Question

The trigonometric equation \( \sin^{-1}x = 2\sin^{-1}a \) has a solution. Find the valid range for \( a \).

• A. \(\text{only when } \frac{1}{\sqrt{2}} < a < \frac{1}{2}\)
• B. \(\text{for all real values of } a\)
• C. \(\text{only when } |a| \leq \frac{1}{\sqrt{2}}\)
• D. \(\text{only when } |a| \geq \frac{1}{\sqrt{2}}\)

Hint:

When solving inverse trigonometric equations, always ensure that the given function values stay within their principal domain and range. This helps in correctly determining valid solutions.

Answer 1

The Correct Option is C

We are given the equation: \[ \sin^{-1} x = 2\sin^{-1} a \] 

Step 1: Define the range of inverse sine function 
For \( \sin^{-1} x \), we know: \[ - \frac{\pi}{2} \leq \sin^{-1} x \leq \frac{\pi}{2} \] Similarly, since \( \sin^{-1} a \) is also defined in the range: \[ - \frac{\pi}{2} \leq \sin^{-1} a \leq \frac{\pi}{2} \] Multiplying both sides of this inequality by 2: \[ - \pi \leq 2\sin^{-1} a \leq \pi \] Since the principal range of \( \sin^{-1} x \) is limited to \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \), for the given equation to have a valid solution, we must satisfy: \[ - \frac{\pi}{2} \leq 2\sin^{-1} a \leq \frac{\pi}{2} \] 

Step 2: Solve for \( a \) 
Dividing the inequality by 2: \[ - \frac{\pi}{4} \leq \sin^{-1} a \leq \frac{\pi}{4} \] Taking sine on both sides: \[ \sin \left( -\frac{\pi}{4} \right) \leq a \leq \sin \left( \frac{\pi}{4} \right) \] Since \( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), we obtain: \[ -\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}} \] 

Step 3: Conclusion 
Thus, the equation has a solution only when: \[ |a| \leq \frac{1}{\sqrt{2}} \] Hence, the correct answer is: \[ \mathbf{|a| \leq \frac{1}{\sqrt{2}}} \]