Question

Let $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{x^4-5x^2+4}{(x-1)(x-2)},& x\ne 1,2\end{array}\\ \begin{array}{cc}6,& x=1\end{array}\\ \begin{array}{cc}12,& x=2\end{array}\end{array}$ then $f\left(x\right)$ is continuous on the set

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

Answer 1

The correct option is A $R$

At $x=1,f\left(x\right)=6$

$\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}\frac{(x-2)(x+2)(x-1)(x+1)}{(x-1)(x-2)}$

$=\underset{x\to 1}{lim}(x+1)(x+2)=6$

$\Rightarrow f\left(x\right)$ is contnuous at $x=1$

${lim}_{x\to 2}f\left(x\right)={lim}_{x\to 2}(x+1)(x+2)=12$

Similarly, it can be shown that it is continuous at $x=2$

$\Rightarrow f\left(x\right)$ is continuous for all on the set $R$.