Given that:
\[ f(m + n) = f(m) + f(n) \]
This implies \( f(x) = kx \).
Now, since \( f(1) = 1 \), we get \( k = 1 \).
Therefore,
\[ f(x) = x \]
Now consider,
\[ \sum_{k=1}^{2022} f(\lambda + k) \leq (2022)^2 \]
Substituting \( f(x) = x \), we have:
\[ \sum_{k=1}^{2022} (\lambda + k) \leq (2022)^2 \]
Simplifying,
\[ 2022\lambda + \frac{2022 \times 2023}{2} \leq (2022)^2 \]
\[ \lambda \leq 2022 - \frac{2023}{2} \]
\[ \lambda \leq 1010.5 \]
Hence, the largest natural number value of \( \lambda \) is:
\[ \boxed{\lambda = 1010} \]
Step 1: Analyze the functional equation The functional equation \( f(m+n) = f(m) + f(n) \) suggests that \( f(x) \) is linear. Assume:
\[ f(x) = kx. \]
Substitute \( f(1) = 1 \):
\[ f(1) = k \cdot 1 \implies k = 1. \]
Thus:
\[ f(x) = x. \]
Step 2: Expand the summation We are given:
\[ \sum_{k=1}^{2022} f(\lambda + k) \leq (2022)^2. \]
Substitute \( f(x) = x \):
\[ \sum_{k=1}^{2022} (\lambda + k) \leq (2022)^2. \]
Split the summation:
\[ \sum_{k=1}^{2022} (\lambda + k) = \sum_{k=1}^{2022} \lambda + \sum_{k=1}^{2022} k. \]
Step 2.1: Simplify each term
Thus:
\[ \sum_{k=1}^{2022} (\lambda + k) = 2022 \cdot \lambda + \frac{2022 \cdot 2023}{2}. \]
Step 3: Solve the inequality Substitute into the inequality:
\[ 2022\lambda + \frac{2022 \cdot 2023}{2} \leq (2022)^2. \]
Simplify:
\[ 2022\lambda \leq (2022)^2 - \frac{2022 \cdot 2023}{2}. \]
Factor 2022 out:
\[ 2022\lambda \leq 2022 \left( 2022 - \frac{2023}{2} \right). \]
Simplify further:
\[ \lambda \leq 2022 - \frac{2023}{2}. \]
Calculate:
\[ \lambda \leq 2022 - 1011.5 = 1010.5. \]
Step 4: Largest natural number Since \( \lambda \) must be a natural number:
\[ \lambda = 1010. \]
Final Answer:- \( 1010. \)
If the domain of the function \( f(x) = \dfrac{1}{\sqrt{10 + 3x - x^2}} + \dfrac{1}{\sqrt{x + |x|}} \) is \( (a, b) \), then \((1 + a)^2 + b^2\) is equal to: