MHTCET 26 Apr 2025 Shift 2 Paper

Concept:A function is continuous at a point if the left-hand limit, right-hand limit, and the function value all exist and are equal.Explanation:The given function can be rewritten using factorization:$$x^{4}-5x^{2}+4 = (x^{2}-1)(x^{2}-4) = (x-1)(x+1)(x-2)(x+2)$$Therefore, for $x \neq 1,2$,$$f(x)=\frac{(x-1)(x+1)(x-2)(x+2)}{|x-1||x-2|}$$Now, the factor $\frac{x-1}{|x-1|}$ behaves differently from the left and right of $x=1$.As $x \to 1^{-}$, $\frac{x-1}{|x-1|} = -1$ and as $x \to 1^{+}$, $\frac{x-1}{|x-1|} = 1$.Hence, $\lim_{x \to 1} \frac{x-1}{|x-1|}$ does not exist.Similarly, $\lim_{x \to 2} \frac{x-2}{|x-2|}$ does not exist.Thus, $f(x)$ is discontinuous at $x=1$ and $x=2$, even though $f(1)=6$ and $f(2)=12$ are defined.For any $x \neq 1,2$, the expression is a quotient of polynomials, and a polynomial is continuous everywhere.Therefore, $f(x)$ is continuous for all real $x$ except at $x=1$ and $x=2$.Answer:$$f(x) \text{ is continuous on } \mathbb{R}-\{1,2\}$$Hence, the correct option is D: $\mathbb{R}-\{1,2\}$.