If a function f satisfies f(m+n)=f(m)+f(n) for all m, n in N and f(1)=1 , then the largest natural number λ such that sum k=1 2022 f(λ+k) leq(2022) 2 is equal to .

Mathematics · Functions

JEE Main 2024 — 9 April, Shift 1 — Question 29

If a function $f$ satisfies $f(m+n)=f(m)+f(n)$ for all $m, n \in N$ 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 $\qquad$ .

Numerical answer — enter this value.

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

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

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

$\Rightarrow \mathrm{k}=1$

$f(x)=x$ Now

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

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

$\Rightarrow 2022 \lambda+\frac{2022 \times 2023}{2} \leq(2022)^{2}$

$\Rightarrow \lambda \leq 2022-\frac{2023}{2}$

$\Rightarrow \lambda \leq 1010.5$

$\therefore$ largest natural no. $\lambda$ is $1010 .$

Answer key and solution verified before publishing.

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