Differential Equations
Linear ODE
MJAT None
Grade 12
Question:
If $f(x)$ is a function satisfying $f(x+y) = f(x) + f(y) + f(x)f(y)$ and $f(x) = xg(x)$ for all $x, y \in \mathbb{R}$, and if $\lim_{x \to 0} g(x) = 1$, then which of the following statements is/are TRUE?
A) $f$ is differentiable at every $x \in \mathbb{R}$
B) If $g(0) = 1$, then $g$ is differentiable at every $x \in \mathbb{R}$
C) The derivative $f'(1)$ is equal to 1
D) The derivative $f'(0)$ is equal to 1
Step-by-Step Solution
Key Concept: Transform the additive‑multiplicative functional equation by adding 1: \(h(x)=f(x)+1\) reduces it to the exponential Cauchy equation \(h(x+y)=h(x)h(y)\), whose continuous solutions are exponentials.
1. Set \(h(x)=f(x)+1\). Then
\[
h(x+y)=f(x+y)+1=f(x)+f(y)+f(x)f(y)+1=(f(x)+1)(f(y)+1)=h(x)h(y).
\]
2. The equation \(h(x+y)=h(x)h(y)\) together with \(h(0)=1\) and continuity at \(0\) (since \(\displaystyle\lim_{x\to0}g(x)=1\) gives \(\displaystyle\lim_{x\to0}\frac{h(x)-1}{x}=1\)) forces
\[
h(x)=e^{cx}\quad\text{for some constant }c.
\]
3. Compute \(c\) from the derivative at \(0\):
\[
h'(0)=\lim_{x\to0}\frac{h(x)-1}{x}= \lim_{x\to0}\frac{f(x)}{x}= \lim_{x\to0}g(x)=1,
\]
so \(c=1\). Hence \(h(x)=e^{x}\).
4. Recover \(f\) and \(g\):
\[
f(x)=h(x)-1=e^{x}-1,\qquad
g(x)=\frac{f(x)}{x}= \frac{e^{x}-1}{x}\;(x\neq0),\; g(0)=1.
\]
All statements that follow from \(f(x)=e^{x}-1\) (e.g., \(f\) is strictly increasing, \(f'(0)=1\), \(\displaystyle\lim_{x\to0}\frac{f(x)}{x}=1\)) are true.
Correct Answer: A, B, D