The function f(x)= x^3+x^2-16x+20 x-2 is not defind for x = 2. in… - Vidyadip

MCQ

The function $\text{f(x)}=\frac{\text{x}^3+\text{x}^2-16\text{x}+20}{\text{x}-2}$ is not defind for $x = 2$. in order to make $f(x)$ continuous at $x = 2$, here $f(2)$ should be defined as:

✓
$0$
B
$1$
C
$2$
D
$3$

✓

Answer

Correct option: A.

$0$

Here,
$x^3+ x^2- 16x + 20$
$= x^3- 2x^2+ 3x^2- 6x - 10x + 20$
$= x^2(x - 2) + 3x(x - 2) - 10(x - 2)$
$= (x - 2)(x^2+ 3x - 10)$
$= (x - 2)(x - 2) (x - 5)$
$= (x - 2)^2(x + 5)$
So, the given function can be rewritten as
$\text{f(x)}=\frac{(\text{x}-2)^2(\text{x}+5)}{\text{x}-2}$
$\Rightarrow\text{f(x)}=(\text{x}-2)(\text{x}+5)$
If $f(x)$ is continuous at $x = 2$, then
$\lim\limits_{\text{x}\rightarrow2}\text{f(x)}=\text{f}(2)$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}(\text{x}-2)(\text{x}+5)=\text{f}(2)$
$\Rightarrow\text{f}(2)=0$
Hence, in order to make $f(x)$ continuous at $x = 2, f(2)$ should be defined as $0$.

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