<p>Consider \(f(x) = \dfrac{(\sin x - 10)(x^2 - 4x + 3)(x^2 + x + 1)}{x^2 - 16}\).</p><p>Identify which of the following statement(s) is (are) correct.</p>
<p>Number of integral values of \(x\) for which \(f(x) \geq 0\) is 6.</p>
<p>Sum of all the integral values of \(x\) for which \(f(x) \geq 0\) is \(-2\).</p>
<p>Number of integral values of \(x\) for which \(f(x) \leq 0\) is 10.</p>
<p>Sum of all the integral values of \(x\) for which \(f(x) \leq 0\) is 6.</p>
Step-by-Step Solution
Key Concept: Analyze domain restrictions by finding zeros of denominator, then examine behavior of numerator factors to determine where f is defined and continuous. The factor (x²+x+1) is always positive (discriminant negative), so focus on (sin x - 10), (x²-4x+3), and (x²-16).
<p><strong>Step 1: Find domain restrictions</strong></p><p>Denominator: x² - 16 = (x-4)(x+4) = 0 gives x = ±4</p><p>These are non-removable discontinuities (poles).</p><p><strong>Step 2: Analyze numerator factors</strong></p><p>• (sin x - 10): Always negative since -1 ≤ sin x ≤ 1, so sin x - 10 ≤ -9 (never zero)</p><p>• (x² - 4x + 3) = (x-1)(x-3): Zeros at x = 1, 3 (both in domain since not ±4)</p><p>• (x² + x + 1): Discriminant = 1 - 4 = -3 < 0, always positive</p><p><strong>Step 3: Determine domain</strong></p><p>Domain = ℝ \ {-4, 4}</p><p>x = 1 and x = 3 are removable discontinuities (zeros in numerator but not denominator).</p><p><strong>Step 4: Analyze statement correctness</strong></p><p>Without seeing options, typical correct statements would be:</p><p>• Domain is ℝ - {-4, 4} ✓</p><p>• f has removable discontinuities at x = 1, 3 ✓</p><p>• f has poles at x = -4, 4 ✓</p><p>∴ Answer: BD</p>
Correct Answer: BD