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ExamsJEE AdvancedMaths

A function f, defined for all real x and y, satisfies f(1) = 2 and f(x + y) - k*x*y = f(x) + 2*y² for some constant k. Find an explicit formula for f(x).

  1. f(x) = 2*x² (with k = 4)
  2. f(x) = x² + 1 (with k = 2)
  3. f(x) = 2*x² + x (with k = 4)
  4. f(x) = 2*x (with k = 0)

Correct answer: f(x) = 2*x² (with k = 4)

Solution

Put x = 1: f(1 + y) - k*y = f(1) + 2*y² = 2 + 2*y², so f(1 + y) = 2 + k*y + 2*y². Let z = 1 + y, i.e. y = z - 1: f(z) = 2 + k(z - 1) + 2(z - 1)² = 2 + k*z - k + 2*z² - 4*z + 2 = 2*z² + (k - 4)*z + (4 - k). For this to be consistent with the original relation for all x, y (a pure quadratic with no linear/constant term forced by symmetry), substituting back shows k = 4, giving f(z) = 2*z². Check: f(1) = 2. So f(x) = 2*x². Then f(x+y)*f(1/(x+y)) = 2(x+y)² * 2/(x+y)² = 4 = k.

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