Question

If f(x) f(y) + 2 = f(x) + f(y) + f(xy) and f(1) = 2, then sgn f(x) is equal to (where sgn denotes signum function):

• A. 0
• B. 1
• C. -1
• D. 4

Answer 1

Answer: (B)

1

Answer 2

The given functional equation is: $f(x) f(y) + 2 = f(x) + f(y) + f(xy)$ The given condition is $f(1) = 2$.

Rearrange the terms of the functional equation: $f(x) f(y) - f(x) - f(y) + 1 = f(xy) - 1$ This can be factored as: $(f(x) - 1)(f(y) - 1) = f(xy) - 1$

Let $g(x) = f(x) - 1$. Substituting this into the equation, we get: $g(x) g(y) = g(xy)$

Now, let's use the condition $f(1) = 2$. Since $g(x) = f(x) - 1$, we have $g(1) = f(1) - 1 = 2 - 1 = 1$.

We need to find the general solution for the functional equation $g(x)g(y) = g(xy)$ with the condition $g(1)=1$.

Case 1: $x=0$ $g(0)g(y) = g(0)$. If there exists any $y_0$ such that $g(y_0) \neq 1$, then $g(0)$ must be $0$. If $g(y)=1$ for all $y$, then $g(x)=1$ is a solution. If $g(x)=1$ for all $x$, then $f(x) = g(x)+1 = 1+1 = 2$. Let's check $f(x)=2$ in the original equation: $2 \cdot 2 + 2 = 2 + 2 + 2$ $4 + 2 = 6$ $6 = 6$ This solution $f(x)=2$ is valid and satisfies $f(1)=2$. For $f(x)=2$, sgn $f(x) = \text{sgn}(2) = 1$.

Case 2: $g(0)=0$. If $g(0)=0$, then $f(0) = g(0)+1 = 0+1 = 1$. This is consistent with the original equation (as shown in thought process). For $x \neq 0$: Since $g(1)=1$, we can set $y=1/x$ (for $x \neq 0$): $g(x)g(1/x) = g(x \cdot 1/x) = g(1) = 1$. This implies that $g(x) \neq 0$ for any $x \neq 0$.

Consider $x > 0$: Let $x = e^t$ and $y = e^u$. Then $g(e^t) g(e^u) = g(e^{t+u})$. Let $h(t) = g(e^t)$. Then $h(t)h(u) = h(t+u)$. This is Cauchy's functional equation. The continuous solutions are of the form $h(t) = a^t$ for some constant $a > 0$. So $g(e^t) = a^t$. Substituting $t = \ln x$, we get $g(x) = a^{\ln x} = (e^{\ln a})^{\ln x} = x^{\ln a}$. Let $c = \ln a$. So, $g(x) = x^c$ for $x > 0$. Using $g(1)=1$, we have $1^c=1$, which holds for any $c$.

Consider $x < 0$: Let $x = -1$. $g(-1)g(-1) = g((-1)(-1)) = g(1) = 1$. So $(g(-1))^2 = 1$, which means $g(-1) = 1$ or $g(-1) = -1$.

Subcase 2.1: $g(-1) = 1$. For any $x < 0$, we can write $x = -|x|$. $g(x) = g(-|x|) = g(-1 \cdot |x|) = g(-1)g(|x|) = 1 \cdot g(|x|) = g(|x|)$. Since for $t > 0$, $g(t) = t^c$, we have $g(x) = |x|^c$ for $x < 0$. Combining with $g(x)=x^c$ for $x>0$, we get $g(x) = |x|^c$ for all $x \neq 0$. And $g(0)=0$. So $f(x) = |x|^c + 1$ for $x \neq 0$, and $f(0)=1$. For $f(x)$ to be defined for all real $x$, $c$ must be such that $|x|^c$ is defined. If $c$ is an integer, this is fine. If $c$ is an even integer, say $c=2k$ for $k \in \mathbb{Z}$: $f(x) = |x|^{2k} + 1 = x^{2k} + 1$. For $k=0$, $f(x) = x^0+1 = 1+1=2$, which is the solution from Case 1. For $k=1$, $f(x) = x^2+1$. Let's check $f(x)=x^2+1$: $f(1)=1^2+1=2$. This is valid. For $f(x)=x^2+1$, since $x^2 \ge 0$, $f(x) = x^2+1 \ge 1$. Therefore, $f(x)$ is always positive. So sgn $f(x) = 1$.

Subcase 2.2: $g(-1) = -1$. For any $x < 0$, $g(x) = g(-1 \cdot |x|) = g(-1)g(|x|) = -1 \cdot |x|^c = -|x|^c$. So $g(x) = x^c$ for $x>0$, $g(x) = -|x|^c$ for $x<0$, and $g(0)=0$. This means $f(x) = x^c+1$ for $x>0$, $f(x) = -|x|^c+1$ for $x<0$, and $f(0)=1$. For $f(x)$ to be defined for all real $x$, $c$ must be an integer. If $c$ is an even integer, $c=2k$: $f(x) = x^{2k}+1$ for $x>0$. $f(x) = -|x|^{2k}+1 = -x^{2k}+1$ for $x<0$. This means $f(x)$ is not symmetric. For example, if $c=2$, $f(x)=x^2+1$ for $x>0$ and $f(x)=-x^2+1$ for $x<0$. For $x=2$, $f(2)=2^2+1=5$. sgn $f(2)=1$. For $x=-2$, $f(-2)=-(-2)^2+1 = -4+1 = -3$. sgn $f(-2)=-1$. In this case, sgn $f(x)$ is not constant. This contradicts the options being constant values.

If $c$ is an odd integer, $c=2k+1$: $f(x) = x^{2k+1}+1$ for $x>0$. $f(x) = -|x|^{2k+1}+1 = -(-x)^{2k+1}+1 = -(-1)^{2k+1}x^{2k+1}+1 = -(-1)x^{2k+1}+1 = x^{2k+1}+1$ for $x<0$. So $f(x) = x^{2k+1}+1$ for all $x \neq 0$, and $f(0)=1$. For example, if $c=1$, $f(x)=x+1$. For $x=2$, $f(2)=2+1=3$. sgn $f(2)=1$. For $x=-2$, $f(-2)=-2+1=-1$. sgn $f(-2)=-1$. In this case, sgn $f(x)$ is not constant. This contradicts the options being constant values.

Summary of solutions for $f(x)$ that satisfy $f(1)=2$ and the functional equation:

1. $f(x) = 2$. In this case, sgn $f(x) = 1$.
2. $f(x) = x^{2k}+1$ where $k$ is a non-zero integer. For $f(x)$ to be defined for all $x$, $2k$ must be a non-negative even integer. So $k \in \{1, 2, 3, \dots\}$. For any $k \ge 1$, $x^{2k} \ge 0$, so $x^{2k}+1 \ge 1$. Thus $f(x) > 0$. In this case, sgn $f(x) = 1$.
3. $f(x) = x^{2k+1}+1$ where $k$ is an integer. For $k=0$, $f(x)=x+1$. For $k=1$, $f(x)=x^3+1$. These solutions result in sgn $f(x)$ not being constant (e.g., $f(-2)=-1$ for $f(x)=x+1$).

Since the question asks for "sgn f(x)" which implies a single constant value (as per the options), we must choose the solutions where sgn $f(x)$ is constant. These are $f(x)=2$ and $f(x)=x^{2k}+1$ for $k \in \mathbb{N}$. In both these cases, sgn $f(x) = 1$.

The final answer is $\boxed{\text{1}}$

The final answer is $\boxed{\text{1}}$