Let \( f(x+y)=f(x) \cdot f(y) \forall x, y \in R \). Suppose that \( f

1.	Let \( f(x+y)=f(x) \cdot f(y) \forall x, y \in R \). Suppose that \( f(k)=4 \), \( k \in R \) and \( f(0)=12 \), then \( f(k) \) is equal to \( (f(0) \neq 0) \)
Answer» \(f(x + y) = f(x) f(y)\) (Given) \(f( 0 + 0) = f(0) .f(0)\) (By taking x = 0 = y) ⇒ \(f(0) = (f(0))^2\) ⇒ \((f(0))^2 - f(0) = 0\) ⇒ \(f(0)\, (f(0) -1) = 0\) ⇒ \(f(0) = 1\) \((\because f(0)\ne 0)\) Given that \(f(1) = 2\) Then \(f(1 + 1) = f(1) f(1) = 2 \times 2 = 4\) ⇒ \(f(2) = 4\) \(\therefore k = 2\)

Let \( f(x+y)=f(x) \cdot f(y) \forall x, y \in R \). Suppose that \( f(k)=4 \), \( k \in R \) and \( f(0)=12 \), then \( f(k) \) is equal to \( (f(0) \neq 0) \)

Answer»

\(f(x + y) = f(x) f(y)\) (Given)

\(f( 0 + 0) = f(0) .f(0)\) (By taking x = 0 = y)

⇒ \(f(0) = (f(0))^2\)

⇒ \((f(0))^2 - f(0) = 0\)

⇒ \(f(0)\, (f(0) -1) = 0\)

⇒ \(f(0) = 1\) \((\because f(0)\ne 0)\)

Given that \(f(1) = 2\)

Then

\(f(1 + 1) = f(1) f(1) = 2 \times 2 = 4\)

⇒ \(f(2) = 4\)

\(\therefore k = 2\)

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Let \( f(x+y)=f(x) \cdot f(y) \forall x, y \in R \). Suppose that \( f(k)=4 \), \( k \in R \) and \( f(0)=12 \), then \( f(k) \) is equal to \( (f(0) \neq 0) \)

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