`Let f(x)={(x^4-5x^2+4)/(|(x-1)(x-2)|`, x!=1,2 & 6, x=1 & 12 , x = 2 ` `then `f(x)` is continuous on the set` (a) R (b) R-{1} (c) R-{2} (d) R-{1,2}

Explanation

To determine where the function \( f(x) \) is continuous, we need to analyze the given piecewise function:

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

### Step 1: Identify the points of discontinuity
The function is defined differently at \( x = 1 \) and \( x = 2 \). We need to check the continuity at these points.

### Step 2: Check continuity at \( x = 1 \)
To check continuity at \( x = 1 \), we need to find the left-hand limit (LHL), right-hand limit (RHL), and the function value at \( x = 1 \).

**Left-hand limit as \( x \to 1^- \)**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x^4 - 5x^2 + 4}{-(x-1)(x-2)} \]
Substituting \( x = 1 \):
\[ = \frac{1^4 - 5(1^2) + 4}{-(1-1)(1-2)} = \frac{1 - 5 + 4}{0} = \frac{0}{0} \text{ (indeterminate)} \]
We need to simplify:
\[ x^4 - 5x^2 + 4 = (x^2 - 4)(x^2 - 1) = (x-2)(x+2)(x-1)(x+1) \]
Thus,
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{(x-2)(x+2)(x-1)(x+1)}{-(x-1)(x-2)} \]
Cancelling \( (x-1)(x-2) \):
\[ = \lim_{x \to 1^-} \frac{(x+2)(x+1)}{-1} = \frac{(1+2)(1+1)}{-1} = \frac{3 \cdot 2}{-1} = -6 \]

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

Since \( \lim_{x \to 1^-} f(x) = -6 \) and \( \lim_{x \to 1^+} f(x) = 6 \), the left-hand limit does not equal the right-hand limit. Thus, \( f(x) \) is discontinuous at \( x = 1 \).

### Step 3: Check continuity at \( x = 2 \)
**Left-hand limit as \( x \to 2^- \)**:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{(x-2)(x+2)(x-1)(x+1)}{-(x-1)(x-2)} \]
Cancelling \( (x-2) \):
\[ = \lim_{x \to 2^-} \frac{(x+2)(x+1)}{-1} = \frac{(2+2)(2+1)}{-1} = \frac{4 \cdot 3}{-1} = -12 \]

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

Since \( \lim_{x \to 2^-} f(x) = -12 \) and \( \lim_{x \to 2^+} f(x) = 12 \), the left-hand limit does not equal the right-hand limit. Thus, \( f(x) \) is discontinuous at \( x = 2 \).

### Step 4: Conclusion
The function \( f(x) \) is continuous everywhere except at \( x = 1 \) and \( x = 2 \). Therefore, the set where \( f(x) \) is continuous is:

\[ \text{Answer: } \mathbb{R} - \{1, 2\} \]