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

Correct Options:
1. There exists a function f ∈ S such that X_f = 0.
2. For every function f ∈ S, we have X_f ≤ 2.
3. There exists a function f ∈ S such that X_f = 2.

Step-by-Step Explanation:

Let $S$ be the set of all twice differentiable functions $f:ℝ\to ℝ$ such that for all $x\in (-1,1)$, the second derivative satisfies:

$\frac{d^{2}f}{d{x}^2}\left(x\right)>0$

This condition implies that $f\left(x\right)$ is strictly convex on the open interval $(-1,1)$.

We define $X_f$ as the number of points $x\in (-1,1)$ for which $f\left(x\right)=x$.

To analyze the number of solutions to $f\left(x\right)=x$, let us define an auxiliary function $g\left(x\right)$ on $(-1,1)$:

$g\left(x\right)=f\left(x\right)-x$

The roots of $g\left(x\right)=0$ in $(-1,1)$ correspond exactly to the points where $f\left(x\right)=x$. Thus, $X_f$ is the number of distinct roots of $g\left(x\right)$ in $(-1,1)$.

Taking the second derivative of $g\left(x\right)$, we get:

$g^{″}\left(x\right)=f^{″}\left(x\right)$

Since $f^{″}\left(x\right)>0$ for all $x\in (-1,1)$, it follows that $g^{″}\left(x\right)>0$ for all $x\in (-1,1)$.

1. Proving $X_f\le 2$ for every $f\in S$:
Suppose, by way of contradiction, that $g\left(x\right)$ has 3 or more distinct roots in $(-1,1)$, say $x_1<x_2<x_3$.

By Rolle's Theorem, since $g\left(x_1\right)=g\left(x_2\right)=0$, there exists at least one point $c_1\in (x_1,x_2)$ such that $g^{\prime }\left(c_1\right)=0$.
Similarly, since $g\left(x_2\right)=g\left(x_3\right)=0$, there exists at least one point $c_2\in (x_2,x_3)$ such that $g^{\prime }\left(c_2\right)=0$.

Now, applying Rolle's Theorem to $g^{\prime }\left(x\right)$ on $(c_1,c_2)$, there must exist some point $d\in (c_1,c_2)$ such that $g^{″}\left(d\right)=0$.
However, this contradicts our condition that $g^{″}\left(x\right)>0$ everywhere on $(-1,1)$.

Therefore, $g\left(x\right)$ can have at most 2 roots in $(-1,1)$, which means:

$X_f\le 2$ for every $f\in S$. (Statement 2 is true)

2. Existence of a function with $X_f=0$:
Consider the function $f\left(x\right)=x^2+2$.
- Second derivative: $f^{″}\left(x\right)=2>0$ for all $x\in (-1,1)$, so $f\in S$.
- Setting $f\left(x\right)=x$ yields $x^2-x+2=0$. The discriminant is $D={(-1)}^2-4\left(1\right)\left(2\right)=-7<0$, so there are no real solutions.
Thus, $X_f=0$. (Statement 1 is true)

3. Existence of a function with $X_f=2$:
Consider the function $f\left(x\right)=2x^2-\frac{1}{8}$.
- Second derivative: $f^{″}\left(x\right)=4>0$ for all $x\in (-1,1)$, so $f\in S$.
- Setting $f\left(x\right)=x$ yields $2x^2-x-\frac{1}{8}=0$, which simplifies to $16x^2-8x-1=0$.
- Using the quadratic formula, the solutions are:

$x=\frac{8\pm \sqrt{64-4\left(16\right)(-1)}}{32}=\frac{8\pm \sqrt{128}}{32}=\frac{1\pm \sqrt{2}}{4}$

Both points $x_1=\frac{1-\sqrt{2}}{4}\approx -0.1035$ and $x_2=\frac{1+\sqrt{2}}{4}\approx 0.6035$ lie within the interval $(-1,1)$.
Thus, $X_f=2$. (Statement 3 is true)