Functional Equation Condition

CAT 2022 Slot 3 · QA · Medium · Algebra

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Let rr be a real number and f(x)={2x−rif x≥rrif x<rf(x) = \begin{cases} 2x - r & \text{if } x \ge r \\ r & \text{if } x < r \end{cases}. Then, the equation f(x)=f(f(x))f(x) = f(f(x)) holds for all real values of xx where

  1. A.

    x≤rx \le r

  2. B.

    x≥rx \ge r

  3. C.

    x>rx > r

  4. D.

    x≠rx \neq r

Answer

A

Explanation

Case 1: x<rx < r f(x)=rf(x) = r. Then f(f(x))=f(r)=2(r)−r=rf(f(x)) = f(r) = 2(r) - r = r. So f(x)=f(f(x))f(x) = f(f(x)) holds for all x<rx < r.

Case 2: x=rx = r f(r)=2r−r=rf(r) = 2r - r = r. f(f(r))=f(r)=rf(f(r)) = f(r) = r. So f(x)=f(f(x))f(x) = f(f(x)) holds for x=rx = r.

Case 3: x>rx > r f(x)=2x−r>rf(x) = 2x - r > r. Then f(f(x))=f(2x−r)=2(2x−r)−r=4x−3rf(f(x)) = f(2x - r) = 2(2x - r) - r = 4x - 3r. For f(x)=f(f(x))f(x) = f(f(x)), we need 2x−r=4x−3r  ⟹  2x=2r  ⟹  x=r2x - r = 4x - 3r \implies 2x = 2r \implies x = r, which contradicts x>rx > r.

Therefore, f(x)=f(f(x))f(x) = f(f(x)) holds if and only if x≤rx \le r.

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