JEE Mains · Maths · STD 12 - 1. relation and function
If \(f( x + y )=f( x ) f( y )\) and \(\sum \limits_{ x =1}^{\infty} f( x )=2, x , y \in N\) where \(N\) is the set of all natural numbers, then the value of \(\frac{f(4)}{f(2)}\) is
- A \(\frac{1}{9}\)
- B \(\frac{4}{9}\)
- C \(\frac{1}{3}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{9}\)
Step-by-step Solution
Detailed explanation
\(f(x+y)=f(x) \cdot f(y)\) \(\sum_{x=1}^{\infty} f(x)=2 \quad\) where \(x, y \in N\) \(f(1)+f(2)+f(3)+\ldots . \infty=2 \ldots .(1)(\) Given \()\) Now for \(f(2)\) put \(x=y=1\) \(f(2)=f(1+1)=f(1) \cdot f(1)=(f(1))^{2}\) \(f(3)=f(2+1)=f(2) \cdot f(1)=(f(1))^{3}\) Now put these…
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