<p>The value of \(\sin\sqrt{x^2 - \frac{\pi^2}{36}}\) lies in the interval</p>
Step-by-Step Solution
Key Concept: The argument of sine function √(x² - π²/36) must satisfy the domain constraint where the expression under the square root is non-negative, meaning |x| ≥ π/6. The output range of sine depends on the range of its argument.
<p><strong>Step 1:</strong> For √(x² - π²/36) to be real, we require x² - π²/36 ≥ 0, which gives |x| ≥ π/6.</p><p><strong>Step 2:</strong> When |x| = π/6 (minimum case), the argument of sine is √(0) = 0, so sin(0) = 0.</p><p><strong>Step 3:</strong> As |x| increases beyond π/6, the argument √(x² - π²/36) increases from 0 towards ∞, meaning the sine function oscillates through all its values.</p><p><strong>Step 4:</strong> Since sine oscillates over the entire interval [-1, 1] as its argument varies over [0, ∞), and the minimum value of the argument is 0 (giving sin(0) = 0), the range is [-1, 1] with the specific point 0 being achieved.</p><p><strong>Step 5:</strong> The most natural answer for such problems is <strong>[-1, 1]</strong> or the interval containing all possible values that sine attains given the domain constraint.</p><p>∴ Answer: D</p>
Correct Answer: D