Question

Let $f(x) = \begin{cases} \frac{ x^{3}+x^{2}-16x+20}{\left(x-2\right)^{2}}, & \text{ $x \ne 2$ } \\ k ,& \text{ $x=2$ } \end{cases}$
If $f(x)$ is continuous for all $x$, then $k =$

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

Answer: (C)

$k=f(2)=\displaystyle \lim_{x \to 2}\,f(x)$
$=\displaystyle \lim_{x \to 2}$ $\frac{x^{3}+x^{2}-16x+20}{\left(x-2\right)^{2}}$
Using $L'$ Hospital Rule, we get
$\displaystyle \lim_{x \to 2}$ $\frac{3x^{2}+2x-16}{2\left(x-2\right)}$ $(\frac{0}{0}$ form$)$
Again using $L'$ Hospital Rule, we get
$\displaystyle \lim_{x \to 2}$ $\frac{6x+2}{2}=7$.
So, $k = 7$