Question

Considering only the principal values of inverse functions, the set
A=\left\\{ x\ge 0 :{{\tan }^{-1}}\left( 2x \right)+{{\tan }^{-1}}\left( 3x \right)=\dfrac{\pi }{4} \right\\} :
A. is an empty set
B. Contains more than two elements
C. Contains two elements
D. is a singleton

Answer 1

We will start the question by applying some trigonometric identity to solve for x. We will use the properties of tan inverse function that is ${{\tan }^{-1}}\left( \theta \right)+{{\tan }^{-1}}\left( \varphi \right)={{\tan }^{-1}}\left( \dfrac{\theta +\varphi }{1-\theta \varphi } \right)$ . After that on obtaining a quadratic equation in x, we will solve it by taking out factors, and whatever value of x that will be greater than or equal to 0 will satisfy the given condition in question.

Complete step by step answer:
We will have to find the value of x for identifying what kind of set is A,
We are given : ${{\tan }^{-1}}\left( 2x \right)+{{\tan }^{-1}}\left( 3x \right)=\dfrac{\pi }{4}$
We will use the following identity to proceed with this question: ${{\tan }^{-1}}\left( \theta \right)+{{\tan }^{-1}}\left( \varphi \right)={{\tan }^{-1}}\left( \dfrac{\theta +\varphi }{1-\theta \varphi } \right)$
Applying the above identity to our function:
${{\tan }^{-1}}\left( 2x \right)+{{\tan }^{-1}}\left( 3x \right)={{\tan }^{-1}}\left( \dfrac{2x+3x}{1-\left( 2x.3x \right)} \right)$
Now, according to the question: ${{\tan }^{-1}}\left( 2x \right)+{{\tan }^{-1}}\left( 3x \right)=\dfrac{\pi }{4}$,
Therefore: ${{\tan }^{-1}}\left( \dfrac{2x+3x}{1-\left( 2x.3x \right)} \right)=\dfrac{\pi }{4}\text{ }..............\text{Equation 1}\text{.}$
We know the standard property for inverse functions that:
${{\tan }^{-1}}x=a\Rightarrow x=\tan a$

We will apply this property in our equation number 1.
$\begin{aligned} & {{\tan }^{-1}}\left( \dfrac{2x+3x}{1-\left( 2x.3x \right)} \right)=\dfrac{\pi }{4} \\\ & \left( \dfrac{2x+3x}{1-\left( 2x.3x \right)} \right)=\tan \left( \dfrac{\pi }{4} \right)\text{ }.............\text{ Equation 2}\text{.} \\\ \end{aligned}$
We will find out the value of $\tan \dfrac{\pi }{4}$ :
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }\Rightarrow \tan \dfrac{\pi }{4}=\dfrac{\sin \left( \dfrac{\pi }{4} \right)}{\cos \left( \dfrac{\pi }{4} \right)}\text{ }...........\text{Equation 3}\text{.}$
We already know that the value of $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ and $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ , so we will putting these values in Equation 3 in order to find the value of $\tan \dfrac{\pi }{4}$:
$\tan \dfrac{\pi }{4}=\dfrac{\sin \left( \dfrac{\pi }{4} \right)}{\cos \left( \dfrac{\pi }{4} \right)}\Rightarrow \tan \dfrac{\pi }{4}=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}}$
After cancelling out $\dfrac{1}{\sqrt{2}}$ , we get $\tan \dfrac{\pi }{4}=1$.
We will put it in our Equation no. 2 :
$\begin{aligned} & \left( \dfrac{2x+3x}{1-\left( 2x.3x \right)} \right)=\tan \left( \dfrac{\pi }{4} \right) \\\ & \left( \dfrac{2x+3x}{1-\left( 2x.3x \right)} \right)=1 \\\ \end{aligned}$
Now we will solve for x from here,
$\begin{aligned} & \left( \dfrac{2x+3x}{1-\left( 2x.3x \right)} \right)=1 \\\ & \left( 5x \right)=1.\left( 1-6{{x}^{2}} \right) \\\ \end{aligned}$
Taking the Right Hand Side to the Left Hand Side;

We will get the following Quadratic Equation, we will find the value of x by solving the obtained quation:
$\begin{aligned} & 6{{x}^{2}}+5x-1=0 \\\ & \Rightarrow 6{{x}^{2}}+\left( 6x-x \right)-1=0 \\\ & \Rightarrow 6{{x}^{2}}+6x-x-1=0 \\\ & \Rightarrow 6x(x+1)-1\left( x+1 \right)=0 \\\ & \Rightarrow \left( x+1 \right)\left( 6x-1 \right)=0 \\\ & \Rightarrow \left( x+1 \right)=0\text{ or }\left( 6x-1 \right)=0 \\\ & \Rightarrow x=-1,x=\dfrac{1}{6} \\\ \end{aligned}$
Since in the question we have been given that the value of x should be greater than or equal to 0, Therefore, we will neglect the negative value of x and hence we are left with only one value of x that is $\dfrac{1}{6}$ .
As there is only one value for set A, the set A will be a singleton.

So, the correct answer is “Option D”.

Note: Students can also directly put the value of $\tan \dfrac{\pi }{4}$ as 1, it is not necessary to show the whole calculation as how you obtained this value. Also, after solving the quadratic equation remember to neglect or reject the negative value as in our question it only demands 0 or greater than 0, therefore it will not qualify for the answer.