Step 1: Understanding the Question:
A function is continuous if the limit at a point equals the value of the function at that point. We check the points of interest $x=1$ and $x=2$.
Step 3: Detailed Explanation:
The numerator is $x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4) = (x-1)(x+1)(x-2)(x+2)$.
So for $x \neq 1, 2$:
$f(x) = \frac{(x-1)(x+1)(x-2)(x+2)}{|(x-1)(x-2)|}$.
- At $x=1$:
Limit depends on the sign of $(x-1)(x-2)$.
As $x \to 1^-$, $(x-1) < 0$ and $(x-2) < 0$, so product $> 0$. Denom $= (x-1)(x-2)$.
LHL $= (1+1)(1+2) = 6$.
As $x \to 1^+$, $(x-1) > 0$ and $(x-2) < 0$, so product $< 0$. Denom $= -(x-1)(x-2)$.
RHL $= -(1+1)(1+2) = -6$.
Since LHL $\neq$ RHL, $f(x)$ is discontinuous at $x=1$.
- Similarly at $x=2$:
As $x \to 2^-$, Denom $= -(x-1)(x-2)$. RHL $= -(2+1)(2+2) = -12$.
As $x \to 2^+$, Denom $= (x-1)(x-2)$. RHL $= (2+1)(2+2) = 12$.
Discontinuous at $x=2$.
- The function is continuous everywhere else on $\mathbb{R}$.
Step 4: Final Answer:
The function is continuous on $\mathbb{R} - \{1, 2\}$.