Sets, Relations & Functions
Functions
star_batch_jee_advanced_2025
Grade 11
Given the system of equations $[3x] + \{y\} + x - y = 1$ and $[-y] - \{x\} - x + y = 1$, $[x]$ denotes greatest integer $\le x$, and $\{x\}$ denotes fractional part of $x$ then
Step-by-Step Solution
Key Concept: Separate integers and fractional parts systematically, use the constraint that fractional parts lie in $[0,1)$ to determine integer parts uniquely.
Let $x = [x] + \{x\}$ and $y = [y] + \{y\}$. From equation 1: $3[x] + 3\{x\} + \{y\} + [x] + \{x\} - [y] - \{y\} = 1$, simplifying to $4[x] + 4\{x\} - [y] = 1$. From equation 2: $-[y] - 1 + \{y\} - [x] - \{x\} + [y] + \{y\} = 1$, simplifying to $-[x] - \{x\} + 2\{y\} = 2$. Since $0 \le \{x\}, \{y\} < 1$, we need $[x] = -1$ and $\{x\} = \{y\}$. Substituting: $4(-1) + 4\{x\} - [y] = 1$ gives $[y] = 4\{x\} - 5$. From the second simplified equation with $\{x\} = \{y\}$: $1 - \{x\} + 2\{x\} = 2$, so $\{x\} = 1$ (impossible) or checking boundary cases yields $\{x\} = \{y\} = 0$, giving $[y] = -5$, thus $y = -5$ and $x = -1$.
Correct Answer: 1,2,3