Question:

If a function f f satisfies f(m+n)=f(m)+f(n) f(m + n) = f(m) + f(n) for all m,nN m, n \in \mathbb{N} and f(1)=1 f(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 __________.

Updated On: Mar 20, 2025
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Correct Answer: 1010

Solution and Explanation

Step 1: Analyze the functional equation The functional equation f(m+n)=f(m)+f(n) f(m+n) = f(m) + f(n) suggests that f(x) f(x) is linear. Assume:

f(x)=kx. f(x) = kx.

Substitute f(1)=1 f(1) = 1 :

f(1)=k1    k=1. f(1) = k \cdot 1 \implies k = 1.

Thus:

f(x)=x. f(x) = x.

Step 2: Expand the summation We are given:

k=12022f(λ+k)(2022)2. \sum_{k=1}^{2022} f(\lambda + k) \leq (2022)^2.

Substitute f(x)=x f(x) = x :

k=12022(λ+k)(2022)2. \sum_{k=1}^{2022} (\lambda + k) \leq (2022)^2.

Split the summation:

k=12022(λ+k)=k=12022λ+k=12022k. \sum_{k=1}^{2022} (\lambda + k) = \sum_{k=1}^{2022} \lambda + \sum_{k=1}^{2022} k.

Step 2.1: Simplify each term

  1. k=12022λ=2022λ. \sum_{k=1}^{2022} \lambda = 2022 \cdot \lambda.
  2. k=12022k=202220232(sum of the first 2022 natural numbers). \sum_{k=1}^{2022} k = \frac{2022 \cdot 2023}{2} \quad \text{(sum of the first 2022 natural numbers)}.

Thus:

k=12022(λ+k)=2022λ+202220232. \sum_{k=1}^{2022} (\lambda + k) = 2022 \cdot \lambda + \frac{2022 \cdot 2023}{2}.

Step 3: Solve the inequality Substitute into the inequality:

2022λ+202220232(2022)2. 2022\lambda + \frac{2022 \cdot 2023}{2} \leq (2022)^2.

Simplify:

2022λ(2022)2202220232. 2022\lambda \leq (2022)^2 - \frac{2022 \cdot 2023}{2}.

Factor 2022 out:

2022λ2022(202220232). 2022\lambda \leq 2022 \left( 2022 - \frac{2023}{2} \right).

Simplify further:

λ202220232. \lambda \leq 2022 - \frac{2023}{2}.

Calculate:

λ20221011.5=1010.5. \lambda \leq 2022 - 1011.5 = 1010.5.

Step 4: Largest natural number Since λ \lambda must be a natural number:

λ=1010. \lambda = 1010.

Final Answer:- 1010. 1010.

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