$f(x) = \begin{cases} \frac{x^{2}-4}{x-\lambda}, & x 2 \end{cases}$ be a continuous function, then
$f(x) =
\begin{cases}
\frac{x^{2}-4}{x-\lambda}, & x < 2 \\
\mu, & x = 2 \\
\frac{x^{3}-8}{x-\mu}, & x > 2
\end{cases}$
be a continuous function, then
$\lambda = 2, \upsilon = 4$
$\lambda \neq 2, \upsilon = 2$
$\lambda \neq 2, \upsilon \neq 2$
$\lambda = 2, \upsilon \neq 2$
Solution
(i) If $\lambda=2$ then $f\left(2^{-}\right)=4$
(ii) If $\lambda\neq2$ then $f\left(2^{-}\right)=0$
(iii) If $\upsilon=2$ then $f\left(2^{+}\right)=12$
(iv) If $\upsilon\neq2$ then $f\left(2^{+}\right)=0$
Therefore, $f\left|x\right|$ is continuous at $x=2$ if $\lambda\neq2$, $\upsilon\neq2$ and $\mu=0$
Therefore, $\lambda\neq2$, $\upsilon\neq2$ is the only correct option.