Discussion

Explanation:

Case 1: x < r
⇒ f(x) = r
⇒ f(f(x)) = f(f(r)) = 2r – r = r
∴ f(x) = f(f(x))

Case 2: x = r
⇒ f(x) = 2r – r = r
⇒ f(f(x)) = f(f(r)) = 2r – r = r
∴ f(x) = f(f(x))

Case 3: x > r
⇒ f(x) = 2x – r > r
⇒ f(f(x)) = f(f(2x - r)) = 2(2x - r) – r = 4x - 3r
∴ f(x) ≠ f(f(x))

⇒ f(x) = f(f(x) when x ≤ r

Hence, option (d).

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