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?
There exists a function $f \in S$ such that $X_f = 0$
For every function $f \in S$, we have $X_f \le 2$
There exists a function $f \in S$ such that $X_f = 2$
There does NOT exist any function $f$ in $S$ such that $X_f = 1$
Step-by-Step Solution
Key Concept: Finding the coordinates of a point using distance along a line, and evaluating the shortest distance between a point and a line.
**Step 1: Define an auxiliary function**
Let $g(x) = f(x) - x$. Since $f$ is twice differentiable, $g$ is twice differentiable. We are given $f''(x) > 0$ for $x \in (-1, 1)$, which implies $g''(x) > 0$. Thus, $g(x)$ is strictly convex on $(-1, 1)$. We are looking for the number of roots of $g(x) = 0$ in $(-1, 1)$, which is $X_f$.
**Step 2: Apply properties of convex functions**
A strictly convex function can intersect a straight line (in this case, the x-axis, $y=0$) at most 2 times. If it intersected 3 times, by Rolle's Theorem, $g'(x)$ would have at least 2 roots, meaning $g''(x)$ would have at least 1 root where it equals 0, contradicting $g''(x) > 0$. Thus, $X_f \le 2$ for all $f \in S$, making (B) true.
**Step 3: Construct examples to test statements**
To check (A), let $f(x) = x^2 + 2$. Then $f''(x) = 2 > 0$. $g(x) = x^2 - x + 2$. The discriminant is $1 - 8 = -7 < 0$, so there are no real roots. Thus $X_f = 0$ is possible. (A) is true.\nTo check (C), let $f(x) = x^2 + x - 1/4$. $g(x) = x^2 - 1/4 = (x-1/2)(x+1/2)$. Both roots are in $(-1, 1)$. Thus $X_f = 2$ is possible. (C) is true.\nTo check (D), let $f(x) = x^2$. Then $f''(x) = 2 > 0$. $g(x) = x^2 - x = x(x-1)$. The roots are $x=0$ and $x=1$. Since $1 \notin (-1, 1)$, $X_f = 1$ here. So such a function DOES exist, making (D) false.
Correct Answer: 1, 2, 3