# let the function :

Question:

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

(a) $R$

(b) $R-\{1\}$

(c) $R-\{2\}$

(d) $R-\{1,2\}$

Solution:

(d) $R-\{1,2\}$

Given: $f(x)=\left\{\begin{array}{c}\frac{x^{4}-5 x^{2}+4}{|(x-1)(x-2)|}, x \neq 1,2 \\ 6, \quad x=1 \\ 12, \quad x=2\end{array}\right.$

Now,

$x^{4}-5 x^{2}+4=x^{4}-x^{2}-4 x^{2}+4=x^{2}\left(x^{2}-1\right)-4\left(x^{2}-1\right)=\left(x^{2}-1\right)\left(x^{2}-4\right)=(x-1)(x+1)(x-2)(x+2)$

$\Rightarrow f(x)=\left\{\begin{array}{c}\frac{(x-1)(x+1)(x-2)(x+2)}{|(x-2)(x-1)|}, x \neq 1,2 \\ 6, x=1 \\ 12, x=2\end{array}\right.$

$\Rightarrow f(x)=\left\{\begin{array}{cl}(x+1)(x+2), & x<1 \\ -(x+1)(x+2), & 12 \\ 6, & x=1 \\ 12, & x=2\end{array}\right.$

$\mathrm{SO}$,

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(1-h+1)(1-h+2)=2 \times 3=6$

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=-\lim _{h \rightarrow 0}(1+h+1)(1+h+2)=-2 \times 3=-6$

Also,

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h)=-\lim _{h \rightarrow 0}(2-h+1)(2-h+2)=-12$

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}(2+h+1)(2+h+2)=12$

Thus, $\lim _{x \rightarrow 1^{+}} f(x) \neq \lim _{x \rightarrow 1^{-}} f(x)$ and $\lim _{x \rightarrow 2^{+}} f(x) \neq \lim _{x \rightarrow 2^{-}} f(x)$

Therefore, the only points of discontinuities of the function $f(x)$ are $x=1$ and $x=2$.

Hence, the given function is continuous on the set $R-\{1,2\}$.