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:
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:
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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.
Updated On: May 28, 2026
- $[\frac{1}{6},\frac{1}{2}]$
- $[\frac{1}{4},\frac{1}{2}]$
- $[2, 4]$
- $[1, 3]$
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The Correct Option is D
Solution and Explanation
Concept:
For an inverse trigonometric equation to have a valid solution, the constant value on one side must fall entirely within the mathematical range of the function expression on the other side. We solve this by finding the domain limits of the inner expression and tracking them through the outer function layers.
Step 1: Find the range of the internal fraction argument.
Let us analyze the range of the denominator term. We know that for all real values of $x$, the squared sine function is bounded by: $$0 \le \sin^2 x \le 1$$ Add 1 across the entire inequality: $$1 \le 1 + \sin^2 x \le 2$$ Since all terms are positive, taking the reciprocal flips the inequality signs: $$\frac{1}{2} \le \frac{1}{1+\sin^2 x} \le 1 \quad \cdots (1)$$
Step 2: Apply the inverse sine layer across the inequality boundaries.
The function $\sin^{-1}(y)$ is strictly increasing for all inputs within its domain $[-1, 1]$. Applying this function directly preserves the inequality direction of equation (1): $$\sin^{-1}\left(\frac{1}{2}\right) \le \sin^{-1}\left(\frac{1}{1+\sin^2 x}\right) \le \sin^{-1}(1)$$ Evaluate the exact known angles at these boundaries: $$\frac{\pi}{6} \le \sin^{-1}\left(\frac{1}{1+\sin^2 x}\right) \le \frac{\pi}{2}$$
Step 3: Solve for the constant parameter $K$.
Substitute our target expression $\frac{K\pi}{6}$ into the middle of the range inequality: $$\frac{\pi}{6} \le \frac{K\pi}{6} \le \frac{\pi}{2}$$ Divide all sections of the inequality by $\pi$: $$\frac{1}{6} \le \frac{K}{6} \le \frac{1}{2}$$ Multiply the entire expression by 6 to isolate $K$: $$1 \le K \le 3 \quad \Rightarrow \quad K \in [1, 3]$$ This matches option (D) perfectly.
Let us analyze the range of the denominator term. We know that for all real values of $x$, the squared sine function is bounded by: $$0 \le \sin^2 x \le 1$$ Add 1 across the entire inequality: $$1 \le 1 + \sin^2 x \le 2$$ Since all terms are positive, taking the reciprocal flips the inequality signs: $$\frac{1}{2} \le \frac{1}{1+\sin^2 x} \le 1 \quad \cdots (1)$$
Step 2: Apply the inverse sine layer across the inequality boundaries.
The function $\sin^{-1}(y)$ is strictly increasing for all inputs within its domain $[-1, 1]$. Applying this function directly preserves the inequality direction of equation (1): $$\sin^{-1}\left(\frac{1}{2}\right) \le \sin^{-1}\left(\frac{1}{1+\sin^2 x}\right) \le \sin^{-1}(1)$$ Evaluate the exact known angles at these boundaries: $$\frac{\pi}{6} \le \sin^{-1}\left(\frac{1}{1+\sin^2 x}\right) \le \frac{\pi}{2}$$
Step 3: Solve for the constant parameter $K$.
Substitute our target expression $\frac{K\pi}{6}$ into the middle of the range inequality: $$\frac{\pi}{6} \le \frac{K\pi}{6} \le \frac{\pi}{2}$$ Divide all sections of the inequality by $\pi$: $$\frac{1}{6} \le \frac{K}{6} \le \frac{1}{2}$$ Multiply the entire expression by 6 to isolate $K$: $$1 \le K \le 3 \quad \Rightarrow \quad K \in [1, 3]$$ This matches option (D) perfectly.
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