Relations & Functions
Fractional part functions and range
Grade 12
Question:
<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>[Note: \(\{k\}\) 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} lies in [0,1), so {-x} = 1 - {x} for non-integer x. Analyze f(x) = 1/2^(1-{x}) - {x} by substituting t = {x} ∈ [0,1) and find the minimum value of g(t) = 2^(t-1) - t.
<p><strong>Step 1:</strong> For any real x, write x = ⌊x⌋ + {x}, where {x} ∈ [0,1). For non-integer x: {-x} = 1 - {x}.</p><p><strong>Step 2:</strong> Substitute t = {x} ∈ [0,1). Then f(x) = 2^(t-1) - t = g(t).</p><p><strong>Step 3:</strong> Find critical points: g'(t) = 2^(t-1)·ln(2) - 1 = 0 ⟹ 2^(t-1) = 1/ln(2) ≈ 1.443. Since 2^(t-1) ranges from 2^(-1) = 0.5 to 2^0 = 1 as t goes from 0 to 1, and 1/ln(2) > 1, there's no critical point in [0,1).</p><p><strong>Step 4:</strong> Since g'(t) = 2^(t-1)·ln(2) - 1 < 0 for all t ∈ [0,1) (as 2^(t-1)·ln(2) < ln(2) ≈ 0.693 < 1), g(t) is strictly decreasing.</p><p><strong>Step 5:</strong> Minimum occurs as t → 1⁻: g(1⁻) = 2^(0) - 1 = 0. Maximum at t = 0: g(0) = 2^(-1) - 0 = 1/2.</p><p><strong>Step 6:</strong> Therefore range is [0, 1/2), so a = 0.</p><p>∴ Answer: C (a = 0)</p>
Correct Answer: C