Question

Let\( S={x∈R:0<x<1 and\ 2 tan−1\frac{(1+x)}{(1−x​)}=cos^{−1}\frac{(1-x^2)}{(1+x^2​)}}\). If n(S) denotes the number of elements in S then :

• A. \(n ( S )=2\) and only one element in\(S\)is less than \(\frac{1}{2}\)
• B. \(n(S)=1\) and the element in \(S\) is less than \(\frac{1}{2}\).
• C. \(n(S)=1\)and the elements in \(S\) is more than \(\frac{1}{2}\).
• D. \(n(S)=0\)

Hint:

For equations involving inverse trigonometric functions, simplify using known identities and consider the domain restrictions carefully. Ensure all possible solutions are within the given interval.

Answer 1

The Correct Option is B

The given equation is:
\[2 \tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right).\]
Step 1: Apply the Domain Condition
From the problem, \(0 < x < 1\).
Step 2: Simplify the Equation
Let:
\[\tan^{-1}\left(\frac{1-x}{1+x}\right) = \theta.\]
Thus:
\[\frac{1-x}{1+x} = \tan \theta \quad \text{and} \quad \theta \in \left(0, \frac{\pi}{4}\right).\]
Substitute into the equation:
\[2 \tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right).\]
\[2\theta = \cos^{-1}(\cos 2\theta).\]
Step 3: Solve for \(\theta\)
Since \(\cos^{-1}(\cos x) = x\) when \(x \in [0, \pi]\), we get:
\[2\theta = 2\theta.\]
The solution is valid, and hence:
\[x = \tan \theta.\]
Step 4: Calculate \(x\)
Let:
\[2\theta = \frac{\pi}{4}.\]
Thus:
\[\theta = \frac{\pi}{8}.\]
\[x = \tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1 \approx 0.414.\]
Step 5: Verify if \(x < \frac{1}{2}\)
Since \(x = 0.414\), we have \(x < \frac{1}{2}\).
Step 6: Determine \(n(S)\)
The solution is unique, so \(n(S) = 1\).
Conclusion: \(n(S) = 1\), and the element in \(S\) is less than \(\frac{1}{2}\) .