Let `f(x) = {{:((x^(4) -5x^(2)+4)/(|(x-1)(x-2)|), (x ne 1,2)),(6, x=1),(12, x=2):}`. Then f(x) is continuous on the set

Options

A. R
B. R-{1}
C. R-{2}
D. R-{1,2}

Answer

To determine the continuity of the function

\[ f(x) = \begin{cases} \frac{x^4 - 5x^2 + 4}{|(x-1)(x-2)|} & \text{if } x \neq 1, 2 \\ 6 & \text{if } x = 1 \\ 12 & \text{if } x = 2 \end{cases} \]

we need to analyze the function at the points \(x = 1\) and \(x = 2\), as these are the points where the function is defined piecewise.

### Step 1: Factor the Polynomial
First, we factor the polynomial in the numerator:

\[ x^4 - 5x^2 + 4 = (x^2 - 4)(x^2 - 1) = (x-2)(x+2)(x-1)(x+1) \]

Thus, we can rewrite \(f(x)\) for \(x \neq 1, 2\) as:

\[ f(x) = \frac{(x-2)(x+2)(x-1)(x+1)}{|(x-1)(x-2)|} \]

### Step 2: Simplify the Function
For \(x \neq 1, 2\), we can simplify \(f(x)\):

\[ f(x) = \frac{(x+2)(x+1)}{|1|} = (x+2)(x+1) \quad \text{for } x > 2 \text{ or } x < 1 \]

For \(1 < x < 2\), the expression becomes:

\[ f(x) = -(x+2)(x+1) \]

### Step 3: Check Continuity at \(x = 1\)
To check continuity at \(x = 1\), we need to find the left-hand limit, right-hand limit, and the function value at \(x = 1\):

1. **Left-hand limit as \(x \to 1^-\)**:
\[ \lim_{x \to 1^-} f(x) = -(1+2)(1+1) = -3 \cdot 2 = -6 \]

2. **Right-hand limit as \(x \to 1^+\)**:
\[ \lim_{x \to 1^+} f(x) = (1+2)(1+1) = 3 \cdot 2 = 6 \]

3. **Function value at \(x = 1\)**:
\[ f(1) = 6 \]

Since the left-hand limit \(-6\) is not equal to the right-hand limit \(6\), \(f(x)\) is **not continuous at \(x = 1\)**.

### Step 4: Check Continuity at \(x = 2\)
Next, we check continuity at \(x = 2\):

1. **Left-hand limit as \(x \to 2^-\)**:
\[ \lim_{x \to 2^-} f(x) = -(2+2)(2+1) = -4 \cdot 3 = -12 \]

2. **Right-hand limit as \(x \to 2^+\)**:
\[ \lim_{x \to 2^+} f(x) = (2+2)(2+1) = 4 \cdot 3 = 12 \]

3. **Function value at \(x = 2\)**:
\[ f(2) = 12 \]

Since the left-hand limit \(-12\) is not equal to the right-hand limit \(12\), \(f(x)\) is **not continuous at \(x = 2\)**.

### Conclusion
The function \(f(x)\) is continuous on the set \(\mathbb{R} \setminus \{1, 2\}\).