<p>If \(f : A \to B\), \(f(x) = \sin^{-1}\left(\dfrac{[x]}{\{x\}}\right)\) and \(g : C \to D\), \(g(x) = \cos^{-1}\left(\dfrac{[x]}{\{x\}}\right)\), then which of the following is always <strong>correct</strong>?</p><p>[Note: \([\cdot]\) and \(\{\cdot\}\) denotes greatest integer and fractional part function respectively.]</p>
<p>(a) \(A = C\)</p>
<p>(b) \(f(x)\) and \(g(x)\) both are injective</p>
<p>(c) \(B\) and \(D\) both are singleton sets</p>
<p>(d) Number of integral solution of the equation \(f(x) + g(x) = \dfrac{\pi}{2}\) is zero.</p>
Step-by-Step Solution
Key Concept: For sin⁻¹ and cos⁻¹ to be defined, the argument must lie in [-1,1]. Since [x] is an integer and {x} ∈ [0,1), the ratio [x]/{x} can only equal specific values when both functions are simultaneously defined. The key is finding when [x]/{x} ∈ [-1,1] AND when this ratio makes sense for BOTH inverse functions together.
<p><strong>Step 1: Domain Analysis</strong></p><p>For f(x) = sin⁻¹([x]/{x}) to be defined: [x]/{x} ∈ [-1,1] and {x} ≠ 0 (so x ∉ ℤ)</p><p>For g(x) = cos⁻¹([x]/{x}) to be defined: [x]/{x} ∈ [-1,1] and {x} ≠ 0 (so x ∉ ℤ)</p><p><strong>Step 2: Finding Valid Range for Ratio</strong></p><p>Let x = n + f where n = [x] ∈ ℤ and f = {x} ∈ (0,1)</p><p>Then [x]/{x} = n/f where f ∈ (0,1)</p><p>For this ratio to be in [-1,1]:</p><p>• If n > 0: n/f ≥ n > 1 (impossible for n ≥ 1)</p><p>• If n = 0: x ∈ [0,1), so 0/f = 0 ✓</p><p>• If n = -1: -1/f ∈ [-∞,-1], only -1/f ∈ [-1,0] when f ≥ 1 (impossible)</p><p>• If n < -1: ratio < -1 (impossible)</p><p><strong>Step 3: Valid Domain</strong></p><p>Only valid case: x ∈ (0,1) where [x] = 0, so [x]/{x} = 0</p><p><strong>Step 4: Function Values</strong></p><p>For all x ∈ (0,1): f(x) = sin⁻¹(0) = 0 and g(x) = cos⁻¹(0) = π/2</p><p>Therefore: <strong>f(x) + g(x) = π/2 for all x in common domain</strong></p><p>∴ Answer: C</p>
Correct Answer: C