Question

Let \(\tan \left( \frac{\pi}{4} + \frac{1}{2} \cos^{-1} \frac{2}{3} \right) + \tan \left( \frac{\pi}{4} - \frac{1}{2} \sin^{-1} \frac{2}{3} \right) = k\). Then number of solution of the equation \(\sin^{-1}(kx - 1) = \sin x - \cos^{-1} x\) is/are :

• A. No solution
• B. exactly one solution
• C. Two solutions
• D. infinite solutions

Hint:

For equations involving \( \sin^{-1} \) or \( \cos^{-1} \), the domain of the function often simplifies the problem by narrowing down the range of \( x \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
First, evaluate the constant \( k \). Then solve the inverse trigonometric equation.
Step 2: Key Formula or Approach:
Let \( \cos^{-1} \frac{2}{3} = \theta \implies \cos \theta = \frac{2}{3} \).
Also \( \sin^{-1} \frac{2}{3} = \frac{\pi}{2} - \theta \).
The expression for \( k \) involves \( \tan(\frac{\pi}{4} \pm \frac{\theta}{2}) \).
Step 3: Detailed Explanation:
Using \( \tan(\frac{\pi}{4} + \alpha) = \frac{1 + \tan \alpha}{1 - \tan \alpha} \):
Evaluating the expression for \( k \) leads to \( k = 2 \).
Now, solve: \( \sin^{-1}(2x - 1) = \sin x - \cos^{-1} x \).
We know \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \implies -\cos^{-1} x = \sin^{-1} x - \frac{\pi}{2} \).
So, \( \sin^{-1}(2x - 1) - \sin^{-1} x = \sin x - \frac{\pi}{2} \).
Domain check: \( -1 \le 2x-1 \le 1 \implies 0 \le x \le 1 \).
By checking the behavior of the LHS and RHS on the interval \( [0, 1] \), we find they intersect at exactly one point.
Step 4: Final Answer:
There is exactly one solution.