The range of the function <span class="math-inline">f(x) = \text{sgn}\left( \frac{\sin^{2} x + 2\sin x + 4}{\sin^{2} x + 2\sin x + 3} \right)</span> is (where <span class="math-inline">\text{sgn}(.)</span> denotes signum function)-
(A) <span class="math-inline">\{-1,0,1\}</span>
(B) <span class="math-inline">\{-1,0\}</span>
(C) <span class="math-inline">\{1\}</span>
(D) <span class="math-inline">\{0,1\}</span>
Step-by-Step Solution
Key Concept: We need to determine the sign of the argument of the signum function by analyzing whether the numerator and denominator are always positive, always negative, or can change sign. The signum function returns 1 for positive arguments, -1 for negative arguments, and 0 for zero arguments.
<p><strong>Step 1:</strong> Let t = sin x. Since sin x ∈ [-1, 1], we have t ∈ [-1, 1].</p><p><strong>Step 2:</strong> Rewrite the argument of sgn as: g(t) = (t² + 2t + 4)/(t² + 2t + 3)</p><p><strong>Step 3:</strong> Analyze the numerator N(t) = t² + 2t + 4. Complete the square: N(t) = (t + 1)² + 3. Since (t + 1)² ≥ 0, we have N(t) ≥ 3 > 0 for all t ∈ [-1, 1].</p><p><strong>Step 4:</strong> Analyze the denominator D(t) = t² + 2t + 3. Complete the square: D(t) = (t + 1)² + 2. Since (t + 1)² ≥ 0, we have D(t) ≥ 2 > 0 for all t ∈ [-1, 1].</p><p><strong>Step 5:</strong> Since both numerator and denominator are always positive for all values of t ∈ [-1, 1], the fraction g(t) is always positive. Therefore, g(t) > 0 for all x in the domain.</p><p><strong>Step 6:</strong> By definition of the signum function: sgn(g(t)) = 1 whenever g(t) > 0.</p><p><strong>Step 7:</strong> The fraction never equals zero (numerator ≠ 0) and never becomes negative, so sgn(g(t)) can only take the value 1.</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C