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if f(x) ={x^(2)}, where {x} denotes t...

if `f(x) ={x^(2)}`, where {x} denotes the fractional part of x , then

A

f(X) is continuous at x=-2 but not at x=2

B

f(x) is continuous at x=2 but not at x=-2

C

f(x) is continuous at x=2 and x=-2

D

f(x) is discontinuous at x=-2 and at x=2

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the function \( f(x) = \{x^2\} \), where \( \{x\} \) denotes the fractional part of \( x \). The fractional part function is defined as \( \{x\} = x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). ### Step-by-Step Solution: 1. **Understanding the Function**: The function \( f(x) = \{x^2\} \) means we will take the square of \( x \) and then find its fractional part. 2. **Identifying Points of Interest**: We need to check the continuity of \( f(x) \) at \( x = -2 \) and \( x = 2 \). The function \( \{x^2\} \) will be discontinuous at points where \( x^2 \) is an integer, which occurs when \( x \) is an integer (since squaring an integer results in an integer). 3. **Evaluating at \( x = 2 \)**: - Calculate \( f(2) \): \[ f(2) = \{2^2\} = \{4\} = 4 - \lfloor 4 \rfloor = 4 - 4 = 0 \] - Check the limit as \( x \) approaches 2 from the left (\( x \to 2^- \)): \[ f(2^-) = \{(2 - \epsilon)^2\} = \{4 - 4\epsilon + \epsilon^2\} \approx \{4 - 0\} = 0 \quad (\text{as } \epsilon \to 0) \] - Check the limit as \( x \) approaches 2 from the right (\( x \to 2^+ \)): \[ f(2^+) = \{(2 + \epsilon)^2\} = \{4 + 4\epsilon + \epsilon^2\} \approx \{4 + 0\} = 0 \quad (\text{as } \epsilon \to 0) \] - Since both limits equal \( f(2) \), the function is continuous at \( x = 2 \). 4. **Evaluating at \( x = -2 \)**: - Calculate \( f(-2) \): \[ f(-2) = \{(-2)^2\} = \{4\} = 0 \] - Check the limit as \( x \) approaches -2 from the left (\( x \to -2^- \)): \[ f(-2^-) = \{(-2 - \epsilon)^2\} = \{4 + 4\epsilon + \epsilon^2\} \approx \{4 + 0\} = 0 \quad (\text{as } \epsilon \to 0) \] - Check the limit as \( x \) approaches -2 from the right (\( x \to -2^+ \)): \[ f(-2^+) = \{(-2 + \epsilon)^2\} = \{4 - 4\epsilon + \epsilon^2\} \approx \{4 - 0\} = 0 \quad (\text{as } \epsilon \to 0) \] - Since both limits equal \( f(-2) \), the function is continuous at \( x = -2 \). 5. **Conclusion**: The function \( f(x) = \{x^2\} \) is continuous at both \( x = -2 \) and \( x = 2 \). ### Final Answer: The function \( f(x) \) is continuous at \( x = -2 \) and \( x = 2 \).
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