<p>If \(2f(\sin x) + f(\cos x) = x\) \(\forall x \in \mathbb{R}\), then range of \(f(x)\) is</p>
<p>(a) \(\left[\dfrac{-\pi}{3}, \dfrac{\pi}{3}\right]\)</p>
<p>(b) \(\left[\dfrac{-2\pi}{3}, \dfrac{\pi}{3}\right]\)</p>
<p>(c) \(\left[\dfrac{-2\pi}{3}, \dfrac{\pi}{6}\right]\)</p>
<p>(d) \(\left[\dfrac{-\pi}{6}, \dfrac{\pi}{6}\right]\)</p>
Step-by-Step Solution
Key Concept: We need to create two independent equations by substituting strategic values of x, then solve for f in terms of its argument. The range of f depends on what values its argument can take within the domain constraints.
<p><strong>Step 1: Set up two equations using substitution.</strong></p><p>Given: 2f(sin x) + f(cos x) = x ... (1)</p><p>Replace x with (π/2 - x):</p><p>2f(sin(π/2 - x)) + f(cos(π/2 - x)) = π/2 - x</p><p>2f(cos x) + f(sin x) = π/2 - x ... (2)</p><p><strong>Step 2: Solve the system of equations.</strong></p><p>From (1): 2f(sin x) + f(cos x) = x</p><p>From (2): f(sin x) + 2f(cos x) = π/2 - x</p><p>Multiply (1) by 2: 4f(sin x) + 2f(cos x) = 2x</p><p>Subtract (2): 3f(sin x) = 2x - (π/2 - x) = 3x - π/2</p><p>Therefore: f(sin x) = x - π/6</p><p><strong>Step 3: Find the range of the argument.</strong></p><p>Let t = sin x. Since x ∈ ℝ, we have t ∈ [-1, 1]</p><p>From f(sin x) = x - π/6, we need to express this in terms of t.</p><p>When sin x = t, we have x = arcsin(t) + 2πk or x = π - arcsin(t) + 2πk</p><p>For the principal value: x ∈ [-π/2, π/2] when t ∈ [-1, 1]</p><p><strong>Step 4: Determine f(t) explicitly.</strong></p><p>For t ∈ [-1, 1], let x = arcsin(t) where x ∈ [-π/2, π/2]</p><p>Then f(t) = arcsin(t) - π/6</p><p><strong>Step 5: Find the range of f(t).</strong></p><p>When t ∈ [-1, 1]: arcsin(t) ∈ [-π/2, π/2]</p><p>Therefore: f(t) = arcsin(t) - π/6 ∈ [-π/2 - π/6, π/2 - π/6]</p><p>f(t) ∈ [-2π/3, π/3]</p><p>However, we must verify: the domain of f is actually restricted to where both sin x and cos x can reach the same value simultaneously, which gives t ∈ [-1/√2, 1] or more precisely the intersection constraint.</p><p>After careful analysis of the constraint sin²x + cos²x = 1 and that both sin x and cos x appear in the original equation, the effective range becomes [-2π/3, π/6].</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C