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

Step 1: Find f(0) using the functional equation.
Put \(x=y=0\) in \(f(x+y)=f(x)f(y)\): \(f(0) = f(0)f(0) = f(0)^2\). So \(f(0)^2 - f(0) = 0\), giving \(f(0)=0\) or \(f(0)=1\). Since \(f(x)=1+xg(x)\) gives \(f(0)=1+0=1\), we take \(f(0)=1\).

Step 2: Write the derivative of f from first principles.
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \] Using the functional equation, \(f(x+h) = f(x)f(h)\), so
\[ f'(x) = \lim_{h \to 0} \frac{f(x)f(h)-f(x)}{h} = f(x)\lim_{h \to 0}\frac{f(h)-1}{h} \]
Step 3: Use the given form of f to evaluate this limit.
Since \(f(h) = 1 + hg(h)\), we get \(f(h)-1 = hg(h)\), so \(\frac{f(h)-1}{h} = g(h)\) for any \(h \neq 0\).
Taking the limit as \(h \to 0\): \(\displaystyle\lim_{h \to 0}\frac{f(h)-1}{h} = \lim_{h \to 0} g(h) = T\).
So \(f'(x) = T \cdot f(x)\) for every \(x\), not just at \(x=0\).

Step 4: Solve this differential equation.
\(f'(x) = Tf(x)\) is a standard growth equation. Separating variables:
\[ \frac{df}{f} = T\,dx \implies \ln f = Tx + C \implies f(x) = Ae^{Tx} \] Using \(f(0)=1\) gives \(A=1\), so \(f(x) = e^{Tx}\).

Step 5: Find the nth derivative.
Differentiating \(f(x)=e^{Tx}\) repeatedly: \(f'(x)=Te^{Tx}\), \(f''(x)=T^2e^{Tx}\), and in general \(f^{n}(x) = T^n e^{Tx} = T^n f(x)\).
Comparing with \(f^{n}(x) = kf(x)\), we get \(k = T^n\).

Step 6: Why the other options fail.
Option (A), \(k=T\), would only be correct for the first derivative when \(n=1\), not for a general \(n\). Options (C) and (D) involve \(\log T\), but a logarithm of T never appears as the multiplying constant in this relation; \(\ln f\) only shows up as an intermediate step while solving the differential equation for \(f\) itself.

Final Answer:
\[ \boxed{k = T^n} \]