UPSC Maths 2025 Paper 1 Q1d — Solution

10 marks · Section A

Question

Given that $f(x + y) = f(x)\,f(y)$ for all real $x, y$ , $f(x) \neq 0$ for any real $x$ and $f'(0) = 2$ . Show that for all real $x$ , $f'(x) = 2 f(x)$ . Hence find $f(x)$ .

Technique

Differentiate the functional equation from first principles (limit definition of the derivative), then solve the resulting separable ODE $f'=2f$ .

Solution

Step 1 — Preliminary: $f(0)=1$ .

Put $x=y=0$ : $f(0)=f(0)^2$ , so $f(0)(f(0)-1)=0$ . Since $f$ is nonzero everywhere, $f(0)\ne 0$ , hence $f(0)=1.$

Step 2 — Derive $f'(x)=2f(x)$ from first principles.

By definition, $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$ Use the functional equation $f(x+h)=f(x)f(h)$ : $f'(x)=\lim_{h\to 0}\frac{f(x)f(h)-f(x)}{h}=f(x)\lim_{h\to 0}\frac{f(h)-1}{h}.$ Since $f(0)=1$ , the remaining limit is exactly $f'(0)$ : $\lim_{h\to 0}\frac{f(h)-1}{h}=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=f'(0)=2.$ Therefore $f'(x)=2f(x)\quad\text{for all real }x.$

Step 3 — Solve the ODE.

$f'(x)=2f(x)$ is separable: $\dfrac{f'(x)}{f(x)}=2$ (valid since $f\ne0$ ), so $\dfrac{d}{dx}\ln|f(x)|=2$ , giving $\ln|f(x)|=2x+C$ , i.e. $f(x)=A e^{2x}$ . Apply $f(0)=1$ : $A=1$ .

$f(x)=e^{2x}.$

Answer

$\boxed{\;f'(x)=2f(x)\ \text{for all }x,\qquad f(x)=e^{2x}.\;}$