R − {1, 2}
Given :
\[f\left( x \right) = \begin{cases}\frac{x^4 - 5 x^2 + 4}{\left| \left( x - 1 \right)\left( x - 2 \right) \right|}, x \neq 1, 2 \\ 6, x = 1 \\ 12, x = 2\end{cases}\]
\[\text{ Now,} \]
\[ x^4 - 5 x^2 + 4 = x^4 - x^2 - 4 x^2 + 4 = x^2 \left( x^2 - 1 \right) - 4\left( x^2 - 1 \right) = \left( x^2 - 1 \right)\left( x^2 - 4 \right) = \left( x - 1 \right)\left( x + 1 \right)\left( x - 2 \right)\left( x + 2 \right)\]
\[ \Rightarrow f\left( x \right) = \begin{cases}\frac{\left( x - 1 \right)\left( x + 1 \right)\left( x - 2 \right)\left( x + 2 \right)}{\left| \left( x - 2 \right)\left( x - 1 \right) \right|}, x \neq 1, 2 \\ 6, x = 1 \\ 12, x = 2\end{cases}\]
\[\Rightarrow f\left( x \right) = \begin{cases}\left( x + 1 \right)\left( x + 2 \right), x < 1 \\ - \left( x + 1 \right)\left( x + 2 \right), 1 < x < 2 \\ \left( x + 1 \right)\left( x + 2 \right), x > 2 \\ 6, x = 1 \\ 12, x = 2\end{cases}\]
So,
\[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right) = \lim_{h \to 0} \left( 1 - h + 1 \right)\left( 1 - h + 2 \right) = 2 \times 3 = 6\]
\[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right) = - \lim_{h \to 0} \left( 1 + h + 1 \right)\left( 1 + h + 2 \right) = - 2 \times 3 = - 6\]
Also,
\[\lim_{x \to 2^-} f\left( x \right) = \lim_{h \to 0} f\left( 2 - h \right) = - \lim_{h \to 0} \left( 2 - h + 1 \right)\left( 2 - h + 2 \right) = - 12\]
\[\lim_{x \to 2^+} f\left( x \right) = \lim_{h \to 0} f\left( 2 + h \right) = \lim_{h \to 0} \left( 2 + h + 1 \right)\left( 2 + h + 2 \right) = 12\]
Thus,
\[\lim_{x \to 1^+} f\left( x \right) \neq \lim_{x \to 1^-} f\left( x \right) \text{ and } \lim_{x \to 2^+} f\left( x \right) \neq \lim_{x \to 2^-} f\left( x \right)\]
Therefore, the only points of discontinuities of the function \[f\left( x \right)\]are \[x = 1 \text{ and }x = 2\]
Hence, the given function is continuous on the set R − {1, 2}.
