If a function \( f \) satisfies \( f(m + n) = f(m) + f(n) \) for all \( m, n \in \mathbb{N} \) and \( f(1) = 1 \), then the largest natural number \( \lambda \) such that \[ \sum_{k=1}^{2022} f(\lambda + k) \leq (2022)^2 \] is equal to __________.
Correct Answer: 1010
Solution and Explanation
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
- \[ \sum_{k=1}^{2022} \lambda = 2022 \cdot \lambda. \]
- \[ \sum_{k=1}^{2022} k = \frac{2022 \cdot 2023}{2} \quad \text{(sum of the first 2022 natural numbers)}. \]
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:- \( 1010. \)
Top Questions on Functions
- The sum of all local minimum values of the function \( f(x) \) as defined below is:
\[ f(x) = \begin{cases} 1 - 2x & \text{if } x < -1, \\[10pt] \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2, \\[10pt] \frac{11}{18}(x-4)(x-5) & \text{if } x > 2. \end{cases} \] - In \( I(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx \), where \( m, n > 0 \), then \( I(9, 14) + I(10, 13) \) is:
- Let \( f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32} \). Then the value of \[ 8 \left( f\left( \frac{1}{15} \right) + f\left( \frac{2}{15} \right) + \dots + f\left( \frac{59}{15} \right) \right) \] is equal to:
- Let \( f: \mathbb{R} \to \mathbb{R} \) be a function defined by \( f(x) = \left( 2 + 3a \right)x^2 + \left( \frac{a+2}{a-1} \right)x + b, a \neq 1 \). If \[ f(x + y) = f(x) + f(y) + 1 - \frac{2}{7}xy, \] then the value of \( 28 \sum_{i=1}^5 f(i) \) is:
- The area of the region enclosed by the curves \( y = e^x \), \( y = |e^x - 1| \), and the y-axis is:
Questions Asked in JEE Main exam
- A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is \( -3 \), then the magnitude of the radius of curvature of the mirror is:
- JEE Main - 2025
- Spherical Mirrors
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to:
- JEE Main - 2025
- Differential Equations
- The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is _____.
- JEE Main - 2025
- Number Systems
- Let \( P_n = \alpha^n + \beta^n \), \( P_{10} = 123 \), \( P_9 = 76 \), \( P_8 = 47 \), and \( P_1 = 1 \). The quadratic equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
- JEE Main - 2025
- Sequences and Series
- If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:
- JEE Main - 2025
- Matrices and Determinants