<p>Let \(f: R \to R\) be continuous function and \(f(x) = f(2x)\) is true \(\forall\, x \in R\) and \(f(1) = 3\), then the value of \(\displaystyle\int_{-1}^{1} f(f(x))\, dx\) is equal to:</p>
Step-by-Step Solution
Key Concept: Since f(x) = f(2x) for all x ∈ ℝ, the function f is constant (applying this recursively: f(x) = f(2x) = f(4x) = f(8x)... = f(2ⁿx) → constant as n→∞). Combined with f(1) = 3, we get f(x) = 3 for all x.
<p><strong>Step 1: Analyze the functional equation</strong></p><p>Given: f(x) = f(2x) for all x ∈ ℝ</p><p>This means f(x) = f(2x) = f(2²x) = f(2³x) = ... = f(2ⁿx) for all positive integers n.</p><p><strong>Step 2: Prove f is constant</strong></p><p>For any x ∈ ℝ, as n → ∞, we have 2ⁿx → ±∞ (or 0 if x = 0). Since f is continuous everywhere and satisfies f(x) = f(2ⁿx) for all n, the function must be constant. If f varied, the sequence f(2ⁿx) couldn't maintain equality with f(x).</p><p><strong>Step 3: Determine the constant value</strong></p><p>Since f is constant and f(1) = 3, we have f(x) = 3 for all x ∈ ℝ.</p><p><strong>Step 4: Evaluate f(f(x))</strong></p><p>f(f(x)) = f(3) = 3</p><p><strong>Step 5: Compute the definite integral</strong></p><p>$$\int_{-1}^{1} f(f(x))\, dx = \int_{-1}^{1} 3\, dx = 3[x]_{-1}^{1} = 3(1-(-1)) = 3 \times 2 = 6$$</p><p>∴ Answer: <strong>6</strong></p>
Correct Answer: C