Verified Solution • 2025–26 Session moderate QJEEM24Ma65ba: JEE Main 2024 Mathematics - Class 11 JEEM24Ma65ba Solution: If a function satisfies for all and , then the largest natural… If a function fff satisfies f( m+n)=f( m)+f(n)f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n})f( m+n)=f( m)+f(n) for all m,n∈N\mathrm{m}, \mathrm{n} \in \mathbf{N}m,n∈N and f(1)=1f(1)=1f(1)=1 , then the largest natural number λ\lambdaλ such that ∑k=12022f(λ+k)≤(2022)2 is equal to _________. Step-by-Step Verified Solution amp;f(m+n)=f(m)+f(n)amp;f(x)=kxamp;∵f(1)=1amp;⇒k=1amp;⇒f(x)=x\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}amp;f(m+n)=f(m)+f(n)amp;f(x)=kxamp;∵f(1)=1amp;⇒k=1amp;⇒f(x)=x amp;∑k=12022f(λ+k)=∑k=12022(λ+k)=λ+λ+…+λ⏟2022+(1+2+…+2022)amp;=2022λ+2022×20232≤(2022)2amp;⇒λ≤20212amp; largest λ=1010\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}amp;k=1∑2022f(λ+k)=k=1∑2022(λ+k)=2022λ+λ+…+λ+(1+2+…+2022)amp;=2022λ+22022×2023≤(2022)2amp;⇒λ≤22021amp; largest λ=1010 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring