Q60 / 90 · JEE Main 2024 (09 Apr Shift 1) · Mathematics
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 $\sum_{k=1}^{2022} f(\lambda+k) \leq(2022)^2$ is equal to _________
Correct answer: 1010
Solution
$\begin{aligned}
& \mathrm{f}(\mathrm{m}+\mathrm{n})=\mathrm{f}(\mathrm{m})+\mathrm{f}(\mathrm{n}) \\
& \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{kx} \\
& \Rightarrow \mathrm{f}(1)=1 \\
& \Rightarrow \mathrm{k}=1 \\
& \mathrm{f}(\mathrm{x})=\mathrm{x}
\end{aligned}$
Now $\sum_{\mathrm{k}=1}^{2022} \mathrm{f}(\lambda+\mathrm{k}) \leq(2022)^2$ $\begin{aligned} & \Rightarrow \sum_{\mathrm{k}=1}^{2022}(\lambda+\mathrm{k}) \leq(2022)^2 \\ & \Rightarrow 2022 \lambda+\frac{2022 \times 2023}{2} \leq(2022)^2\end{aligned}$ $\begin{aligned} & \Rightarrow \lambda \leq 2022-\frac{2023}{2} \\ & \Rightarrow \lambda \leq 1010.5 \end{aligned}$ $\therefore$ largest natural no. $\lambda$ is 1010 .
Now $\sum_{\mathrm{k}=1}^{2022} \mathrm{f}(\lambda+\mathrm{k}) \leq(2022)^2$ $\begin{aligned} & \Rightarrow \sum_{\mathrm{k}=1}^{2022}(\lambda+\mathrm{k}) \leq(2022)^2 \\ & \Rightarrow 2022 \lambda+\frac{2022 \times 2023}{2} \leq(2022)^2\end{aligned}$ $\begin{aligned} & \Rightarrow \lambda \leq 2022-\frac{2023}{2} \\ & \Rightarrow \lambda \leq 1010.5 \end{aligned}$ $\therefore$ largest natural no. $\lambda$ is 1010 .
Hence, the correct answer is 1010.