If a function f satisfies f ( m + n ) = f ( m ) + f ( n ) for all m , n ? N and f ( 1 ) = 1 , then the largest natural number ? such that ? k = 1 2022 f ( ? + k ) ? ( 2022 ) 2 is equal to ___________ . [2024]

Question

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 ___________ . [2024]

Smart Achievers · Accepted Answer

(1010)

We have, $f (m + n) = f (m) + f (n)$

So, $f (x) = k x$

$∵$ $f (1) = 1 \Rightarrow k = 1$

Hence, $f (x) = x$

Now, $\sum_{k = 1}^{2022} f (λ + k) = \sum_{k = 1}^{2022} (λ + k)$

$= \underset{2022}{\underset{⏟}{λ + λ + . . . + λ}}$ $+ (1 + 2 + . . . . + 2022)$

$= 2022 λ + \frac{2022 \times 2023}{2} \leq {(2022)}^{2}$ (Given)

$\Rightarrow λ \leq \frac{2021}{2}$

So, largest $λ = 1010$

Solution

Q.
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 ___________ . [2024]

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Solution

Q. If a function f satisfies f(m+n)=f(m)+f(n) for all m, n∈N and f(1)=1, then the largest natural number λ such that ∑k=12022f(λ+k)≤(2022)2 is equal to ___________ . [2024]

Ans. (1010) We have, f(m+n)=f(m)+f(n) So, f(x)=kx ∵ f(1)=1⇒k=1 Hence, f(x)=x Now, ∑k=12022f(λ+k)=∑k=12022(λ+k) =λ+λ+...+λ⏟2022+(1+2+....+2022) =2022λ+2022×20232≤(2022)2 (Given) ⇒λ≤20212 So, largest λ=1010

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Q.
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 ___________ . [2024]