JEE Main 2024 (Online) 9th April Morning ShiftQuestion 1
If a function $f$ satisfies $f(m+n)=f(m)+f(n)$ for all $m,n\in\mathbf{N}$ and $f(1)=1$, then the largest natural number $\lambda$ such that $\sum_{k=1}^{2022}f(\lambda+k)\le(2022)^{2}$ is equal to $\_\_\_\_\_\_$ .
Functions
Answer$1010$
Solution
$f(n)=n$. $\sum_{k=1}^{2022}(\lambda+k)=2022\lambda+\frac{2022\cdot2023}{2}\le2022^2\Rightarrow\lambda\le2022-1011.5=1010.5$. Largest natural $\lambda=1010$.