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Given that f(x+y)=f(x)+2y²+kxy, and that f(a)=2 and f(b)=8, the function f(x) must be of the form

1. 2x²
2. 2x²+1
3. 2x²−1
4. x²

Correct answer: 2x²

Solution

Put x=0: f(y)=f(0)+2y^2, so f(x)=2x^2+c, and matching the cross term gives k=4. The leading coefficient is therefore 2, so f(x)=x^2 cannot be correct; the standard form is f(x)=2x^2.

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