Question

Let $f(x)=\begin{cases}\frac{x^{4}-5x^{2}+4}{|(x-1)(x-2)|}&,x\ne1,2\\ 6&,x=1
12&,x=2\end{cases}$. Then $f(x)$ is continuous on the set}

• A. $\mathbb{R}-\{1\}$
• B. $\mathbb{R}-\{2\}$
• C. $\mathbb{R}$
• D. $\mathbb{R}-\{1,2\}$

Hint:

The presence of $|x-a|$ in the denominator often indicates a jump discontinuity if the factor $(x-a)$ cancels out.

Answer 1

The Correct Option is D

Step 1: Concept
For continuity at $x=a$, $\lim_{x\to a} f(x)$ must exist and equal $f(a)$.

Step 2: Analysis
- The numerator $x^4-5x^2+4$ factors into $(x^2-1)(x^2-4) = (x-1)(x+1)(x-2)(x+2)$. - The denominator contains $|(x-1)(x-2)|$. - When $x \to 1$ or $x \to 2$, the limit depends on the direction because of the absolute value, leading to different signs for LHL and RHL.

Step 3: Calculation
- At $x=1$, LHL and RHL will involve $\pm (x+1)(x-2)(x+2)$. The values will be $\pm (2)(-1)(3) = \pm 6$. Since LHL $\neq$ RHL, it is discontinuous. - Similar behavior occurs at $x=2$.

Step 4: Conclusion
The function is discontinuous at both $x=1$ and $x=2$. Final Answer: (D)