1. Reframe the condition using a new function

Define $g(x)=f(x)-x.$ Then:

• $g$ is twice differentiable on $\mathbb{R}$,
• and $g''(x)=f''(x)>0 \quad \text{for all } x\in(-1,1).$

So $g$ is strictly convex on $(-1,1)$.

Also, the points where $f(x)=x$ are exactly the points where $g(x)=0.$ Hence $X_f$ is the number of zeros of the strictly convex function $g$ in $(-1,1)$.

2. Key fact: a strictly convex function can have at most two zeros

Suppose, for contradiction, that $g$ has three distinct zeros $x_1<x_2<x_3, \qquad g(x_1)=g(x_2)=g(x_3)=0.$

For a strictly convex function, the graph lies strictly below the chord joining any two distinct points of the graph. Consider the chord joining $(x_1,0)$ and $(x_3,0)$. This chord is the line $y=0$.

Then for every $x\in(x_1,x_3)$, $g(x)<0.$ In particular, $g(x_2)<0,$ which contradicts $g(x_2)=0$.

Therefore, $g$ can have at most two zeros in $(-1,1)$. So for every $f\in S$, $X_f\le 2.$

Hence Option B is true.

3. Check Option A: Can $X_f=0$ happen?

Yes. Take $f(x)=x^2+1.$ Then $f''(x)=2>0$ for all $x$, so $f\in S$.

Now solve $f(x)=x \iff x^2+1=x \iff x^2-x+1=0.$ The discriminant is $\Delta = (-1)^2-4(1)(1)=1-4=-3<0.$ So there is no real solution, hence no solution in $(-1,1)$. Thus $X_f=0.$

Hence Option A is true.

4. Check Option C: Can $X_f=2$ happen?

Yes. Take $f(x)=x^2.$ Then $f''(x)=2>0$ for all $x$, so $f\in S$.

Now solve $f(x)=x \iff x^2=x \iff x(x-1)=0.$ So the solutions are $x=0,1.$ Among these, only $x=0$ lies in $(-1,1)$, so this gives only one solution, not two. So this example does not work.

Let us choose a better example: $f(x)=x^2-\frac14.$ Then $f''(x)=2>0,$ so $f\in S$.

Now solve $f(x)=x \iff x^2-\frac14=x \iff x^2-x-\frac14=0.$ Using the quadratic formula, $x=\frac{1\pm\sqrt{1+1}}{2}=\frac{1\pm\sqrt2}{2}.$ These are approximately $\frac{1-\sqrt2}{2}\approx -0.207, \qquad \frac{1+\sqrt2}{2}\approx 1.207.$ Only one lies in $(-1,1)$, so still not enough.

Take instead $f(x)=x^2-\frac12.$ Then $f''(x)=2>0,$ so $f\in S$.

Solve $x^2-\frac12=x \iff x^2-x-\frac12=0.$ So $x=\frac{1\pm\sqrt{1+2}}{2}=\frac{1\pm\sqrt3}{2}.$ These are approximately $\frac{1-\sqrt3}{2}\approx -0.366, \qquad \frac{1+\sqrt3}{2}\approx 1.366,$ again only one in $(-1,1)$.

So let us shift differently. We want $g(x)=f(x)-x$ to be a strictly convex function with two zeros in $(-1,1)$. The simplest choice is $g(x)=x^2-\frac14,$ which has zeros at $x=\pm \frac12$. Then define $f(x)=g(x)+x=x^2+x-\frac14.$ Now $f''(x)=2>0,$ so $f\in S$.

And $f(x)=x \iff x^2+x-\frac14=x \iff x^2-\frac14=0,$ so $x=\pm \frac12,$ both of which lie in $(-1,1)$. Thus $X_f=2.$

Hence Option C is true.

5. Check Option D: Is it impossible to have $X_f=1$?

We only need one counterexample.

Take $f(x)=x+x^2.$ Then $f''(x)=2>0,$ so $f\in S$.

Now solve $f(x)=x \iff x+x^2=x \iff x^2=0 \iff x=0.$ This gives exactly one solution in $(-1,1)$. So $X_f=1.$

Therefore, the statement “There does NOT exist any function $f\in S$ such that $X_f=1$” is false.

Hence Option D is false.

6. Final conclusion

The true statements are: $\boxed{A,\ B,\ C}$

This matches the stored correct answer.