Exams Joint Entrance Examination Mathematics Functions Question 6 Question 6 of 162 Medium NUMERICAL
If a function f f f satisfies f ( m + n ) = f ( m ) + f ( n ) f(m+n)=f(m)+f(n) f ( m + n ) = f ( m ) + f ( n ) for all m , n ∈ N m,n\in N m , n ∈ N and f ( 1 ) = 1 f(1)=1 f ( 1 ) = 1 , then the largest natural number λ \lambda λ such that ∑ k = 1 2022 f ( λ + k ) ≤ ( 2022 ) 2 \sum _{k=1}^{2022}f(\lambda +k)\le (2022{)}^{2} ∑ k = 1 2022 f ( λ + k ) ≤ ( 2022 ) 2 is equal to \_\_\_\_\_\_\_\_\_.
Answer: 1010
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f ( m + n ) = f ( m ) + f ( n ) f ( x ) = k x ∵ f ( 1 ) = 1 ⇒ k = 1 ⇒ f ( x ) = x \begin{gathered} f(m+n)=f(m)+f(n) \\ f(x)=kx \\ \because f(1)=1 \\ \Rightarrow k=1 \\ \Rightarrow f(x)=x \end{gathered} f ( m + n ) = f ( m ) + f ( n ) f ( x ) = k x ∵ f ( 1 ) = 1 ⇒ k = 1 ⇒ f ( x ) = x
∑ k = 1 2022 f ( λ + k ) = ∑ k = 1 2022 ( λ + k ) = λ + λ + … + λ 2022 + ( 1 + 2 + … + 2022 ) = 2022 λ + 2022 × 2023 2 ≤ ( 2022 ) 2 ⇒ λ ≤ 2021 2 λ = 1010 \begin{gathered} \sum _{k=1}^{2022}f(\lambda +k)=\sum _{k=1}^{2022}(\lambda +k)=\underset{2022}{\underset{}{\lambda +\lambda +\ldots +\lambda }}+(1+2+\ldots +2022) \\ =2022\lambda +\frac{2022\times 2023}{2}\le (2022{)}^{2} \\ \Rightarrow \lambda \le \frac{2021}{2} \\ \lambda =1010 \end{gathered} k = 1 ∑ 2022 f ( λ + k ) = k = 1 ∑ 2022 ( λ + k ) = 2022 λ + λ + … + λ + ( 1 + 2 + … + 2022 ) = 2022 λ + 2 2022 × 2023 ≤ ( 2022 ) 2 ⇒ λ ≤ 2 2021 λ = 1010 Answer the question to unlock