Question

Suppose the function \(f\) satisfies the equation \(f(x+y) = f(x)f(y)\) for all \(x\) and \(y\). Here \(f(x) = 1 + xg(x)\), where \(\displaystyle\lim_{x \to 0} g(x) = T\), and \(T\) is a positive integer. If \(f^{n}(x) = kf(x)\), where \(f^{n}(x)\) denotes the \(n\)th derivative of \(f\), then \(k\) is equal to:

• A. \(T\)
• B. \(T^n\)
• C. \(\log T\)
• D. \((\log T)^n\)

Hint:

Differentiate the functional equation to show f'(x) = Tf(x), then repeat this n times to find the nth derivative in terms of f(x).

Answer 1

The Correct Option is B

Here is a different way to reach the same result, using induction instead of solving the differential equation outright.

1. Find f(0): putting $x=y=0$ into $f(x+y)=f(x)f(y)$ gives $f(0)=f(0)^2$, so $f(0)=0$ or $f(0)=1$. Since $f(x)=1+xg(x)$ forces $f(0)=1$, we keep $f(0)=1$.
2. Find f'(0): from $f(x)=1+xg(x)$, for small $h$ we get $f(h)-1=hg(h)$, so $\frac{f(h)-1}{h}=g(h)$. Taking $h \to 0$ gives $f'(0)=\lim_{h\to0}g(h)=T$. This is the instantaneous rate of change of $f$ right at $x=0$.
3. Show f'(x) = Tf(x) everywhere: differentiate $f(x+h)=f(x)f(h)$ with respect to $h$ and then set $h=0$. This gives $f'(x)=f(x)f'(0)=Tf(x)$ for every $x$, not only at $x=0$.
4. Induction base case: for $n=1$, $f'(x)=Tf(x)$ is already shown above, so $k=T^1=T$ works for $n=1$.
5. Induction step: assume $f^{(n)}(x)=T^nf(x)$ holds for some $n$. Differentiating both sides gives $f^{(n+1)}(x)=T^nf'(x)=T^n \cdot Tf(x)=T^{n+1}f(x)$, so the formula also holds for $n+1$.

By induction, $f^{(n)}(x)=T^nf(x)$ for every positive integer $n$. Comparing with the given relation $f^{n}(x)=kf(x)$, we read off $k=T^n$.

Let's summarize:

• The functional equation plus the given form of $f$ pin down $f'(0)=T$.
• The same functional equation forces $f'(x)=Tf(x)$ at every point, not just at $0$.
• Repeating differentiation $n$ times multiplies in a factor of $T$ each time, giving $k=T^n$.

So $k=T^n$, matching option (B). A logarithm of $T$ never enters this relation; that only appears if you try to solve for $f(x)$ itself, which is a different quantity from $k$.