Question

$f(x)$ is a twice differentiable function such that

$f(x)+f''(x) = -xg(x)f'(x), \quad g(x) \ge 0 \quad \forall \quad x \ge 0$ then:

• A. If $f(x)$ is an increasing function for $x \ge 0$, then $f''(1) \le -f(1)$
• B. $|f(x)| \le \sqrt{f(0)^2 + f'(0)^2}$ for $x \ge 0$
• C. $f'(a)(f(1) + f(0)) + f''(b) (f'(0) + f'(1)) > 0$ for some $a, b \in (0,1)$

Answer 1

Answer: (A), (B)

(1), (2)

Answer 2

The problem asks us to determine which of the given statements are true for a twice differentiable function $f(x)$ satisfying the differential equation $f(x)+f''(x) = -xg(x)f'(x)$, where $g(x) \ge 0$ for all $x \ge 0$.

Let's analyze each statement:

Statement (1): If $f(x)$ is an increasing function for $x \ge 0$, then $f''(1) \le -f(1)$.

Given that $f(x)$ is an increasing function for $x \ge 0$, it implies that $f'(x) \ge 0$ for all $x \ge 0$. The given differential equation is $f(x)+f''(x) = -xg(x)f'(x)$. We are given $x \ge 0$ and $g(x) \ge 0$. Since $f'(x) \ge 0$, the product $xg(x)f'(x) \ge 0$. Therefore, $-xg(x)f'(x) \le 0$. Substituting this into the differential equation, we get: $f(x)+f''(x) \le 0$ for all $x \ge 0$. This can be rewritten as $f''(x) \le -f(x)$ for all $x \ge 0$. Now, substitute $x=1$ into this inequality: $f''(1) \le -f(1)$. Thus, statement (1) is TRUE.

Statement (2): $|f(x)| \le \sqrt{f(0)^2 + f'(0)^2}$ for $x \ge 0$.

To prove this, let's consider the function $h(x) = f(x)^2 + f'(x)^2$. We want to show that $h(x) \le h(0)$ for $x \ge 0$. Let's find the derivative of $h(x)$ with respect to $x$: $h'(x) = \frac{d}{dx}(f(x)^2 + f'(x)^2)$ $h'(x) = 2f(x)f'(x) + 2f'(x)f''(x)$ Factor out $2f'(x)$: $h'(x) = 2f'(x)(f(x) + f''(x))$ From the given differential equation, we know $f(x)+f''(x) = -xg(x)f'(x)$. Substitute this into the expression for $h'(x)$: $h'(x) = 2f'(x)(-xg(x)f'(x))$ $h'(x) = -2xg(x)(f'(x))^2$ We are given that $x \ge 0$ and $g(x) \ge 0$. Also, $(f'(x))^2 \ge 0$ since it's a square. Therefore, the product $-2xg(x)(f'(x))^2$ must be less than or equal to zero. So, $h'(x) \le 0$ for all $x \ge 0$. Since $h'(x) \le 0$, the function $h(x)$ is a decreasing function for $x \ge 0$. This means that for any $x \ge 0$, $h(x) \le h(0)$. Substituting back the definition of $h(x)$: $f(x)^2 + f'(x)^2 \le f(0)^2 + f'(0)^2$ for $x \ge 0$. Since $f(x)^2 \le f(x)^2 + f'(x)^2$, it follows that: $f(x)^2 \le f(0)^2 + f'(0)^2$. Taking the square root of both sides (and remembering that $\sqrt{y^2} = |y|$): $|f(x)| \le \sqrt{f(0)^2 + f'(0)^2}$ for $x \ge 0$. Thus, statement (2) is TRUE.

Statement (3): $f'(a)(f(1) + f(0)) + f''(b) (f'(0) + f'(1)) > 0$ for some $a, b \in (0,1)$.

