Question

The function $f\left(x\right)=\frac{x^3+x^2-16x+20}{x-2}$ is not defined for x = 2. In order to make f (x) continuous at x = 2, Here f (2) should be defined as
(a) 0
(b) 1
(c) 2
(d) 3

Answer 1

Here,

$$\begin{gathered} x^3+x^2-16x+20 \\ =x^3-2x^2+3x^2-6x-10x+20 \\ =x^2\left(x-2\right)+3x\left(x-2\right)-10\left(x-2\right) \\ =\left(x-2\right)\left(x^2+3x-10\right) \\ =\left(x-2\right)\left(x-2\right)\left(x+5\right) \\ ={\left(x-2\right)}^2\left(x+5\right) \end{gathered}$$

So, the given function can be rewritten as

$f\left(x\right)=\frac{{\left(x-2\right)}^2\left(x+5\right)}{x-2}$

$\Rightarrow f\left(x\right)=\left(x-2\right)\left(x+5\right)$

If $f\left(x\right)$ is continuous at $x=2$, then
$\underset{x\to 2}{\lim }f\left(x\right)=f\left(2\right)$

$$\begin{gathered} \Rightarrow \underset{x\to 2}{\lim }\left(x-2\right)\left(x+5\right)=f\left(2\right) \\ \Rightarrow f\left(2\right)=0 \end{gathered}$$

Hence, in order to make $f\left(x\right)$ continuous at $x=2,f\left(2\right)$ should be defined as 0.