Number of Fixed Points for Twice Differentiable Convex Functions

To determine the correct statements, let us define an auxiliary function $g: \mathbb{R} \to \mathbb{R}$ given by: $g(x) = f(x) - x$

Since $f \in S$, $f$ is twice differentiable on $\mathbb{R}$, which implies $g$ is also twice differentiable on $\mathbb{R}$. Taking derivatives, we get: $g'(x) = f'(x) - 1$ $g''(x) = f''(x)$

We are given that $f''(x) > 0$ for all $x \in (-1,1)$. Therefore, $g''(x) > 0$ for all $x \in (-1,1)$.

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Analysis of Option (B):

Since $g''(x) > 0$ for all $x \in (-1,1)$, the derivative $g'(x)$ is strictly increasing on $(-1,1)$.

Suppose $g(x) = 0$ has $3$ or more distinct roots in $(-1,1)$, say $x_1 < x_2 < x_3$. By Rolle's Theorem:

• There exists $c_1 \in (x_1, x_2)$ such that $g'(c_1) = 0$.
• There exists $c_2 \in (x_2, x_3)$ such that $g'(c_2) = 0$.

Since $c_1 < c_2$ and $g'(c_1) = g'(c_2) = 0$, this contradicts the fact that $g'(x)$ is strictly increasing on $(-1,1)$.

Thus, $g(x) = 0$ can have at most $2$ distinct roots in $(-1,1)$. Hence, for every function $f \in S$, we have $X_f \le 2$.
Option (B) is TRUE.

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Analysis of Option (A):

Consider the function $f(x) = x^2 + 2$.

• $f$ is twice differentiable on $\mathbb{R}$ and $f''(x) = 2 > 0$ for all $x \in (-1,1)$, so $f \in S$.
• Setting $f(x) = x$, we get: $x^2 - x + 2 = 0$ The discriminant of this quadratic equation is $\Delta = (-1)^2 - 4(1)(2) = -7 < 0$, which means it has no real roots.
• Therefore, $X_f = 0$.

Option (A) is TRUE.

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Analysis of Option (C):

Consider the function $f(x) = 2x^2 - \frac{1}{8}$.

• $f$ is twice differentiable on $\mathbb{R}$ and $f''(x) = 4 > 0$ for all $x \in (-1,1)$, so $f \in S$.
• Setting $f(x) = x$, we get: $2x^2 - x - \frac{1}{8} = 0$ Solving for $x$: $x = \frac{1 \pm \sqrt{1 - 4(2)(-1/8)}}{4} = \frac{1 \pm \sqrt{2}}{4}$
• The two solutions are: $x_1 = \frac{1 - \sqrt{2}}{4} \approx -0.1035 \in (-1,1)$ $x_2 = \frac{1 + \sqrt{2}}{4} \approx 0.6035 \in (-1,1)$ Both roots lie in $(-1,1)$, which means $X_f = 2$.
  Option (C) is TRUE.

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Analysis of Option (D):

Consider the function $f(x) = x^2 + x$.

• $f$ is twice differentiable on $\mathbb{R}$ and $f''(x) = 2 > 0$ for all $x \in (-1,1)$, so $f \in S$.
• Setting $f(x) = x$, we get: $x^2 + x = x \implies x^2 = 0 \implies x = 0$
• Since $x = 0 \in (-1,1)$, $X_f = 1$.
• This shows that there does exist a function $f \in S$ such that $X_f = 1$.

Option (D) is FALSE.

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Conclusion:

The correct statements are A, B, and C.