Since $f(x)$ is twice differentiable, both $f(x)$ and $f'(x)$ are continuous on $[0,1]$ and differentiable on $(0,1)$. By the Mean Value Theorem (MVT) for $f(x)$ on $[0,1]$, there exists some $a \in (0,1)$ such that: $f'(a) = \frac{f(1) - f(0)}{1 - 0} = f(1) - f(0)$. By the Mean Value Theorem (MVT) for $f'(x)$ on $[0,1]$, there exists some $b \in (0,1)$ such that: $f''(b) = \frac{f'(1) - f'(0)}{1 - 0} = f'(1) - f'(0)$. Now, substitute these expressions for $f'(a)$ and $f''(b)$ into the given statement's expression: $f'(a)(f(1) + f(0)) + f''(b) (f'(0) + f'(1))$ $= (f(1) - f(0))(f(1) + f(0)) + (f'(1) - f'(0))(f'(0) + f'(1))$ Using the difference of squares formula, $(A-B)(A+B) = A^2-B^2$: $= (f(1)^2 - f(0)^2) + (f'(1)^2 - f'(0)^2)$ Rearrange the terms: $= (f(1)^2 + f'(1)^2) - (f(0)^2 + f'(0)^2)$ Recall the function $h(x) = f(x)^2 + f'(x)^2$ from statement (2). So, the expression simplifies to $h(1) - h(0)$. From the analysis of statement (2), we established that $h'(x) \le 0$ for $x \ge 0$, which means $h(x)$ is a decreasing function. Therefore, for $x=1$ and $x=0$, we must have $h(1) \le h(0)$. This implies $h(1) - h(0) \le 0$. So, the expression $f'(a)(f(1) + f(0)) + f''(b) (f'(0) + f'(1))$ is less than or equal to zero. The statement claims that this expression is strictly greater than zero ($>0$). This contradicts our finding that it is $\le 0$. Thus, statement (3) is FALSE.

Based on the analysis, statements (1) and (2) are true, while statement (3) is false.

The final answer is $\boxed{(1), (2)}$.

Explanation of the solution:

1. For statement (1): If $f(x)$ is increasing, $f'(x) \ge 0$. Given $x \ge 0$ and $g(x) \ge 0$, the term $-xg(x)f'(x)$ in the differential equation $f(x)+f''(x) = -xg(x)f'(x)$ must be $\le 0$. This implies $f(x)+f''(x) \le 0$, or $f''(x) \le -f(x)$. Setting $x=1$ gives $f''(1) \le -f(1)$.
2. For statement (2): Define $h(x) = f(x)^2 + f'(x)^2$. Differentiate $h(x)$ to get $h'(x) = 2f'(x)(f(x)+f''(x))$. Substitute the given differential equation: $h'(x) = 2f'(x)(-xg(x)f'(x)) = -2xg(x)(f'(x))^2$. Since $x \ge 0$, $g(x) \ge 0$, and $(f'(x))^2 \ge 0$, it follows that $h'(x) \le 0$. This means $h(x)$ is a decreasing function. Therefore, $h(x) \le h(0)$ for $x \ge 0$, which translates to $f(x)^2 + f'(x)^2 \le f(0)^2 + f'(0)^2$. Since $f(x)^2 \le f(x)^2 + f'(x)^2$, we have $f(x)^2 \le f(0)^2 + f'(0)^2$. Taking the square root gives $|f(x)| \le \sqrt{f(0)^2 + f'(0)^2}$.
3. For statement (3): Apply the Mean Value Theorem. There exist $a \in (0,1)$ such that $f'(a) = f(1)-f(0)$, and $b \in (0,1)$ such that $f''(b) = f'(1)-f'(0)$. Substitute these into the expression: $f'(a)(f(1) + f(0)) + f''(b) (f'(0) + f'(1)) = (f(1)-f(0))(f(1)+f(0)) + (f'(1)-f'(0))(f'(0)+f'(1))$. This simplifies to $(f(1)^2 - f(0)^2) + (f'(1)^2 - f'(0)^2) = (f(1)^2+f'(1)^2) - (f(0)^2+f'(0)^2)$, which is $h(1)-h(0)$. Since $h(x)$ is a decreasing function, $h(1) \le h(0)$, implying $h(1)-h(0) \le 0$. Thus, the expression is $\le 0$, contradicting the statement that it is $>0$.