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A differentiable function y = f(x) satisfies f(x) + f(y) = f(x)*f(y) + f(xy) for all real x, y, where f(1) = 0 and f'(1) = -2. Find the number of solutions of the equation f(x) = 2^x.

1. 1
2. 2
3. 3
4. 4

Correct answer: 1

Solution

From the functional equation with y=1: f(x)+f(1)=f(x)*f(1)+f(x*1) => f(x)+0=0+f(x). Consistent but no new info. Differentiate the functional equation w.r.t. y: f'(y) = f(x)*f'(y) + x*f'(xy). At y=1: f'(1)=f(x)*f'(1)+x*f'(x) => -2 = -2*f(x)+x*f'(x) => x*f'(x)-2f(x) = -2. This is a linear ODE: f'(x) - (2/x)*f(x) = -2/x. Integrating factor: e^(-integral(2/x)dx) = 1/x². Multiply: (f/x²)' = -2/x³. Integrate: f/x² = 1/x² + C => f(x) = 1 + Cx². Apply f(1)=0: 0=1+C => C=-1. So f(x) = 1-x². Check f'(1): f'(x)=-2x => f'(1)=-2. Consistent! Equation f(x)=2^x => 1-x²=2^x. At x=0: 1=1 yes. At x=-1: 0=1/2 no. At x=1: 0=2 no. For x<-1: 1-x²<0, 2^x>0: no intersection. For large x: 1-x² -> -inf, 2^x -> +inf: no intersection. The function 1-x² and 2^x intersect only at x=0. Number of solutions = 1.

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