Let f : R to R be a twice differentiable func | vedclass

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(A) Given $f'(x) = f^2(x)$ and $f(0) = 0$. If $f(x)$ is not identically zero,then $\frac{f'(x)}{f^2(x)} = 1$. Integrating both sides,we get $-\frac{1}

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$2f'(2) + 4f''(2)$

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(A) Given $f'(x) = f^2(x)$ and $f(0) = 0$. If $f(x)$ is not identically zero,then $\frac{f'(x)}{f^2(x)} = 1$. Integrating both sides,we get $-\frac{1}{f(x)} = x + c$,which implies $f(x) = -\frac{1}{x+c}$. However,$f(0) = 0$ implies $-\frac{1}{c} = 0$,which is impossible. Thus,the only twice differentiable function satisfying $f(0) = 0$ and $f'(x) = f^2(x)$ is $f(x) = 0$ for all $x \in R$. If $f(x) = 0$,then $f'(x) = 0$ and $f''(x) = 0$. Therefore,$\lim_{x 	o 2} (f(x) + xf'(x) + x^2f''(x)) = 0 + 2(0) + 4(0) = 0$. Checking the options for $f(x) = 0$: $A: 2(0) + 4(0) = 0$. $B: 2(0) + 4(0) - 0 = 0$. $C: -1 + 0 + 0 = -1$. $D: 1 + 0 + 0 = 1$. Since $f(x) = 0$ is the solution,the expression evaluates to $0$. Option $A$ and $B$ both result in $0$. Given the structure,$A$ is the most simplified form.

Let $f : R \to R$ be a twice differentiable function satisfying $f(0) = f(1) = 0$ and $f'(x) = f^2(x)$ for all $x \in R$. Then $\lim_{x \to 2} (f(x) + xf'(x) + x^2f''(x))$ is equal to:

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