Question

Let f(x) be a continuous and differentiable function satisfying f(x + y) = f(x)f(y) ∀ x, y ∈ R if f(x) can be expressed as f(x) = 1 + x P(x) + x²Q(x) where $\lim_{x\to 0} P(x) = a$ and $\lim_{x\to 0} Q(x) = b$, then

f'(x) is equal to :

• A. a f(x)
• B. b f(x)
• C. (a + b) f(x)
• D. (a+2b) f(x)

Answer 1

Answer: (A)

a f(x)

Answer 2

The given functional equation is $f(x + y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$. A continuous function satisfying this equation must be of the form $f(x) = c^x$ for some constant $c$. Since $f(x)$ is also differentiable, this form is valid.

Let's evaluate $f(0)$ using the given expression $f(x) = 1 + x P(x) + x^2 Q(x)$. Since $f$ is continuous, $\lim_{x\to 0} f(x) = f(0)$. $\lim_{x\to 0} f(x) = \lim_{x\to 0} (1 + x P(x) + x^2 Q(x))$ $= 1 + \lim_{x\to 0} (x P(x)) + \lim_{x\to 0} (x^2 Q(x))$ $= 1 + (\lim_{x\to 0} x) (\lim_{x\to 0} P(x)) + (\lim_{x\to 0} x^2) (\lim_{x\to 0} Q(x))$ $= 1 + 0 \cdot a + 0 \cdot b = 1$. So, $f(0) = 1$. This is consistent with $f(x) = c^x$, since $f(0) = c^0 = 1$.

Now, let's find the derivative $f'(x)$ using the definition: $f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$ Using the functional equation $f(x+h) = f(x)f(h)$, we have: $f'(x) = \lim_{h\to 0} \frac{f(x)f(h) - f(x)}{h} = \lim_{h\to 0} \frac{f(x)(f(h) - 1)}{h}$ Since $f(x)$ does not depend on $h$, we can take it out of the limit: $f'(x) = f(x) \lim_{h\to 0} \frac{f(h) - 1}{h}$

The limit $\lim_{h\to 0} \frac{f(h) - 1}{h}$ is the definition of the derivative of $f$ at $x=0$, since $f(0)=1$. $f'(0) = \lim_{h\to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h\to 0} \frac{f(h) - 1}{h}$. So, $f'(x) = f(x) f'(0)$.

Now we need to find $f'(0)$. We use the given expression for $f(x)$ near $x=0$: $f(x) = 1 + x P(x) + x^2 Q(x)$ for $x \neq 0$. For $h \neq 0$, we have $f(h) = 1 + h P(h) + h^2 Q(h)$. So, $\frac{f(h) - 1}{h} = \frac{(1 + h P(h) + h^2 Q(h)) - 1}{h} = \frac{h P(h) + h^2 Q(h)}{h} = P(h) + h Q(h)$.

Now, we evaluate the limit as $h \to 0$: $f'(0) = \lim_{h\to 0} (P(h) + h Q(h))$ Using the properties of limits: $f'(0) = \lim_{h\to 0} P(h) + \lim_{h\to 0} (h Q(h))$ We are given $\lim_{h\to 0} P(h) = a$ and $\lim_{h\to 0} Q(h) = b$. $\lim_{h\to 0} (h Q(h)) = (\lim_{h\to 0} h) (\lim_{h\to 0} Q(h)) = 0 \cdot b = 0$. So, $f'(0) = a + 0 = a$.

Substituting this back into the expression for $f'(x)$: $f'(x) = f(x) \cdot a = a f(x)$.

The derivative of $f(x)$ is $a f(x)$.