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JEE Main 20249 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Functions

If a function f satisfies f(~m+n)=f(~m)+f(n) for all m, n ∈ N and...

If a function ff satisfies f( m+n)=f( m)+f(n)f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n}) for all m,n∈N\mathrm{m}, \mathrm{n} \in \mathbf{N} and f(1)=1f(1)=1, then the largest natural number λ\lambda such that ∑k=12022f(λ+k)≤(2022)2\sum_{k=1}^{2022} f(\lambda+k) \leq(2022)^2 is equal to _________

Correct answer

0

Step-by-step solution

f(m+n)=f(m)+f(n)⇒f(x)=kx⇒f(1)=1⇒k=1f(x)=x\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 ∑k=12022f(λ+k)≤(2022)2\sum_{\mathrm{k}=1}^{2022} \mathrm{f}(\lambda+\mathrm{k}) \leq(2022)^2⇒∑k=12022(λ+k)≤(2022)2⇒2022λ+2022×20232≤(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}⇒λ≤2022−20232⇒λ≤1010.5\begin{aligned} & \Rightarrow \lambda \leq 2022-\frac{2023}{2} \\ & \Rightarrow \lambda \leq 1010.5 \end{aligned}∴\therefore largest natural no. λ\lambda is 1010 .
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