If a function f satisfies f(m + n) = f(m) + f(n) for all m, n ∈ N and f(1) = 1, then the largest natural

Question

If a function f satisfies \(f(m+n)=f(m)+f(n)\) for all \(\mathrm{m}, \mathrm{n} \in \mathrm{N}\) and \(\mathrm{f}(1)=1\), then the largest natural number \(\lambda\) such that \(\sum\limits_{k=1}^{2022} f(\lambda+k) \leq(2022)^{2}\) is equal to ________ . 

Answer 1

Correct answer is : 1010 

\(f(m+n)=f(m)+f(n)\)

\(\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{kx}\)

\(\Rightarrow \mathrm{f}(1)=1\)

\(\Rightarrow \mathrm{k}=1\)

\(\mathrm{f}(\mathrm{x})=\mathrm{x}\)

Now

\( \begin{aligned} \sum_{\mathrm{k}=1}^{2022} \mathrm{f}(\lambda+\mathrm{k}) \leq(2022)^{2} \end{aligned} \) 

\( \begin{aligned} \Rightarrow \sum_{\mathrm{k}=1}^{2022}(\lambda+\mathrm{k}) \leq(2022)^{2} \end{aligned} \) 

\( \begin{aligned} \Rightarrow 2022 \lambda+\frac{2022 \times 2023}{2} \leq(2022)^{2} \end{aligned} \)

\( \begin{aligned} \Rightarrow \lambda \leq 2022-\frac{2023}{2} \\ \end{aligned} \) 

\( \begin{aligned} \Rightarrow \lambda \leq 1010.5 \end{aligned} \) 

\(\therefore\)  largest natural no. \(\lambda\) is 1010.