cot }^{-1}\left(|\ln x|\right)={\displaystyle \frac{\pi }{2}}$$
We have a standard result of inverse trigonometric ratios that is if $${\tan }^{-1}x+{\cot }^{-1}y={\displaystyle \frac{\pi }{2}}$$ happens if and only if $$x=y$$
By using the above result we get
$$\Rightarrow e^{-x}=|\ln x|$$