Question

If \(f(x) \cdot f(y)=f(x)+f(y)+f(x y)-2 \forall x, y \in R\) and if \(f(x)\) is not a constant function, then the value of \(f(1)\) is (A) 1 (B) 2 (C) 0 (D) \(-1\)

Short answer

#Answer# The value of \(f(1)\) is \(2\).

Step-by-step solution

set x = 1 and y = 1

We will set \(x = 1\) and \(y = 1\) to get more information about the equation: $$ f(1) \cdot f(1)=f(1)+f(1)+f(1 \cdot 1)-2 $$

Simplify

Simplify the equation obtained in Step 1: $$ f(1)^2 = 3f(1)-2 $$

set x = y

Now we will set \(x = y\) to get more information about the equation: $$ f(x) \cdot f(x)=f(x)+f(x)+f(x^2)-2 $$

Simplify

Simplify the equation obtained in Step 3: $$ f(x)^2 = 2f(x) + f(x^2) - 2 $$

set x = 1

Now we will set \(x = 1\) in the simplified equation obtained in Step 4: $$ f(1)^2 = 2f(1) + f(1) - 2 $$

Compare

Compare the equations from Step 2 and Step 5: $$ f(1)^2 = 3f(1)-2 $$ $$ f(1)^2 = 2f(1) + f(1) - 2 $$

Solve

Both equations are equal, we will solve for \(f(1)\): $$ 3f(1) - 2 = 2f(1) + f(1) - 2 $$ $$ f(1) = 2 $$ Therefore, the value of \(f(1)\) is \(2\). This corresponds to the option (B) in the given choices.