Question

The true set of values of \(K\) for which \(\sin^{-1}\left(\frac{1}{1+\sin^{2}x}\right)=\frac{K\pi}{6}\) may have a solution is:

• A. $[\frac{1}{6},\frac{1}{2}]$
• B. $[\frac{1}{4},\frac{1}{2}]$
• C. $[2, 4]$
• D. $[1, 3]$

Hint:

When computing ranges for inverse functions, always verify if the inner expression can cause a division-by-zero error or step outside the domain of the outer function. Here, since the denominator sits safely between 1 and 2, the fraction stays well within the regular $[-1, 1]$ domain of the $\sin^{-1}$ function.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To find the set of values of \( K \), we must find the range of the function \( f(x) = \sin^{-1} \left( \frac{1}{1+\sin^2 x} \right) \). The equation will have a solution only if the right-hand side \( \frac{K\pi}{6} \) falls within this range.
Step 2: Key Formula or Approach:
1. Find the range of \( \sin^2 x \).
2. Find the range of \( \frac{1}{1+\sin^2 x} \).
3. Apply the \( \sin^{-1} \) function to these bounds.
Step 3: Detailed Explanation:
- We know that for any real \( x \), \( 0 \le \sin^2 x \le 1 \).
- Adding 1 to all sides: \( 1 \le 1 + \sin^2 x \le 2 \).
- Taking the reciprocal (which reverses the inequality signs): \[ \frac{1}{2} \le \frac{1}{1 + \sin^2 x} \le 1 \] - Now apply \( \sin^{-1} \) (an increasing function) to the interval: \[ \sin^{-1} \left( \frac{1}{2} \right) \le \sin^{-1} \left( \frac{1}{1 + \sin^2 x} \right) \le \sin^{-1}(1) \] - Evaluating the inverse sine values: \[ \frac{\pi}{6} \le \text{Range} \le \frac{\pi}{2} \] The given equation is \( \frac{K\pi}{6} = \text{Range} \). So we substitute the bounds: \[ \frac{\pi}{6} \le \frac{K\pi}{6} \le \frac{\pi}{2} \] Multiply the entire inequality by \( \frac{6}{\pi} \): \[ 1 \le K \le \frac{\pi}{2} \cdot \frac{6}{\pi} \] \[ 1 \le K \le 3 \] Step 4: Final Answer:
The value of \( K \) must be in the closed interval \( [1, 3] \) for a solution to exist.