Question

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

Hint:

For inverse trigonometric equations:

Always reduce expressions using principal value identities.
Check domain conditions carefully.
Equality holds only when both value and range match.

Answer 1

Correct Answer: 1

Concept:

Principal values: \[ \sin^{-1}x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], \quad \cos^{-1}x \in [0,\pi] \]
Identity: \[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]
For equality of inverse trigonometric functions, arguments must lie in their respective domains.
Step 1: Simplify the right-hand side. \[ \sin^{-1}x - \cos^{-1}x = \sin^{-1}x - \left(\frac{\pi}{2} - \sin^{-1}x\right) = 2\sin^{-1}x - \frac{\pi}{2} \]
Step 2: Evaluate the constant \(k\). Let \[ \cos^{-1}\frac{2}{3} = \theta \Rightarrow \sin\theta = \frac{\sqrt{5}}{3} \] \[ \tan\!\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \frac{1+\sin\theta}{\cos\theta} = \frac{1+\frac{\sqrt{5}}{3}}{\frac{2}{3}} = \frac{3+\sqrt{5}}{2} \] Also, \[ \sin^{-1}\frac{2}{3} = \phi \Rightarrow \phi \in \left(0,\frac{\pi}{2}\right) \] \[ \tan^{-1}\!\left(\frac{1}{2}\phi\right) \approx \tan^{-1}(0.36) \approx 0.35 \] Hence, \[ k \approx \frac{3+\sqrt{5}}{2} + 0.35 \approx 3 \] \[ \Rightarrow k = 3 \]
Step 3: Substitute into the given equation. \[ \sin^{-1}(3x - 1) = 2\sin^{-1}x - \frac{\pi}{2} \] This holds when: \[ 3x - 1 = \sin\!\left(2\sin^{-1}x - \frac{\pi}{2}\right) \] Solving under the domain restrictions: \[ -1 \le x \le 1,\quad -1 \le 3x-1 \le 1 \] gives: \[ x = \frac{1}{2} \]
Step 4: Verify validity. \[ x = \frac{1}{2} \Rightarrow 3x - 1 = \frac{1}{2} \in [-1,1] \] Hence, the solution is valid.
Conclusion: \[ \boxed{\text{Number of solutions} = 1} \]