Let $S$ be the set of all twice differentiable functions $f$ from $\mathbb{R}$ to $\mathbb{R}$ such that $\frac{d^{2}f}{dx^{2}}(x)>0$ for all $x\in(-1,1)$. For $f\in S$, let $X_{f}$ be the number of points $x\in(-1,1)$ for which $f(x)=x$. Then which of the following statements is(are) true?

$S=\{f:\ f''(x)>0\ \forall x\in(-1,1)\}$ $f \text{ is strictly convex on } (-1,1)$ $X_f=\#\{x\in(-1,1): f(x)=x\}$ $\text{(1) } \exists f:\ X_f=0 \qquad \text{(2) } \forall f:\ X_f\le 2$ $\text{(3) } \exists f:\ X_f=2 \qquad \text{(4) } \nexists f:\ X_f=1$ $1.\ \text{set } g(x)=f(x)-x$ $2.\ X_f = \#\{\text{zeros of } g \text{ in } (-1,1)\}$ $3.\ \text{check each statement with examples}$ $g(x)=f(x)-x, \quad f(x)=x \iff g(x)=0$ $g''(x)=f''(x)>0 \Rightarrow g \text{ strictly convex on } (-1,1)$ $\text{Claim: a strictly convex } g \text{ has} \le 2 \text{ zeros}$ $a<b<c,\quad g(a)=g(b)=g(c)=0$ $\text{convexity} \Rightarrow g(b)<0, \text{ a contradiction}$ $X_f\le 2 \text{ for every } f\in S \Rightarrow \text{(2) TRUE}$ $g(x)=x^2+1>0 \Rightarrow X_f=0$ $f(x)=x+x^2+1,\ f''=2>0 \Rightarrow \text{(1) TRUE}$ $g(x)=x^2-\dfrac14=0 \Rightarrow x=\pm\dfrac12$ $f(x)=x+x^2-\dfrac14 \Rightarrow X_f=2 \Rightarrow \text{(3) TRUE}$ $g(x)=x^2 \Rightarrow g(x)=0 \text{ only at } x=0$ $f(x)=x+x^2 \Rightarrow X_f=1 \Rightarrow \text{(4) FALSE}$ $\text{(1) TRUE},\quad \text{(2) TRUE},\quad \text{(3) TRUE},\quad \text{(4) FALSE}$ $\boxed{\text{Answer: (1), (2), (3)}}$