Key Concepts and Formulas

• Greatest Integer Function $[x]$: The greatest integer less than or equal to $x$.
• Absolute Value Function $|y|$: The distance of $y$ from zero. $|y| = y$ if $y \ge 0$, and $|y| = -y$ if $y < 0$.
• Fractional Part $\{x\}$: Defined as $x - [x]$.
• Continuity at a point $c$: A function $f(x)$ is continuous at $x=c$ if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
  • Left-Hand Limit (LHL): $\lim_{x \to c^-} f(x)$
  • Right-Hand Limit (RHL): $\lim_{x \to c^+} f(x)$
  • Function Value: $f(c)$

Step-by-Step Solution

The function is given by $f(x) = [x^2 - x] + |-x + [x]|$. We need to check the continuity of $f(x)$ at $x=0$ and $x=1$.

Step 1: Simplify the expression for $f(x)$ by considering different intervals.

We can rewrite the term $|-x + [x]|$ using the definition of the fractional part function. We know that $\{x\} = x - [x]$, so $-x + [x] = -(x - [x]) = -\{x\}$. Therefore, $|-x + [x]| = |-\{x\}|$. Since $0 \le \{x\} < 1$, we have $-1 < -\{x\} \le 0$. Thus, $|-\{x\}| = -(-\{x\}) = \{x\}$ for $-1 < -\{x\} \le 0$, which is always true since $\{x\} \ge 0$. So, $|-x + [x]| = \{x\}$.

Now, the function becomes $f(x) = [x^2 - x] + \{x\}$.

We need to analyze the behavior of $f(x)$ around the points where the terms $[x^2-x]$ and $\{x\}$ might change their values abruptly, which are usually integers for $[x^2-x]$ and integers for $x$ for $\{x\}$. The points of interest are $x=0$ and $x=1$.

Let's analyze the behavior of $[x^2 - x]$ and $\{x\}$ in the vicinity of $x=0$ and $x=1$.

Case 1: Analyzing continuity at $x=0$.

We need to calculate LHL, RHL, and $f(0)$.

• Calculate $f(0)$: Substitute $x=0$ into the simplified expression: $f(0) = [0^2 - 0] + \{0\} = [0] + 0 = 0 + 0 = 0$.

• Calculate LHL at $x=0$: $\lim_{x \to 0^-} f(x)$ As $x \to 0^-$, $x$ is a small negative number (e.g., $x = -0.1, -0.01$). For $x$ close to $0$ and $x < 0$, we have $[x] = -1$. Also, $x^2 - x = x(x-1)$. If $x$ is a small negative number, then $x-1$ is close to $-1$. So $x(x-1)$ will be a small positive number. For example, if $x = -0.1$, $x^2 - x = (-0.1)^2 - (-0.1) = 0.01 + 0.1 = 0.11$. So, $[x^2 - x] = 0$ for $x$ slightly less than $0$. And $\{x\} = x - [x] = x - (-1) = x+1$. As $x \to 0^-$, $\{x\} \to 0+1 = 1$. Therefore, $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} ([x^2 - x] + \{x\}) = 0 + 1 = 1$.

• Calculate RHL at $x=0$: $\lim_{x \to 0^+} f(x)$ As $x \to 0^+$, $x$ is a small positive number (e.g., $x = 0.1, 0.01$). For $x$ close to $0$ and $x > 0$, we have $[x] = 0$. Also, $x^2 - x = x(x-1)$. If $x$ is a small positive number, then $x-1$ is close to $-1$. So $x(x-1)$ will be a small negative number. For example, if $x = 0.1$, $x^2 - x = (0.1)^2 - 0.1 = 0.01 - 0.1 = -0.09$. So, $[x^2 - x] = -1$ for $x$ slightly greater than $0$. And $\{x\} = x - [x] = x - 0 = x$. As $x \to 0^+$, $\{x\} \to 0$. Therefore, $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} ([x^2 - x] + \{x\}) = -1 + 0 = -1$.

• Conclusion for $x=0$: LHL = 1, RHL = -1, $f(0) = 0$. Since LHL $\ne$ RHL, the function $f(x)$ is not continuous at $x=0$.

