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(D) Given $f(x) = x^2 + g'(1)x + g''(2)$. Let $g'(1) = a$ and $g''(2) = b$. Then $f(x) = x^2 + ax + b$. $f'(x) = 2x + a$ and $f''(x) = 2$. Given $g(x)

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$x^2 - 2$

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(D) Given $f(x) = x^2 + g'(1)x + g''(2)$. Let $g'(1) = a$ and $g''(2) = b$. Then $f(x) = x^2 + ax + b$. $f'(x) = 2x + a$ and $f''(x) = 2$. Given $g(x) = f(1)x^2 + xf'(x) + f''(x)$. Substituting $f(1) = 1 + a + b$, $f'(x) = 2x + a$, and $f''(x) = 2$: $g(x) = (1 + a + b)x^2 + x(2x + a) + 2 = (1 + a + b + 2)x^2 + ax + 2 = (3 + a + b)x^2 + ax + 2$. Now, $g'(x) = 2(3 + a + b)x + a$ and $g''(x) = 2(3 + a + b)$. Using $g'(1) = a$: $2(3 + a + b) + a = a \implies 6 + 2a + 2b = 0 \implies a + b = -3$. Using $g''(2) = b$: $2(3 + a + b) = b \implies 6 + 2a + 2b = b \implies 2a + b = -6$. Solving the system: $(2a + b) - (a + b) = -6 - (-3) \implies a = -3$. Then $-3 + b = -3 \implies b = 0$. Thus, $f(x) = x^2 - 3x$ and $g(x) = (3 - 3 + 0)x^2 - 3x + 2 = -3x + 2$. $f(x) - g(x) = (x^2 - 3x) - (-3x + 2) = x^2 - 2$.

Let $f(x)$ and $g(x)$ be twice differentiable functions such that $f(x) = x^2 + g'(1)x + g''(2)$ and $g(x) = f(1)x^2 + xf'(x) + f''(x)$. Then $f(x) - g(x) =$

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