JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 25)
If a function $$f$$ satisfies $$f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n})$$ for all $$\mathrm{m}, \mathrm{n} \in \mathbf{N}$$ and $$f(1)=1$$, then the largest natural number $$\lambda$$ such that $$\sum_\limits{\mathrm{k}=1}^{2022} f(\lambda+\mathrm{k}) \leq(2022)^2$$ is equal to _________.
Answer
1010
Explanation
$$\begin{aligned} & f(m+n)=f(m)+f(n) \\ & f(x)=k x \\ & \because f(1)=1 \\ & \Rightarrow k=1 \\ & \Rightarrow f(x)=x \end{aligned}$$
$$\begin{aligned} & \sum_{k=1}^{2022} f(\lambda+k)=\sum_{k=1}^{2022}(\lambda+k)=\underbrace{\lambda+\lambda+\ldots+\lambda}_{2022}+(1+2+\ldots+2022) \\ & =2022 \lambda+\frac{2022 \times 2023}{2} \leq(2022)^2 \\ & \Rightarrow \lambda \leq \frac{2021}{2} \\ & \text { largest } \lambda=1010 \end{aligned}$$
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