Step 2: Analyzing continuity at $x=1$.

We need to calculate LHL, RHL, and $f(1)$.

• Calculate $f(1)$: Substitute $x=1$ into the simplified expression: $f(1) = [1^2 - 1] + \{1\} = [0] + 0 = 0 + 0 = 0$.

• Calculate LHL at $x=1$: $\lim_{x \to 1^-} f(x)$ As $x \to 1^-$, $x$ is slightly less than 1 (e.g., $x = 0.9, 0.99$). For $x$ close to $1$ and $x < 1$, we have $[x] = 0$. Also, $x^2 - x = x(x-1)$. If $x$ is slightly less than 1, then $x-1$ is a small negative number. So $x(x-1)$ will be a small negative number. For example, if $x = 0.9$, $x^2 - x = (0.9)^2 - 0.9 = 0.81 - 0.9 = -0.09$. So, $[x^2 - x] = -1$ for $x$ slightly less than $1$. And $\{x\} = x - [x] = x - 0 = x$. As $x \to 1^-$, $\{x\} \to 1$. Therefore, $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} ([x^2 - x] + \{x\}) = -1 + 1 = 0$.

• Calculate RHL at $x=1$: $\lim_{x \to 1^+} f(x)$ As $x \to 1^+$, $x$ is slightly greater than 1 (e.g., $x = 1.1, 1.01$). For $x$ close to $1$ and $x > 1$, we have $[x] = 1$. Also, $x^2 - x = x(x-1)$. If $x$ is slightly greater than 1, then $x-1$ is a small positive number. So $x(x-1)$ will be a small positive number. For example, if $x = 1.1$, $x^2 - x = (1.1)^2 - 1.1 = 1.21 - 1.1 = 0.11$. So, $[x^2 - x] = 0$ for $x$ slightly greater than $1$. And $\{x\} = x - [x] = x - 1$. As $x \to 1^+$, $\{x\} \to 1 - 1 = 0$. Therefore, $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} ([x^2 - x] + \{x\}) = 0 + 0 = 0$.

• Conclusion for $x=1$: LHL = 0, RHL = 0, $f(1) = 0$. Since LHL = RHL = $f(1)$, the function $f(x)$ is continuous at $x=1$.

Step 3: Combine the conclusions.

We found that $f(x)$ is not continuous at $x=0$ and continuous at $x=1$.

Common Mistakes & Tips

• Incorrectly handling the absolute value: Remember that $|y|$ is always non-negative. In this case, $|-x + [x]| = |-\{x\}|$, and since $0 \le \{x\} < 1$, we have $-1 < -\{x\} \le 0$. Thus, $|-\{x\}| = -(-\{x\}) = \{x\}$.
• Approximation errors for $[x^2-x]$: Be careful when evaluating $[x^2-x]$ for values slightly above or below an integer. For example, when $x \to 0^+$, $x^2-x$ is a small negative number, so its floor is $-1$. When $x \to 1^-$, $x^2-x$ is a small negative number, so its floor is $-1$.
• Understanding the behavior of $\{x\}$: The fractional part $\{x\}$ is $x$ for $x \in [0, 1)$, and $x-1$ for $x \in [1, 2)$, and so on. For $x \to 0^+$, $\{x\} \to 0$. For $x \to 0^-$, $[x]=-1$, so $\{x\} = x - (-1) = x+1 \to 1$. For $x \to 1^-$, $[x]=0$, so $\{x\} = x - 0 = x \to 1$. For $x \to 1^+$, $[x]=1$, so $\{x\} = x - 1 \to 0$.

Summary

We analyzed the function $f(x) = [x^2 - x] + |-x + [x]|$ by first simplifying it to $f(x) = [x^2 - x] + \{x\}$. We then checked the continuity at $x=0$ and $x=1$ by comparing the left-hand limit, the right-hand limit, and the function value at these points. At $x=0$, the LHL was 1, the RHL was -1, and $f(0)$ was 0, indicating discontinuity. At $x=1$, the LHL was 0, the RHL was 0, and $f(1)$ was 0, indicating continuity. Therefore, the function is continuous at $x=1$ but not at $x=0$.

The final answer is \boxed{A}.