Question

Let $f(x): R \rightarrow [-1, 1]$ be twice differentiable and $f^2(0) + (f'(0))^2 = 4$, then which of the following(s) is/are TRUE?

• A. There exist $\alpha, \beta \in R$ where $\alpha < \beta$, such that $f$ is one-one on the interval $(\alpha, \beta)$
• B. There exists $c \in (0, 2)$ such that $|f'(c)| \leq 1$
• C. $\lim_{x \rightarrow \infty} f(x) = 1$
• D. There exists some $x_0$ such that $f(x_0) + f''(x_0) = 0$ but $f'(x_0) \neq 0$

Answer 1

Answer: (A), (B), (D)

(1), (2), (4)

Answer 2

The function $f(x): R \rightarrow [-1, 1]$ is twice differentiable, and $f^2(0) + (f'(0))^2 = 4$. Since the range of $f(x)$ is $[-1, 1]$, we have $|f(x)| \leq 1$ for all $x \in R$. In particular, $|f(0)| \leq 1$, so $f^2(0) \leq 1$. From the given condition, $(f'(0))^2 = 4 - f^2(0)$. Since $f^2(0) \leq 1$, $(f'(0))^2 \geq 4 - 1 = 3$. Thus, $|f'(0)| \geq \sqrt{3}$.

(1) There exist $\alpha, \beta \in R$ where $\alpha < \beta$, such that $f$ is one-one on the interval $(\alpha, \beta)$.

A function is one-one on an interval if it is strictly monotonic on that interval. This means its derivative must be strictly positive or strictly negative on that interval. We know $|f'(0)| \geq \sqrt{3}$, so $f'(0) \geq \sqrt{3}$ or $f'(0) \leq -\sqrt{3}$. Since $f$ is twice differentiable, $f'$ is continuous. If $f'(0) > 0$, by the continuity of $f'$, there exists an interval $(-\delta, \delta)$ around 0 for some $\delta > 0$ such that $f'(x) > 0$ for all $x \in (-\delta, \delta)$. On this interval, $f(x)$ is strictly increasing and hence one-one. If $f'(0) < 0$, similarly, there exists $(-\delta, \delta)$ such that $f'(x) < 0$ for all $x \in (-\delta, \delta)$. On this interval, $f(x)$ is strictly decreasing and hence one-one. In either case, such an interval $(\alpha, \beta)$ containing 0 exists. So, option (1) is TRUE.

(2) There exists $c \in (0, 2)$ such that $|f'(c)| \leq 1$.

Consider the interval $[0, 2]$. Since $f$ is differentiable on $R$, it is continuous on $[0, 2]$ and differentiable on $(0, 2)$. By the Mean Value Theorem, there exists $c \in (0, 2)$ such that $f'(c) = \frac{f(2) - f(0)}{2 - 0} = \frac{f(2) - f(0)}{2}$. Since the range of $f(x)$ is $[-1, 1]$, we have $-1 \leq f(2) \leq 1$ and $-1 \leq f(0) \leq 1$. Therefore, $-1 - 1 \leq f(2) - f(0) \leq 1 - (-1)$, which means $-2 \leq f(2) - f(0) \leq 2$. Dividing by 2, we get $-1 \leq \frac{f(2) - f(0)}{2} \leq 1$. So, $-1 \leq f'(c) \leq 1$, which implies $|f'(c)| \leq 1$. This holds for some $c \in (0, 2)$. So, option (2) is TRUE.

(3) $\lim_{x \rightarrow \infty} f(x) = 1$.

The function $f(x) = \sin(2x)$ has range $[-1, 1]$. $f(0) = \sin(0) = 0$. $f'(x) = 2 \cos(2x)$. $f'(0) = 2 \cos(0) = 2$. $f^2(0) + (f'(0))^2 = 0^2 + 2^2 = 4$. This function $f(x) = \sin(2x)$ satisfies the given conditions. However, $\lim_{x \rightarrow \infty} \sin(2x)$ does not exist. Therefore, option (3) is FALSE.

(4) There exists some $x_0$ such that $f(x_0) + f''(x_0) = 0$ but $f'(x_0) \neq 0$.

