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Let f be a differentiable function defined for all real x satisfying 3*f(x + 2y) = f(2x) + f(y) + 2x + 10y + 1 for all real x, y, and f'(0) = 2*f(0). Which of the following is/are correct?

1. f(x) is an odd function
2. f^(-1)(3) = 1
3. f(0.5) = f'(5.0)
4. sin(f(x)) is a periodic function with period pi

Correct answer: f^(-1)(3) = 1

Solution

Put x=y=0: 3f(0) = f(0) + f(0) + 1 => 3f(0) = 2f(0) + 1 => f(0) = 1. From f'(0) = 2f(0) = 2. Put y = 0: 3f(x) = f(2x) + f(0) + 2x + 1 = f(2x) + 2x + 2. Put x = 0: 3f(2y) = f(0) + f(y) + 10y + 1 = f(y) + 10y + 2. So 3f(2y) = f(y) + 10y + 2... (A). Assume f(x) = ax + b. f(0) = b = 1. f'(x) = a, f'(0) = a = 2f(0) = 2. So f(x) = 2x + 1. Check in original: 3f(x+2y) = 3(2(x+2y)+1) = 3(2x+4y+1) = 6x+12y+3. f(2x)+f(y)+2x+10y+1 = (4x+1)+(2y+1)+2x+10y+1 = 6x+12y+3. Checks out! So f(x) = 2x+1. (A) f(x) is odd: f(-x) = -2x+1 ≠ -f(x) = -(2x+1). Not odd. (B) f⁻¹(3) = 1: f(1) = 3, so f⁻¹(3) = 1. Correct! (C) f(0.5) = 2(0.5)+1 = 2; f'(5.0) = 2. So f(0.5) = f'(5.0) = 2. Also correct! (D) sin(f(x)) = sin(2x+1). Period of sin(2x+1) = 2pi/2 = pi. So period = pi. Also correct! Multiple options are correct. Answer is B, C, D.

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