JEE Mains · Maths · STD 12 - 1. relation and function
Let \(\mathrm{f}: N \rightarrow N\) be a function such that \(\mathrm{f}(\mathrm{m}+\mathrm{n})=\mathrm{f}(\mathrm{m})+\mathrm{f}(\mathrm{n})\) for every \(\mathrm{m}, \mathrm{n} \in N\). If \(\mathrm{f}(6)=18\) then \(\mathrm{f}(2) \cdot \mathrm{f}(3)\) is equal to :
- A \(6\)
- B \(54\)
- C \(18\)
- D \(36\)
Answer & Solution
Correct Answer
(B) \(54\)
Step-by-step Solution
Detailed explanation
\(f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n})\) \(\text { Put } \mathrm{m}=1, \mathrm{n}=1\) \(f(2)=2 f(1)\) \(\text { Put } \mathrm{m}=2, \mathrm{n}=1\) \(f(3)=f(2)+f(1)=3 f(1)\) \(\text { Put } \mathrm{m}=3, \mathrm{n}=3\) \(f(6)=2 f(3) \Rightarrow f(3)=9\)…
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