Question

f(x+y)=f(x)×f(y) x belongs to all real numbers then f(x) is

• A. 0
• B. a^x (where a is a positive constant)
• C. 1
• D. x^a (where a is a constant)

Answer 1

Answer: (A), (B)

The possible forms of $f(x)$ are $f(x) = 0$ and $f(x) = a^x$ where $a$ is a positive constant.

Answer 2

The given functional equation is $f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$.

1. Find $f(0)$: Set $x=0$ and $y=0$. $f(0+0) = f(0) \times f(0)$ $f(0) = (f(0))^2$ This implies $f(0) = 0$ or $f(0) = 1$.

2. Case 1: $f(0) = 0$. Set $y=0$. $f(x+0) = f(x) \times f(0)$ $f(x) = f(x) \times 0$ $f(x) = 0$ for all $x \in \mathbb{R}$. This is a valid solution, as $f(x+y)=0$ and $f(x)f(y)=0 \times 0 = 0$.

3. Case 2: $f(0) = 1$.

   • Show $f(x) \neq 0$ for any $x$: Set $y = -x$. $f(x+(-x)) = f(x) \times f(-x)$ $f(0) = f(x)f(-x)$ $1 = f(x)f(-x)$ This implies that $f(x)$ cannot be zero for any $x$, otherwise $1=0$, which is a contradiction.
   • Show $f(x) > 0$ for all $x$: $f(x) = f\left(\frac{x}{2} + \frac{x}{2}\right) = f\left(\frac{x}{2}\right)f\left(\frac{x}{2}\right) = \left(f\left(\frac{x}{2}\right)\right)^2$. Since the square of any real number is non-negative, $f(x) \ge 0$. As we've shown $f(x) \neq 0$, it must be that $f(x) > 0$ for all $x \in \mathbb{R}$.
   • Determine the form of $f(x)$: Let $f(1) = a$. Since $f(1)>0$, we have $a>0$. For any positive integer $n$: $f(n) = f(1+1+...+1) = f(1)f(1)...f(1) = (f(1))^n = a^n$. For any rational number $q = \frac{m}{n}$ (where $m, n$ are integers, $n \neq 0$): $f(nq) = f(q+...+q) = (f(q))^n$. Also, $f(nq) = f(n \times \frac{m}{n}) = f(m) = a^m$. So, $(f(q))^n = a^m$, which means $f(q) = (a^m)^{1/n} = a^{m/n} = a^q$. Thus, $f(x) = a^x$ for all rational numbers $x$. Assuming continuity or other regularity conditions for real numbers, $f(x)=a^x$ for all real numbers $x$. The function $f(x) = a^x$ with $a>0$ satisfies the original equation: $f(x+y) = a^{x+y}$ $f(x)f(y) = a^x a^y = a^{x+y}$ This holds true. The case $a=1$ gives $f(x)=1^x=1$, which is a valid solution and is a specific instance of $a^x$.

Therefore, the possible forms of $f(x)$ are $f(x) = 0$ and $f(x) = a^x$ where $a$ is a positive constant.