Consider the function $g(x) = f^2(x) + (f'(x))^2$. Since $|f(x)| \leq 1$ for all $x$, $f^2(x) \leq 1$. Also, we know that $f$ is bounded, so its derivative $f'$ must also be bounded. If $f'$ were unbounded, say $f'(x) > M$ on an interval of length $L$, then $f$ would increase by more than $ML$. If $ML > 2$, this would contradict $|f(x)| \leq 1$. A more rigorous argument using MVT shows that $|f'(x)|$ must be bounded. Let $M = \sup_{x \in R} |f'(x)|$. Then $g(x) = f^2(x) + (f'(x))^2 \leq 1 + M^2$. So $g(x)$ is bounded. Since $f$ is twice differentiable, $g(x)$ is differentiable. $g'(x) = 2f(x)f'(x) + 2f'(x)f''(x) = 2f'(x) [f(x) + f''(x)]$. Since $g(x)$ is differentiable and bounded, its derivative $g'(x)$ cannot be strictly positive or strictly negative everywhere. If $g'(x) > \epsilon > 0$ for all $x > a$, then $g(x) > g(a) + \epsilon(x-a)$ for $x>a$, which implies $g(x) \rightarrow \infty$ as $x \rightarrow \infty$, contradicting the boundedness of $g(x)$. Similarly, $g'(x)$ cannot be strictly negative everywhere. Thus, there must exist some $x_0 \in R$ such that $g'(x_0) = 0$. $g'(x_0) = 2f'(x_0) [f(x_0) + f''(x_0)] = 0$. This implies that either $f'(x_0) = 0$ or $f(x_0) + f''(x_0) = 0$. So, for any point $x_0$ where $g'(x_0)=0$, either $f'(x_0)=0$ or $f(x_0) + f''(x_0)=0$. The question asks if there exists $x_0$ such that $f(x_0) + f''(x_0) = 0$ but $f'(x_0) \neq 0$. This is equivalent to asking if there exists a point $x_0$ such that $g'(x_0)=0$ and $f'(x_0) \neq 0$. Since $g'(x_0)=0$ for some $x_0$, and the condition $g'(x_0)=0$ implies ($f'(x_0)=0$ or $f(x_0) + f''(x_0)=0$), the only way the statement "there exists $x_0$ such that $f(x_0) + f''(x_0) = 0$ but $f'(x_0) \neq 0$" is FALSE is if for every $x_0$ such that $g'(x_0)=0$, we must have $f'(x_0)=0$. In this case, $f(x_0) + f''(x_0) = 0$ would also be true. So, if the statement in (4) is false, it means that the set $\{x \mid g'(x)=0\}$ is a subset of $\{x \mid f'(x)=0\}$. Let $S = \{x \mid g'(x)=0\}$. We know $S$ is non-empty. If $x_0 \in S$, then $f'(x_0)=0$. Consider the function $f(x) = \sin(2x)$. We found this function satisfies the initial condition. $f(x) = \sin(2x)$. $f'(x) = 2 \cos(2x)$. $f''(x) = -4 \sin(2x)$. We want to check if there exists $x_0$ such that $f(x_0) + f''(x_0) = 0$ and $f'(x_0) \neq 0$. $f(x_0) + f''(x_0) = \sin(2x_0) - 4 \sin(2x_0) = -3 \sin(2x_0)$. Setting $f(x_0) + f''(x_0) = 0$, we get $-3 \sin(2x_0) = 0$, so $\sin(2x_0) = 0$. This happens when $2x_0 = n\pi$ for some integer $n$, i.e., $x_0 = \frac{n\pi}{2}$. Now check $f'(x_0)$ at these points: $f'(x_0) = 2 \cos(2x_0) = 2 \cos(n\pi) = 2 (-1)^n$. $f'(x_0) = 2$ if $n$ is even, and $f'(x_0) = -2$ if $n$ is odd. In either case, $f'(x_0) \neq 0$ for $x_0 = \frac{n\pi}{2}$. So, for the function $f(x) = \sin(2x)$, there exist points $x_0 = \frac{n\pi}{2}$ where $f(x_0) + f''(x_0) = 0$ and $f'(x_0) \neq 0$. For example, at $x_0 = \frac{\pi}{2}$, $f(\frac{\pi}{2}) = \sin(\pi) = 0$, $f''(\frac{\pi}{2}) = -4 \sin(\pi) = 0$, so $f(\frac{\pi}{2}) + f''(\frac{\pi}{2}) = 0$. $f'(\frac{\pi}{2}) = 2 \cos(\pi) = -2 \neq 0$. So, option (4) is TRUE.