<p>The value of \(a\) for which the function \(f(x) = (4a-3)(x + \log 5) + 2(a-7)\cot\dfrac{x}{2}\sin^2\dfrac{x}{2}\) does not possess critical points is</p>
Step-by-Step Solution
Key Concept: A function has no critical points when its derivative is never zero. Simplify f'(x) first, then find the condition on 'a' such that f'(x) ≠ 0 for all x in the domain.
<p><strong>Step 1: Find the derivative of f(x).</strong></p><p>f(x) = (4a-3)(x + log 5) + 2(a-7)cot(x/2)sin²(x/2)</p><p>f'(x) = (4a-3) + 2(a-7)·d/dx[cot(x/2)sin²(x/2)]</p><p><strong>Step 2: Simplify the second term.</strong></p><p>Note that cot(x/2)sin²(x/2) = (cos(x/2)/sin(x/2))·sin²(x/2) = cos(x/2)sin(x/2) = ½sin(x)</p><p>Therefore: d/dx[cot(x/2)sin²(x/2)] = ½·cos(x)</p><p><strong>Step 3: Express f'(x) completely.</strong></p><p>f'(x) = (4a-3) + 2(a-7)·½cos(x)</p><p>f'(x) = (4a-3) + (a-7)cos(x)</p><p><strong>Step 4: Condition for no critical points.</strong></p><p>For f(x) to have no critical points, f'(x) ≠ 0 for all x.</p><p>This means: (4a-3) + (a-7)cos(x) ≠ 0 for all x ∈ ℝ</p><p>Since cos(x) ∈ [-1, 1], the range of (a-7)cos(x) is [(a-7)(-1), (a-7)(1)] when a > 7, or [(a-7)(1), (a-7)(-1)] when a < 7.</p><p><strong>Step 5: Analyze the range condition.</strong></p><p>For f'(x) to never equal zero:</p><p>• Minimum of f'(x) = (4a-3) + (a-7)(-1) = (4a-3) - (a-7) = 3a + 4</p><p>• Maximum of f'(x) = (4a-3) + (a-7)(1) = 5a - 10</p><p>We need either: (3a+4) > 0 or (5a-10) < 0</p><p>From (3a+4) > 0: a > -4/3</p><p>From (5a-10) < 0: a < 2</p><p>For NO critical points, we need the derivative to maintain constant sign, so BOTH conditions cannot be true simultaneously.</p><p>We need: a < -4/3 OR a > 2</p><p>However, checking a > 2: both 3a+4 > 0 and 5a-10 > 0, so f'(x) > 0 always (no critical points). ✓</p><p><strong>∴ Answer:</strong> D</p>
Correct Answer: D