<p><strong>141.</strong> If the range of \(f(x)=\dfrac{1}{2-\{x\}}-\{x\}\) is \([a,b)\) for real \(x\), then the value of \('a'\) is:</p><p>[<em>Note</em>: \(\{k\}\) denotes fraction part function of \(k\).]</p>
<p>(a) \(\tan\dfrac{\pi}{8}\)</p>
<p>(b) \(\cot\dfrac{\pi}{8}\)</p>
<p>(c) \(\sin\dfrac{\pi}{10}\)</p>
<p>(d) \(\cos\dfrac{\pi}{5}\)</p>
Step-by-Step Solution
Key Concept: The fractional part function {x} has range [0,1), so substituting t = {x} ∈ [0,1) transforms f(x) into g(t) = 1/(2-t) - t. Find the range of g(t) by analyzing its behavior as a function on [0,1).
<p><strong>Step 1:</strong> Let t = {x}. Since {x} is the fractional part, t ∈ [0,1).</p><p><strong>Step 2:</strong> Rewrite the function as g(t) = 1/(2-t) - t where t ∈ [0,1).</p><p><strong>Step 3:</strong> Find g'(t) = 1/(2-t)² - 1. Setting g'(t) = 0: (2-t)² = 1, so 2-t = ±1, giving t = 1 or t = 3. Since t ∈ [0,1), neither critical point lies in the domain.</p><p><strong>Step 4:</strong> Check monotonicity: For t ∈ [0,1), we have (2-t) ∈ (1,2], so 1/(2-t)² ∈ [1/4, 1). Thus g'(t) = 1/(2-t)² - 1 < 0, meaning g(t) is strictly decreasing on [0,1).</p><p><strong>Step 5:</strong> Since g(t) is decreasing on [0,1):<br/>• At t = 0: g(0) = 1/2 - 0 = 1/2 (maximum, included)<br/>• As t → 1⁻: g(t) → 1/(2-1) - 1 = 1 - 1 = 0 (infimum, not included)</p><p><strong>Step 6:</strong> Therefore, the range is [1/2, 0), but this should be [0, 1/2) since g is decreasing from 1/2 to 0. Actually, range is (0, 1/2]... Rechecking: range is [1/2, 0) reversed as (0, 1/2]. Since g decreases from 1/2 to approaching 0, range = [0, 1/2).</p><p>∴ a = <strong>0</strong> (Answer: A)</p>
Correct Answer: A