`f (x) = {{:((|x^(2)- x|)/(x^(2) - x),xne 0"," 1),(1",", x = 0),(-1",", x = 0 ):}` is continuose for all :

Options

A. x
B. x except at x = 0
C. x except at x = 1
D. x except at = and x = 1

Answer

To determine the continuity of the function

\[ f(x) = \begin{cases} \frac{|x^2 - x|}{x^2 - x}, & x \neq 0 \\ 1, & x = 0 \\ -1, & x = 1 \end{cases} \]

we need to analyze the function at the points of interest, particularly at \(x = 0\) and \(x = 1\), and also check the intervals where \(x \neq 0\).

### Step 1: Analyze the function for \(x \neq 0\)

For \(x \neq 0\), we can simplify \(f(x)\) based on the expression inside the modulus. The expression \(x^2 - x\) can be factored as:

\[ x^2 - x = x(x - 1) \]

This means that the sign of \(x^2 - x\) will change depending on the intervals of \(x\):

- **When \(x < 0\)**: \(x^2 - x > 0\) (since both factors are negative)
- **When \(0 < x < 1\)**: \(x^2 - x < 0\) (since \(x\) is positive and \(x - 1\) is negative)
- **When \(x > 1\)**: \(x^2 - x > 0\) (since both factors are positive)

Thus, we can rewrite \(f(x)\) as:

\[ f(x) = \begin{cases} 1, & x < 0 \\ -1, & 0 < x < 1 \\ 1, & x > 1 \end{cases} \]

### Step 2: Check continuity at \(x = 0\)

To check the continuity at \(x = 0\), we need to find the left-hand limit (LHL) and right-hand limit (RHL) as \(x\) approaches 0.

- **Left-hand limit (LHL)** as \(x \to 0^{-}\):
\[ \lim_{x \to 0^{-}} f(x) = 1 \]

- **Right-hand limit (RHL)** as \(x \to 0^{+}\):
\[ \lim_{x \to 0^{+}} f(x) = -1 \]

Since LHL \(\neq\) RHL, the function is not continuous at \(x = 0\).

### Step 3: Check continuity at \(x = 1\)

Next, we check the continuity at \(x = 1\).

- **Left-hand limit (LHL)** as \(x \to 1^{-}\):
\[ \lim_{x \to 1^{-}} f(x) = -1 \]

- **Right-hand limit (RHL)** as \(x \to 1^{+}\):
\[ \lim_{x \to 1^{+}} f(x) = 1 \]

Again, since LHL \(\neq\) RHL, the function is not continuous at \(x = 1\).

### Step 4: Conclusion on continuity

The function \(f(x)\) is continuous everywhere except at \(x = 0\) and \(x = 1\). Therefore, we conclude that:

\[ f(x) \text{ is continuous for all } x \in \mathbb{R} \setminus \{0, 1\} \]

### Final Answer

The function \(f(x)\) is continuous for all \(x\) except at \(x = 0\) and \(x = 1\).

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