Question

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

Answer 1

(d) R − {1, 2}

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

$$\begin{gathered} \mathrm{Now}, \\ {x}^4-5x^2+4=x^4-x^2-4x^2+4=x^2\left(x^2-1\right)-4\left(x^2-1\right)=\left(x^2-1\right)\left(x^2-4\right)=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right) \\ \Rightarrow f\left(x\right)=\left\{\begin{array}{c}\frac{\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left|\left(x-2\right)\left(x-1\right)\right|},x\ne 1,2\\ 6,x=1\\ 12,x=2\end{array}\right. \end{gathered}$$


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


So,

$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)=\underset{h\to 0}{\lim }\left(1-h+1\right)\left(1-h+2\right)=2\times 3=6$

$\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1+h\right)=-\underset{h\to 0}{\lim }\left(1+h+1\right)\left(1+h+2\right)=-2\times 3=-6$

Also,

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(2-h\right)=-\underset{h\to 0}{\lim }\left(2-h+1\right)\left(2-h+2\right)=-12$

$\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(2+h\right)=\underset{h\to 0}{\lim }\left(2+h+1\right)\left(2+h+2\right)=12$

Thus, $\underset{x\to 1^{+}}{\lim }f\left(x\right)\ne \underset{x\to 1^{-}}{\lim }f\left(x\right)\mathrm{and}\underset{x\to 2^{+}}{\lim }f\left(x\right)\ne \underset{x\to 2^{-}}{\lim }f\left(x\right)$

Therefore, the only points of discontinuities of the function $f\left(x\right)$ are $x=1\mathrm{and}x=2$ .

Hence, the given function is continuous on the set R − {1, 2}.
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