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(A) Let $S$ be the set of all twice differentiable functions $f: R \rightarrow R$ such that $\frac{d^2 f}{dx^2} > 0$ for all $x \in (-1, 1)$.This impl

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$A, B, C$

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(A) Let $S$ be the set of all twice differentiable functions $f: R \rightarrow R$ such that $\frac{d^2 f}{dx^2} > 0$ for all $x \in (-1, 1)$.This implies that the graph of $f$ is strictly concave upward (convex) on the interval $(-1, 1)$.Let $\phi(x) = f(x) - x$. Then $\phi''(x) = f''(x) - 0 = f''(x) > 0$ for all $x \in (-1, 1)$.Since $\phi''(x) > 0$,the function $\phi(x)$ is strictly convex.$A$ strictly convex function can intersect the $x$-axis at most at two points.Therefore,the equation $\phi(x) = 0$,which is equivalent to $f(x) = x$,can have at most two solutions in the interval $(-1, 1)$.Thus,$X_f \leq 2$ for every $f \in S$,which makes statement $(B)$ true.By choosing appropriate convex functions,we can construct examples where $X_f = 0$ (e.g.,$f(x) = x^2 + 2$),$X_f = 1$ (e.g.,$f(x) = x^2 + x + 0.1$),and $X_f = 2$ (e.g.,$f(x) = x^2 + 0.1$).Since $X_f$ can be $0, 1,$ or $2$,statement $(A)$ is true,statement $(C)$ is true,and statement $(D)$ is false.Therefore,the correct statements are $(A)$,$(B)$,and $(C)$.

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